Haritha

Equilibrium

Question 1. In liquid-gas equilibrium, the pressure of vapours above the liquid is constant at

1. Constant temperature
2. Low temperature
3. High temperature
4. None of these.

Answer: 1. Constant temperature

Vapour pressure is directly related to temperature. The greater is the temperature, the greater will be the vapour pressure. So to keep it constant, the temperature should be constant.

Question 2. \(3 \mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{O}_{3(g)}\) For the above reaction at 298 K, K_c is found to be 3.0 x 10^-59. If the concentration of O₂ at equilibrium is 0.040 M then the concentration of O₃ in M is

1. \(4.38 \times 10^{-32}\)
2. \(1.9 \times 10^{-63}\)
3. \(2.4 \times 10^{31}\)
4. \(1.2 \times 10^{21}\)

Answer: 1. \(4.38 \times 10^{-32}\)

∴ \(K_c=\frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]^3}=3 \times 10^{-59}\)

Given: \(\left[\mathrm{O}_2\right]=0.040 \mathrm{M}\)

⇒ \(K_c=\frac{\left[\mathrm{O}_3\right]^2}{(0.040)^3}=3 \times 10^{-59}\)

⇒ \({\left[\mathrm{O}_3\right]^2=1.92 \times 10^{-63}}\)

⇒ \({\left[\mathrm{O}_3\right]=4.38 \times 10^{-32} \mathrm{M}}\)

Read and Learn More NEET MCQs with Answers

Question 3. The equilibrium constants of the following are \(\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 ; K_1\); \(\mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} ; K_2\); \(\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{H}_2 \mathrm{O} ; K_3\)

The equilibrium constant (K) of the reaction: \(2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \stackrel{\mathrm{K}}{\rightleftharpoons} 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O}\) will be

1. \(K_2 K_3^3 / K_1\)
2. \(K_2 K_3 / K_1\)
3. \(K_2^3 K_3 / K_1\)
4. \(K_1 K_3^3 / K_2\)

Asnwer: 1. \(K_2 K_3^3 / K_1\)

From the given equations, \(2 \mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2+3 \mathrm{H}_2 ; \frac{1}{K_1}\)….(1)

⇒ \(\mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} ; K_2\)…..(2)

⇒ \(3 \mathrm{H}_2+\frac{3}{2} \mathrm{O}_2 \rightleftharpoons 3 \mathrm{H}_2 \mathrm{O} ; K_3^3\)….(3)

By adding equations (1), (2) and (3), we get \(2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \stackrel{K}{\rightleftharpoons} 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O}, K=\frac{K_2 K_3^3}{K_1}\)

Question 4. If the equilibrium constant for \(\mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{(\mathrm{g})}\) is K, the equilibrium constant for \(\frac{1}{2} \mathrm{~N}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{NO}_{(g)}\) will be

1. 1/2 K
2. K
3. K²
4. K½

Answer: 4. K½

If the reaction is multiplied by 1/2, then the equilibrium constant, K’ = K^1/2

Question 5. Given that the equilibrium constant for the reaction, \(2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{SO}_{3(g)}\) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature; \(\mathrm{SO}_{3(g)} \rightleftharpoons \mathrm{SO}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)}\)

1. \(1.8 \times 10^{-3}\)
2. \(3.6 \times 10^{-3}\)
3. \(6.0 \times 10^{-2}\)
4. \(1.3 \times 10^{-5}\)

Answer: 3. \(6.0 \times 10^{-2}\)

⇒ \(2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{SO}_{3(g)}, K=278\)…..(1)

By reversing the equation (1), we get \(2 \mathrm{SO}_{3(g)} \rightleftharpoons 2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)}\)…..(2)

Equilibrium constant for this reaction is, \(K^{\prime}=\frac{1}{K}=\frac{1}{278}\)

By dividing the equation (2) by 2, we get the desired equation, \(\mathrm{SO}_{3(g)} \rightleftharpoons \mathrm{SO}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)}\)….(3)

Equilibrium constant for this reaction, \(K^{\prime \prime}=\sqrt{K^{\prime}}=\sqrt{\frac{1}{K}}=\sqrt{\frac{1}{278}}=0.0599=0.06 \text { or } 6 \times 10^{-2}\)

Question 6. Given the reaction between 2 gases represented by A₂ and B₂ to give the compound AB_(g). \(A_{2(g)}+B_{2(g)} \rightleftharpoons 2 A B_{(g)}\) At equilibrium, the concentration of \(A_2=3.0 \times 10^{-3} \mathrm{M}\), of \(B_2=4.2 \times 10^{-3} \mathrm{M}\), of \(A B=2.8 \times 10^{-3} \mathrm{M}\)

If the reaction takes place in a sealed vessel at 527°C, then the value of Kc will be

1. 2.0
2. 1.9
3. 0.62
4. 4.5

Answer: 3. 0.62

⇒ \(A_{2(g)}+B_{2(g)} \rightleftharpoons 2 A B_{(g)}\)

∴ \(K_c=\frac{[A B]^2}{\left[A_2\right]\left[B_2\right]}=\frac{\left(2.8 \times 10^{-3}\right)^2}{\left(3.0 \times 10^{-3}\right)\left(4.2 \times 10^{-3}\right)}=\frac{2.8 \times 2.8}{3.0 \times 4.2}=0.62\)

Question 7. For the reaction, \(\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}\), the equilibrium constant is K₁. The equilibrium constant is K₂ for the reaction, \(2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(\mathrm{~g})}\) What is K for the reaction, \(\mathrm{NO}_{2(g)} \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2(g)}+\mathrm{O}_{2(g)} \text {? }\)

1. \(\frac{1}{2 K_1 K_2}\)
2. \(\frac{1}{4 K_1 K_2}\)
3. \(\left[\frac{1}{K_1 K_2}\right]^{1 / 2}\)
4. \(\frac{1}{K_1 K_2}\)

Answer: 3. \(\left[\frac{1}{K_1 K_2}\right]^{1 / 2}\)

⇒ \(\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} ; K_1\)

⇒ \(2 \mathrm{NO}+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 ; K_2\)

⇒ \(\mathrm{NO}_2 \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\mathrm{O}_2 ; K\)

∴ \(K_1=\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} ; K_2=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{NO}^2\left[\mathrm{O}_2\right]\right.}\)

K = \(\frac{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]}{\left[\mathrm{NO}_2\right]}=\sqrt{\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right] \times[\mathrm{NO}]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{NO}^2 \times\left[\mathrm{NO}_2\right]^2\right.}} \Rightarrow K=\sqrt{\frac{1}{K_1 K_2}}\)

Question 8. The dissociation constants for acetic acid and HCN at 25°C are 1.5 x 10^-5 and 4.5 x 10^-10 respectively. The equilibrium constant for the equilibrium, \(\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-}\) would be

1. \(3.0 \times 10^{-5}\)
2. \(3.0 \times 10^{-4}\)
3. \(3.0 \times 10^4\)
4. \(3.0 \times 10^5\)

Answer: 3. \(3.0 \times 10^4\)

Given, \(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\)

⇒ \(K_1=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=1.5 \times 10^{-5}\)

HCN \(\rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-}\)

⇒ \(K_2=\frac{\left[\mathrm{CN}^{-}\right]\left[\mathrm{H}^{+}\right]}{[\mathrm{HCN}]}=4.5 \times 10^{-10}\)

⇒ \(\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-}\)

K = \(\frac{\left[\mathrm{HCN}^{-}\right]\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CN}^{-}\right]\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\)

K = \(\frac{K_1}{K_2}=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=0.3 \times 10^5 \text { or } K=3 \times 10^4\)

Question 9. The value of equilibrium constant of the reaction, \(\mathrm{HI}_{(\mathrm{g})} \rightleftharpoons \frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{I}_{2(\mathrm{~g})}\) is 8.0. The equilibrium constant of the reaction \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}\) will be

1. 16
2. 1/8
3. 1/16
4. 1/64

Answer: 4. 1/64

⇒ \(\mathrm{HI}_{(\mathrm{g})} \rightleftharpoons 1 / 2 \mathrm{H}_{2(g)}+1 / 2 \mathrm{I}_{2(g)}\)

i.e. \(\mathrm{K}=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]}=8\)

⇒ \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(g)}\)

K’ \(=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\left(\frac{1}{8}\right)^2 \Rightarrow K^{\prime}=\frac{1}{64}\)

Question 10. Equilibrium constants K₁ and K₂ for the following equilibrium: \(\mathrm{NO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \stackrel{K_1}{\rightleftharpoons} \mathrm{NO}_{2(g)}\) and \(2 \mathrm{NO}_{2(\mathrm{~g})} \stackrel{\mathrm{K}_2}{\rightleftharpoons} 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})}\) are related as

1. \(K_2=1 / K_1^2\)
2. \(K_2=K_1^2\)
3. \(K_2=1 / K_1\)
4. \(K_2=K_1 / 2\)

Answer: 1. \(K_2=1 / K_1^2\)

⇒ \(K_1=\frac{p_{\mathrm{NO}_2}}{p_{\mathrm{NO}} \cdot\left(p_{\mathrm{O}_2}\right)^{1 / 2}}\)…..(1)

⇒ \(K_2=\frac{\left(p_{\mathrm{NO}}\right)^2 \cdot p_{\mathrm{O}_2}}{\left(p_{\mathrm{NO}_2}\right)^2}\)….(2)

Taking the square root on both sides in Equation 2,

⇒ \( \sqrt{K_2}=\frac{p_{\mathrm{NO}} \cdot\left(p_{\mathrm{O}_2}\right)^{1 / 2}}{p_{\mathrm{NO}_2}} \Rightarrow \sqrt{K_2}=\frac{1}{K_1} \Rightarrow K_2=\frac{1}{K_1^2}\)

Question 11. If K₁ and K₂ are the respective equilibrium constants for the two reactions, \(\mathrm{XeF}_{6(g)}+\mathrm{H}_2 \mathrm{O}_{(g)} \rightarrow \mathrm{XeOF}_{4(g)}+2 \mathrm{HF}_{(g)}\);
\(\mathrm{XeO}_{4(g)}+\mathrm{XeF}_{6(g)} \rightarrow \mathrm{XeOF}_{4(g)}+\mathrm{XeO}_3 \mathrm{~F}_{2(g)^{\prime}}\) the equilibrium constant of the reaction, \(\mathrm{XeO}_{4(g)}+2 \mathrm{HF}_{(g)} \rightarrow \mathrm{XeO}_3 \mathrm{~F}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(g)} \text {, }\) will be

1. \(K_1 / K_2\)
2. \(K_1 \cdot K_2\)
3. \(K_1 /\left(K_2\right)^2\)
4. \(K_2 / K_1\)

Answer: 4. \(K_2 / K_1\)

Given, \(\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{XeOF}_4+2 \mathrm{HF}, K_{\text {eq }}=K_1\)

⇒ \(\mathrm{XeOF}_4+2 \mathrm{HF} \rightleftharpoons \mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O}, K_{\text {eq }}=1 / K_1\)….(1) and

⇒ \(\mathrm{XeO}_4+\mathrm{XeF}_6 \rightleftharpoons \mathrm{XeOF}_4+\mathrm{XeO}_3 \mathrm{~F}_2, K_{\text {eq }}=K\)….(2)

The reaction, \(\mathrm{XeO}_4+2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_3 \mathrm{~F}_2+\mathrm{H}_2 \mathrm{O}\), can be obtained by adding equation (1) and equation (2).

So, the equilibrium constant for the above reaction can be obtained by multiplying the equilibrium constants of equation (1) and equation (2).

Hence, the value is \(\frac{K_2}{K_1}\)

Question 12. The equilibrium constant for the reaction \(\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3\) is K, then the equilibrium constant for the equilibrium \(2 \mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2+3 \mathrm{H}_2\) is

1. \(\sqrt{K}\)
2. \(\sqrt{\frac{1}{K}}\)
3. \(\frac{1}{K}\)
4. \(\frac{1}{K^2}\)

Answer: 3. \(\sqrt{K}\)

The equilibrium constant for the reverse reaction will be 1/K.

Question 13. K₁ and K₂ are equilibrium constants for reactions (1) and (2) respectively. \(\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons \mathrm{NO}_{(g)}\)…..(1); \(\mathrm{NO}_{(g)} \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)}\)

1. \(K_1=\left(\frac{1}{K_2}\right)^2\)
2. \(K_1=K_2{ }^2\)
3. \(K_1=\frac{1}{K_2}\)
4. \(K_1=\left(K_2\right)^0\)

Answer: 1. \(K_1=\left(\frac{1}{K_2}\right)^2\)

Reaction (2) is the reversible reaction of (1) and is half of the reaction (1). Thus, rate constant can be given as \(K_2=\sqrt{\frac{1}{K_1}} \text { or } K_1=\left[\frac{1}{K_2}\right]^2\)

Question 14. The reaction, \(2 A_{(g)}+B_{(g)} \rightleftharpoons 3 C_{(g)}+D_{(g)}\) begins with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

1. [(0.75)³ (0.25)] ÷ [(1.00)² (1.00)]
2. [(0.75)³ (0.25)] ÷ [(0.50)² (0.75)]
3. [(0.75)³ (0.25)] ÷ [(0.50)² (0.25)]
4. [(0.75)³ (0.25)] ÷ [(0.75)² (0.25)]

Answer: 2. [(0.75)³ (0.25)] + [(0.50)² (0.75)]

Equilibrium constant, \(K=\frac{[C]^3[D]}{[A]^2[B]}\)

∴ K= \(\frac{(0.75)^3(0.25)}{(0.5)^2(0.75)}\)

Question 15. The dissociation equilibrium of a gas AB₂ can be represented as \(2 A B_{2(g)} \rightleftharpoons 2 A B_{(g)}+B_{2(g)}\) The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K_p and total pressure P is

1. \(\left(2 K_{/} / P\right)^{1 / 2}\)
2. \(\left(K_p / P\right)\)
3. \(\left(2 K_p / P\right)\)
4. \(\left(2 K_p / P\right)^{1 / 3}\)

Answer: 4. \(\left(2 K_p / P\right)^{1 / 3}\)

Amount of moles at equilibrium = 2(1 – x) + 2x + x = 2 + x

⇒ \(K_p=\frac{\left[p_{A B}\right]^2\left[p_{B_2}\right]}{\left[p_{A B_2}\right]^2}\)

⇒ \(K_p=\frac{\left(\frac{2 x}{2+x} \times P\right)^2 \times\left(\frac{x}{2+x} \times P\right)}{\left(\frac{2(1-x)}{2+x} \times P\right)^2}=\frac{\frac{4 x^3}{2+x} \times P}{4(1-x)^2}\)

⇒ \(K_p=\frac{4 x^3 \times P}{2} \times \frac{1}{4}\) (because \(1-x \approx 1\) and \(2+x \approx 2\))

x = \(\left(\frac{8 K_p}{4 P}\right)^{1 / 3} \Rightarrow x=\left(\frac{2 K_p}{P}\right)^{1 / 3}\)

Question 16. The values of \(K_{p_1}\) and \(K_{p_2}\) for the reactions,

Y \(\rightleftharpoons\) Z ……(1)

A \(\rightleftharpoons\) 2B……….(2)

are in the ratio 9: 1. If the degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio

1. 36:1
2. 1:1
3. 3: 1
4. 1: 9

Answer: 1. 36:1

X \(\rightleftharpoons\) +Z…(1)

A \(\rightleftharpoons 2 B\) ……(2)

X \(\rightleftharpoons\) Y+Z

Total no. of moles at equilibrium =1-α+2α =1+α

Similarly,

Total Number of moles at equilibrium = 1-α+2α=1+α

∴ \(K_{p_1}=\frac{p_Y \times p_Z}{p_X}=\frac{\frac{\alpha}{1+\alpha} \times P_1 \times \frac{\alpha}{1+\alpha} \times P_1}{\frac{1-\alpha}{1+\alpha} \times P_1}=\frac{\alpha^2 P_1}{(1+\alpha)(1-\alpha)}\)

∴ \(K_{p_2}=\frac{\left(p_B\right)^2}{p_A}=\frac{\left(\frac{2 \alpha}{1+\alpha} \times P_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times P_2}=\frac{(2 \alpha)^2 P_2}{(1+\alpha)(1-\alpha)}\)

Now \(\frac{K_{p_1}}{K_{p_2}}=\frac{P_1}{4 P_2} \Rightarrow \frac{K_{p_1}}{K_{p_2}}=\frac{9}{1}=\frac{P_1}{4 P_2} \Rightarrow \frac{P_1}{P_2}=\frac{36}{1}=36: 1\)

Question 17. A 20-litre container at 400 K contains CO_2(g) at pressure 0.4 atm and an excess of SrO (neglecting the volume of solid S₂O). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of CO₂ attains its maximum value, will be (Given that: \(\mathrm{SrCO}_{3(s)} \rightleftharpoons \mathrm{SrO}_{(s)}+\mathrm{CO}_{2(g)}\), K_p = 1.6 atm)

1. 10 litre
2. 4 litre
3. 2 litre
4. 5 litre

Answer: 4. 5 litre

⇒ \(\mathrm{SrCO}_{3(s)} \rightleftharpoons \mathrm{SrO}_{(s)}+\mathrm{CO}_{2(g)} ; K_p=1.6 \mathrm{~atm}\)

⇒ \(K_p=\frac{p_{\mathrm{CO}_2} \times p_{\mathrm{SrO}}}{p_{\mathrm{SrCO}_3}} \Rightarrow 1.6=p_{\mathrm{CO}_2}\) (because \(p_{\mathrm{SrO}}=p_{\mathrm{SrCO}_3}=1\))

∴ Maximum pressure of CO₂ = 1.6 atm

Let the maximum volume of the container when the pressure of CO₂ is 1.6 atm be V L

During the process, PV = constant

∴ 0.4 x 20 = 1.6 X V

⇒ V = \(\frac{0.4 \times 20}{1.6}=5 \mathrm{~L}\)

Question 18. In which of the following equilibriums K_c and K_p are not equal?

1. \(2 \mathrm{NO}_{(g)} \rightleftharpoons \mathrm{N}_{2(g)}+\mathrm{O}_{2(g)}\)
2. \(\mathrm{SO}_{2(g)}+\mathrm{NO}_{2(g)} \rightleftharpoons \mathrm{SO}_{3(g)}+\mathrm{NO}_{(g)}\)
3. \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(g)} \rightleftharpoons 2 \mathrm{HI}_{(g)}\)
4. \(2 \mathrm{C}_{(s)}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{CO}_{2(\mathrm{~g})}\)

Answer: 4. \(2 \mathrm{C}_{(s)}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{CO}_{2(\mathrm{~g})}\)

K_c and K_p are related by the equation, \(K_p=K_c(R T)^{\Delta n_g}\)

where \({\Delta n_g}\) = difference in the no. of moles of products and reactants in the gaseous state.

for \(2 \mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{CO}_{2(g)}\)

∴ \(\Delta n_g=2-1=1 \neq 0\)

Question 19. If the concentration of OH- ions in the reaction \(\mathrm{Fe}(\mathrm{OH})_{3(s)} \rightleftharpoons \mathrm{Fe}_{(a q)}^{3+}+3 \mathrm{OH}_{(a q)}^{-}\) is decreased by 1/4 times, then equilibrium concentration of Fe³⁺ will increase by

1. 64 times
2. 4 times
3. 8 times
4. 16 times.

Answer: 1. 64 times

Fe\((\mathrm{OH})_{3(s)} \rightleftharpoons \mathrm{Fe}_{(a q)}^{3+}+3 \mathrm{OH}_{(a q)}^{-}\)

K = \(\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3}{\left[\mathrm{Fe}(\mathrm{OH})_3\right]}\)

K = \(\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3\) (activity of solid is taken unity)

The concentration of OH^– ion in the reaction is decreased by 1/4 times then the equilibrium concentration of Fe³⁺ will be increased by 64 times in order to keep the value of K constant.

Question 20. Equilibrium constant K_p for the following reaction \(\mathrm{MgCO}_{3(s)} \rightleftharpoons \mathrm{MgO}_{(s)}+\mathrm{CO}_{2(g)}\)

1. \(K_p=p_{\mathrm{CO}_2}\)
2. \(K_p=p_{\mathrm{CO}_2} \times \frac{p_{\mathrm{CO}_2} \times p_{\mathrm{MgO}}}{p_{\mathrm{MgCO}_3}}\)
3. \(K_p=\frac{p_{\mathrm{CO}_2}+p_{\mathrm{MgO}}}{p_{\mathrm{MgCO}_3}}\)
4. \(K_p=\frac{p_{\mathrm{MgCO}_3}}{p_{\mathrm{CO}_2} \times p_{\mathrm{MgO}}}\)

Answer: 1. \(K_p=p_{\mathrm{CO}_2}\)

K_p = PCO₂.

Solids do not exert pressure, so their partial pressure is taken as unity.

Question 21. If the value of the equilibrium constant for a particular reaction is 1.6 x 10¹², then at equilibrium the system will contain

1. Mostly products
2. Similar amounts of reactants and products
3. All reactants
4. Mostly reactants.

Answer: 1. Mostly products

The value of K is high which means the reaction proceeds almost to completion i.e., the system will contain mostly products.

Question 22. In the Haber process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for the reaction which yielded only 50% of the expected product. What will be the composition of a gaseous mixture under the aforesaid condition in the end?

1. 20 litres of ammonia, 20 litres of nitrogen, 20 litres of hydrogen
2. 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
3. 20 litres of ammonia, 10 litres of nitrogen, and 30 litres of hydrogen
4. 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

Answer: 2. 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

⇒ \(\begin{array}{ccc}
3 \mathrm{H}_2 & +\mathrm{N}_2 & \rightarrow 2 \mathrm{NH}_3 \\
3 & 1 & 2 \\
3 / 2 & 1 / 2 & 1 \\
10 \times \frac{3}{2} & 10 \times \frac{1}{2} & 10 \times 1 \\
15 & 5 & 10
\end{array}\)

The composition of a gaseous mixture under the aforesaid condition in the end will be

H₂ =30-15=15 litres

N₂ = 30 – 5 =25 litres; NH₃ = l0 litres

Question 23. The reaction quotient (Q) for the reaction \(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}\) is given by Q = \(\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\). The reaction will proceed from right to left if

1. Q = K_c
2. Q < K_c
3. Q>K_c
4. Q = 0

where K_c is the equilibrium constant.

Answer: 3. Q>K_c

⇒ \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\)

⇒ \(K_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} ; \Delta n_{(\mathrm{g})}=2-4=-2\)

Thus, the reaction will go from right to left when Q > K_c

Question 24. The equilibrium concentrations of the species in the reaction A + B \(\rightleftharpoons\) C + D are 2, 3, 10 and 6 mol L^-1, respectively at 300 K. ΔG° for the reaction is (R = 2 cal/mol K)

1. -1381.80 cal
2. -13.73 cal
3. 1372.60 cal
4. -137.26 cal

Answer: 1. -1381.80 cal

A + \(B \rightleftharpoons C+D\)

K = \(\frac{[C][D]}{[A][B]}=\frac{10 \times 6}{2 \times 3}=10\)

Δ\(G^{\circ}\)=-R T ln K

Δ\(G^{\circ}\)=-2.303 R T log K

Δ\(G^{\circ}=-2.303 \times 2 \times 300 \times \log 10\)

= \(-2.303 \times 2 \times 300=-1381.80 \mathrm{cal}\)

Question 25. Hydrolysis of sucrose is given by the following reaction : Sucrose + H₂O \(\rightleftharpoons\) Glucose + Fructose If the equilibrium constant (K_C) is 2 x 10¹³ at 300 K, the value of Δ_rG° at the same temperature will be

1. -8.314 J mol^-1 K^-1 x 300 K x ln(2 x 10¹³)
2. 8.314 J mol^-1 K^-1 x 300 K x ln(2 x 10¹³)
3. 8.314 J mol^-1 K^-1 x 300 K x ln(3 x 10¹³)
4. -8.314 J mol^-1 K^-1 x 300 K x ln(4 x 10¹³)

Answer: 1. -8.314 J mol^-1 K^-1 x 300 K x ln(2 x 10¹³)

ΔG = ΔG° + RT In Q

At equilibrium, ΔG = 0 and Q = K_C.

∴ 0 = ΔG°+ RT In K_C.

⇒ ΔG° = -RT ln K_C

= -8.314 J mol^-1 K^-1 x 300 K x ln(2 x 10¹³)

Question 26. Which of the following statements is correct for a reversible process in a state of equilibrium?

1. ΔG° = -2.30 RT log K
2. ΔG° = 2.30 RT log K
3. ΔG = -2.30 RT log K
4. ΔG = 2.30 RT log K

Answer: 1. ΔG° = -2.30 RT log K

Question 27. Match List 1 (Equations) with List 2 (Type of Processes) and select the correct option.

1. 1-(A), 2-(B), 3-(C), 4-(D)
2. 1 -(C), 2-(D), 3-(B), 4-(A)
3. 1-(D), 2-(A), 3-(B), 4-(C)
4. 1-(B), 2-(A), 3-(D), 4-(C)

Answer: 3. 1-(D), 2-(A), 3-(B), 4-(C)

When K_p>Q, rate of forward reaction > rate of backward reaction.

∴ Reaction is spontaneous.

When ΔG° < RT ln Q, ΔG° is positive, the reverse reaction is feasible, thus reaction is non-spontaneous.

When K_p = Q, rate of forward reaction = rate of backward reaction.

∴ The reaction is in equilibrium.

When TΔS > ΔH, ΔG will be negative only when ΔH = +ve.

∴ The reaction is spontaneous and endothermic.

Question 28. Which one of the following conditions will favour the maximum formation of the product in the reaction \(A_{2(g)}+B_{2(g)} \rightleftharpoons X_{2(g)}, \Delta_r H=-X \mathrm{~kJ}?\)

1. Low temperature and high pressure
2. Low temperature and low pressure
3. High temperature and high pressure
4. High temperature and low pressure

Answer: 1. Low temperature and high pressure

On increasing the pressure and decreasing the temperature, equilibrium will shift in the forward direction.

Question 29. For the reversible reaction, \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\text { heat }\) The equilibrium shifts in forward direction

1. By increasing the concentration of NH_3(g)
2. By decreasing the pressure
3. By decreasing the concentrations of N_2(g) and H_2(g)
4. By increasing pressure and decreasing temperature.

Answer: 4. By increasing pressure and decreasing temperature.

As the forward reaction is exothermic and leads to a lowering of pressure (produces a lesser number of gaseous moles) hence, according to Le Chatelier’s principle, at high pressure and low temperature, the given reversible reaction will shift in the forward direction to form more product.

Question 30. For a given exothermic reaction, K_p and K’_p are the equilibrium constants at temperatures T₁ and T₂, respectively. Assuming that the heat of the reaction is constant in the temperature range between T₁ and T₂ it is readily observed that

1. \(K_p>K_p^{\prime}\)
2. \(K_p<K_p^{\prime}\)
3. \(K_p=K_p^{\prime}\)
4. \(K_p=\frac{1}{K_p^{\prime}}\)

Answer: 1. \(K_p>K_p^{\prime}\)

log\(\frac{K_p^{\prime}}{K_p}=-\frac{\Delta H}{2.303 R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]\)

For an exothermic reaction, ΔH = -ve i.e., heal is evolved. The temperature T₂ is higher than T₁

Thus, \(\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) is negative.

so, \(\log K_p^{\prime}-\log K_p=- \text { ve } \quad \text { or } \quad \log K_p>\log K_p^{\prime}\)

or \(K_p>K_p^{\prime}\)

Question 31. KMnO₄ can be prepared from K₂MnO₄ as per the region, \(3 \mathrm{MnO}_4^{2-}+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+4 \mathrm{OH}^{-}\) The reaction can go to completion by removing OH^– ions by adding

1. CO₂
2. SO₂
3. HCl
4. KOH

Answer: 1. CO₂

HCI and SO₂ are reducing agents, So, they can reduce MnO₄^–.

CO₂ is neither an oxidising nor a reducing agent, it will provide only an acidic medium. It can shift the reaction in the forward direction and the reaction can go to completion.

Question 32. The value of ΔH for the reaction \(X_{2(g)}+4 Y_{2(g)} \rightleftharpoons 2 X Y_{4(g)}\) Formation of XY_4(g) will be favoured at

1. High temperature and high pressure
2. Low pressure and low temperature
3. High temperature and low pressure
4. High pressure and low temperature.

Answer: High pressure and low temperature.

⇒ \(X_{2(g)}+4 Y_{2(g)} \rightleftharpoons 2 X Y_{4(g)}\)

Δn_g = -ve and ΔH = -ve

The reaction is favoured in the forward direction at low temperature and high pressure.

Question 33. For the reaction \(\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(l)}\), ΔH_r = -170.8 kJ mol^-1. Which of the following statements is not true?

1. The reaction is exothermic.
2. At equilibrium, the concentrations of CO_2(g) and H₂O_(l) are not equal.
3. The equilibrium constant for the reaction is given by \(K_p=\frac{\left[\mathrm{CO}_2\right]}{\left[\mathrm{CH}_4\right]\left[\mathrm{O}_2\right]}\)
4. Addition of CH_4(g) or O_2(g) at equilibrium mm will cause a shift to the right.

Answer: 3. The equilibrium constant for the reaction is given by \(K_p=\frac{\left[\mathrm{CO}_2\right]}{\left[\mathrm{CH}_4\right]\left[\mathrm{O}_2\right]}\)

⇒ \(\mathrm{CH}_{4(g)}+2 \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{CO}_{2(g)}+2 \mathrm{H}_2 \mathrm{O}_{(l)}\)

⇒ \(K_p=\frac{p_{\mathrm{CO}_2}}{p_{\mathrm{CH}_4} \cdot p_{\mathrm{O}_2}^2}\)

Question 34. Reaction \(\mathrm{BaO}_{2(s)} \rightleftharpoons \mathrm{BaO}_{(s)}+\mathrm{O}_{2(g)} ; \Delta H=+\mathrm{ve}\) In equilibrium condition, pressure of O₂ depends on

1. Increase mass of BaO₂
2. Increase the mass of BaO
3. Increase the temperature on equilibrium
4. Increase the mass of BaO₂ and BaO.

Answer: 3. Increase the temperature on equilibrium

The pressure of O₂ does not depend on concentration terms of other reactants (because both are in solid-state), Since this is an endothermic reaction if the temperature is raised, dissociation of BaO₂ would occur, and more O₂ is produced at equilibrium, the pressure of O₂ increases.

Question 35. For any reversible reaction, if we increase the concentration of the reactants, then the effect on the equilibrium constant

1. Depends on the amount of concentration
2. Unchanged
3. Decrease
4. Increase.

Answer: 2. Unchanged

For a reaction, \(A+B \rightleftharpoons C+D\),

⇒ \(K_{\mathrm{eq}}\)=\(\frac{[C][D]}{[A][B]}\)

Increase in conc. of reactants will proceed the equilibrium in the forward direction giving more products so that the equilibrium constant value remains constant and independent of concentration.

Question 36. According to Le Chatelier’s principle, adding heat to a solid and liquid in equilibrium will cause the

1. Temperature to increase
2. Temperature to decrease
3. Amount of liquid to decrease
4. Amount of solid to decrease.

Answer: 4. Amount of solid to decrease.

When solid and liquid are in equilibrium, the increase in temperature results in an increase in the volume of liquid or a decrease in the amount of solid

Solid \(\rightleftharpoons\) Liquid

With the increase in temperature equilibrium shifts in the forward direction.

Question 37. Which one of the following information can be obtained on the basis of the Le Chatelier principle?

1. Dissociation constant of a weak acid
2. Entropy change in a reaction
3. The equilibrium constant of a chemical reaction
4. Shift in equilibrium position on changing the value of a constraint

Answer: 4. Shift in equilibrium position on changing the value of a constraint

According to Le Chatelier’s principle, if an equilibrium is subjected to a change in concentration, pressure temperature, etc. equilibrium shifts in such a way so as to undo the effect of a change imposed.

Question 38. Aqueous solution which of the following compounds is the best conductor of electric current?

1. Hydrochloric acid, HCl
2. Ammonia, NH₃
3. Fructose, C₆H₁₂O₆
4. Acetic acid, C₂H₄O₂

Answer: 1. Hydrochloric acid, HCl

HCl is a strong acid and dissociates completely into in aqueous solution.

Question 39. Aqueous solution of acetic acid contains

1. \(\mathrm{CH}_3 \mathrm{COO}^{-}\) and \(\mathrm{H}^{+}\)
2. \(\mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{H}_3 \mathrm{O}^{+}\) and \(\mathrm{CH}_3 \mathrm{COOH}\)
3. \(\mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{H}_3 \mathrm{O}^{+}\) and \(\mathrm{H}^{+}\)
4. \(\mathrm{CH}_3 \mathrm{COOH}, \mathrm{CH}_3 \mathrm{COO}^{-}\) and \(\mathrm{H}^{+}\)

Answer: 2. \(\mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{H}_3 \mathrm{O}^{+}\) and \(\mathrm{CH}_3 \mathrm{COOH}\)

∴ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}\)

As acetic acid is a weak acid so, it also contains some undissociated CH₃COOH along with CH₃COO^– and H₃O⁺ ions.

Question 40. Amongst the given options which of the following molecules/ion acts as a Lewis acid?

1. BF₃
2. OH^–
3. NH₃
4. H₂O

Answer: 1. BF₃

Among the given molecules/ions, BF₃ acts as a Lewis acid as it is an electron-deficient species, it can accept a lone pair of electrons.

Question 41. The conjugate base for Bronsted acids H₂O and HF are

1. \(\mathrm{H}_3 \mathrm{O}^{+}\) and \(\mathrm{H}_2 \mathrm{~F}^{+}\) respectively
2. \(\mathrm{OH}^{-}\) and \(\mathrm{H}_2 \mathrm{~F}^{+}\), respectively
3. \(\mathrm{H}_3 \mathrm{O}^{+}\) and \(\mathrm{F}^{-}\), respectively
4. \(\mathrm{OH}^{-}\) and \(\mathrm{F}^{-}\), respectively.

Answer: 4. \(\mathrm{OH}^{-}\) and \(\mathrm{F}^{-}\), respectively.

⇒ \(\begin{array}{cc}
\text { Bronsted acid } & \text { Conjugate base } \\
\mathrm{H}_2 \mathrm{O} & \mathrm{OH}^{-} \\
\mathrm{HF} & \mathrm{F}^{-}
\end{array}\)

Question 42. Which of the following cannot act both as Bronsted acid and as Bronsted base?

1. \(\mathrm{HCO}_3^{-}\)
2. \(\mathrm{NH}_3\)
3. \(\mathrm{HCl}\)
4. \(\mathrm{HSO}_4^{-}\)

Answer: 3. \(\mathrm{HCl}\)

HCI cannot accept H⁺ ions and, therefore cannot act as Bronsted Base.

Question 43. Which of the following fluoro-compounds is most likely to behave as a Lewis base?

1. BF₃
2. PF₃
3. CF₄
4. SiF₄

Answer: 2. PF₃

BF₃ → Lewis acid (incomplete octet)

PF₃ → Lewis base (presence of lone pair on P atom)

CF₄ → Complete octet

SiF₄ → Lewis acid (empty d-orbital in Si-atom)

Question 44. Which of these is least likely to act as a Lewis base?

1. BF₃
2. PF₃
3. CO
4. F^–

Answer: 1. BF₃

BF₃ is Lewis acid (e^– pair acceptor).

Question 45. Which is the strongest acid in the following?

1. HClO₄
2. H₂SO₃
3. H₂SO₄
4. HClO₃

Answer: 1. HClO₄

⇒ \(\stackrel{+7}{\mathrm{HClO}}_4\) with the highest oxidation number and its conjugate base is resonance stabilised, hence it is the most acidic. CI is more electronegative than S.

Question 46. Which one of the following molecular hydrides acts as a Lewis acid?

1. NH₃
2. H₂O
3. B₂H₆
4. CH₄

Answer: 3. B₂H₆

Compounds that are electron deficient act as Lewis acids. Out of the given hydrides, B₂H₆ satisfies this condition and is, therefore, a Lewis acid.

Question 47. Which of the following molecules acts as a Lewis acid?

1. (CH₃)₂O
2. (CH₃)₃P
3. (CH₃)₃N
4. (CH₃)₃B

Answer: 4. (CH₃)₃B

Lewis acids are electron-deficient compounds since (CH₃)₃ B is electron-deficient (due to an incomplete octet of B); it acts as a Lewis acid.

Question 48. Which one of the following statements is not true?

1. Among halide ions, iodide is the most powerful reducing agent.
2. Fluorine is the only halogen that does not show a variable oxidation state.
3. HOCl is a stronger acid than HOBr.
4. HF is a stronger acid than HCl

Answer: 4. HF is a stronger acid than HCl

Due to the strong hydrogen-fluorine bond, proton is not given off easily and hence, HF is the weakest acid.

Question 49. Which one of the following compounds is not a protonic acid?

1. B(OH)₃
2. PO(OH)₃
3. SO(OH)₂
4. SO₂(OH)₂

Asnwer: 1. B(OH)₃

B(OH)₃ in an aqueous medium coordinates a molecule of water to form the hydrated species

In this species, B³⁺ ion, because of its small size, has high polarizing power thereby pulling the sigma electron charge of the coordinated O atom towards itself. The coordinated oxygen, in turn, pulls the sigma electron charge of the OH bond of the attached water molecule towards itself. This facilitates the removal of H⁺ ions from the O – H bond.

Thus, the solution of B(OH)₃ in water acts as a weak acid, and it is not a protonic acid.

Question 50. In \(\mathrm{HS}^{-}, \mathrm{I}^{-}, R-\mathrm{NH}_2, \mathrm{NH}_3\) order of proton accepting tendency will be

1. \(\mathrm{I}^{-}>\mathrm{NH}_3>R-\mathrm{NH}_2>\mathrm{HS}^{-}\)
2. \(\mathrm{NH}_3>\mathrm{R}-\mathrm{NH}_2>\mathrm{HS}^{-}>\mathrm{I}^{-}\)
3. \(\mathrm{R}-\mathrm{NH}_2>\mathrm{NH}_3>\mathrm{HS}^{-}>\mathrm{I}^{-}\)
4. \(\mathrm{HS}^{-}>\mathrm{R}-\mathrm{NH}_2>\mathrm{NH}_3>\mathrm{I}^{-}\)

Answer: 3. \(\mathrm{R}-\mathrm{NH}_2>\mathrm{NH}_3>\mathrm{HS}^{-}>\mathrm{I}^{-}\)

Proton accepting tendency is known as the strength of basicity.

In \(R-\ddot{\mathrm{N}}_2\), N has a lone pair of electrons which intensifies due to electron releasing R-group and increases the tendency to donate a lone pair of electrons to H⁺.

Secondly, as the size of the ion increases, there is less attraction for H⁺ to form a bond with the H-atom and are is basic. Thus the order of proton accepting tendency: \(\mathrm{R}-\mathrm{NH}_2>\mathrm{NH}_3>\mathrm{HS}^{-}>\mathrm{I}^{-}\)

Question 51. The conjugate acid of NH^–₂ is

1. \(\mathrm{NH}_4 \mathrm{OH}\)
2. \(\mathrm{NH}_2^{-}\)
3. \(\mathrm{NH}_4^{+}\)
4. \(\mathrm{NH}_3\)

Answer: 4. \(\mathrm{NH}_3\)

NH^–₂ + H⁺ → NH₃ (conjugate acid)

Substance + H⁺ → conjugate acid

Substance – H⁺ → conjugate base

Question 52. Which compound is electron deficient?

1. \(\mathrm{BeCl}_2\)
2. \(\mathrm{BCl}_3\)
3. \(\mathrm{CCl}_4\)
4. \(\mathrm{PCl}_5^3\)

Answer: 2. \(\mathrm{BCl}_3\)

In BCl₃ the central atom ‘B’ is sp² hybridised and contains only ‘six’-electrons in its valence shell. Therefore, it is electron deficient.

Question 53. The strongest conjugate base is

1. \(\mathrm{SO}_4^{2-}\)
2. \(\mathrm{Cl}^{-}\)
3. \(\mathrm{NO}_3^{-}\)
4. \(\mathrm{CH}_3 \mathrm{COO}^{-}\)

Answer: 4. \(\mathrm{CH}_3 \mathrm{COO}^{-}\)

⇒ \(\underset{\text { Weak acid }}{\mathrm{CH}_3 \mathrm{COOH}} \rightleftharpoons \underset{\text { Strong conjugate base }}{\mathrm{CH}_3 \mathrm{COO}^{-}}+\mathrm{H}^{+}\)

As CH₃COOH is the weakest acid, its conjugate base (CH₃COO^–) is the strongest base, H₂SO₄, HCl, and HNO₃ are strong acids, so their conjugate bases are weak.

Question 54. Which of the following is not a Lewis acid?

1. \(\mathrm{SiF}_4\)
2. \(\mathrm{C}_2 \mathrm{H}_4\)
3. \(\mathrm{BF}_3\)
4. \(\mathrm{FeCl}_3\)

Answer: 2. \(\mathrm{C}_2 \mathrm{H}_4\)

In BF₃ and FeCI₃ molecules, the central atoms have incomplete octets and in SiF₄ the central atom has empty d-orbitals. Hence, according to Lewis’s concept, these are Lewis acids

Question 55. Repeated use of which one of the following fertilizers would increase the acidity of the soil?

1. Ammonium sulphate
2. Superphosphate of lime
3. Urea
4. Potassium nitrate

Answer: 1. Ammonium sulphate

Ammonium sulphate is a salt of strong acid (H₂SO₄ and weak base (NH₄OH). Therefore, repeated use of ammonium sulphate would increase the concentration of sulphuric acid, while ammonia from NH₄OH is used up by the plant. Hence, the acidity of the soil will increase.

Question 56. The pK_b of dimethylamine and pK_a of acetic acid are 3.27 and 4.77 respectively at T(K). The correct option for the pH of dimethylammonium acetate solution is

1. 6.25
2. 8.50
3. 5.50
4. 7.75

Answer: 4. 7.75

For a salt of weak acid and weak base

pH = \(7+\frac{1}{2}\left(\mathrm{p}_a-\mathrm{p} K_b\right)\)

Given, \(\mathrm{p} K_a=4.77, \mathrm{p}_b=3.27\)

pH = \(7+\frac{1}{2}(4.77-3.27)=7.75\)

Question 57. Find out the solubility of Ni(OH)₂ in 0.1 M NaOH. Given that the ionic product of Ni(OH)₂ is 2 x 10^-15.

1. \(2 \times 10^{-13} \mathrm{M}\)
2. \(2 \times 10^{-8} \mathrm{M}\)
3. \(1 \times 10^{-13} \mathrm{M}\)
4. \(1 \times 10^8 \mathrm{M}\)

Answer: 1. \(2 \times 10^{-13} \mathrm{M}\)

⇒ \(\mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}_s^{2+}+\underset{2 s}{2 \mathrm{OH}^{-}}\) where s is the solubility of \(\mathrm{Ni}(\mathrm{OH})_2\)

⇒ \(\underset{0.1 \mathrm{M}}{\mathrm{NaOH}} \rightleftharpoons \underset{0.1 \mathrm{M}}{\mathrm{Na}^{+}}+\underset{0.1 \mathrm{M}}{\mathrm{OH}^{-}}\)

⇒ \(\left[\mathrm{OH}^{-}\right]=2 s+0.1\) approx 0.1(because 2 s<<<0.1)

Ionic product of \(\mathrm{Ni}(\mathrm{OH})_2=\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2\) \(2 \times 10^{-15}=s(0.1)^2\)

s = \(\frac{2 \times 10^{-15}}{0.1 \times 0.1}=2 \times 10^{-13} \mathrm{M}\)

Question 58. The pH of 0.01 M NaOH_(aq) solution will be

1. 7.01
2. 2
3. 12
4. 9

Answer: 3. 12

⇒ \(\underset{0.01 M}{\mathrm{NaOH}} \rightleftharpoons \mathrm{Na}^{+}+ \underset{0.1 M}{\mathrm{OH}^{-}}\)

∴ [OH^–] = 0.01 M

∴ pOH = -log [OH^–] = -log(0.01) = 2

∴ pH = 14 – pOH = 14-2 = 12

Question 59. The following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:

1. \(60 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+40 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}\)
2. \(55 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+45 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}\)
3. \(75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}\)
4. \(100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}\)

pH of which one of them will be equal to 1?

1. B
2. A
3. D
4. C

Answer: 4. C

pH = 1, so [H⁺] = 10^-1

For the acid-base mixture: N₁V₁ – N₂V₂ = N₃V₃.

(For NaOH and HCl, Normaiity = Molarity)

1. \(M_1\left(\mathrm{H}^{+}\right)=\frac{60 \times \frac{1}{10}-40 \times \frac{1}{10}}{100}=2 \times 10^{-2} \mathrm{M}\) i.e. \(\mathrm{pH}=1.698 \approx 1.7\)
2. \(M_2\left(\mathrm{H}^{+}\right)=\frac{55 \times \frac{1}{10}-45 \times \frac{1}{10}}{100}=\frac{1}{100}=10^{-2} \mathrm{M}\) i.e. \(\mathrm{pH}=2\)
3. \(M_3\left(\mathrm{H}^{+}\right)=\frac{75 \times \frac{1}{5}-25 \times \frac{1}{5}}{100}=10^{-1} \mathrm{M}\) i.e. \(\mathrm{pH}=1\)
4. \(M_4\left(\mathrm{H}^{+}\right)=\frac{100 \times \frac{1}{10}-100 \times \frac{1}{10}}{200}=0\) i.e. \(\mathrm{pH}=7\)

Question 60. The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅N⁺H) in a 0.10 M aqueous pyridine solution (Kb for C₅H₅N = 1.7 x 10^-9) is

1. 0.0060%
2. 0.013%
3. 0.77%
4. 1.6%

Answer: 2. 0.013%

⇒ \(\underset{0.10\mathrm{M}}{\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_5 \mathrm{H}_5 \stackrel{+}{\mathrm{N}} \mathrm{H}+\mathrm{OH}^{-}\)

α \(=\sqrt{\frac{K_b}{C}}=\sqrt{\frac{1.7 \times 10^{-9}}{0.10}}=1.30 \times 10^{-4}\)

∴ Percentage of pyridine that forms pyridinium ion = 1.30 x 10^-4 x 100 = 0.013%

Question 61. What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

1. 2.0
2. 7.0
3. 1.04
4. 12.65

Answer: 4. 12.65

One mole of NaOH is completely neutralised by one mole of HCl

Hence, 0.01 mole of NaOH will be completely neutralised by 0.01 mole of HCl.

NaOH left unneutralised = 0.1 – 0.01 = 0.09 mol

As equal volumes of two solutions are mixed, \([\mathrm{OH}]^{-}=\frac{0.09}{2}=0.045 \mathrm{M}\)

⇒ pOH =2 -log(0.04s) = 1.35 ∴ pH = 14 – 1.35 = 12.65

Question 62. Which of the following salts will give the highest pH in water?

1. KCI
2. NaCl
3. Na₂CO₃
4. CuSO₄

Answer: 3. Na₂CO₃

Na₂CO₃ which is a salt of NaOH (strong base) and H₂CO₃ (weak acid) will produce a basic solution with pH greater than 7.

Question 63. Accumulation of lactic acid (HC₃H₅O₃), a monobasic acid in tissues leads to pain and a feeling of fatigue. In a 0.10 M aqueous solution, lactic acid is 3.7% dissociated. The value of dissociation constant, K_a, for this acid, will be

1. \(1.4 \times 10^{-5}\)
2. \(1.4 \times 10^{-4}\)
3. \(3.7 \times 10^{-4}\)
4. \(2.8 \times 10^{-4}\)

Answer: 2. \(1.4 \times 10^{-4}\)

Degree of dissociation, \(\alpha=\frac{3.7}{100}=0.037\)

According to Ostwald’s formula, \(K_a=\alpha^2 C=(0.037)^2 \times 0.10=1.369 \times 10^{-4} \approx 1.4 \times 10^{-4}\)

Question 64. At 100°C the K_w of water is 55 times its value at 25°C. What will be the pH of the neutral solution? (log 55 = 1.74)

1. 7.00
2. 7.87
3. 5.13
4. 6.13

Answer: 4. 6.13

We know that, at \(25^{\circ} \mathrm{C}, K_w=1 \times 10^{-14}\)

At \(100^{\circ} \mathrm{C}, K_w=55 \times 10^{-14}\)

(because \(K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\))

⇒ \(K_w=\left[\mathrm{H}^{+}\right]^2\)

⇒ \(\mathrm{H}^{+}=\sqrt{K_w}\)

⇒ \(\mathrm{H}^{+}=\sqrt{55 \times 10^{-14}} \Rightarrow \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\)

pH = \(-\log \left[\sqrt{55 \times 10^{-14}}\right]\)

= \(\frac{1}{2}\left[-\log \left(55 \times 10^{-14}\right)\right]=\frac{1}{2}[-\log 55+14 \log 10]\)

= \(\frac{1}{2}[-1.74+14]=\frac{1}{2}[12.26]=6.13\)

Question 65. Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value?

1. BaCl₂
2. AlCl₃
3. LiCl
4. BeCl₂

Answer: 1. BaCl₂

BaCl₂ is made up of Ba(OH)₂ and HCl.

AlCl₃ is made up of Al(OH)₃ and HCL

LiCl is made up of LiOH and HCl.

BeCI₂ is made up of Be(OH)₂ and HCl.

Ba(OH)₂ is the strongest base among the given options and thus has the maximum pH

Question 66. What is [H⁺] in mol/L of a solution that is 0.20 M in CH₃COONa and 0.10 M in CH₃COOH? (K_a for CH₃COOH = 1.8 X 10^-5)

1. \(3.5 \times 10^{-4}\)
2. \(1.1 \times 10^{-5}\)
3. \(1.8 \times 10^{-5}\)
4. \(9.0 \times 10^{-6}\)

Answer: 4. \(9.0 \times 10^{-6}\)

Question 67. The ionization constant of ammonium hydroxide is 1.77 x 10^-5 at 298 K. Hydrolysis constant of ammonium chloride is

1. \(6.50 \times 10^{-12}\)
2. \(5.65 \times 10^{-13}\)
3. \(5.65 \times 10^{-12}\)
4. \(5.65 \times 10^{-10}\)

Answer: 4. \(5.65 \times 10^{-10}\)

NH₄CI is a salt of strong acid and weak base, so the hydrolysis constant is \(K_h=\frac{K_w}{K_b}\)

Given, \(K_b\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5}\)

∴ \(K_w=10^{-14}\)

∴ \(K_h=\frac{10^{-14}}{1.77 \times 10^{-5}}=0.565 \times 10^{-9} \text { or } K_h=5.65 \times 10^{-10}\)

Question 68. What is the [OH^–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂?

1. 0.40 M
2. 0.0050 M
3. 0.12 M
4. 0.10 M

Answer: 4. 0.10 M

Millimoles of H⁺ produced = 20 x 0.05 = 1

Millimoles of OH^– produced = 30 x 0.1 x 2 = 6

(Each Ba(OH)₂ gives 2OH^–.)

∴ Millimoles of OH^– remaining in solution = 6 – 1 = 5

Total volume of solution = 20 + 30 = 50 mL

∴ \(\left[\mathrm{OH}^{-}\right]=\frac{5}{50}=0.1 \mathrm{M}\)

Question 69. Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?

1. \(3.7 \times 10^{-3} \mathrm{M}\)
2. \(1.11 \times 10^{-3} \mathrm{M}\)
3. \(1.11 \times 10^{-4} \mathrm{M}\)
4. \(3.7 \times 10^{-4} \mathrm{M}\)

Answer: 4. \(3.7 \times 10^{-4} \mathrm{M}\)

pH = \(-\log \left[\mathrm{H}^{+}\right]\)

or \(\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}} ;\left[\mathrm{H}^{+}\right]\) of solution  1 = \(10^{-3}\)

∴ \(\left[\mathrm{H}^{+}\right]\) of soln. \(2=10^{-4} ;\left[\mathrm{H}^{+}\right]\) of solution. \(3=10^{-5}\)

Total concentration of \(\left[\mathrm{H}^{+}\right]=10^{-3}\left(1+1 \times 10^{-1}+1 \times 10^{-2}\right)\)

⇒ \(10^{-3}\left(\frac{1}{1}+\frac{1}{10}+\frac{1}{100}\right) \Rightarrow 10^{-3}\left(\frac{100+10+1}{100}\right)\)

⇒ \(10^{-3}\left(\frac{111}{100}\right)=1.11 \times 10^{-3}\)

So, \(\mathrm{H}^{+}\) ion concentration in mixture of equal volumes of these acid solution = \(\frac{1.11 \times 10^{-3}}{3}=3.7 \times 10^{-4} \mathrm{M}\)

Question 70. A weak acid, H_A, has a K_a of 1.00 x 10^-5. If 0.100 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to

1. 1.00%
2. 99.9%
3. 0.100%
4. 99.0%

Answer: 1. 1.00%

For a weak acid, the degree of dissociation,

α \(=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{1 \times 10^{-5}}{0.1}}=10^{-2}\) i.e. 1.00%

Question 71. Calculate the pOH of a solution at 25°C that contains 1 x 10^-10 M of hydronium ions, i.e. \(\mathrm{H}_3 \mathrm{O}^{+}\).

1. 4.000
2. 9.000
3. 1.000
4. 7.000

Answer: 1. 4.000

Given, [H₃O⁺] = 1 x 10^-10 or, pH = 10

Now at 25°C, pH + POH = PK_w = 14

or, pOH = 14-PH = 14-10=4

Question 72. The hydrogen ion concentration of a 10^-8 M HCl aqueous solution at 298 K (K_w = 10^-14) is

1. \(1.0 \times 10^{-8} \mathrm{M}\)
2. \(1.0 \times 10^{-6} \mathrm{M}\)
3. \(1.0525 \times 10^{-7} \mathrm{M}\)
4. \(9.525 \times 10^{-8} \mathrm{M}\)

Answer: 3. \(1.0525 \times 10^{-7} \mathrm{M}\)

10^-8 M HCl = 10^-8 MH⁺

Also from water, [H⁺] = 10^-7

Total [H⁺] = 10^-7 + 0.10 x 10^-7 = 1.1 x 10^-7 M

Question 73. At 25°C, the dissociation constant of a base, BOH, is 1.0 x 10^-12. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be

1. \(1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
2. \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\)
3. \(2.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\)
4. \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

Answer: 4. \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

C = \(0.01 \mathrm{M}\)

⇒ \(K_b=1 \times 10^{-12} \text { at } 25^{\circ} \mathrm{C}\)

⇒ \(\begin{array}{lcccc}
& \mathrm{BOH} & & B^{+}+\mathrm{OH}^{-} \\
\text {Initially } & \mathrm{C} & & 0 & 0 \\
\text { At eq. } & \mathrm{C}-\mathrm{C} \alpha & & \mathrm{C} \alpha & \mathrm{C} \alpha
\end{array}\)

⇒ \({\left[\mathrm{OH}^{-}\right]=\mathrm{C} \alpha}\)

⇒ \({\left[\mathrm{OH}^{-}\right]=\sqrt{K_b \mathrm{C}}=\sqrt{1 \times 10^{-12} \times 10^{-2}}}\)

⇒ \({\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}}\)

Question 74. Which has the highest pH?

1. CH₃COOK
2. Na₂CO₃
3. NH₄Cl
4. NaNO₃

Answer: 2. Na₂CO₃

NH₄OH is a weak base but HCl is a strong acid in solution, so the pH of NH₄Cl solution is comparable. NaNO₃ is a salt of a strong base and strong acid, so the pH of the solution will be 7.

Hydrolysis of potassium acetate (a salt of a weak acid and a strong alkali) gives a weakly alkaline solution since the acetate ion acts as a weak base.

⇒ \(\mathrm{CH}_3 \mathrm{COOK}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{COOH}+\mathrm{K}^{+}+\mathrm{OH}^{-}\)

The pH of this solution = 8.8.

Hydrolysis of sodium carbonate (a salt of strong alkali and a weak acid) gives an alkaline solution.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2\left(\mathrm{Na}^{+}+\mathrm{OH}^{-}\right)+\mathrm{H}_2 \mathrm{CO}_3\)

The pH of this solution is > 10

Question 75. Ionisation constant of CH₃COOH is 1.7 x 10^-5 and concentration of H⁺ ions is 3.4 x 10^-4. Then find out the initial concentration of CH₃COOH molecules,

1. \(3.4 \times 10^{-4}\)
2. \(3.4 \times 10^{-3}\)
3. \(6.8 \times 10^{-4}\)
4. \(6.8 \times 10^{-3}\)

Answer: 4. \(6.8 \times 10^{-3}\)

⇒ \(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\)

⇒ \(K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\)

⇒ \({\left[\mathrm{CH}_3 \mathrm{COOH}\right]=\frac{3.4 \times 10^{-4} \times 3.4 \times 10^{-4}}{1.7 \times 10^{-5}}=6.8 \times 10^{-3}}\)

Question 76. The correct relation between dissociation constants of a dibasic acid is

1. \(K_{a_1}=K_{a_2}\)
2. \(K_{a_1}>K_{a_2}\)
3. \(K_{a_1}<K_{a_2}\)
4. \(K_{a_1}=\frac{1}{K_{a_2}}\)

Answer: 2. \(K_{a_1}>K_{a_2}\)

1. \(\mathrm{H}_2 A \stackrel{K_{a_1}}{\rightleftharpoons} \mathrm{HA}^{-}+\mathrm{H}^{+}\)
2. \(\mathrm{HA}^{-} \stackrel{K_{a_2}}{\rightleftharpoons} A^{2-}+\mathrm{H}^{+}\)

In the 1st step, the H⁺ ion comes from the neutral molecule, while in the 2nd step, the H⁺ ion comes from negatively charged ions. The presence of the -ve charge makes the removal of the H⁺ ion difficult.

Thus, \(K_{a_1}>K_{a_2}\)

Question 77. Which statement is wrong about pH and H⁺?

1. pH of neutral water is not zero.
2. Adding 1 N solution of CH₃COOH and 1 N solution of NaOH, the pH will be seven.
3. [H⁺] concentrated and cold H₂SO₄.
4. Mixing solution of CH₃COOH and HCl, the pH will be less than 7.

Answer: 2. Adding 1 N solution of CH₃COOH and 1 N solution of NaOH, the pH will be seven.

After mixing 1 N solution of CH₃COOH (weak acid) and 1 N NaOH (strong base), the resulting solution will have free OH^– ions. Thus, pH will be higher than 7.

Question 78. The concentration of [H⁺] and concentration of [OH^–] of a 0.1 aqueous solution of 2% ionised weak acid is [ionic product of water = 1 x 10^-14]

1. \(2 \times 10^{-3} \mathrm{M}\) and \(5 \times 10^{-12} \mathrm{M}\)
2. \(1 \times 10^{-3} \mathrm{M}\) and \(3 \times 10^{-11} \mathrm{M}\)
3. \(0.02 \times 10^{-3} \mathrm{M}\) and \(5 \times 10^{-11} \mathrm{M}\)
4. \(3 \times 10^{-2} \mathrm{M}\) and \(4 \times 10^{-13} \mathrm{M}\)

Answer: 1. \(2 \times 10^{-3} \mathrm{M}\) and \(5 \times 10^{-12} \mathrm{M}\)

[H⁺] = Cα = 0.1 x 0.02 = 2 x 10^-3 M

(As degree of dissociation = 2% = 0.02)

Hence, \(\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{2 \times 10^{-3}}=5 \times 10^{-12} \mathrm{M}\)

Question 79. The hydride ion H^– is a stronger base than its hydroxide ion OH^–. Which of the following reactions will occur if sodium hydride (NaH) is dissolved in water?

1. \(\mathrm{H}^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow\) no reaction
2. \(\mathrm{H}_{(a q)}^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{H}_{(a q)}^{-}+\mathrm{H}_2 \mathrm{O}_{(b)} \rightarrow \mathrm{OH}^{-}+\mathrm{H}_2\)
4. None of these.

Answer: 3. \(\mathrm{H}_{(a q)}^{-}+\mathrm{H}_2 \mathrm{O}_{(b)} \rightarrow \mathrm{OH}^{-}+\mathrm{H}_2\)

⇒ \(\mathrm{NaH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{NaOH}+\mathrm{H}_2\)

or, \(\mathrm{H}_{(a q)}^{-}+\mathrm{H}_2 \mathrm{O}_{(l)} \rightarrow \mathrm{OH}^{-}+\mathrm{H}_2 \uparrow\)

Question 80. The ionic product of water at 25°C is 10^-14. Its ionic product at 90°C will be,

1. \(1 \times 10^{-14}\)
2. \(1 \times 10^{-16}\)
3. \(1 \times 10^{-20}\)
4. \(1 \times 10^{-12}\)

Answer: 4. \(1 \times 10^{-12}\)

At high temperatures, the value of ionic products increases.

Question 81. If α is the dissociation constant, then the total number of moles for the reaction, 2HI → H₂ + I₂ will be

1. 1
2. 1 – α
3. 2
4. 2 – α

Answer: 3. 2

Total number of moles = 2(1 – α) + 2α = 2

Question 82. The pH value of N/10 NaOH solution is

1. 12
2. 13
3. 10
4. 11

Answer: 2. 13

Since NaOH is a strong base, therefore it completely ionises. Thus, the hydroxyl ion concentration is equal to that of the base itself. We know that the concentration of OH^– is N/ 10. NaOH=0.1 =10^-1

Therefore value of \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]}=\frac{1 \times 10^{-14}}{10^{-1}}=1 \times 10^{-13}\)

pH = – log [H₃O⁺] = – log [1 x 10^-13] = 13

Question 83. The pH value of a 10 M solution of HCl is

1. Equal to 1
2. Equal to 2
3. Less than 0
4. Equal to 0

Answer: 3. Less than 0

Since HCl is a strong acid and it completely ionises, therefore H₃O⁺ ions concentration is equal to that of the acid itself i.e., [H₃O⁺] = [HCl] = 10 M.

Therefore, pH = – log [H₃O⁺] = – log [10] = – 1

Question 84. At 80°C, distilled water has [H₃O⁺] concentration equal to 1 x 10^-6 mole/litre. The value of K_w at this temperature will be

1. 1 x 10^-12
2. 1 x 10^-15
3. 1 x 10^-6
4. 1 x 10^-9

Answer: 1. 1 x 10^-12

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=1 \times 10^{-6}\) mole/litre

⇒\(K_w=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\left[1 \times 10^{-6}\right] \times\left[1 \times 10^{-6}\right]=1 \times 10^{-12}\)

Question 85. 0.1 M solution of which one of these substances will act basic?

1. Sodium borate
2. Ammonium chloride
3. Calcium nitrate
4. Sodium sulphate

Answer: 1. Sodium borate

Sodium borate is a salt formed from a strong base (NaOH) and weak acid (H₃BO₃). Hence, sodium borate will act as a basic solution.

Question 86. The compound whose water solution has the highest pH is

1. NaCl
2. NaHCO₃
3. Na₂CO₃
4. NH₄Cl

Answer: 3. Na₂CO₃

NH₄CI and NaHCO₃ are acidic in nature and NaCl is neutral. Only Na₂CO₃ is basic and thus, has the highest pH.

Question 87. The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given: pK_a of CH₃COOH = 4.57]

1. 5.57
2. 3.57
3. 4.57
4. 2.57

Answer: 1. 5.57

It is an acidic buffer.

For acidic buffer, pH = \(\mathrm{p} K_a+\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

= \(4.57+\log \frac{0.1}{0.01}=4.57+\log 10=4.57+1=5.57\)

Question 88. Which will make a basic buffer?

1. 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
2. 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH₃COOH
3. 100 mL of 0.1 M CH₃COOH + 100 mL of 0.1M NaOH
4. 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH

Answer: 4. 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH

Acid-base titration: \(\underset{10 \mathrm{mmol}}{\mathrm{HCl}}+\underset{20 \mathrm{mmol}}{\mathrm{NH}_4 \mathrm{OH}} \longrightarrow \mathrm{NH}_4 \mathrm{Cl}\)

∴ HCI is the limiting reagent.

Solution contains NH₄OH (weak base) and NH₄CI (sait of strong acid and weak base). Therefore, a basic buffer will be formed.

Question 89. Which one of the following pairs of solutions is not an acidic buffer?

1. CH₃COOH and CH₃COONa
2. H₂CO₃ and Na₂CO₃
3. H₃PO₄ and Na₃PO₄
4. HClO₄ and NaClO₄

Answer: 4. HClO₄ and NaClO₄

An acidic buffer is a mixture of a weak acid and its salt with a strong base. HCIO₄ is a strong acid.

Question 90. The dissociation constant of a weak acid is 1 x 10^-4. In order to prepare a buffer solution with a pH = 5, the [Salt]/[Acid] ratio should be

1. 4:5
2. 10:1
3. 5: 4
4. 1: 10

Answer: 2. 10:1

pH = \(\mathrm{p} K_a+\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

5 = \(-\log K_a+\log \frac{[\text { Salt }]}{[\text { Acid }]}\) (because \(p K_a=-\log K_a\))

5 = \(-\log \left[1 \times 10^{-4}\right]+\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

5 = \(4+\log \frac{[\text { Salt }]}{[\text { Acid }]}, 5-4=\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

1 = \(\log \frac{[\text { Salt }]}{[\text { Acid }]}, \frac{[\text { Salt }]}{[\text { Acid }]}=10\) i.e. 10: 1

Question 91. Buffer solutions have constant acidity and alkalinity because

1. These give unionised acid or base on reaction with added acid or alkali
2. Acids and alkalies in these solutions are shielded from attack by other ions
3. They have a large excess of H⁺ or OH^– ions
4. They have a fixed value of pH.

Answer: 1. These give unionised acid or base on reaction with added acid or alkali

Question 92. A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of NH⁺₄ is 0.20 M. If the equilibrium constant, K_b for NH₃ equals 1.8 x 10^-5, what is the pH of this solution? (log 2.7 = 0.43)

1. 9.08
2. 9.43
3. 11.72
4. 8.73

Answer: 2. 9.43

⇒ \(\left[\mathrm{NH}_3\right]=0.30 \mathrm{M}_3 K_b=1.8 \times 10^{-5}\)

⇒ \({\left[\mathrm{NH}_4^{+}\right]=0.20 \mathrm{M}}\)

⇒ \(\mathrm{pK}_b=-\log \left(1.8 \times 10^{-5}\right)=4.74\)

pOH = \(\mathrm{p} K_b+\log \frac{[\text { salt }]}{[\text { base }]}=4.74+\log \frac{0.2}{0.3}=4.56\)

pH = (14-4.56)=9.44

Question 93. In a buffer solution containing equal concentrations of B^– and HB, the K_b for B^– is 10^-10. The pH of the buffer solution is

1. 10
2. 7
3. 6
4. 4

Answer: 4. 4

We know, pOH = \(p K_b+\log \frac{\left[B^{-}\right]}{[\mathrm{HB}]}\)

Since, [B^–] = [HB] (given)

∴ pOH = pK_b  ⇒pOH = 10

∴ pH = 44- 10=4

Question 94. Which of the following pairs constitutes a buffer?

1. HCl and KCl
2. HNO₂ and NaNO₂
3. NaOH and NaCl
4. HNO₃ and NH₂NO₃

Answer: 2. HNO₂ and NaNO₂

HNO₂ (weak acid) and NaNO₂ (salt of conjugate base) is an example of acidic buffer

Question 95. The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to the ratio of the concentrations of the conjugate acid (H In) and base (In^–) forms of the indicator by the expression

1. \(\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{p} K_{\text {In }}-\mathrm{pH}\)
2. \(\log \frac{[\mathrm{HIn}]}{\left[\mathrm{In}^{-}\right]}=\mathrm{p} K_{\text {In }}-\mathrm{pH}\)
3. \(\log \frac{[\mathrm{HIn}]}{\left[\mathrm{In}^{-}\right]}=\mathrm{pH}-\mathrm{p} K_{\text {In }}\)
4. \(\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{p} K_{\text {In }}\)

Answer: 4. \(\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{p} K_{\text {In }}\)

Let us consider the formation of a salt of a weak acid and a strong base.

In\(^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{HIn}+\mathrm{OH}^{-}\)

⇒ \(K_h=\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]}\)….(1)

Other equations present in the solution are

⇒ \(\mathrm{HIn} \rightleftharpoons \mathrm{H}^{+}+\mathrm{In}^{-}\)

⇒ \(\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}\)

⇒ \(K_{\mathrm{In}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}\)….(2)

⇒ \(K_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\)….(3)

From (2) and (3), \(\frac{K_w}{K_{\text {In }}}=\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]}=K_h\)……(4)

⇒ \({\left[\mathrm{OH}^{-}\right]=\frac{K_w}{K_{\text {In }}} \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}}\)

⇒ \(\log \left[\mathrm{OH}^{-}\right]=\log K_w-\log K_{\text {In }}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}\)

⇒ \(-\mathrm{pOH}=-\mathrm{p} K_w+\mathrm{p} K_{\text {In }}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \)

⇒ \(\mathrm{p} K_w-\mathrm{pOH}=\mathrm{p} K_{\text {In }}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \text { or, } \mathrm{pH}=\mathrm{p} K_{\text {In }}+\log \frac{\left[\mathrm{In}^{-}\right]}{\left[\mathrm{HIn}^{-}\right]}\)

i.e. \(\log \frac{[\mathrm{In}]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{p} K_{\text {In }}\)

Question 96. The solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has a pH of 9.25. Then find out pK_b of NH₄OH.

1. 9.25
2. 4.75
3. 3.75
4. 8.25

Answer: 2. 4.75

A solution of 0.1 N NH₄OH and 0.1 N NH₄CI is a buffer solution.

According to Henderson’s equation, the pH of a basic buffer

pH = \(14-\mathrm{p} K_b-\log \frac{[\text { Salt }]}{[\text { Base }]} \Rightarrow \mathrm{p} K_b=14-\mathrm{pH}-\log \frac{[\text { Salt }]}{[\text { Base }]}\)

⇒ \(\mathrm{p} K_b=14-9.25-\log \frac{0.1}{0.1}\)

⇒ \(\mathrm{p} K_b=14-9.25=4.75\)

∴ \(\mathrm{pK}_b \text { of } \mathrm{NH}_4 \mathrm{OH}=4.75\)

Question 97. A physician wishes to prepare a buffer solution at pH = 3.85 that efficiently resists changes in pH yet contains only a small concentration of the buffering agents. Which of the following weak acids together with its sodium salt would be best to use?

1. 2, 5-Dihydroxybenzoic acid (pK_a= 2.97)
2. Acetoacetic acid (pK_a = 3.58)
3. m-Chlorobenzoic acid (pK_a = 3.98)
4. p-Chlorocinnamic acid (pK_a = 4.41)

Answer: 2. Acetoacetic acid (pK_a = 3.58)

pH of an acidic buffer solution is given by Henderson equation: pH = \(\mathrm{pH}=\mathrm{pK}_a+\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

Its buffer capacity = pK_a±1

Since a buffer solution is more effective in the pH range pK_a±1 therefore, the weak acid having pK_a = 3.58 together with its sodium salt is chosen. Acetoacetic acid is, therefore, a suitable weak acid.

Question 98. The pH value of blood does not appreciably change by a small addition of an acid or a base, because the blood

1. Can be easily coagulated
2. Contains iron as a part of the molecule
3. Is a body fluid
4. Contains serum protein which acts as a buffer.

Answer: 4. Contains serum protein which acts as a buffer.

The pH value of the blood is maintained constant by the buffer solution present in the blood itself. Buffer solutions resist the change in pH values.

Question 99. pH of a saturated solution of Ca(OH)₂ is 9. The solubility product (K_sp) of Ca(OH)₂ is

1. 0.5 x 10^-10
2. 0.5 x 10^-15
3. 0.25 x 10^-10
4. 0.125 x 10^-15

Answer: 2. 0.5 x 10^-15

pH of the saturated solution of Ca(OH)₂ = 9

∴ pOH of the saturated solution of Ca(OH)₂ = 14 – 9 = 5

⇒ [OH^–] = 10^-5 (pH + pOH = 14)

⇒ \(\mathrm{Ca}(\mathrm{OH})_2 \rightleftharpoons \underset{s}{\mathrm{Ca}^{2+}}+\underset{2s}{2 \mathrm{OH}^{-}}\)

⇒ \(K_{\text {sp }}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=\left[1 / 2 \times 10^{-5}\right]\left[10^{-5}\right]^2\)

= \(0.5 \times 10^{-15}\)

Question 100. The molar solubility of CaF₂ (K_sp = 5.3 x 10^-11) in 0.1 M solution of NaF will be

1. 5.3 x 10^-11 mol L^-1
2. 5.3 x 10^-8 mol L^-1
3. 5.3 x 10^-9 mol L^-1
4. 5.3 x 10^-10 mol L^-1

Answer: 3. 5.3 x 10^-9 mol L^-1

⇒ \(\mathrm{CaF}_2 \longrightarrow \underset{s}{\mathrm{Ca}^{2+}}+\underset{2 s}{2 \mathrm{~F}^{-}}\)

⇒ \(\mathrm{NaF} \longrightarrow \underset{0.1 M}{\mathrm{Na}^{+}}+\underset{0.1 M}{\mathrm{~F}^{-}}\)

⇒ \({\left[\mathrm{Ca}^{2+}\right]=s,\left[\mathrm{~F}^{-}\right]=(2 s+0.1)=0.1 \mathrm{M}}\)

⇒ \(K_{s p}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2\)

⇒ \(5.3 \times 10^{-11}=(s)(0.1)^2\)

s = \(\frac{5.3 \times 10^{-11}}{(0.1)^2}=5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\)

∴ Molar solubility is \(5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\)

Question 101. The solubility of BaSO₄ in water is 2.42 x 10^-3 g L^-1 at 298 K. The value of its solubility product (K_sp) will be (Given molar mass of BaSO₄ = 233 g mol^-1)

1. 1.08 x 10^-10 mol² L^-2
2. 1.08 x 10^-12 mol² L^-2
3. 1.08 x 10^-14 mol² L^-2
4. 1.08 x 10^-8 mol² L^-2

Answer: 1. 1.08 x 10^-10 mol² L^-2

Solubility of \(\mathrm{BaSO}_4\),

s = \(\frac{2.42 \times 10^{-3}}{233} \mathrm{~mol} \mathrm{~L}^{-1}=1.04 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

⇒ \(\mathrm{BaSO}_4\) ionizes completely in the solution as

⇒ \(\mathrm{BaSO}_{4(s)} \rightleftharpoons \mathrm{Ba}_{(a q)}^{2+}+\mathrm{SO}_{4(a q)}^{2-}\)

⇒ \(K_{s p}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]=s^2\)

= \(\left(1.04 \times 10^{-5}\right)^2=1.08 \times 10^{-10} \mathrm{~mol}^2 \mathrm{~L}^{-2}\)

Question 102. The concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 x 10^-4 mol L^-1. The Solubility product of Ag₂C₂O₄ is

1. 2.66 x 10^-12
2. 4.5 x 10^-11
3. 5.3 x 10^-12
4. 2.42 x 10^-8

Answer: 3. 5.3 x 10^-12

Let solubility of Ag₂C₂O₄ be s mol L^-1

⇒ \(\underset{s}{\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_{4(s)}} \rightleftharpoons \underset{2 s}{2 \mathrm{Ag}_{(a q)}^{+}}+\underset{s}{\mathrm{C}_2 \mathrm{O}_{4(a q)}^{2-}}\)

K_sp = [Ag⁺]² [C₂O^2-₄]

K_sp = (2s)²(s)=4s³

K_sp = 4x (1.1x 10^-4)³ ([Ag⁺] =2s=2.2x 10^-4)

K_sp =5.3×10^-12

Question 103. The solubility of AgCl_(s) with solubility product 1.6 x 10^-10 in 0.1 M NaCl solution would be

1. 1.26 x 10^-5 M
2. 1.6x 10^-9 M
3. 1.6x 10^-11 M
4. Zero.

Answer: 2. 1.6x 10^-9 M

Let solubility of AgCl in moles per litre.

⇒ \(\underset{s}{\mathrm{AgCl}_{(a q)}} \rightleftharpoons \underset{s}{\mathrm{Ag}_{(a q)}^{+}}+\underset{(s+0.1)}{\mathrm{Cl}_{(a q)}^{-}}\)

(0.1 M NaCl solution also provides 0.1 M Cl^– ion).

K_sp = [Ag⁺] [CI^–]; 1.6 x 10^-10 = s(s + 0.1)

1.6 x 10^-10 = s(0.1) (‘s<<<<0.1)

1.6 x 10^-10 = s(0.1)

s = \(\frac{1.6 \times 10^{-10}}{0.1}=1.6 \times 10^{-9} \mathrm{M}\)

Question 104. MY and NY₃ two nearly insoluble salts, have the same K_sp values of 6.2 x 10^-13 at room temperature. Which statement would be true in regard to MY and NY₃?

1. The salts MY and NY₃ are more soluble in 0.5 M KY than in pure water.
2. The addition of the salt of KY to the solution of MY and NY₃ will have no effect on their solubilities.
3. The molar solubilities of MY and NY₃ in water are identical.
4. The molar solubility of MY in water is less than that of NY₃

Answer: 4. The molar solubility of MY in water is less than that of NY₃

For \(M Y: K_{s p}=s_1^2\)

⇒ \(s_1=\sqrt{K_{s p}}=\sqrt{6.2 \times 10^{-13}}=7.87 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

For \(N Y_3: K_{s p}=\left(s_2\right)\left(3 s_2\right)^3=27 s_2^4\)

⇒ \(s_2=\sqrt[4]{\frac{6.2 \times 10^{-13}}{27}}=3.89 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)

Hence, the molar solubility of MY in water is less than that of NY₃.

Question 105. The K_sp of Ag₂CrO₄, AgCl, AgBr and Agl are respectively, 1.1 x 10^-12, 1.8 x 10^-10, 5.0 x 10^-13, 8.3 x 10^-17. Which one of the following salts will precipitate last if AgNO₃ solution is added to the solution containing equal moles of NaCl, NaBr, Nal and Na₂CrO₄?

1. AgBr
2. Ag₂CrO₄
3. Agl
4. AgCl

Answer: 2. Ag₂CrO₄

⇒ \(\begin{array}{lcc}
Salt & {1}{c}{K_{s p}} & Solubility \\
\mathrm{Ag}_2 \mathrm{CrO}_4 & 1.1 \times 10^{-12}=4 s^3 & s=\sqrt[3]{\frac{K_{s p}}{4}}=0.65 \times 10^{-4} \\
\mathrm{AgCl} & 1.8 \times 10^{-10}=s^2 & s=\sqrt{K_{s p}}=1.34 \times 10^{-5} \\
\mathrm{AgBr} & 5 \times 10^{-13}=s^2 & s=\sqrt{K_{s p}}=0.71 \times 10^{-6} \\
\mathrm{AgI} & 8.3 \times 10^{-17}=s^2 & s=\sqrt{K_{s p}}=0.9 \times 10^{-8}
\end{array}\)

The solubility of Ag₂CrO₄ is highest thus, it will be precipitated at last.

Question 106. Using the Gibbs’ energy change, ΔG° = +63.3 kJ, for the following reaction, \(\mathrm{Ag}_2 \mathrm{CO}_{3(s)} \rightleftharpoons 2 \mathrm{Ag}_{(a q)}^{+}+\mathrm{CO}_3^{2-}{ }_{(a q)}\) the K_sp of Ag₂CO_3(s) in water at 25 °C is

(R = 8.314 J K^-1 mol^-1)

1. 3.2 x10^-26
2. 8.0 x 10^-12
3. 2.9 x10^-3
4. 7.9 x 10^-2

Answer: 2. 8.0 x 10^-12

ΔG° = -2.303 RT logK_sp

63.3 x 10³ J = – 2.303 x 8.314 x 298 log K_sp

63.3 x 10³ J = -5705.84 logK_sp

⇒ \(\log K_{s p}=-\frac{63.3 \times 10^3}{5705.84}=-11.09\)

⇒ \(K_{s p}={antilog}(-11.09)=8.128 \times 10^{-12}\)

Question 107. The values of K of CaCO₃ and CaC₂O₄ are 4.7 x 10^-9 and 1.3 x 10^-9 respectively at 25°C. If the mixture of these two is washed with water, what is the concentration of Ca²⁺ ions in water?

1. 5.831 x 10^-5 M
2. 6.856 x 10^-5 M
3. 3.606 x 10^-5 M
4. 7.746 x 10^-5 M

Answer: 4. 7.746 x 10^-5 M

⇒ \(\mathrm{CaCO}_3 \rightarrow \underset{x}{\mathrm{Ca}^{2+}}+\underset{x}{\mathrm{CO}_3^{2-}}\)

⇒ \(\mathrm{CaC}_2 \mathrm{O}_4 \rightarrow \underset{y}{\mathrm{Ca}^{2+}}+\underset{y}{\mathrm{C}_2 \mathrm{O}_4^{2-}}\)

Now, \(\left[\mathrm{Ca}^{2+}\right]=x+y\)

and \(x(x+y)=4.7 \times 10^{-9}\)…….(1)

y(x+y)=1.3\(\times 10^{-9}\)…..(2)

Dividing equations (1) and (2) we get \(\frac{x}{y}=3.6\)

∴ x = 3.6 y

Putting this value in equation (2), we get \(y(3.6 y+y)=1.3 \times 10^{-9}\)

On solving, we get \(y=1.68 \times 10^{-5}\)

and \(x=3.6 \times 1.68 \times 10^{-5}=6.05 \times 10^{-5}\)

∴ \(\left[\mathrm{Ca}^{2+}\right]=(x+y)=\left(1.68 \times 10^{-5}\right)+\left(6.05 \times 10^{-5}\right)\)

∴ \(\left[\mathrm{Ca}^{2+}\right]=7.73 \times 10^{-5} \mathrm{M}\)

Question 108. Identify the correct order of solubility in aqueous medium.

1. Na₂S > CuS > ZnS
2. Na₂S > ZnS > CuS
3. CuS > ZnS > Na₂S
4. ZnS > Na₂S > CuS

Answer: 2. Na₂S > ZnS > CuS

Sodium sulphide is soluble in water. The solubility product (and hence solubility) of ZnS is larger than that of CuS

Question 109. pH of a saturated solution of Ba(OH)₂ is 12. The value of the solubility product (K ) of Ba(OH)₂ is

1. 3.3 x 10^-7
2. 5.0 x 10^-7
3. 4.0 x 10^-6
4. 5.0 x 10^-6

Answer: 2. 5.0 x 10^-7

pH of solution =12

⇒ \({\left[\mathrm{H}^{+}\right]=10^{-12}}\)

⇒ \({\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{10^{-12}}=10^{-2}}\)

⇒ \(\mathrm{Ba}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}\)

2s = \(10^{-2} \Rightarrow s=\frac{10^{-2}}{2}\)

⇒ \(K_{s p}=(s)(2 s)^2=4 s^3\)

= \(4 \times\left(\frac{10^{-2}}{2}\right)^3=\frac{4}{8} \times 10^{-6}=5 \times 10^{-7}\)

Question 110. In qualitative analysis, the metals of group 1 can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag⁺ and Pb²⁺ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl^– concentration is 0.10 M. What will the concentrations of Ag⁺ and Pb²⁺ be at equilibrium?

(\(K_{s p} \text { for } \mathrm{AgCl}=1.8 \times 10^{-10}, K_{s p} \text { for } \mathrm{PbCl}_2=1.7 \times 10^{-5}\))

1. \(\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}\)
   
2. \(\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-5} \mathrm{M}\)
   
3. \(\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}\)
   
4. \(\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-4} \mathrm{M}\)

Answer: 3. \(\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}\)

⇒ \(K_{s p}[\mathrm{AgCl}]=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\)

⇒ \({\left[\mathrm{Ag}^{+}\right]=\frac{1.8 \times 10^{-10}}{10^{-1}}=1.8 \times 10^{-9} \mathrm{M}}\)

⇒ \(K_{s p}\left[\mathrm{PbCl}_2\right]=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2\)

⇒ \({\left[\mathrm{~Pb}^{2+}\right]=\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}}=1.7 \times 10^{-3} \mathrm{M}}\)

Question 111. H₂S gas when passed through a solution of cations containing HCl precipitates the cations of the second group of qualitative analysis but not those belonging to the fourth group. It is because

1. The presence of HCl decreases the sulphide ion concentration
2. The solubility product of group 2 sulphides is more than that of group 4 sulphides
3. The presence of HCl increases the sulphide ion concentration
4. Sulphides of group 4 cations are unstable in HCl.

Answer: 1. Presence of HCl decreases the sulphide ion concentration

The cations of group 2 are precipitated as their sulphides.

The solubility product of sulphide of group 2 radicals is very low. Therefore, even with a low conc. of S^2- ions, the ionic product exceeds the value of their solubility product and the radicals of group 2 get precipitated. The low. of S^2- ions is obtained by passing H₂S gas through the solution of the salt in the presence of oil. HCI suppresses the degree of ionisation of H₂S by the common ion effect.

Note that the solubility product of group 4 radicals is quite high. It is necessary to suppress the conc. of S^2- ions, otherwise radicals of group 4 will also get precipitated along with group 2 radicals.

Question 112. The solubility product of a sparingly soluble salt AX₂ is 3.2 x 10^-11. Its solubility (in moles/L) is

1. 5.6 x 10^-6
2. 3.1 x 10^-4
3. 2 x 10^-4
4. 4 x 10^-4

Answer: 3. 2 x 10^-4

⇒ \(K_{s p}=3.2 \times 10^{-11}\)

⇒ \(A X_2 \rightleftharpoons \underset{s}{A^{2+}}+\underset{2 s}{2 X^{-}}\)

⇒ \(K_{s p}=s \times(2 s)^2=4 s^3 ; \text { i.e., } 3.2 \times 10^{-11}=4 s^3\)

or, \(s^3=0.8 \times 10^{-11}=8 \times 10^{-12}\)

s = \(2 \times 10^{-4}\)

Question 113. The solubility product of Agl at 25°C is 1.0 x 10^-16 mol² L^-2. The solubility of Agl in 10^-4 N solution of KI at 25°C is approximately (in mol L^-1)

1. 1.0 x 10^-16
2. 1.0 x 10^-12
3. 10 x 10^-10
4. 1.0 x 10^-8

Answer: 2. 1.0 x 10^-12

⇒ \(AgI \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{I}^{-}\)

⇒ \(\mathrm{KI} \rightleftharpoons \underset{10^{-4} \mathrm{M}}{\mathrm{K}^{+}}+\underset{10^{-4} \mathrm{M}}{\mathrm{I}^{-}}\)

(For \(KI, 1 \mathrm{~N}=1 \mathrm{M}\))

⇒ \({\left[\mathrm{I}^{-}\right]=s+10^{-4}}\)

⇒ \(K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]\)

⇒ \(1 \times 10^{-16}=s\left(s+10^{-4}\right) \Rightarrow 1 \times 10^{-16}=s^2+10^{-4} s\)

⇒ \(1 \times 10^{-16}=10^{-4} s\)

s = \(\frac{1 \times 10^{-16}}{10^{-4}}=1 \times 10^{-12} \mathrm{~mol} \mathrm{~L}^{-1}\) (because \(s^2<<<10^{-4} s\))

Question 114. The solubility of MX₂ type electrolytes is 0.5 x 10^-4 mol/lit., then find out K_sp of electrolytes.

1. 5 x 10^-12
2. 25 x 10^-10
3. 1 x 10^-13
4. 5 x 10^-13

Answer: 4. 5 x 10^-13

Let s be the solubility of the electrolyte MX₂.

⇒ \(\left[M^{2+}\right]=s,\left[X^{-}\right]=2 s\)

Solubility product, \(K_{s p}=s \times(2 s)^2=4 s^3;s=0.5 \times 10^{-4} \mathrm{~mol} /\) litre

∴ \(K_{s p}=4 \times\left(0.5 \times 10^{-4}\right)^3;K_{s p}=5 \times 10^{-13}\)

Question 115. The solubility of M₂S salt is 3.5 x 10^-6 then find out the solubility product.

1. 1.7 x 10^-6
2. 1.7 x 10^-16
3. 1.7 x 10^-18
4. 1.7 x 10^-12

Answer: 2. 1.7 x 10^-16

For reaction, \(M_2 \mathrm{~S} \rightleftharpoons \underset{2 s}{2 M^{+}}+\underset{s}{\mathrm{~S}^{2-}}\)

Solubility \(=3.5 \times 10^{-6}\)

Solubility product, \(K_{s p}=\left[M^{+}\right]^2\left[\mathrm{~S}^{2-}\right]\)

= \((2 s)^2 s=4 s^3=4 \times\left(3.5 \times 10^{-6}\right)^3=1.7 \times 10^{-16}\)

Question 116. The solubility of a saturated solution of calcium fluoride is 2 x 10^-4 moles per litre. Its solubility product is

1. 22 x 10^-11
2. 14 x 10^-4
3. 2 x 10^-2
4. 32 x 10^-12

Answer: 4. 32 x 10^-12

For \(\mathrm{CaF}_2\), decomposition is as follows:

⇒ \(\mathrm{CaF}_2 \rightarrow \underset{s}{ } \mathrm{Ca}^{2+}+\underset{2 s}{2 \mathrm{~F}^{-}}\)

⇒ \(K_{s p}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2=s \times(2 s)^2\)

or \(K_{s p}=4 s^3 \Rightarrow K_{s p}=4 \times\left(2 \times 10^{-4}\right)^3 \Rightarrow K_{s p}=32 \times 10^{-12}\)

Question 117. The solubility products of CuS, Ag₂S and HgS are 10^-31, 10^-44 and 10^-54 respectively. The solubilities of these sulphides are in the order

1. HgS > Ag₂S > CuS
2. CuS > Ag₂S > HgS
3. Ag₂S > CuS > HgS
4. Ag₂S > HgS > CuS

Answer: 2. CuS > Ag₂S > HgS

The greater the solubility product, the greater is the solubility.

Question 118. The solubility of AgCl will be minimal in

1. 0.01 M CaCl₂
2. Pure water
3. 0.001 M AgNO₃
4. 0.01M NaCl

Answer: 1.0.01 M CaCl₂

There are a greater number of Cl^– ions in CaCl₂ compared to others. Hence, the solubility of AgCl will be minimal in 0.01M CaCl₂ due to the common ion effect.

Question 119. Which one of the following is most soluble?

1. \(\mathrm{Bi}_2 \mathrm{~S}_3\left(K_{s p}=1 \times 10^{-70}\right)\)
2. \(\mathrm{Ag}_2 \mathrm{~S}\left(K_{s p}=6 \times 10^{-51}\right)\)
3. \({CuS}\left(K_{s p}=8 \times 10^{-37}\right)\)
4. \({MnS}\left(K_{s p}=7 \times 10^{-16}\right)\)

Answer: 4. \({MnS}\left(K_{s p}=7 \times 10^{-16}\right)\)

The higher the value of the solubility product, the greater the solubility.

 

 

Hydrogen

Question 1. One would expect a proton to have a very large

1. Charge
2. Ionization potential
3. Hydration energy
4. Radius.

Answer: 3. Hydration energy

Proton (H⁺) ion being very small in size would have very large hydration energy

Question 2. The ionization of hydrogen atoms would give rise to

1. Hydride ion
2. Hydronium ion
3. Proton
4. Hydroxyl ion.

Answer: 3. Proton

It gives rise to Proton, \(\mathrm{H}_{(g)} \rightarrow \underset{\text { Tritium }}{\mathrm{H}_{(a q)}^{+}}+e^{-}\)

Question 3. Tritium a radioactive isotope of hydrogen, emits which of the following particles?

1. Neutron(n)
2. Beta(β^–)
3. Alpha (α)
4. Gamma(γ)

Answer: 2. Beta(β^–)

Tritium is a beta particle emitting radioactive isotope of hydrogen.

⇒ \(\underset{\text { Tritium }}{{ }_1^3 \mathrm{H}} \longrightarrow{ }_2^3 \mathrm{He}+\underset{\beta-\text {-particle }}{-1}{ }^0 e+0\)

Question 4. Which one of the following pairs of substances on reaction will not evolve H₂ gas?

1. Copper and HCl (aqueous)
2. Iron and steam
3. Iron and H₂SO₄ (aqueous)
4. Sodium and ethyl alcohol

Answer: 1. Copper and HCl (aqueous)

Copper is a noble metal, as it lies below hydrogen in the electrochemical series. Therefore, it cannot displace hydrogen from dilute HCl. Iron and sodium lie above hydrogen in the electrochemical series, so they can liberate H₂ either from steam or H₂SO₄ solution.

⇒ \(\mathrm{C}_2 \mathrm{H}_5-\mathrm{OH}+\mathrm{Na} \rightarrow \mathrm{C}_2 \mathrm{H}_5-\mathrm{ONa}+1 / 2 \mathrm{H}_2\)

⇒ \(\mathrm{Fe}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{FeSO}_4+\mathrm{H}_2\)

⇒ \(3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2\)

Read and Learn More NEET MCQs with Answers

Question 5. Water gas is produced by

1. Passing steam through a red-hot coke
2. Saturating hydrogen with moisture
3. Mixing oxygen and hydrogen in a ratio of 1: 2
4. Heating a mixture of CO₂ and CH₄ in petroleum refineries.

Answer: 1. Passing steam through a red-hot coke

⇒ \(\underset{\text { Steam }}{\mathrm{H}_2 \mathrm{O}}+\underset{\text { Red hot }}{\mathrm{C}} \longrightarrow \underbrace{\mathrm{H}_2+\mathrm{CO}}_{\text {Water gas }}\)

Question 6. Which of the following metals evolves hydrogen on reacting with cold dilute HNO₃?

1. Mg
2. Al
3. Fe
4. Cu

Answer: 1. Mg

Mg reacts with nitric acid to give Mg(NO₃)₂ and evolves H₂ gas \(\mathrm{Mg}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2\)

Question 7. Which of the following statements about hydrogen is incorrect?

1. Hydronium ion, H₃O⁺ exists freely in solution.
2. Dihydrogen does not act as a reducing agent.
3. Hydrogen has three isotopes of which tritium is the most common.
4. Hydrogen never acts as a cation in ionic salts.

Answer: 2. Dihydrogen does not act as a reducing agent. and 3. Hydrogen has three isotopes of which tritium is the most common.

⇒ \(\mathrm{CuO}+\mathrm{H}_2 \rightarrow \mathrm{Cu}+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{ZnO}+\mathrm{H}_2 \rightarrow \mathrm{Zn}+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2 \rightarrow 3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O}\)

Hydrogen has three isotopes of which protium is the most common and tritium is radioactive.

Question 8. Match List 1 with List 2.

Choose the correct answer from the options given below:

1. (1) -(D), (2) – (A), (3) – (B), (4) – (C)
2. (1) -(C), (2) – (A), (3) – (B), (4) – (D)
3. (1) -(A), (2) – (B), (3) – (D), (4) – (C)
4. (1) -(B), (2) – (C), (3) – (D), (4) – (A)

Answer: 1. (1) -(D), (2) – (A), (3) – (B), (4) – (C)

MgH₂ – Ionic hydride

GeH₄ – Electron Precise hydride

B₂H₆ – Electron-deficient hydride

HF – Eiectron rich hydride

Question 9. Which of the following is electron-deficient? 

1. (BH₃)₂
2. PH₃
3. (CH₃)₂
4. (SiH₃)₂

Answer: 1. (BH₃)₂

Boron hydrides are electron-def,cient compounds.

Question 10. The method used to remove the temporary hardness of water is

1. Synthetic resins method
2. Calgon’s method
3. Clark’s method
4. Ion-exchange method.

Answer: 3. Clark’s method

Clarks process is used to remove the temporary hardness of the water. In this method, quick lime is added. The bicarbonates present in temporary hard water react with lime water to form insoluble calcium and magnesium carbonates which can be easily filtered off.

⇒ \(\underset{\text{Quick lime}}{\mathrm{CaO}}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text{Lime water}}{{\mathrm{Ca}{\mathrm{OH}}_2}}\)

⇒ \(\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow 2 \mathrm{CaCO}_3 \downarrow+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Mg}\left(\mathrm{HCO}_3\right)_2+2 \mathrm{Ca}(\mathrm{OH})_2 \longrightarrow 2 \mathrm{CaCO}_3 \downarrow+\mathrm{Mg}(\mathrm{OH})_2 \downarrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 11. The number of hydrogen-bonded water molecule(s) associated with CuSO₄. 5H₂O is

1. 3
2. 1
3. 2
4. 5

Answer: 2. 1

The ionic formulation of CuSO₄ · 5H₂O is [Cu(H₂O)₄]H₂O · SO₄, in which four H₂O molecules are coordinated to a central Cu²⁺ ion while the fifth H₂O molecule is hydrogen bonded to the sulfate group.

Question 12. Which of the following groups of ions makes the water hard?

1. Sodium and bicarbonate
2. Magnesium and chloride
3. Potassium and sulphate
4. Ammonium and chloride

Answer: 2. Magnesium and chloride

The hardness of water is due to the presence of bicarbonates, chlorides, and sulfates of Ca and Mg. Hence, hard water will consist of Mg²⁺ and Cl^– ions.

Question 13. At its melting point, ice is lighter than water because

1. H₂O molecules are more closely packed in a solid state
2. Ice crystals have a hollow hexagonal arrangement of H₂O molecules
3. On melting of ice, the H₂O molecules shrink in size
4. Ice forms mostly heavy water on first melting.

Answer: 2. Ice crystals have a hollow hexagonal arrangement of H₂O molecules

In ice crystals, water molecules are linked through H-bonds in a hollow hexagonal arrangement so, the volume is large and the density is less. In a liquid state, this hollow arrangement breaks into closer arrangements of molecules. Consequently, the density is increased in a liquid state.

Question 14. Match the following and identify the correct option.

1. 1-C; 2-A; 3-B; 4-A
2. 1-C; 2-B; 3-A; 4-D
3. 1-C; 2-D; 3-B; 4-A
4. 1-A; 2-C; 3-B; 4-D

Answer: 1. 1-C; 2-A; 3-B; 4-A

Question 15. The structure of H₂O₂ is

1. Spherical
2. Non-planar
3. Planar
4. Linear.

Answer: 2. Non-planar

Hydrogen peroxide has a non-planar structure

Question 16. The volume strength of 1.5 N H₂O₂ solution is

1. 8.8
2. 8.4
3. 4.8
4. 5.2

Answer: 2. 8.4

Normality (N) = 1.5

We know that the equivalent weight of H₂O₂ is 17 and the strength of H₂O₂= Normality x Equivalent weight

= 1.5 x 17 = 25.5

⇒ \(\underset{(2 \times 34=68 \mathrm{~g})}{2 \mathrm{H}_2 \mathrm{O}_2} \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\underset{(22.4 \mathrm{~L})}{\mathrm{O}_2}\)

Since 68 grams of H₂O₂, produces 22.4 litres of oxygen at NTP therefore, 25.5 grams of H₂O₂, will produce = 22.4/68 x 25.5 = 8.4 litre of hydrogen

Thus, the volume strength of the given H₂O₂, solution is 8.4

Question 17. The O – O – H bond angle in H₂O₂ is

1. 106°
2. 109°28′
3. 120°
4. 97°

Answer: 4. 97°

The bond angle of O – O – H in H₂O₂, is 97°

Question 18. Hydrogen peroxide molecules are

1. Monoatomic and form X^2-₂ ions
2. Diatomic and form X^– ions
3. Diatomic and form X₂^– ions
4. Monoatomic and form X^– ions

Answer: 2. Diatomic and form X^– ions

H₂O₂ is diatomic and forms H⁺ + HO₂^– (X^–) (hydroperoxide ion).

Question 19. Which of the following is the true structure of H₂O₂?

Answer: 2

is the true structure of H₂O₂

Question 20. The reaction of H₂O₂ with H₂S is an example of a reaction.

1. Addition
2. Oxidation
3. Reduction
4. Acidic

Answer: 2. Oxidation

It is an example of an oxidation reaction.

⇒ \(\mathrm{H}_2 \mathrm{~S}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{S}\)

⇒ \(\mathrm{H}_2 \mathrm{O}_2\) oxidises \(\mathrm{H}_2 \mathrm{~S}\) into \(\mathrm{S}\).

Question 21. Which of the following statements is not correct?

1. Hydrogen is used to reduce heavy metal oxides to metals.
2. Heavy water is used to study the reaction mechanism.
3. Hydrogen is used to make saturated fats from oils.
4. The H-H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any element.
5. Hydrogen reduces oxides of metals that are more active than iron.

Choose the most appropriate answer from the options given below:

1. 4, 5 only
2. 1, 2, 3 only
3. 2, 3, 4, 5 only
4. 2, 4 only

Answer: 1. 4, 5 only

Statements 4 and 5 are incorrect.

The H-H bond dissociation enthalpy is the highest for a single bond between two atoms of any element. Hydrogen reduces oxides of metals that are less reactive than iron.

Question 22. Some statements about heavy water are given below,

1. Heavy water is used as a moderator in nuclear reactors.
2. Heavy water is more associated than ordinary water.
3. Heavy water is a more effective solvent than ordinary water.

Which of the above statements is correct?

1. (1) and (2)
2. (1), (2), and (3)
3. (2) and (3)
4. (1) and (3)

Answer: 1. (1) and (2)

Heavy water is used for slowing down the speed of neutrons in nuclear reactors, hence used as a moderator. The boiling point of heavy water is greater (37 4.42 K) than that of ordinary water (373 K), hence heat water is more associated. The Dielectric constant of ordinary water is greater than that of heavy water, hence ordinary water is a better solvent.

Environmental Chemistry

Question 1. The pollution due to oxides of sulphur gets enhanced due to the presence of

1. Particulate matter
2. Ozone
3. Hydrocarbons
4. Hydrogen peroxide

Choose the most appropriate answer from the options given below:

1. (1), (4) only
2. (1), (2), (4) only
3. (2), (3), (4) only
4. (1), (3), (4) only

Answer: 2. (1), (2), (4) only

The presence of particulate matter in polluted air catalyses the oxidation of sulphur dioxide to sulphur trioxide.

⇒ \(2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{SO}_{3(g)}\)

The reaction can also be promoted by ozone and hydrogen peroxide.

⇒ \(\mathrm{SO}_{2(g)}+\mathrm{O}_{3(g)} \rightarrow \mathrm{SO}_{3(g)}+\mathrm{O}_{2(g)}\)

⇒ \(\mathrm{SO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{2(l)} \rightarrow \mathrm{H}_2 \mathrm{SO}_{4(a q)}\)

Question 2. Match List-1 with List-2

Choose the correct answer from the options given below.

1. (1) – (C), (2) – (B), (3) – (D), (4) – (A)
2. (1) – (A), (2) – (B), (3) – (C), (4) – (D)
3. (1) – (B), (2) – (C), (3) – (D), (4) – (A)
4. (1) – (D), (2) – (C), (3) – (A), (4) – (B)

Answer: 4. (1) – (D), (2) – (C), (3) – (A), (4) – (B)

Photochemical smog: \(\mathrm{NO}_{2(g)}\) \(\longrightarrow{h v}\) \(\mathrm{NO}_{(g)}+\mathrm{O}_{(g)}\)

Ozone depletion: \(\mathrm{HOCl}_{(g)}\) \(\longrightarrow{h \mathrm{v}}\) \(\dot{\mathrm{O}} \mathrm{H}+\dot{\mathrm{Cl}}\)

Acid rain: \(\mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CaSO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

Tropospheric pollution: \(2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{SO}_{3(\mathrm{~g})}\)

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Question 3. Which of the following is not correct about carbon monoxide?

1. It forms carboxyhaemoglobin.
2. It reduces oxygen oxygen-carrying ability of blood.
3. The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin.
4. It is produced due to incomplete combustion.

Answer: 3. It is produced due to incomplete combustion.

The carboxyhaemoglobin is about 300 times more stable than oxyhaemoglobin.

Question 4. Among the following, the one that is not a greenhouse gas is

1. Sulphur dioxide
2. Nitrous oxide
3. Methane
4. Ozone.

Answer: 1. Sulphur dioxide

Besides carbon dioxide, other greenhouse gases are methane, water vapours, nitrous oxide, CFCs and ozone.

Question 5. Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?

1. N₂O₅
2. NO₂
3. N₂O
4. NO

Answer: 1. N₂O₅

Question 6. Which of the following is a sink for CO?

1. Microorganisms present in the soil
2. Oceans
3. Plants
4. Haemoglobin

Answer: 1. Microorganisms present in the soil

Microorganisms present in the soil consume atmospheric CO

Question 7. Which one of the following is not a common component of photochemical smog?

1. Ozone
2. Acrolein
3. Peroxyacetyl nitrate
4. Chlorofluorocarbons

Answer: 4. Chlorofluorocarbons

Question 8. Which one of the following statements regarding photochemical smog is not correct?

1. Carbon monoxide does not play any role in photochemical smog formation.
2. Photochemical smog is an oxidising agent in character.
3. Photochemical smog is formed through a photochemical reaction involving solar energy.
4. Photochemical smog does not cause

Answer: 4. Photochemical smog does not cause

Photochemical smog causes irritation in the eyes and throat.

Question 9. Which one of the following is responsible for the depletion of the ozone layer in the upper strata of the atmosphere?

1. Polyhalogens
2. Ferrocene
3. Fullerenes
4. Freons

Answer: 4. Freons

Chlorofluorocarbons such as freon-11 (CFCI₃) and freon-12 (CF₂Cl₂) emitted as propellants in aerosol spray cans, refrigerators, fire fighting reagents etc. are stable compounds and chemically inert.

They do not react with, any substance with which they come in contact and thus float through the atmosphere unchanged and eventually enter the stratosphere.

There they absorb UV radiation and break down liberating free atomic chlorine which causes the decomposition of ozone. This results in the depletion of the ozone layer.

⇒ \(\dot{\mathrm{Cl}}+\mathrm{O}_3 \rightarrow \mathrm{ClO}+\mathrm{O}_2 ; \mathrm{ClO}+\mathrm{O}_3 \rightarrow \dot{\mathrm{Cl}}+2 \mathrm{O}_2\)

Question 10. About 20 km above the earth, there is an ozone layer. Which one of the following statements about ozone and the ozone layer is true?

1. It is beneficial to us as it stops UV radiation.
2. Conversion of O₃ to O₂ is an endothermic reaction.
3. Ozone is a triatomic linear molecule.
4. It is harmful as it stops useful radiation. (1995)

Answer: 1. It is beneficial to us as it stops UV radiation.

The ozone layer is very beneficial to us because it stops harmful ultraviolet radiation from reaching the Earth.

Question 11. Given below are two statements.

Statement 1: The nutrient-deficient water bodies lead to eutrophication.

Statement 2: Eutrophication leads to a decrease in the level of oxygen in the water bodies.

In the light of the above statements, choose the correct answer from the options given below:

1. Statement 1 is correct but Statement 2 is false.
2. Statement 1 is incorrect but Statement 2 is true.
3. Both Statement 1 and Statement 2 are true.
4. Both Statement 1 and Statement 2 are false.

Answer: 2. Statement 1 is incorrect but Statement 2 is true.

The process in which nutrient-enriched water bodies support a dense plant population, which kills animal litter by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication.

Question 12. Which one of the following statements is not true?

1. Clean water would have a BOD value of 5 ppm.
2. Fluoride deficiency in drinking water is harmful. Soluble fluoride is often used to bring its concentration up to 1 ppm.
3. When the pH of rainwater is higher than 6.5, it is called acid rain.
4. Dissolved Oxygen (DO) in cold water can reach a concentration of up to 10 ppm.

Answer: 3. When the pH of rainwater is higher than 6.5, it is called acid rain.

When the pH of rainwater drops below 5.6 it is called acid rain.

Question 13. Which one of the following statements is not true?

1. pH of drinking water should be between 5.5 and 9.5.
2. The concentration of DO below 6 ppm is good for the growth of fish.
3. Clean water would have a BOD value of less than 5 ppm.
4. Oxides of sulphur, nitrogen and carbon, are the most widespread air pollutant.

Answer: 2. The concentration of DO below 6 ppm is good for the growth of fish.

Fish flies in water bodies polluted by sewage due to a decrease in dissolved oxygen (D.O.)

Question 14. Green chemistry means such reactions which

1. Are related to the depletion of the ozone layer
2. Study the reactions in plants
3. Produce colour during reactions
4. Reduce the use and production of hazardous chemicals.

Answer: 4. Reduce the use and production of hazardous chemicals.

Green chemistry is the design, development, and implementation of chemical products and processes to reduce or eliminate the use and generation of substances hazardous to human health and the environment.

Green chemistry also refers to the redesign of chemical products and processes with the goal of reducing or eliminating any negative environmental or health effects.

Alcohols Phenols And Ethers

Question 1. The general molecular formula, which represents the homologous series of alkanols is

1. \(\mathrm{C}_n \mathrm{H}_{2 n+2} \mathrm{O}\)
2. \(\mathrm{C}_n \mathrm{H}_{2 n} \mathrm{O}_2\)
3. \(\mathrm{C}_n \mathrm{H}_{2 n} \mathrm{O}\)
4. \(\mathrm{C}_n \mathrm{H}_{2 n+1} \mathrm{O}\)

Answer: 1. \(\mathrm{C}_n \mathrm{H}_{2 n+2} \mathrm{O}\)

All alcohols follow the general formula \(\mathrm{C}_n \mathrm{H}_{2 \pi+2} \mathrm{O}\).

⇒ \(\begin{gathered}
\mathrm{CH}_3 \mathrm{OH}\left[\mathrm{CH}_{2+2} \mathrm{O}\right] ; \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\left[\mathrm{C}_2 \mathrm{H}_{(2 \times 2)+2} \mathrm{O}\right] \\
n=1, \quad n=2
\end{gathered}
\)

Question 2. Consider the following reaction and identify the product (P).

Answer: 3

Question 3. Which of the following will be most readily dehydrated under acidic conditions?

Answer: 4

Question 4. Given below are two statements:

Statement-1: The acidic strength of monosubstituted nitrophenol is higher than phenol because of the electron-withdrawing nitro group.

Statement 2: 0-nitrophenol, m-nitrophenol, and p-nitrophenol will have the same acidic strength as they have one nitro group attached to the phenolic ring. In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both statement-1 and statement-2 are correct.
2. Both statement-1 and statement-2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but Statement 2 is

Answer: 3. Statement 1 is correct but statement 2 is incorrect.

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Electron withdrawing groups (for example, -NO₂) stabilize the phenoxide ion more by dispersing the negative charge relative to phenol (i.e. release of proton becomes easy) and thus, increase the acidic strength of phenols.

The particular effect is more when the substituent is present on o- and p-positions than in re-position to the phenolic group.

Thus, the acidic strength of nitrophenols decreases in the order: p-nitrophenol > o-nitrophenol > z-nitrophenol > phenol

Question 5. Given below are two statements:

Statement 1: In the Lucas test, primary, secondary, and tertiary alcohols are distinguished on the basis of their reactivity with the cone. HCl + ZnCl₂, known as Lucas Reagent.

Statement 2: Primary alcohols are most reactive and immediately produced turbidity at room temperature on reaction with Lucas Reagent.

In the light of the above statements, choose the most appropriate answer from the options given below.

1. Both statement 1 and statement 2 are correct.
2. Both statement 1 and statement 2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but statement 2 is correct.

Answer: 3. Statement 1 is correct but statement 2 is incorrect.

Tertiary alcohols are most reactive and immediately produce turbidity at room temperature while primary alcohols do not react with Lucas reagent at room temperature.

Question 6. The reaction between acetone and methyl magnesium chloride followed by hydrolysis will give

1. isopropyl alcohol
2. sec-butyl alcohol
3. fert-butyl alcohol
4. iso-butyl alcohol.

Answer: 3. fert-butyl alcohol

Question 7. The structure of intermediate A in the following reaction is

Answer: 3

Question 8. When vapors of a secondary alcohol is passed over heated copper at 573 K, the product formed is

1. A carboxylic acid
2. An aldehyde
3. A ketone
4. An alkene.

Answer: 3. A ketone

Question 9. In the reaction,

the electrophile involved is

1. Dichloromethyl cation \(\left({ }_{\mathrm{C}}^{+} \mathrm{HCl}_2\right)\)
2. Formyl cation \((\stackrel{\rightharpoonup}{\mathrm{C}} \mathrm{HO})\)
3. Dichloromethyl anion \(\left(\overline{\mathrm{C}} \mathrm{HCl}_2\right)\)
4. Dichlorocarbene \((:\left.\mathrm{CCl}_2\right)\)

Answer: 4. Dichlorocarbene \((:\left.\mathrm{CCl}_2\right)\)

It is Reimer-Tiemann reaction. The electrophile formed is dichlorocarbene (CCl) which is formed according to the following mechanism

Question 10. Compound A, C₈H₁₀O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with a characteristic smell. A and Y are respectively

Answer: 3

As the compound is giving yellow precipitate with NaOI that shows it is undergoing a haloform reaction. Haloform reaction is shown by the compounds having

Question 11. Identify the major products P, Q, and R in the following sequence of reactions

Answer: 4

Question 12. Which one is the most acidic compound?

Answer: 3

Electron withdrawing groups increase the acidity while electron donating groups decrease the acidity of phenol.

Question 13. The reaction of phenol with chloroform in the presence of dilute sodium hydroxide finally introduces which one of the following functional groups?

1. -COOH
2. -CHCl₂
3. -CHO
4. -CH₂Cl

Answer: 3. -CHO

This is Reimer-Tiemann reaction.

Question 14. Which of the following reaction(s) can be used for the preparation of alkyl halides?

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HCl}\) \(\underrightarrow{\mathrm{Anh} . \mathrm{ZnCl} l_2}\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HCl} \longrightarrow\)
3. \(\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{HCl} \longrightarrow\)
4. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH}+\mathrm{HCl}\) \(\underrightarrow{\text { Anh. } \mathrm{ZnCl}_2}\)

1. 1 and 2 only
2. 4 only
3. 3 and 4 only
4. 1, 3 and 4 only

Answer: 4. 1, 3 and 4 only

1° and 2° alcohols react with HCI in the presence of an anhydrous ZICI₂ as a catalyst while in the case of 3° alcohols, ZnCI₂ is not required.

Quetsion 15. Which of the following will not be soluble in sodium hydrogen carbonate?

1. 2,4,6-Trinitrophenol
2. Benzoic acid
3. o-Nitropbenol
4. Benzene sulphonic acid

Answer: 3. o-Nitropbenol

The reaction is as follows: Acid + \(\mathrm{NaHCO}_3 \rightarrow\) Sodium salt of acid+ \(\mathrm{H}_2 \mathrm{CO}_3\)

Among all the given compounds, o-nitrophenol is a weaker acid than \(\mathrm{HCO}_3^{-}\). Hence, it does not react with \(\mathrm{NaHCO}_3\).

Question 16. The number of isomeric alcohols of molecular formula C₆H₁₄O which give positive iodoform test is

1. Three
2. Four
3. Five
4. Two.

Answer: 2. Four

The iodoform test is positive for alcohols with the formula R – CHOH – CH₃.

Among C₆H₁₄O isomers, the ones with positive iodoform tests are:

Question 17. In the following sequence of reactions, the end product (C) is

1. Acetone
2. Methane
3. Acetaldehyde
4. Ethyl alcohol.

Answer: 4. Ethyl alcohol.

Quetsion 18. In the following reactions, the major products (A) and (C) are respectively

Answer: 2

Quetsion 19. Given are cyclohexanol(1), acetic acid(2), 2,4,6-trinitrophenol (3), and phenol (4). In these, the order of decreasing acidic character will be

1. 3>2>4>1
2. 2>3>1>4
3. 2>3>4>1
4. 3>4>2>1

Answer: 1. 3>2>4>1

Since phenols and carboxylic acids are more acidic than aliphatic alcohols, we find that cyclohexanol (1) is the least acidic. Out of the two given phenols, 3 is more acidic than 4.

This is because of the presence of three highly electron-withdrawing -NO₂ groups on the benzene ring which makes the O-H bond extremely polarized. This facilitates the release of H as H⁺.

Thus, 3 and 4.

ln acetic acid, the electron-withdrawing in the -COOH group polarises the O-H bond and increases the acidic strength. Acetic acid is therefore more acidic than phenol or cyclohexanol.

∴ The order of acidic character is 3 > 2 > 4 > 1

Question 20. Which of the following compounds has the most acidic nature?

Answer: 2

Phenol is the most acidic of all the given compounds. In phenol, the electron-withdrawing phenyl ring polarizes the O-H bond, thereby facilitating the release of H as H⁺ and hence, phenol is most acidic.

The electron-withdrawing effect of the phenyl ring is somewhat diminished by the – CH₂ group and it is therefore, less acidic than phenol. In (3) and (4), -OH group is attached to alkyl groups which, due to their +1 effect reduces the polarity of -OH bond and so, the acidic strength is low.

Question 21. Among the following four compounds

1. Phenol
2. Methyl phenol
3. Mefa-nitrophenol
4. Para-nitrophenol

The acidity order is

1. 4>3>1>2
2. 3>4>1>2
3. 1>4>3>2
4. 2>1>3>4

Answer: 1. 4>3>1>2

In phenols, the presence of electron-releasing groups decreases the acidity, whereas the presence of electron-withdrawing groups increases the acidity, compared to phenol.

Among the meta and para-nitrophenols, the latter is more acidic as the presence of – NO₂ in the group at the para position stabilizes the phenoxide ion to a greater extent than when it is present at the meta position.

Thus, correct order of acidity is: para -nitrophenol(4) > meta-nitrophenol(3) > phenol(1) > methyl phenol(2)

Quetsion 22. When glycerol is treated with excess of HI, it produces

1. 2-Iodopropane
2. Allyl Iodide
3. Propene
4. Glycerol Triiodide.

Answer: 1. 2-Iodopropane

Question 23. Consider the following reaction:

the product Z is

1. \(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3\)
2. \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}-\mathrm{SO}_3 \mathrm{H}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)
4. \(\mathrm{CH}_2=\mathrm{CH}_2\)

Answer: 3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)

Quetsion 24. HOCH₂CH₂OH on heating with periodic acid gives

Answer: 3

When 1,2-diol-like ethylene glycol is treated with HIO₂ each alcoholic group is oxidized to a carbonyl group by HIO₂. Since in glycol, both the -OH groups are terminal, so oxidation would yield two formaldehyde molecules.

Question 25. Consider the following reaction product Z is

1. Benzaldehyde
2. Benzoic acid
3. Benzene
4. Toluene.

Answer: 2. Benzoic acid

Quetsion 26. Ethylene oxide when treated with Grignard reagent yields

1. Primary alcohol
2. Secondary alcohol
3. Tertiary alcohol
4. Cyclopropyl alcohol.

Answer: 1. Primary alcohol

Quetsion 27. Which one of the following compounds is most acidic?

Answer: 3

Phenols are much more acidic than alcohols, due to the stabilization of phenoxide ion by resonance.

-NO₂ is the electron-withdrawing group and helps in stabilizing the negative charge on the oxygen hence equilibrium shifts in the forward direction, and more H⁺ are removed easily Hence, it is the most acidic.

CH₃ is the electron-donating group. Hence, electron density increases on the oxygen and destabilizes the product. Thus, equilibrium shifts in the backward direction.

Question 28. Which one of the following will not form a yellow precipitate on heating with an alkaline solution of iodine?

1. \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3\)
3. \(\mathrm{CH}_3 \mathrm{OH}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)

Answer: 3. \(\mathrm{CH}_3 \mathrm{OH}\)

The formation of a yellow precipitate on heating a compound with an alkaline solution of iodine is known as an iodoform reaction. Methyl alcohol does not respond to this test. Iodoform test is exhibited by ethyl alcohol, acetaldehyde, acetone, methyl ketones, and those alcohols that possess the CH₃CH(OH)- group.

Quetsion 29. n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?

1. PCl₅
2. Reduction
3. Oxidation with potassium dichromate
4. Ozonolysis

Answer: 3. Oxidation with potassium dichromate

n-Propyl alcohol on oxidation with potassium dichromate gives an aldehyde which on further oxidation gives an acid. Both aldehydes and acids contain the same number of C atoms as the original alcohol.

Isopropyl alcohol on oxidation gives a ketone with the same number of C atoms as the original alcohol.

Question 30. When phenol is treated with CHCl₃ and NaOH, the product formed is

1. Benzaldehyde
2. Salicylaldehyde
3. Salicylic acid
4. Benzoic acid.

Answer: 2. Salicylaldehyde

This reaction is called the Reimer-Tiemann reaction

Question 31. Which of the following is correct?

1. On reduction, any aldehyde gives secondary alcohol.
2. The reaction of vegetable oil with H₂SO₄ gives glycerine.
3. Alcoholic iodine with NaOH gives iodoform.
4. Sucrose in reaction with NaCl gives inverted sugar.

Answer: 3. Alcoholic iodine with NaOH gives iodoform.

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+4 \mathrm{I}_2+\mathrm{NaOH} \longrightarrow \mathrm{CHI}_3+\mathrm{NaI}\)

+ \(\mathrm{HCOONa}+\mathrm{H}_2 \mathrm{O}\)

Iodoform is a pale yellow solid which crystallizes in hexagonal plates.

Question 32. The correct acidic order of the following is

1. 1>2>3
2. 3>1>2
3. 2>3>1
4. 1>3>2

Answer: 2. 3>1>2

Phenol exists as a resonance hybrid of the following structures.

Thus, due to resonance, the oxygen atom of the – OH group acquires a positive charge and hence attracts the electron pair of the O-H bond leading to the release of a hydrogen atom as a proton.

Once the phenoxide ion is formed it stabilises itself by resonance which is more stable than the parent phenol as there is no charge separation.

Effect of substituent: The presence of electron withdrawing groups (-NO₂, -X, -CN) increases the acidity of phenols while the presence of electron releasing groups (-NH₂ -CH₃) decreases the acidity of phenols.

This explains the following order of acidity: p-nitrophenol > phenol >p-cresol.

Question 33. The reaction of   with RMgX leads to the formation of

Answer: 1

Question 34. When 3,3-dimethyl-2-butanol is heated with H₂SO₄, the major product obtained is

1. 2,3-dimethyl-2-butene
2. cis and trans isomers of 2,3-dimethyl-2-butene
3. 2,3-dimethyl-1-butene
4. 3,3-dimethyl 1-butene

Answer: 1. 2,3-dimethyl-2-butene

Question 35. The alkene R – CH = CH₂ reacts readily with B₂H₆ and the product on oxidation with alkaline hydrogen peroxides produces

Answer: 4

Question 36. On heating glycerol with cone. H₂SO₄, a compound is obtained which has a bad odor. The compound is

1. Acrolein
2. Formic acid
3. Allyl alcohol
4. Glycerol sulfate.

Answer: 1. Acrolein

Question 37. Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether, due to the presence of

1. H-bonding in ethanol
2. H-bonding in dimethyl ether
3. CH₃ group in ethanol
4. CH₃ group in dimethyl ether.

Answer: 1. H-bonding in ethanol

Question 38. Increasing order of acid strength among p-methoxyphenyl,p-methyl phenol and p-nitrophenol is

1. p-nitrophenol, p-methoxyphenyl, p-methyl phenol
2. p-methyl phenol, p-methoxyphenyl, p-nitrophenol
3. p-nitrophenol, p-methyl phenol, p -methoxyphenyl
4. p-methoxyphenyl, p-methylphenol, p-nitrophenol.

Answer: 4. p-methoxyphenyl, p-methylphenol, p-nitrophenol.

– OCH₃, – CH₃ are electron donating groups and decrease the acidic character of phenols. -NO₂, is an electron-withdrawing group and tends to increase the acidic character. The electron-donating effect of -OCH₃ group (+R effect) is more than that of – CH₃ group (+1 effect). Thus, the order is p-methoxyphenol < p-methylphenol < p-nitrophenol.

Quetsion 39. Which one of the following on oxidation gives a ketone?

1. Primary alcohol
2. Secondary alcohol
3. Tertiary alcohol
4. All of these

Answer: 2. Secondary alcohol

2° alcohols on oxidation give ketones, 1° alcohols form aldehydes.

Question 40. What is formed when a primary alcohol undergoes catalytic dehydrogenation?

1. Aldehyde
2. Ketone
3. Alkene
4. Acid

Answer: 1. Aldehyde

Primary alcohol undergoes catall.tic dehydrogenation to give aldehyde.

Question 41. How many isomers of C₅H₁₁OH will be primary alcohols?

1. 5
2. 4
3. 2
4. 3

Answer: 2. 4

4-isomers are possible for C₅H₁₁OH.

Question 42. HBr reacts fastest with

1. 2-methyl propan-1-ol
2. methyl propan-2-ol
3. propan-2-ol
4. propan-l-ol.

Answer: 2. methylpropan-2-ol

Generates 3° carbocation which is a very stable intermediate, thus it will react more rapidly with HBr.

Question 43. When phenol is treated with excess bromine water. It gives

1. m-bromophenol
2. o- and p-bromophenols
3. 2,4-bromophenol
4. 2,4,6-tribromophenol.

Answer: 4. 2,4,6-tribromophenol.

Phenol in reaction with excess bromine water gives 2,4,6 – tribromophenol.

Question 44. The compound which reacts fastest with Lucas reagent at room temperature is

1. Butan-1-ol
2. Butan-2-ol
3. 2-Methylpropan-1-ol
4. 2-Methylpropan-2-ol.

Asnwer: 4. 2-Methylpropan-2-ol.

2-Methylpropan-2-ol reacts rapidly with Lucas reagent at room temperature.

Quetsion 45. Which one of the following compounds will be most readily attacked by an electrophile?

1. Chlorobenzene
2. Benzene
3. Phenol
4. Toluene

Answer: 3. Phenol

-OH group being an electron donor increases the electron density in phenol. Thus, the electron density in phenol is higher than that of toluene, benzene, and chlorobenzene.

Question 46. Propene, CH₃CH=CH₂ can be converted into 1-propanol by oxidation. Indicate which set of reagents amongst the following is ideal for the above conversion.

1. KMnO₄ (alkaline)
2. Osmium tetroxide (OSO₄/CH₂Cl₂)
3. B₂H₆ and alk.H₂O₂
4. O₃/Zn

Answer: 3. B₂H₆ and alk.H₂O₂

Question 47. Phenol is heated with CHCl₃ and aqueous KOH when salicylaldehyde is produced. This reaction is known as

1. Rosenmund’s reaction
2. Reimer-Tiemann reaction
3. Friedel-Crafts reaction
4. Sommelet reaction.

Answer: 2. Reimer-Tiemann reaction

Treatment of phenol with CHCI₃ and aqueous hydroxide introduces the – CHO group, onto the aromatic ring generally ortho to the – OH group. This reaction is known as Reimer-Tiemann reaction.

Question 48. Lucas’s reagent is

1. cone. HCl and anhydrous ZnCl₂
2. cone. HNO₃ and hydrous ZnCl₂
3. cone. HCl and hydrous ZnCl₂
4. cone. HNO₃ and anhydrous ZnCl₂

Answer: 1. cone. HCl and anhydrous ZnCl₂

Question 49. Methanol is industrially prepared by

1. Oxidation of CH₄ by steam at 900°C
2. Reduction of HCHO using LiAIH₄
3. The reaction of HCHO with a solution of NaOH
4. Reduction of CO using H₂ and ZnO-Cr₂O₃.

Answer: 4. Reduction of CO using H₂ and ZnO-Cr₂O₃.

⇒ \(\mathrm{CO}+2 \mathrm{H}_2\) \(\underrightarrow{\mathrm{ZnO}-\mathrm{Cr}_2}\) \({\mathrm{O}_3} \mathrm{CH}_3 \mathrm{OH}\)

Question 50. Consider the following reaction

Answer: 1

Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to a more stable aryl-oxygen bond. The reaction yields phenol and alkyl halide.

Question 51. Anisole on cleavage with HI gives

Answer: 1

Question 52. The compound that is most difficult to protonate is

Answer: 1

The lone pair of oxygen is in conjugation with the phenyl group so, it is the least basic among the given compounds and is most difficult to protonate.

Question 53. The major products C and D formed in the following reactions respectively are \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\) \(\underrightarrow{\text { excess } \mathrm{HI}}\) C+D

1. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I}\) and \(\mathrm{I}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)
2. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}\) and \(\mathrm{I}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)
3. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I}\) and \(\mathrm{HO}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)
4. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}\) and \(\mathrm{HO}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)

Answer: 1. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I}\) and \(\mathrm{I}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)

Ethers are readily attacked by HI to give an alkyl halide and alcohol. But when heated with excess of HI, the product alcohol first formed reacts further with HI to form the corresponding alkyl iodide.

⇒ \(\mathrm{R}-\mathrm{O}-\mathrm{R}^{\prime}+\underset{\text { (excess) }}{2 \mathrm{HI}}\) \(\underrightarrow{\text { (Heat) }}\) \(\mathrm{RI}+\mathrm{R}^{\prime} \mathrm{I}+\mathrm{H}_2 \mathrm{O}\)

Question 54. The heating of phenyl methyl ether with HI produces

1. Iodobenzene
2. Phenol
3. Benzene
4. Ethyl chloride.

Answer: 2. Phenol

In case of phenyl methyl ether, methyl phenyl oxonium ion is formed by protonation of ether. The O-CH₃ bond is weaker than O-C₆H₅ bond as O-C₆H₅ has a partial double bond character. Therefore, the attack by I^– ion breaks O-CH₃ bond to form CH₃l.

Question 55. The reaction can be classified as

1. Dehydration reaction
2. Williamson alcohol synthesis reaction
3. Williamson ether synthesis reaction
4. Alcohol formation reaction.

Answer: 3. Williamson ether synthesis reaction

Williamson ether synthesis reaction involves the treatment of sodium alkoxide with a suitable alkyl halide to form an ether.

Question 56. The reaction is called

1. Etard reaction
2. Gattermann-Koch reaction
3. Williamson synthesis
4. Williamson continuous etherification process.

Answer: 3. Williamson synthesis

Williamson synthesis is the best method for the preparation of ether.

Question 57. Among the following sets of reactants which one produces anisole?

1. \(\mathrm{CH}_3 \mathrm{CHO} ; \mathrm{RMgX}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} ; \mathrm{NaOH} ; \mathrm{CH}_3 \mathrm{I}\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\); neutral \(\mathrm{FeCl}_3\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3 ; \mathrm{CH}_3 \mathrm{COCl} ; \mathrm{AlCl}_3\)

Answer: 2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} ; \mathrm{NaOH} ; \mathrm{CH}_3 \mathrm{I}\)

Question 58. Identify Z in the sequence of reactions: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2\) \(\underrightarrow{\mathrm{HBr} / \mathrm{H}_2 \mathrm{O}_2}\) Y \(\underrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}} \) Z

1. \(\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_3-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)
2. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)
3. \(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4-\mathrm{O}-\mathrm{CH}_3\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)

Answer: 1. \(\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_3-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)

Question 59. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?

Answer: 1

Question 60. In the reaction, which of the following compounds will be formed?

Answer: 3

The alkyl iodide produced depends on the nature of the alkyl groups. If one group is Me and the other a primary or secondary alkyl group, it is methyl iodide that is produced. This can be explained by the assumption that the mechanism is S_N2, and because of the steric effect of the larger group, I^– attack the smaller methyl group.

When the substrate is a methyl t-alkyl ether, the products are t-RI and MeOH. This can be explained by S_N1 mechanism, the carbonium ion produced being the t-alkyl since a tertiary carbonium ion is more stable than a primary or secondary carbonium ion.

Question 61. The major organic product in the reaction is \(\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}\left(\mathrm{CH}_3\right)_2+\mathrm{HI} \longrightarrow\) products

Answer: 1

With cold HI, a mixture of alkyl iodide and alcohol is formed. In the case of mixed ethers, the halogen atom attaches to a smaller and less complex alkyl group.

CH₃OCH(CH₃)₂ + HI → CH₃I + (CH₃)₂CHOH

Question 62. Ethyl chloride is converted into diethyl ether by

1. Perkins reaction
2. Grignard reaction
3. Wurtz synthesis
4. Williamson synthesis.

Answer: 4. Williamson synthesis.

⇒ \(\mathrm{C}_2 \mathrm{H}_5-\mathrm{Cl}+\mathrm{Na}-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5 \longrightarrow \mathrm{C}_2 \mathrm{H}_5-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5+\mathrm{NaCl}\)

Question 63. Which one of the following compounds is resistant to nucleophilic attack by hydroxyl ions?

1. Diethyl ether
2. Acetonitrile
3. Acetamide
4. Methyl acetate

Answer: 1. Diethyl ether

Diethyl ether is a saturated compound, so it is resistant to nucleophilic attack by a hydroxyl ion (OH^–). Other compounds have unsaturation and the unsaturated ‘C’ atom bears partial +ve charge, therefore they undergo easy nucleophilic attack by OH^– ion.

Question 64. The compound which does not react with sodium is

1. \(\mathrm{CH}_3 \mathrm{COOH}\)
2. \(\mathrm{CH}_3 \mathrm{CHOHCH}_3\)
3. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
4. \(\mathrm{CH}_3 \mathrm{OCH}_3\)

Answer: 4. \(\mathrm{CH}_3 \mathrm{OCH}_3\)

Ethers are very inert. The chemical inertness of others is due to the absence of an active group in their molecules’ Since CH₃ – O – CH₃ is inert and does not contain an active group, therefore it does not react with sodium

Question 65. Which one is formed when sodium phenoxide is heated with ethyl iodide?

1. Phenetole
2. Ethyl phenyl alcohol
3. Phenol
4. None of these

Answer: 1. Phenetole

Phenetole is formed when sodium phenoxide is heated with ethyl iodide.

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{ONa}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}\) \(\underrightarrow{\Delta}\) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5\)

 

The Solid State

Question 1. The pure crystalline substance on being heated gradually first forms a turbid liquid at constant temperature and still at higher temperature turbidity completely disappears. The behaviour is a characteristic of substance-forming

1. Allotropic crystals
2. Liquid crystals
3. Isomeric crystals
4. Isomorphous crystals.

Answer: 2. Liquid crystals

Liquid crystals on heating first become turbid and then on further heating turbidity completely disappears.

Question 2. Glass is a

1. Liquid
2. Solid
3. Supercooled liquid
4. Transparent organic polymer.

Answer: 3. Supercooled liquid

Glass is a supercooled liquid which forms a noncrystalline solid without a regular lattice.

Question 3. Most crystals show good cleavage because their atoms, ions or molecules are

1. Weakly bonded together
2. Strongly bonded together
3. Spherically symmetrical
4. Arranged in planes.

Answer: 4. Arranged in planes.

Crystals show good cleavage because their constituent particles are arranged in planes.

Question 4. The ability of a substance to assume two or more crystalline structures is called

1. Isomerism
2. Polymorphism
3. Isomorphism
4. Amorphism.

Answer: 2. Polymorphism

The phenomenon of the existence of a substance in two or more crystalline structures is called polymorphism.

Question 5. Cation and anion combine in a crystal to form the following type of compound

1. Ionic
2. Metallic
3. Covalent
4. Dipole-dipole.

Answer: 1. The electrostatic force of attraction which exists between oppositely charged ions is called an ionic bond.

Question 6. For two ionic solids CaO and KI, identify the wrong statement among the following.

1. CaO has a high melting point.
2. The lattice energy of CaO is much larger than that of KI.
3. KI has a high melting point.
4. KI is soluble in benzene.

Answer: 4. KI is soluble in benzene.

KI is an ionic compound while benzene is not.

Read and Learn More NEET MCQs with Answers

Question 7. The correct option for the number of body-centred unit cells in all 14 types of Bravais lattice unit cells is

1. 3
2. 7
3. 5
4. 2

Answer: 1. 3

Out of 14 types of Bravaisiattice, three body-centred units crisis are there which are: orthorhombic, tetragonal and cubic

Question 8. For the orthorhombic system, axial ratios are α = β = γ = 90° and the axial angles are

1. α = β = γ ≠ 90°
2. α = β = γ = 90°
3. α = γ = 90°, β ≠ 90°
4. α ≠ β ≠ γ ≠ 90°

Answer: 2. α = β = γ = 90°

For orthorhombic system, α = β = γ = 90°

Question 9. The number of carbon atoms per unit cell of the diamond unit cell is

1. 6
2. 1
3. 4
4. 8

Answer: 4. 8

Diamond is like ZnS (Zincblende).

Carbon forms ccp (fcc) and also occupies half of the tetrahedral voids.

Total no. of carbon atoms per unit cell = 8 x 1/8 + 6 x 1/2 + 4 = 8

Question 10. In a face-centred cubic lattice, a unit cell is shared equally by how many unit cells?

1. 2
2. 4
3. 6
4. 8

Answer: 3. 6

Here given unit cell is shared equally by six faces in the fcc which is shared equally by six different unit cells.

Question 11. When Zn converts from its melted state to its solid state, it has a hcp structure and then finds the number of nearest atoms.

1. 6
2. 8
3. 12
4. 4

Answer: 3. 12

hcp is a closed-packed arrangement in which the unit cell is hexagonal and the coordination number is 12

Question 12. The fee crystal contains how many atoms in each unit cell?

1. 6
2. 8
3. 4
4. 5

Answer: 3. 4

The contribution of eight atoms of face-centred cubic unit cell = 8 x 1/8 = 1 atom. There is one atom at each of the six faces, which is shared by 2 unit cells each.

The contribution of 6 face-centred atoms = 6 x 1/2 = 3

Therefore n = L + 3 = 4

Question 13. The number of atoms contained in a fee unit cell of a monatomic substance is

1. 1
2. 2
3. 4
4. 6

Answer: 3. 4

fcc crystal contains = 8 x 1/8 + 6 x 1/2 = 4 atoms in a unit cell

Question 14. A compound is formed by two elements A and B. The element B forms a cubic close-packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A_x B_y, then the value of x + y is in the option

1. 3
2. 2
3. 5
4. 4

Answer: 3. 5

Let, the number of atoms in ccp unit cell = N

Thus, the number of B atoms = N

Number of tetrahedral voids = 2N

Number of A atoms = \(\frac{2 N}{3}\)

Molecular ratio A:B = \(\frac{2 N}{3}\)

The formula of the compound is A₂B₂

Hence, x+y=2+3=5

Question 15. What fraction of one edge-centred octahedral void lies in one unit cell of fcc?

1. 1/4
2. 1/12
3. 1/2
4. 1/3

Answer: 1. 1/4

There is an octahedral void at each edge of the unit cell such that 1/4th of the void is in one cell.

Question 16. The right option for the number of tetrahedral and octahedral voids in the hexagonal primitive unit cells are

1. 12,6
2. 8,4
3. 6,12
4. 2, 1

Answer: 1. 12,6

Number of atoms in hexagonal unit cell = N = 6

Number of octahedral void = N = 6

Number of tetrahedral void = 2N = 12

Question 17. A compound is formed by cation C and anion A. Hie anions form hexagonal close-packed (hep) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is

1. \(C_4 A_3\)
2. \(C_2 A_3\)
3. \(C_3 A_2\)
4. \(C_3 A_4\)

Answer: 4. \(C_3 A_4\)

Number of atoms per unit cell in hcp = 6

Number of octahedral voids in hcp = 6

Number of anions per unit cell = 6

Number of octahedral voids occupied by cations = 6 x \(\frac{75}{100}\) = \(\frac{9}{2}\)

∴ Formula of compound = C_9/2A₆ = C₃A₄

Question 18. In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca²⁺) and fluoride ion (F^–) are

1. 4 and 2
2. 6 and 6
3. 8 and 4
4. 4 and 8

Answer: 3. 8 and 4

In fluorite structure, Ca²⁺ ions are in the face-centred cubic arrangement. Each Ca²⁺. is connected to 4 F^– ions below it and to another set of 4 F^– ions above it i.e. Ca²⁺ has a coordination number of 8 and each F¹ ion has a coordination number 4.

Question 19. The ionic radii of A⁺ and B^– ions are 0.98 x 10^-10 m and 1.81 x 10^-10 m. The coordination number of each ion in AB is

1. 8
2. 2
3. 6
4. 4

Answer: 3. 6

Radius ration, \(\frac{r_{+}}{r_{-}}=\frac{0.98 \times 10^{-10}}{1.81 \times 10^{-10}}=0.54\)

It lies in the range of 0.414 to 0.732 hence, the coordination number of each ion will be 6 as the compound will have NaCl type structure i.e., octahedral arrangement.

Question 20. The number of octahedral voids (s) per atom present in a cubic close-packed structure is

1. 1
2. 3
3. 2
4. 4

Answer: 1. 1

The number of octahedral voids is the same as the number of atoms

Question 21. The structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is

1. \(\mathrm{ABO}_2\)
2. \(A_2 B_2\)
3. \(A_2 B_3 \mathrm{O}_4\)
4. \(A B_2 \mathrm{O}_2\)

Answer: 4. \(A B_2 \mathrm{O}_2\)

Number of atoms in ccp = 4= O^2-

Number of tetrahedral voids = 2 x N = 2 x 4 = B

Number of A²⁺ ions = 8 x 1/4 = 2

Number of octahedral voids = Number of B⁺ ions = N = 4

Ratio, O^2-: A²⁺ : Bn = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB₂A₂

Question 22. A solid compound XY has a NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y^–) will be

1. 275.1 pm
2. 322.5 pm
3. 241.5 pm
4. 165.7 pm

Answer: 3. 241.5 pm

For NaCl, \(\frac{r_{+}}{r_{-}}=0.414\)

Given: radius of cation =100pm

⇒ \(\frac{100}{r_{-}}=0.414 \Rightarrow \frac{100}{0.414}=r_{-} \Rightarrow r_{-}=241.5 \mathrm{pm}\)

Question 23. A compound formed by elements X and Y crystallises in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face- centres. The formula of the compound is

1. XY₃
2. X₃Y
3. XY
4. XY₂

Answer: 1. XY₃

In a unit cell, X atoms at the corners = \(\frac{1}{8}\) = 1

Y atoms at the face centres = \(\frac{1}{2}\) x 6 = 3

The ratio of X and Y = 1 : 3. Hence formula is XY

Question 24. In a cube of any crystal, an A-atom is placed at every corner and a B-atom is placed at every centre of the face. The formula of a compound is

1. AB
2. AB₃
3. A₂B₂
4. A₂B₃

Answer: 2. AB₃

‘A’ atoms are at ‘8’ corners of the cube.

Thus, no. of ‘A’atoms per unit cell = 8 x \(\frac{1}{8}\) = 1

‘B’ atoms are at the face centre of six faces. Thus, no. of ‘B’ atoms per unit cell = 6 x \(\frac{1}{2}\) = 3

The formula is AB₃

Question 25. In crystals of which one of the following ionic compounds would you expect the maximum distance between centres of cations and anions?

1. Csl
2. CsF
3. LiF
4. Lil

Answer: 1. Csl

As Cs⁺ ion has a larger size than Li⁺ and I^– has a larger size than F^–, so maximum distance between centres of cations and anions is in CsI.

Question 26. The second-order Bragg diffraction of X-rays with λ = 1.00 Å from a set of parallel planes in a metal occurs at an angle of 60°. Die distance between the scattering planes in the crystal is

1. 2.00 Å
2. 1.00 Å
3. 0.575 Å
4. 1.15 Å

Answer: 4. 1.15 Å

According to Bragg’s equation, nλ = 2d sin θ

As, n = 2λ = 1.00Å, θ = 60°, d = ?

2dsinθ = nλ

2dsin60°=2×1Å

2d x \(\frac{\sqrt{3}}{2}=2 \Rightarrow d=\frac{2}{\sqrt{3}}=1.15\) Å (because \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\))

Question 27. The intermetallic compound LiAg crystallizes in the cubic lattice in which both lithium and silver have a coordination number of eight. The crystal class is

1. Face-centred cube
2. Simple cube
3. Body-centred, cube
4. None of these.

Answer: 3. Body-centred, cube

A body-centred cubic unit cell consists of 8 atoms at the corners and one atom at the centre.

Question 28. In the fluorite structure, the coordination number of Ca²⁺ ions is

1. 4
2. 6
3. 8
4. 3

Answer: 3. 8

In fluorite (CaF₂) structure, C.N. of Ca²⁺ =8, C.N. of F^– = 4

Question 29. An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is

1. \(\frac{\sqrt{3}}{4} \times 288 \mathrm{pm}\)
2. \(\frac{\sqrt{2}}{4} \times 288 \mathrm{pm}\)
3. \(\frac{4}{\sqrt{3}} \times 288 \mathrm{pm}\)
4. \(\frac{4}{\sqrt{2}} \times 288 \mathrm{pm}\)

Answer: 1. \(\frac{\sqrt{3}}{4} \times 288 \mathrm{pm}\)

For bcc structure, \(r=\frac{\sqrt{3}}{4} a\), where a is the unit cell edge length and r is the radius of the sphere (atom).

r = \(\frac{\sqrt{3}}{4} \times 288 \mathrm{pm}\)

Question 30. The Vacant space in bcc lattice unit cell is

1. 48%
2. 23%
3. 32%
4. 26%

Answer: 3. 32%

Packing efficiency of bcc lattice = 68%

Hence, empathy space = 32%

Question 31. If a is the length of the side of a cube, the distance between the body-centred atom and one corner atom in the cube will be

1. \(\frac{2}{\sqrt{3}} a\)
2. \(\frac{4}{\sqrt{3}} a\)
3. \(\frac{\sqrt{3}}{4} a\)
4. \(\frac{\sqrt{3}}{2} a\)

Answer: 4. \(\frac{\sqrt{3}}{2} a\)

The distance between the body-centred atom and one corner atom is \(\frac{\sqrt{3} a}{2}\) i.e. half of the body diagonal.

Question 32. A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is

1. 288 pm
2. 408 pm
3. 144 pm
4. 204 pm

Answer: 1. 288 pm

For a face-centred cubic (fcc) structure.

r = \(\frac{a}{2 \sqrt{2}}, a=408 \mathrm{pm}, r=\frac{408}{2 \sqrt{2}}=144 \mathrm{pm}\)

Diameter = 2r = 2 x 144 = 288 Pm

Question 33. AB crystallizes in a body-centred cubic lattice with edge length ‘a’ equal to 387 pm. The distance between two oppositely charged ions in the lattice is

1. 335 pm
2. 250 pm
3. 200 pm
4. 300 pm

Answer: 1. 335 pm

For a bcc latlice, 2(r₊+r_–)= √3a

where r₊ = radius of cation, r_– = radius of anion

a = edge length

∴ \(\left(r_{+}+r_{-}\right)=\frac{\sqrt{3} \times 387}{2}=335.15 \mathrm{pm}\) = 335 Pm

Question 34. Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of lithium will be

1. 151.8 pm
2. 75.5 pm
3. 300.5 pm
4. 240.8 pm

Answer: 1. 151.8 pm

Since Li crystallises in body-centred cubic crystal atomic radius,

r = \(\frac{\sqrt{3} a}{4} \quad(a=\text { edge length })\)

∴ r = \(\frac{\sqrt{3}}{4} \times 351=151.8 \mathrm{pm}\)

Given: a=351 pm

Question 35. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of a copper atom in pm?

1. 157
2. 181
3. 108
4. 128

Answer: 4. 128

Since Cu crystallises in a face-centred cubic lattice.

Atomic radius r = \(\frac{a}{2 \sqrt{2}}\) (a = edge length = 361 pm)

∴ r = \(\frac{361}{2 \sqrt{2}}=127.6=128 \mathrm{pm}\)

Question 36. Which of the following statements is not correct?

1. The number of carbon atoms in a unit cell of diamond is 8.
2. The number of Bravais lattices in which a crystal can be categorized is 14.
3. The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.
4. Molecular solids are generally volatile.

Answer: 3. The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.

The packing fraction for a cubic unit cell is given by f = \(\frac{Z \times \frac{4}{3} \pi r^3}{a^3}\).

Question 37. If a stands for the edge length of the cubic systems: simple cubic, body-centred cubic and face-centred cubic, then the ratio of radii of the spheres in these systems will be respectively

1. \(\frac{1}{2} a: \frac{\sqrt{3}}{2} a: \frac{\sqrt{2}}{2} a\)
2. \(1 a: \sqrt{3} a: \sqrt{2} a\)
3. \(\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a\)
4. \(\frac{1}{2} a: \sqrt{3} a: \frac{1}{\sqrt{2}} a\)

Answer: 3. \(\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a\)

For simple cubic: r = a/2

For body centred: r = \(a \sqrt{3} / 4\)

For face-centred: r = \(\frac{a}{2 \sqrt{2}}\)

where a = edge length, r = radius

∴ The ratio of radii of the three will be \(\frac{a}{2}: \frac{a \sqrt{3}}{4}: \frac{a}{2 \sqrt{2}}\)

Question 38. The fraction of the total volume occupied by the atoms present in a simple cube is

1. \(\frac{\pi}{3 \sqrt{2}}\)
2. \(\frac{\pi}{4 \sqrt{2}}\)
3. \(\frac{\pi}{4}\)
4. \(\frac{\pi}{6}\)

Answer: 4. \(\frac{\pi}{6}\)

Question 39. The pyknometric density of sodium chloride crystal is 2.165 x 10³ kg m^-3 while its X-ray density is 2.178 x 10³ kg m^-3. The fraction of unoccupied sites in sodium chloride crystals is

1. 5.96
2. 5.96 x 10^-2
3. 5.96 x 10^-1
4. 5.96 x 10^-3

Answer: 4. 5.96 x 10^-3

Molar volume from psychometric density = \(\frac{M}{2.165 \times 10^3} \mathrm{~m}^3\)

Molar volume from X-ray density = \(\frac{M}{2.178 \times 10^3} \mathrm{~m}^3\)

Volume occupied = \(\frac{M}{10^3}\left(\frac{1}{2.165}-\frac{1}{2.178}\right) \mathrm{m}^3\)

Fraction unoccupied

= \(\left(\frac{0.013 M \times 10^{-3}}{2.165 \times 2.178}\right) /\left(\frac{M \times 10^{-3}}{2.165}\right)=5.96 \times 10^{-3}\)

Question 40. The edge length of face-centred unit cubic cells is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is

1. 144 pm
2. 398 pm
3. 288 pm
4. 618 pm

Answer: 1. 144 pm

In the face-centred cubic lattice, the unit cell, a = r + 2R + r

where r = Radius of cation, R = Radius of anion

⇒ 508 = 2 x 110 + 2R

∴ R = 144Pm

Question 41. Copper crystallises in a fcc unit cell with a cell edge length of 3.608 x 10^-8 cm. The density of copper is 8.92 g cm^-3. Calculate the atomic mass of copper.

1. 63.1 u
2. 31.55 u
3. 60 u
4. 65 u

Answer: 1. 63.1 u

Density of unit cell

d = \(\frac{Z \times M}{N_0 \times a^3}\)

Given, a = \(3.608 \times 10^{-8} \mathrm{~cm}\)

d = \(8.92 \mathrm{~g} / \mathrm{cm}^3\)

Z = 4(for fcc)

M = \(\frac{N_0 \times a^3 \times d}{Z}=\frac{6.023 \times 10^{23} \times\left(3.608 \times 10^{-8}\right)^3 \times 8.92}{4}\)

= \(63.08 \mathrm{u} \approx 63.1 \mathrm{u}^4\)

Question 42. Iron exhibits bee structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of the density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remain constant with temperature) is

1. \(\frac{\sqrt{3}}{\sqrt{2}}\)
2. \(\frac{4 \sqrt{3}}{3 \sqrt{2}}\)
3. \(\frac{3 \sqrt{3}}{4 \sqrt{2}}\)
4. \(\frac{1}{2}\)

Answer: 3. \(\frac{3 \sqrt{3}}{4 \sqrt{2}}\)

For bcc lattice: \(Z=2, a=\frac{4 r}{\sqrt{3}}\)

For fcc lattice : Z=4, a = \(2 \sqrt{2}r\)

∴ \(\frac{d_{R T}}{d_{900^{\circ} \mathrm{C}}}=\frac{\left(\frac{Z M}{N_A a^3}\right)_{b c c}}{\left(\frac{Z M}{N_A a^3}\right)_{f c c}}\)

Given, molar mass and atomic radii are constant.

= \(\frac{2}{4}\left(\frac{2 \sqrt{2} r}{\frac{4 r}{\sqrt{3}}}\right)^3=\frac{3 \sqrt{3}}{4 \sqrt{2}}\)

Question 43. Lithium has a bcc structure. Its density is 530 kg m^-3 and its atomic mass is 6.94 g mol^-1. Calculate the edge length of a unit cell of lithium metal. (N_A = 6.02 x 10²³ mol^-1)

1. 527 pm
2. 264 pm
3. 154 pm
4. 352 pm

Answer: 4. 352 pm

For bcc, Z = 2, ρ = 530 kg m^-3,

at, mass of Li = 6.94 g mol^-1, N_A = 6.02 x 10²³ mol^-1

⇒ \(\rho=530 \mathrm{~kg} \mathrm{~m}^{-3}=\frac{530 \times 1000 \mathrm{~g}}{1 \times(100)^3 \mathrm{~cm}^3}=0.53 \mathrm{~g} \mathrm{~cm}^{-3}\)

⇒ \(\rho=\frac{Z \times \text { At. mass }}{N_A \times a^3}\)

⇒ \(a^3=\frac{Z \times \text { At. mass }}{N_A \times \rho}=\frac{2 \times 6.94}{6.02 \times 10^{23} \times 0.53}\)

= \(43.5 \times 10^{-24} \mathrm{~cm}^3\)

a = \(352 \times 10^{-10} \mathrm{~cm}=352 \mathrm{pm}\)

Question 44. A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^-3. The molar mass of the metal is (N_A Avogadro’s constant = 6.02 x 10²³ mol^-1)

1. 27 g mol^-1
2. 20 g mol^-1
3. 40 g mol^-1
4. 30 g mol^-1

Answer: 1. 27 g mol^-1

d = \(\frac{Z M}{N_A a^3}(Z=4 \text { for } f c c)\)

M = \(\frac{d \times N_A \times a^3}{Z}=\frac{2.72 \times 6.023 \times 10^{23} \times\left(404 \times 10^{-10}\right)^3}{4}\)

M = \(27 \mathrm{~g} \mathrm{~mol}^{-1}\)

Question 45. CsBr crystallises in a body-centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 x 10²³ mol^-1, the density of CsBr is

1. 4.25 g/cm³
2. 42.5 g/cm³
3. 0.425 g/cm³
4. 8.25 g/cm³

Answer: 1. 4.25 g/cm³

Density of \(\mathrm{CsBr}=\frac{Z \times M}{a^3 \times N_A}\)

= \(\frac{1 \times 213}{\left(436.6 \times 10^{-10}\right)^3 \times 6.023 \times 10^{23}}=4.25 \mathrm{~g} / \mathrm{cm}^3\)

(It has one formula unit in the unit cell, so Z=1.)

Question 46. An element (atomic mass =100 g/mol) having bcc structure has a unit cell edge of 400 pm. The density of the element is

1. 7.289 g/cm³
2. 2.144 g/cm³
3. 10.376 g/cm³
4. 5.188 g/cm³

Answer: 4. 5.188 g/cm³

Cell edge = 400 pm; number of atoms in bcc (Z) = 2

and atomic mass = 100 g/mol.

Since atomic mass is 100 g/mol, therefore the mass of each atom (m) = \(\frac{100}{6.023 \times 10^{23}}=16.6 \times 10^{-23} \mathrm{~g}\)

We know that volume of unit cell = (400 pm)³

= (64 x 10⁶)pm³ = 64 x 10^-24 cm³ and

mass of unit cell= Z x m =2x (16.6 x 10^-23) = 33.2 x 10^-23 g

Therefore, density = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)

= \(\frac{33.2 \times 10^{-23}}{64 \times 10^{-24}}=5.188 \mathrm{~g} / \mathrm{cm}^3\)

Question 47. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): In a particular point defect, an ionic solid is electrically neutral, even if a few of its cations are missing from its unit cell.

Reason (R): In an ionic solid, Frenkel defect arises due to the dislocation of a cation from its lattice site to the interstitial site, maintaining overall electrical neutrality.

In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both (A) and (R) are correct and (R) is the correct explanation of (A).
2. Both (A) and (R) are correct but (R) is not the correct explanation of (A).
3. (A) is correct but (R) is not correct.
4. (A) is not correct but (R) is correct

Answer: 2. Both (A) and (R) are correct but (R) is not the correct explanation of (A).

Frenkel defect is shown by ionic solids. The smaller ion (usually a cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.

Question 48. The formula of nickel oxide with metal deficiency defect in its crystal is Ni_0-98O. The crystal contains Ni²⁺ and Ni³⁺ ions. The fraction of nickel existing as Ni²⁺ ions in the crystal is

1. 0.96
2. 0.04
3. 0.50
4. 0.3

Answer: 1. 0.96

Let the fraction of metal which exists as Ni²⁺ ion be x.

Then the fraction of metal as Ni³⁺ = 0.98 – x

∴ 2x + 3(0.98 – x) = 2

⇒ 2x+2.94-3x=2

⇒  x=0.94=0.96

Question 49. The correct statement regarding defects in crystalline solids is

1. Frenkel defects decrease the density of crystalline solids
2. Frenkel defect is a dislocation defect
3. Frenkel defect is found in halides of alkaline metals

Answer: 2. Frenkel defect is a dislocation defect

Frenkel defect is a dislocation defect as smaller ions (usually cations) are dislocated from normal sites to interstitial sites. Frenkel defect is shown by compounds having large differences in the size of cations and anions hence, alkali metal halides do not show Frenkel defect.

Also, Schottky’s defect decreases the density of the crystal while Frenkel’s defect has no effect on the density of crystals.

Question 50. The appearance of colour in solid alkali metal halides is generally due to

1. Interstitial positions
2. F-centres
3. Schottky defect
4. Frenkel defect.

Answer: 2. F-centres

F-centres are the sites where anions are missing and instead, electrons are present. They are responsible for colours.

Question 51. Schottky defect in crystals is observed when

1. The density of the crystal is increased
2. Unequal number of cations and anions are missing from the lattice
3. An ion leaves its normal site and occupies an interstitial site
4. An equal number of cations and anions are missing from the lattice.

Answer: 4. Equal number of cations and anions are missing from the lattice.

In the Schottky defect, an equal no. of cations and anions are missing from the lattice. So, the crystal remains neutral. Such defect is more common in highly ionic compounds of similar cationic and anionic size, i.e. NaCl

Question 52. Ionic solids, with Schottky defects, contain in their structure

1. Cation vacancies only
2. Cation vacancies and interstitial cations
3. An equal number of cation and anion vacancies
4. Anion vacancies and interstitial anions.

Answer: 3. Equal number of cation and anion vacancies

When an atom is missing from its normal lattice site, a lattice vacancy is created. Such a defect, which involves an equal number of cation and anion vacancies in the crystal lattice is called a Schottky defect

Question 53. Which is the incorrect statement?

1. Density decreases in the case of crystals with Schottky defect.
2. NaCl_(s) is an insulator, silicon is a semiconductor, silver is a conductor, and quartz is a piezoelectric crystal.
3. Frenkel defect is favoured in those ionic compounds in which sizes of cations and anions are almost equal.
4. FeO_0.98 has a non-stoichiometric metal deficiency defect.

Answer: 3. Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal. and

4. FeO_0.98 has a non-stoichiometric metal deficiency defect.

Frenkel defect is favoured in those ionic compounds in which there is a large difference in the size of cations and anions.

Non-stoichiometric defects due to metal deficiency are shown by  Fe_xO where x = 0.93 to 0.96.

Question 54. With which one of the following elements silicon should be doped so as to give a p-type of semiconductor?

1. Selenium
2. Boron
3. Germanium
4. Arsenic

Answer: 2. Boron

If silicon is doped with any of the elements of group 13 (B, Al, Ga, In, Tl) of the periodic table, the p-type of the semiconductor will be obtained.

Question 55. If NaCl is doped with 10^-4 mol % of SrCl₂, the concentration of cation vacancies will be (N_A = 6.02 x 10²³ mol^-1)

1. 6.02 x 10¹⁶ mol^-1
2. 6.02 x 10¹⁷ mol^-1
3. 6.02 x 10¹⁴ mol^-1
4. 6.02 x 10¹⁵ mol^-1

Answer: 2. 6.02 x 10¹⁷ mol^-1

As each Sr²⁺ ion introduces one cation vacancy, therefore, the concentration of cation vacancies = mole % of SrCI₂ added.

∴ The concentration of cation vacancies = 10^-4 mole%

= \(\frac{10^{-4}}{100} \times 6.023 \times 10^{23}=6.023 \times 10^{17}p\)

Question 56. If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor formation will occur?

1. n-type semiconductor
2. p-type semiconductor
3. Both (1) and (2)
4. None of these

Answer: 1. n-type semiconductor

When an impurity atom with 5 valence electrons (as arsenic) is introduced in a germanium crystal, it replaces one of the germanium atoms. Four of the five valence electrons of the impurity atom form covalent bonds with each valence electron of four geranium atoms and the fifth valence electron becomes free to move in the crystal structure.

This free electron acts as a charge carrier. Such as an impure germanium crystal is called an n-type semiconductor because in it charge carriers are negative (free electrons)

Question 57. On doping Ge metal with a little of In or Ga, one gets

1. p-type semiconductor
2. n-type semiconductor
3. Insulator
4. Rectifier.

Answer: 1. p-type semiconductor

p-type of semiconductors are produced

Due to metal deficiency defects

By adding impurities containing fewer electrons (i.e. atoms of group 13). Ge belongs to Group 14 and In or Ga to Group 13. Hence on doping-type semiconductor is obtained. This doping of Ge with In increases the electrical conductivity of the Ge crystal.

 

Solutions

Question 1. In one molal solution that contains 0.5 moles of a solute, there is

1. 500 mL of solvent
2. 500 g of solvent
3. 100 mL of solvent
4. 1000 g of solvent.

Answer: 2. 500 g of solvent

Molality (m) = \(\frac{\text { Moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)

Let the amount of solvent be \(x \mathrm{~g}\).

1 = \(\frac{0.5}{\frac{x}{1000}}\)

x = \(500 \mathrm{~g}\)

Question 2. Which of the following is dependent on temperature?

1. Molarity
2. Mole fraction
3. Weight percentage
4. Molality

Answer: 1. Molarity

Molarity is a function of temperature as volume depends on temperature.

Question 3. What is the mole fraction of the solute in a 1.00 m aqueous solution?

1. 1.770
2. 0.0354
3. 0.0177
4. 0.177

Answer: 3. 0.0177

1 molal aqueous solution means 1 mole of solute is present in 1000 g of water.

∴ \(x_{\text {solute }}=\frac{1}{1+\frac{1000}{18}}=\frac{1}{56.5}=0.0177\)

Question 4. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO₃? The concentrated acid is 70% HNO₃.

1. 70.0 g cone. HNO₃
2. 54.0 g cone. HNO₃
3. 45.0 g cone. HNO₃
4. 90.0 g cone. HNO₃

Answer: 3. 45.0 g cone. HNO₃

Molarity = \(\frac{w_{\mathrm{HNO}_3} \times 1000}{M_{\mathrm{HNO}_3} \times V_{\mathrm{sol}(\mathrm{mL})}}\)

or \(2=\frac{w_{\mathrm{HNO}_3}}{63} \times \frac{1000}{250} \Rightarrow w_{\mathrm{HNO}_3}=\frac{63}{2} \mathrm{~g}\)

Mass of acid \(\times \frac{70}{100}=\frac{63}{2}\)

Mass of acid \(=45 \mathrm{~g}\)

Read and Learn More NEET MCQs with Answers

Question 5. Which of the following compounds can be used as antifreeze in automobile radiators?

1. Methyl alcohol
2. Glycol
3. Nitrophenol
4. Ethyl alcohol

Answer: 2. Glycol

A 35% (V/V) solution of ethylene glycol is used as an antifreeze in cars for cooling the engine’ At this concentration’ the antifreeze lowers the freezing point of water to 255°4 K (-17.6°C).

Question 6. Concentrated aqueous sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.80 g mol^-1. The volume of acid required to make one litre of 0.1 M H₂SO₄ solution is

1. 16.65 mL
2. 22.20 mL
3. 5.55 mL
4. 11.10 mL

Answer: 3. .55 mL

⇒ \(\mathrm{H}_2 \mathrm{SO}_4\) is \(98 \%\) by weight.

Weight of \(\mathrm{H}_2 \mathrm{SO}_4=98 \mathrm{~g}\), Weight of solution \(=100 \mathrm{~g}\)

∴ Volume of solution = \(\frac{\text { mass }}{\text { density }}=\frac{100}{1.80} \mathrm{~mL}\)

= \(55.55 \mathrm{~mL}=0.0555 \mathrm{~L}\)

Molarity of solution = \(\frac{98}{98 \times 0.0555}=18.02 \mathrm{M}\)

Let \(V \mathrm{~mL}\) of this \(\mathrm{H}_2 \mathrm{SO}_4\) is used to prepare 1 litre of 0.1 M \(\mathrm{H}_2 \mathrm{SO}_4\).

∴ mM of concentrated \(\mathrm{H}_2 \mathrm{SO}_4=\mathrm{mM}\) of dilute \(\mathrm{H}_2 \mathrm{SO}_4\)

or, \(V \times 18.02=1000 \times 0.1\)

⇒ V = \(\frac{1000 \times 0.1}{18.02}=5.55 \mathrm{~mL}\)

Question 7. The mole fraction of the solute in one molal aqueous solution is

1. 0.009
2. 0.018
3. 0.027
4. 0.036

Answer: 2. 0.018

1 molal aqueous solution means I mole of solute present in 1 kg of H₂O.

1 mole of solute present \(\frac{1000}{18}\) mole of H₂O

⇒ \(x_{\text {solute }}=\frac{1}{\frac{1000}{18}+1}=\frac{18}{1018}=0.01768=0.018\)

Question 8. 2.5 litres of 1 M NaOH solution is mixed with another 3 litres of 0.5 M NaOH solution. Then find out the arity of the resultant solution.

1. 0.80 M
2. 1.0 M
3. 0.73 M
4. 0.50 M

Answer: 3. 0.73 M

Molecular weight of NaOH =40 g mol^-1

2.5 litre of 1 M NaOH solution contain 40 g mol^-1 x 1 mol L^-1 x 2.5 L = 40 x 2.5 g of NaOH

3 litre of 0.5 M NaOH solution contain

40 g mol^-1 x 0.5 mol L^-1 x 3 L= 40 x 0.5 x 3 g of NaOH

If these two solutions are mixed, the volume of the resultant solution = (2.5 + 3) = 5.5 litre.

5.51itre of the resultant solution contain = 40(2.5 + 1.5) g of NaOH

1 litre of the resultant solution contain \(\frac{40 \times 4}{5.5} \text { g of } \mathrm{NaOH}=\frac{40 \times 4}{5.5 \times 40} \text { mole of } \mathrm{NaOH}\)

The molarity of the resultant solution = 0.727 = 0.73 M

Question 9. How many grams of dibasic acid (mol. weight 200) should be present in 100 mL of the aqueous solution to give a strength of 0.1 N?

1. 10 g
2. 2 g
3. 1 g
4. 20 g

Answer: 3. 1 g

The strength of the solution is 0.1 N.

⇒ \(\frac{w}{E}=\frac{V \times N}{1000}\) (Equivalent weight = \(\frac{200}{2}=100\))

⇒ w = \(\frac{100 \times 0.1 \times 100}{1000}=1 \mathrm{~g}\)

Question 10. What is the molarity of H₂SO₄ solution, which has a density 1.84 g/cc at 35°C and contains 98% by weight?

1. 18.4 M
2. 18 M
3. 4.18 M
4. 8.14 M

Answer: 1. 18.4 M

We know that 98% H₂SO₄ by weight means 98 g of H₂SO₄ is present in 100 g of solution

Therefore, its weight is 98 and moles of \(\mathrm{H}_2 \mathrm{SO}_4\)

= \(\frac{\text { Weight of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Molecular weight }}=\frac{98}{98}=1\)

and volume of solution = \(\frac{\text { Mass }}{\text { Density }}\)

= \(\frac{100}{1.84}=54.35 \mathrm{~mL}=\frac{54.35}{1000} \mathrm{~L}\)

Therefore, molarity of \(\mathrm{H}_2 \mathrm{SO}_4\)

= \(\frac{\text { Moles of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Volume (in litres) }}=\frac{1 \times 1000}{54.35}=18.4 \mathrm{M}\)

Question 11. The concentration unit, independent of temperature, would be

1. Normality
2. Weight volume per cent
3. Molality
4. Molarity.

Answer: 3. Molality

The molality involves the weights of the solute and the solvent. Since the weight does not change with the temperature. therefore molality does not depend upon the temperature.

Question 12. How many grams of CH₃OH should be added to water to prepare a 150 mL solution of 2 M CH₃OH?

1. 9.6 x 10³
2. 2.4 x 10³
3. 9.6
4. 2.4

Answer: 3. 9.6

Since the molecular mass of CH₃OH is 32 therefore the quantity of CH₃OH to prepare 150 mL solution of 2 M CH₃OH

= \(\left(\frac{2}{1000}\right) \times 150 \times 32=9.6 \mathrm{~g}\)

Question 13. The correct option for the value of vapour pressure of a solution at 45° C with benzene to octane in a molar ratio 3: 2 is

[At 45° C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume ideal gas]

1. 350 mm of Hg
2. 160 mm of Hg
3. 168 mm of Hg
4. 336 mm of Hg

Answer: 4. 336 mm of Hg

⇒ \(P_{\text {total }}=P_{\text {bemone }} x_{\text {benzene }}+P_{\text {ocane }} x_{\text {octane }}\)

= \(280 \times \frac{3}{5}+420 \times \frac{2}{5}=168+168\)

= \(336 \mathrm{~mm} \text { of } \mathrm{Hg}\)

Question 14. In water-saturated air, the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is

1. 1.18 atm
2. 1.76 atm
3. 1.176 atm
4. 0.98 atm.

Answer: 3. 1.176 atm

p_{water vapour} = x_{water vapour} x P_{water vapour}

= 0.02 x 1.2 = 0.024atm

P_total = P_{water vapour}+ P_{dry air}

1.2 = 0.024 + p_{dry air}

Pdry air= 1.2 – 0.024 = 1.176 atm

Partial vapour pressure is directly proportional to mole fraction, p ∝ x.

Question 15. p_A and p_B are the vapour pressures of pure liquid components, A and B, respectively of an ideal binary solution. If x_A represents the mole fraction of component A, the total pressure of the solution will be

1. p_A + x_A(p_B – p_A)
2. p_A + x_A(p_A – p_B)
3. p_B + x_A(p_B – p_A)
4. p_B + x_A(p_A– p_B)

Answer: 4. p_B + x_A(p_A– p_B)

According to Raoult’s law, \(p=x_A p_A+x_B p_B\)….(1)

For binary solutions, \(x_A+x_B=1, x_B=1-x_A\)….(2)

Putting the value of \(x_B\) from eqn. (2) to eqn. (1)

P = \(x_A p_A+\left(1-x_A\right) p_B=x_A p_A+p_B-x_A p_B\)

P = \(p_B+x_A\left(p_A-p_B\right)\)

Question 16. The vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 25°C is 200 mm Hg and 41.5 mm Hg respectively. The vapour pressure of the solution obtained by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at the same temperature will be

(Molecular mass of CHCl₂ = 119.5 u and molecular mass of CH₂Cl₂ = 85 u)

1. 173.9 mm Hg
2. 615.0 mm Hg
3. 347.9 mm Hg
4. 285.5 mm Hg

Answer: None

⇒ \(p_{\mathrm{CHCl}_5}^{\circ}=200 \mathrm{~mm} \mathrm{Hg}, p^{\mathrm{a}} \mathrm{CH}_2 \mathrm{Cl}_2=41.5 \mathrm{~mm} \mathrm{Hg}\)

Moles of \(\mathrm{CHCl}_3=\frac{\text { Weight }}{\text { Molecular weight }}=\frac{25.5}{119.5}=0.2132111\)

Moles of \(\mathrm{CH}_2 \mathrm{Cl}_2=\frac{40}{85}=0.470\)

⇒ \(x_{\mathrm{CHCl}_3}=\frac{0.213}{0.213+0.470}=0.31\)

⇒ \(x_{\mathrm{CH}_2 \mathrm{Cl}_2}=\frac{0.470}{0.213+0.470}=0.69\)

⇒ \(P_{\text {tocal }}=p^{\circ}{ }_{\mathrm{CHC}_3} x_{\mathrm{CHCl}_3}+p^{\circ}{ }_{\mathrm{CH}_2 \mathrm{Cl}_2} x_{\mathrm{CH}_2 \mathrm{C}_2}\)

= \(200 \times 0.31+41.5 \times 0.69=62+28.63=90.63 \mathrm{~mm} \mathrm{Hg}\)

Question 17. A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20 °C are 440 mm Hg for pentane and 120 mm Hg for hexane. Tire mole fraction of pentane in the vapour phase would be

1. 0.200
2. 0.549
3. 0.786
4. 0.478

Answer: 4. 0.478

⇒ \(\frac{n_{C_6} H_{12}}{n_{C_6} H_{14}}=\frac{1}{4}\)

⇒ \(x_{\mathrm{C}_5 \mathrm{H}_{12}}=\frac{1}{5} \text { and } x_{\mathrm{C}_6 \mathrm{H}_{14}}=\frac{4}{5}\)

⇒ \(p_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ}=440 \mathrm{~mm} \mathrm{Hg} ; P_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ}=120 \mathrm{~mm} \mathrm{Hg}\)

⇒ \(P_{\text {total }}=p^{\circ}{ }_{\mathrm{C}_5 \mathrm{H}_{12}} x_{\mathrm{C}_5 \mathrm{H}_{12}}+p^{\circ}{ }_{\mathrm{C}_6 \mathrm{H}_{14}} x_{\mathrm{C}_6 \mathrm{H}_{14}}^5\)

= \(440 \times \frac{1}{5}+120 \times \frac{4}{5}=88+96=184 \mathrm{~mm} \text { of } \mathrm{Hg}\)

By Raoult’s law, \(p_{\mathrm{C}_5 \mathrm{H}_{12}}=p^0 \mathrm{C}_5 \mathrm{H}_{12} x_{\mathrm{C}_5 \mathrm{H}_{12}}\)….(1)

⇒ \(x_{\mathrm{C}_5 \mathrm{H}_{12}} \rightarrow\) mole fraction of pentane in solution.

By Dalton’s law, \(P_{\mathrm{C}_3 \mathrm{H}_{12}}=x_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} P_{\text {total }}\)……(2)

⇒ \(x^{\prime}{ }_{\mathrm{C}_5 \mathrm{H}_{12}} \rightarrow\) mole fraction of pentane above the solution.

From (1) and (2), \(P_{\mathrm{C}_5 \mathrm{H}_{12}}=440 \times \frac{1}{5}=88 \mathrm{~mm} \text { of } \mathrm{Hg}\)

⇒ 88 = \(x^{\prime}{ }_{\mathrm{C}_5 \mathrm{H}_{12}} \times 184 \Rightarrow x^{\prime}{ }_{\mathrm{C}_5 \mathrm{H}_{12}}=\frac{88}{184}=0.478\)

Question 18. The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of the solution obtained by mixing 3 moles of P and 2 mol of Q would be

1. 72 torr
2. 140 torr
3. 68 torr
4. 20 torr

Answer: 1. 72 torr

By Raoult’s Law, \(P_T=p_p^0 x_p^0+p_Q^{\circ} x_Q\) where \(p_P^{\circ}=80\) torr, \(p_Q^{\circ}=60\) torr, \(x_P=\frac{3}{5} ; x_Q=\frac{2}{5}\)

⇒ \(P_T=80 \times \frac{3}{5}+60 \times \frac{2}{5}=48+24=72 \text { torr }\)

⇒ \(P_T=80 \times \frac{3}{5}+60 \times \frac{2}{5}=48+24=72 \text { torr }\)

Question 19. The mixture which shows positive deviation from Raoult’s law is

1. Ethanol + acetone
2. Benzene + toluene
3. Acetone + chloroform
4. Chloroethane + bromoethane.

Answer: 1. Ethanol + acetone

A mixture of ethanol and acetone shows a positive deviation from Raoult’s law.

In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of, the hydrogen bonds between them. Due to the weakening of interactions, the solution shows a positive deviation from Raoult’s law.

Question 20. For an ideal solution, the correct option is

1. Δ_mix G = 0 at constant T and P
2. Δ_mix S = 0 at constant T and P
3. Δ_mix V ≠ 0 at constant Tand P
4. Δ_mix H = 0 at constant T and P.

Answer: 4. Δ_mixH = 0 at constant T and P.

For an ideal solution, Δ_mix H = 0 and Δ_mix V = 0 at constant T and P.

Question 21. The mixture that forms maximum boiling azeotrope is

1. Heptane + octane
2. Water + nitric acid
3. Ethanol + water
4. Acetone + carbon disulphide.

Answer: 2. Water + nitric acid

Maximum boiling azeotropes are formed by those solutions which show negative deviations from Raoult’s law. H₂O and HNO₃ mixture show negative deviations.

Question 22. Which of the following statements is correct regarding a solution of two components A and B exhibiting positive deviation from ideal behaviour?

1. Intermolecular attractive forces between A-A and B-B are stronger than those between A-B.
2. Δ_mix H = 0 at constant T and p.
3. Δ_mix V= 0 at constant T and P.
4. Intermolecular attractive forces between A-A and B-B are equal to those between A-B

Answer: 1. Intermolecular attractive forces between A-A and B-B are stronger than those between A-B.

In case of positive deviation from Raoult’s Law, A-B interactions are weaker than those between A-A or B-B, i.e., in this case, the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules.

This means that in such a solution, molecules of A (or B) will find it easier to escape than in a pure state. This will increase the vapour pressure and result in positive deviation.

Question 23. Which one of the following is incorrect for an ideal solution?

1. ΔH_mix= 0
2. ΔU_mix = 0
3. ΔP = P_obs — P_{calculated by Raoults’s law-} = 0
4. ΔG_mix= 0

Answer: 4. ΔH_mix= 0

For an ideal solution, \(\Delta H_{\text {mix }}=0, \Delta V_{\mathrm{mix}}=0 \text {, }\)

Now, \(\Delta U_{\text {mix }}=\Delta H_{\text {mix }}-P \Delta V_{\text {mix }}\)

∴ \(\Delta U_{\mathrm{mix}}=0\)

Also, for an ideal solution, \(P_A=x_A P_A^{\circ}, P_B=x_B P_B^{\circ}\)

∴ \(\Delta P=P_{\text {cbs }}-P_{\text {calculaned by Rooult’s law }}=0\)

∴ \(\Delta G_{\text {mix }}= \Delta H_{\text {mix }}-T \Delta S_{\text {mix }}\)

For an ideal solution, \(\Delta S_{\text {mis }} \neq 0\)

∴ \(\Delta G_{\text {mix }} \neq 0\)

Question 24. Which of the following statements about the composition of the vapour over an ideal 1: 1 molar mixture of benzene and toluene is correct?

Assume that the temperature is constant at 25°C. (Given, vapour pressure data at 25°C, benzene = 12.8 kPa, toluene = 3.85 kPa)

1. The vapour will contain equal amounts of benzene and toluene.
2. Not enough information is given to make a prediction.
3. The vapour will contain a higher percentage of benzene
4. The vapour will contain a higher percentage of toluene.

Answer: 3. The vapour will contain a higher percentage of benzene

⇒ \(p_{\text {Bewaene }}\) = \(x_{\text {Benzene }}p_{\text {Benaene }}^{\circ}\)

⇒ \(P_{\text {Taluene }}\) = \(x_{\text {Talsene }}p_{\text {Toluene }}^{\circ}\)

For an ideal 1: 1 molar mixture of benzene and toluene, \(x_{\text {Benzene }}=\frac{1}{2} \text { and } x_{\text {Toluene }}=\frac{1}{2}\)

⇒ \(P_{\text {Benzene }}=\frac{1}{2} p_{\text {Benzene }}^{\text {o }}=\frac{1}{2} \times 12.8 \mathrm{kPa}=6.4 \mathrm{kPa}\)

⇒ \(P_{\text {Toluene }}=\frac{1}{2} \rho_{\text {Toluene }}^{\mathrm{a}}=\frac{1}{2} \times 3.85 \mathrm{kPa}=1.925 \mathrm{kPa}\)

Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene.

Question 25. Which condition is not satisfied by an ideal solution?

1. Δ_mixV = 0
2. Δ_mixS = 0
3. Obeyance to Raoulfs Taw
4. Δ_mixH = 0

Answer: 2. Δ_mixS = 0

For an ideal solution:

• Volume change (ΔV) on mixing should be zero.
• Heat change (ΔH) on mixing should be zero.
• Obeys Raouitt law at every range of concentration.
• Entropy change ((ΔS) on mixing ≠ 0.

Question 26. A solution of acetone in ethanol

1. Obeys Raoult’s law
2. Shows a negative deviation from Raoult’s law
3. Shows a positive deviation from Raoult’s law
4. Behaves like a near-ideal solution.

Answer: 3. Shows a positive deviation from Raoult’s law

Both the components escape easily showing higher vapour pressure than the expected value. This is due to the breaking of some hydrogen bonds between ethanol molecules.

Question 27. A solution containing components A and B follows Raoulf’s law

1. A – B attraction force is greater than A – A and B-B
2. The a-B attraction force is less than A-A and B-B
3. A-B attraction force remains the same as A – A and B-B
4. The volume of the solution is different from the sum of the volume of the solute and solvent.

Answer: 3. A-B attraction force remains the same as A – A and B-B

Raoultt’s law is valid for ideal solutions only. The element of non-ideality enters into the picture when the molecules of the solute and solvent affect each other’s intermolecular forces. A solution containing components of A and B behaves as an ideal solution when the A – B attraction force remains the same as the A – A and B – B attraction forces.

Question 28. All form ideal solutions except

1. C₆H₆ and C₆H₅CH₃
2. C₂H₅Br and C₂H₅I
3. C₆H₅Cl and C₆H₅Br
4. C₂H₅I and C₂H₅OH

Answer: 4. C₂H₅I and C₂H₅OH

C₂H₅OH is associated through H-bonding while C₂H₅I does not show H-bonding.

Question 29. An ideal solution is formed when its components

1. Have no volume change on mixing
2. Have no enthalpy change on mixing
3. Have both the above characteristics
4. Have high solubility.

Answer: 3. Have both of the above characteristics

For the ideal solution, ΔV_mixing = 0 and ΔH_mixing = 0

Question 30. The following solutions were prepared by dissolving 10 g of glucose (C₆H₁₂O₆) in 250 ml of water (P₁). 10 g of urea (CH₄N₂O) in 250 ml of water (P₂) and 10 g of sucrose (C₁₂H₂₂O₁₁) in 250 ml of water (P₃). The right option for the decreasing order of osmotic pressure of these solutions is

1. P₃ > P₁ > P₂
2. P₂ > P₁ > P₃
3. P₁ >P₂> P₃
4. P₂ > P3 > P₁

Answer: 2. P₂ > P₁ > P₃

Osmotic pressure \((\pi)=C R T\)

∴ \(\pi \propto \mathrm{C}\)

For glucose solution, \(C_1=\frac{10}{180} \times \frac{1000}{250}=0.22 M\)

For urea solution, \(C_2=\frac{10}{60} \times \frac{1000}{250}=0.66 \mathrm{M}\)

For sucrose solution, \(C_3=\frac{10}{342} \times \frac{1000}{250}=0.117 \mathrm{M}\)

Hence, the order of osmotic pressure is \(P_2>P_1>P_3\).

Question 31. The freezing point depression constant (K_f) of benzene is 5.12 K kg mol^-1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places)

1. 0.20 K
2. 0.80 K
3. 0.40 K
4. 0.60 K

Answer: 3. 0.40 K

Given K_f = 5.12 K kg mol^-1, m =0.078m

ΔT_f = K_f x m = 5.12 x 0.078 = 0.39936 ≈ 0.40 K

Question 32. If the molality of the dilute solution is doubled, the value of the molal depression constant K_f will be

1. Halved
2. Tripled
3. Unchanged
4. Doubled.

Answer: 3. Unchanged

The value of molal depression-constant, K_f is constant for a particular solvent, thus, it will be unchanged when the molality of the dilut. the solution is doubled.

Question 33. At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be

1. 102 °C
2. 103 °C
3. 101 °C
4. 100 °C

Answer: 3. 101 °C

Given: \(W_B=6.5 \mathrm{~g}, W_A=100 \mathrm{~g}\), \(p_{\mathrm{s}}=732 \mathrm{~mm}, K_b=0.52, T_b^{\circ}=100^{\circ} \mathrm{C}, p^{\circ}=760 \mathrm{~mm}\)

∴ \(\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n_2}{n_1} \Rightarrow \frac{760-732}{760}=\frac{n_2}{100 / 18}\)

⇒ \(n_2=\frac{28 \times 100}{760 \times 18}=0.2046 \mathrm{~mol}\)

⇒ \(\Delta T_b=K_b \times m\)

⇒ \(T_b-T_b^{\circ}=K_b \times \frac{n_2 \times 1000}{W_A(\mathrm{~g})}\)

⇒ \(T_b-100^{\circ} \mathrm{C}=\frac{0.52 \times 0.2046 \times 1000}{100}=1.06\)

⇒ \(T_b=100+1.06=101.06^{\circ} \mathrm{C}\)

Question 34. 200 mL of an aqueous solution of a protein contains 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 x 10^-3 bar. The molar mass of the protein will be (R = 0.083 L bar mol^-1 K^-1)

1. 51022 g mol^-1
2. 122044 g mol^-1
3. 31011 g mol^-1
4. 61038 g mol^-1

Answer: 4. 61038 g mol^-1

We know that \(p V=n R T\), where\(n=\frac{w}{M}\)

πV = \(\frac{w}{M} R T\)

M = \(\frac{w R T}{\pi V}=\frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times \frac{200}{1000}}\)

= \(\frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.2}=61038 \mathrm{~g} \mathrm{~mol}^{-1}\)

Question 35. A solution of sucrose (molar mass = 342 g mol^-1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (K_f for water = 1.86 K kg mol^-1)

1. -0.372 °C
2. -0.520 °C
3. +0.372 °C
4. -0.570 °C

Answer: 1. -0.372 °C

We know, \(\Delta T_f=K_f m\)

m = \(\frac{w_B}{M_B} \times \frac{1000}{W_A}=\frac{68.5 \times 1000}{342 \times 1000}=\frac{68.5}{342}\)

⇒ \(\Delta T_f=1.86 \times \frac{68.5}{342}=0.372^{\circ} \mathrm{C}\)

⇒ \(T_f=0-0.372^{\circ} \mathrm{C}=-0.372^{\circ} \mathrm{C}\)

Question 36. During osmosis, the flow of water through a semipermeable membrane is

1. From solution having lower concentration only
2. From solution having higher concentration only
3. From both sides of the semipermeable membrane with equal flow rates
4. From both sides of the semipermeable membrane with unequal flow rates.

Answer: 1. From solution having lower concentration only

Question 37. 1.00 g of a non-electrolyte solute (molar mass 250 g mol^-1) was dissolved in 51.2 g of benzene. If the freezing point constant, Kf of benzene is 5.12 K_f kg mol^-1, the freezing point of benzene will be lowered by

1. 0.2 K
2. 0.4 K
3. 0.3 K
4. 0.5 K

Answer: 2. 0.4 K

= \(M_B=\frac{1000 \times K_f \times w_B}{W_A \times \Delta T_f}\)

or, \(250=\frac{1000 \times 5.12 \times 1}{51.2 \times \Delta T_f}\)

∴ \(\Delta T_f=\frac{1000 \times 5.12 \times 1}{51.2 \times 250}=0.4 \mathrm{~K}\)

Question 38. A solution containing 10 g per dm³ of urea (molecular mass = 60 g mol^-1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is

1. 200 g mol^-1
2. 250 g mol^-1
3. 300 g mol^-1
4. 350 g mol^-1

Answer: 3. 300 g mol^-1

Molar concentration of urea = \(\frac{10}{60}\) dm3

Molar concentration of non-volatile solution = \(\frac{50}{M_B} \mathrm{~L}^{-1}=\frac{50}{M_B} \mathrm{dm}^{-3}\)

For isotonic solutions, \(\frac{10}{60}=\frac{50}{M_B}\)

M_B =300 g mol^-1

Question 39. A solution of urea (mol. mass 56 g mol^-1) boils at 100.18°C at the atmospheric pressure. If K_f and K_b for water are 1.86 and 0.512 K kg mol^-1 respectively, the above solution will freeze at

1. 0.654°C
2. – 0.654°C
3. 6.54°C
4. -6.54°C

Answer: 2. – 0.654°C

⇒ \(\Delta T_f=K_f m\)….(1)

⇒ \(\Delta T_b=K_b m\)….(2)

⇒ \(\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}\)….(3)

⇒ \(\Delta T_f\) depression in freezing point

⇒ \(\Delta T_b \rightarrow\) elevation in boiling point

⇒ \(K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)

⇒ \(K_b=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \Delta T_b=100.18-100=0.18\)

From eq. (3), \(\frac{\Delta T_f}{0.18}=\frac{1.86}{0.512}\)

⇒ \(\Delta T_f=0.654=T_f^{\circ}-T_f=0-T_f \Rightarrow T_f=-0.654^{\circ} \mathrm{C}\)

Freezing point of urea in water \(=-0.654^{\circ} \mathrm{C}\)

Question 40. A solution contains a non-volatile solute of molecular mass M₂. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure? Pure water can be obtained from seawater by

1. \(M_2=\left(\frac{m_2}{\pi}\right) V R T\)
2. \(M_2=\left(\frac{m_2}{V}\right) \frac{R T}{\pi}\)
3. \(M_2=\left(\frac{m_2}{V}\right) \pi R T\)
4. \(M_2=\left(\frac{m_2}{V}\right) \frac{\pi}{R T}\)

Answer: 2. \(M_2=\left(\frac{m_2}{V}\right) \frac{R T}{\pi}\)

For dilute solution, \(\pi=\frac{n}{V} R T\)

⇒ \(\pi V=\frac{m_2}{M_2} R T \Rightarrow M_2=\frac{m_2 R T}{\pi V}\)

Question 41. Pure water can be obtained from seawater by

1. Centrifugation
2. Plasmolysis
3. Reverse osmosis
4. Sedimentation

Answer: 3. Reverse osmosis

Question 42. From the colligative properties of the solution, which one is the best method for the determination of the molecular weight of proteins and polymers?

1. Osmotic pressure
2. Lowering in vapour pressure
3. Lowering in freezing point
4. Elevation in boiling point

Answer: 1. Osmotic pressure

Question 43. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte solid, weighing 2.175 g is added to 39.08 of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of a solid substance?

1. 69.5
2. 59.6
3. 49.50
4. 79.8

Answer: 1. 69.5

p°_benzene = 640 mm Hg, p_s = 600 mm Hg

w_B = 2.175 g, W_benzene = 39.08 g

From Raoultt law \(\frac{p^{\circ}-p_8}{p^{\circ}}=\frac{w_B \times M_{\text {benzene }}}{W_{\text {benzene }} \times M_B} \Rightarrow \frac{640-600}{640}=\frac{2.175 \times 78}{39.08 \times M_B}\)

⇒ M_B = 69.5

Question 44. If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by 0.216°C, than that of the pure solvent. The molecular weight of the substance (Molal elevation constant for the solvent is 2.16°C) is

1. 10.1
2. 100
3. 1.01
4. 1000

Answer: 1. 10.1

⇒ \(w_B=0.15 \mathrm{~g}, W_A=15 \mathrm{~g}, \Delta T_b=0.216^{\circ} \mathrm{C}\)

⇒ \(K_b=2.16, m=? \)

As \(\Delta T_b=\frac{1000 \times K_b \times w_B}{M_B \times W_A}\)

⇒ \(M_B=\frac{1000 \times 2.16 \times 0.15}{0.216 \times 15}=100\)

Question 45. A 5% solution of cane sugar (mol. wt. = 342) is isotonic with 1% solution of a substance X. The molecular weight of X is

1. 68.4
2. 171.2
3. 34.2
4. 136.8

Answer: 1. 68.4

For isotonic solutions, \(C_1=C_2\) \(\frac{W_1}{M_1 V_1}=\frac{W_2}{M_2 V_2} \Rightarrow \frac{5}{342 \times 0.1}=\frac{1}{M_2 \times 0.1}\)

⇒ \(M_2=\frac{342}{5}=68.4\)

Question 46. The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?

1. 0.4
2. 0.6
3. 0.8
4. 0.2

Answer: 2. 0.6

x₂ (mole fraction of solute) = 0.2

From Raoults law \(\frac{p^a-p_8}{p^{\circ}}=x_2 \Rightarrow \frac{10}{p^{\circ}}=0.2 \Rightarrow p^{\circ}=50 \mathrm{~mm} \mathrm{Hg}\)

Again, when \(p^{\circ}-p_s=20 \mathrm{~mm} \mathrm{Hg}\), then \(\frac{p^{\circ}-p_S}{p^{\circ}}=\) mole fraction of solute \(=\frac{20}{50}=0.4\)

⇒ mole fraction of solvent \(=1-0.4=0.6\)

Question 47. The vapour pressure of CCl₄ at 25°C is 143 mm Hg If 0.5 g of a non-volatile solute (mol. weight = 65) is dissolved in 100 g CCl₄, the vapour pressure of the solution will be

1. 199.34 mm Hg
2. 143.99 mm Hg
3. 141.43 mm Hg
4. 94.39 mm Hg.

Answer: 3. 141.43 mm Hg

Vapour pressure of pure solvent (p°_A) = 143 mm Hg, weight of solute (w_B) = 0.5 g, weight of solvent (W_A) = 100 g, the molecular weight of solute M_B = 65 and molecular weight of solvent (M_A) = 154.

⇒ \(\frac{p_A^{\circ}-p_s}{p_A^a}=\frac{w_B M_A}{M_B W_A} \text { or } \frac{143-p_s}{143}=\frac{0.5 \times 154}{65 \times 100}\)

or \(p_s=141.31 \mathrm{~mm} \mathrm{Hg}\)

Question 48. The relationship between osmotic pressure at 273 K when 10 g glucose (p₁), 10 g urea (p₂), and 10 g sucrose (p₃) are dissolved in 250 mL of water is

1. p₂>p₁ > p₃
2. p₂>p₃>p₁
3. p₁>p₂>p₃
4. p₃>p₁>p₂

Answer: 1. p₂>p₁> p₃

Weight of glucose = 10 g,

Weight of urea = 10 g and weight of sucrose = 10 g

The number of moles of glucose, \(\left(n_1\right)=\frac{\text { Weight }}{\text { Molecular weight }}=\frac{10}{180}=0.05\)

Similarly, number of moles of urea \(\left(n_2\right)=\frac{10}{60}=0.16\) and the number of moles of sucrose \(\left(n_3\right)=\frac{10}{342}=0.03\)

Question 49. According to Raoult’s law, the relative lowering of vapour pressure for a solution is equal to

1. Mole fraction of solute
2. Mole fraction of solvent
3. Moles of solute
4. Moles of solvent.

Answer: 1. Mole fraction of solute

Question 50. If 0.1 M solution of glucose and 0.1 M solution of urea are placed on two sides of the semipermeable membrane to equal heights, then it will be correct to say that

1. There will be no net movement across the membrane
2. Glucose will flow towards the glucose solution
3. Urea will flow towards the glucose solution
4. Water will flow from the urea solution to glucose.

Answer: 1. There will be no net movement across the membrane

There is no net movement of the solvent through the semipermeable membrane between two solutions of equal concentration

Question 51. Which one is a colligative property?

1. Boiling point
2. Vapour pressure
3. Osmotic pressure
4. Freezing point

Answer: 3. Osmotic pressure

The properties which depend only upon the number of solute particles present in the solution irrespective of their nature are called colligative properties. Lowering in vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure are colligative properties.

Question 52. Blood cells retain their normal shape in solution which are

1. Hypotonic to blood
2. Isotonic to blood
3. Hypertonic to blood
4. Equinormal to blood.

Answer: 2. Isotonic to blood

Blood cells neither swell nor shrink in isotonic solution. The solutions having the same osmotic pressure are called isotonic solutions.

Question 53. The relative lowering of the vapour pressure is equal to the ratio between the number of

1. Solute molecules to the solvent molecules
2. Solute molecules to the total molecules in the solution
3. Solvent molecules to the total molecules in the solution
4. Solvent molecules to the total number of ions of the solute.

Answer: 2. Solute molecules to the total molecules in the solution

Relative lowering of vapour pressure is equal to the mole fraction of solute which is the ratio of solute molecules to the total molecules in the solution

Question 54. The van’t Hoff factor (1) for a dilute aqueous solution of the strong electrolyte barium hydroxide is

1. 0
2. 1
3. 2
4. 3

Answer: 4. 3

Being a strong electrolyte, Ba(OH)₂ undergoes 100% dissociation in a dilute aqueous solution, \(\mathrm{Ba}(\mathrm{OH})_{2(a q)} \rightarrow \mathrm{Ba}^{2+}{ }_{\langle(q)}+2 \mathrm{OH}^{-}{ }_{(a q)}\)

Thus, vant Hoff factor i = 3

Question 55. The boiling point of 0.2 mol kg^-1 solution of X in water is greater than the equimolal solution of Y in water. Which one of the following statements is true in this case?

1. The molecular mass of X is less than the molecular mass of Y.
2. Y is undergoing dissociation in water while X undergoes no change.
3. X is undergoing dissociation in water.
4. The molecular mass of X is greater than the molecular mass of Y.

Answer: 3. X is undergoing dissociation in water.

ΔT_b = i x K_b m
For equimolal solutions, elevation in boiling point will be higher if the solution undergoes dissociation i.e., i > l.

Question 56. Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?

1. KCl
2. C₆H₁₂O₆
3. AI₂(SO₄)₃
4. K₂SO₄

Answer: 3. AI₂(SO₄)₃

⇒ \(\Delta T_f=i \times K_f \times m\)

So, \(\Delta T_f \propto i\) (van’t Hoff factor)

Hence, I am the maximum i.e., 5 for AI₂(SO₄)₃

Question 57. The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in another solvent is respectively

1. Less than one and greater than one
2. Less than one and less than one
3. Greater than one and less than one
4. Greater than one and greater than one.

Answer: 3. Greater than one and less than one

From the value of van t Hoff factor I it is possible to determine the degree of dissociation or association. In the case of dissociation, i is greater than I and in the case of an association, i is less than 1.

Question 58. The freezing point depression constant for water is -1.86 °Cm^-1, If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by -3.82 °C. Calculate the van’t Hoff factor for Na₂SO₄.

1. 2.05
2. 2.63
3. 3.11
4. 0.381

Answer: 2. 2.63

We know that, \(\Delta T_f=i \times K_f \times \frac{w_B \times 1000}{M_B \times W_A}\)

Given: \(\Delta T_f=3.82, K_f=1.86,\)

⇒ \(w_B=5, M_B=142, W_A=45\)

i = \(\frac{\Delta T_f \times M_B \times W_A}{K_f \times w_B \times 1000}=\frac{3.82 \times 142 \times 45}{1.86 \times 5 \times 1000}=2.63\)

Question 59. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K_f for water is 1.86°C/m, the freezing point of the solution will be

1. – 0.18 °C
2. – 0.54 °C
3. – 0.36 °C
4. – 0.24 °C

Answer: 4. – 0.24 °C

We know that ΔT_f =i x K_f x m

Here i is vant Hoff factor.

i for weak acid is 1 + α.

Here α is the degree of dissociation i,e., 30/100 = 0.3

∴ i = 1 + α =1+0.3=1.3

ΔT_f = i x K_f x m= 1.3 x 1.86 x 0.1 = 0.24

∴ Freezing point of the solution, T_f= T°_f – ΔT_f =0-0.24=-0.24°C

Question 60. An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?

1. Addition of NaCl
2. Addition of Na₂SO₄ m
3. Addition of 1.00 molal KI
4. Addition of water

Answer: 4. Addition of water

The addition of water to an aqueous solution of KI causes the concentration of the solution to decrease thereby increasing the vapour pressure. In the other three options, the electrolytes undergo ionization, which leads to a lowering of vapour pressure.

Question 61. A 0.0020 m aqueous solution of an ionic compound [CO(NH₃)₅(NO₂)]Cl freezes at -0.00732 °C. The number of moles of ions that 1 mol of ionic compound produces on being dissolved in water will be (K_f = -1.86 °C/m)

1. 3
2. 4
3. 1
4. 2

Answer: 4. 2

The number of moles of ions produced by 1 mol of ionic compound = i

Applying, ΔT_f =i x K_f x m

0.00732 = i x 1.86 x 0.002

⇒ i = \(\frac{0.00732}{1.86 \times 0.002}=1.96 \div 2\)

Question 62. 0. 5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K_f for water is 1.86 K kg mol^-1, the lowering in freezing point of the solution is

1. 0.56 K
2. 1.12 K
3. -0.56 K
4. -1.12 K

Answer: 2. 1.12 K

HX ⇔ H⁺ + x^–

Total=1+α

∴ i=1 + α = 1+0.2 = 1.2

ΔT_f = i x K_f x m = 1.2 x 1.86 x 0.5 = 1.116 K≈ 1.12 K

Question 63. Which of the following 0.10m aqueous solution will have the lowest freezing point?

1. KI
2. C₁₂H₂₂O₁₁
3. Al₂(SO₄)₃
4. C₅H₁₀O₅

Answer: 3. Al₂(SO₄)₃

Since Al₂ (SO₄)₃ gives a maximum number of ions on dissociation, therefore it will have the lowest freezing point.

ΔT_f = i K_f m

Question 64. Which of the following salts has the same value of van’t Hoff factor (1) as that of K₃[Fe(CN)₆]?

1. Na₂SO₄
2. Al(NO₃)₃
3. Al₂(SO₄)₄
4. NaCl

Answer: 2. Al(NO₃)₃

⇒ \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 3 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\) and \(\mathrm{Al}\left(\mathrm{NO}_3\right)_3 \rightleftharpoons \mathrm{Al}^{3+}+3 \mathrm{NO}_3^{-}\)

Since both AI(NO₃)₃ and K₃[Fe(CN)₆] give the same number of ions, therefore they have the same van t Hoff factor.

Question 65. At 25°C, the highest osmotic pressure is exhibited by 0.1 M solution of

1. Glucose
2. Urea
3. CaCl₂
4. KCl.

Answer: 3. CaCl₂

In solution, CaCl_f gives three ions, KCl gives two ions while glucose and urea are covalent molecules so they do not undergo ionisation. Since osmotic pressure is a colligative property and depends upon the number of solute particles (ions), therefore, 0.1 M solution of CaCI₂ exhibits the highest osmotic pressure.

Question 66. Which of the following aqueous solutions has a minimum freezing point?

1. 0.01 m NaCl
2. 0.005 m C₂H₅OH
3. 0.005 mMgI₂
4. 0.005 mMgSO₄

Answer: 1. 0.01 m NaCl

Here, ΔT_f = i x K_f x m

vant Hoff factor, i = 2 for NaCl, so conc. = 0.02, which is the maximum in the present case.

Hence, ΔT_f is the maximum or freezing point is minimum in 0.01 m NaCI.

Chemical Kinetics

Question 1. For the chemical reaction, \(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}\) the correct option is

1. \(3 \frac{d\left[\mathrm{H}_2\right]}{d t}=2 \frac{d\left[\mathrm{NH}_3\right]}{d t}\)
2. \(-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}=-\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)
3. \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=2 \frac{d\left[\mathrm{NH}_3\right]}{d t}\)
4. \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)

Answer: 4. \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)

For the given chemical reaction,

Rate of reaction = \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)

Question 2. The rate of the reaction \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\) can be written in three ways.

• \(\frac{-d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=k\left[\mathrm{~N}_2 \mathrm{O}_5\right]\)
• \(\frac{d\left[\mathrm{NO}_2\right]}{d t}=k^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right]\); \(\frac{d\left[\mathrm{O}_2\right]}{d t}=k^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right]\)

The relationship between k and k’ and between k and k” are

1. k’ = 2k, k” = k
2. k’ = 2k, k” = k/2
3. k’ = 2k, k” = 2k
4. k’ = k, k” = k

Answer: 2. k’ = 2k, k” = k/2

For the reaction, \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\)

⇒ \(-\frac{1}{2} (k) \frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t} (k’) =+\frac{1}{4} \frac{d\left[\mathrm{NO}_2\right]}{d t}=+\frac{d\left[\mathrm{O}_2\right]}{d t}(k”)\)

⇒ \(\frac{1}{2} k=\frac{1}{4} k^{\prime}=k^{\prime \prime}, k^{\prime}=2 k ; k^{\prime \prime}=\frac{1}{2} k\)

Question 3. For the reaction \(\mathrm{N}_2 \mathrm{O}_{5(g)} \rightarrow 2 \mathrm{NO}_{2(g)}+1 / 2 \mathrm{O}_{2(g)}\) the value of rate of disappearance of N₂O₅ is given as 6.25 x 10^-3 mol L^-1s^-1 The rate of formation of N02 and 02 is given respectively as

1. \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
2. \(1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
3. \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
4. \(1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(6.25 \times 10^{-3} \mathrm{moll}^{-1} \mathrm{~s}^{-4}\)

Answer: 2. \(1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

⇒ \(\mathrm{N}_2 \mathrm{O}_{5(g)} \rightarrow 2 \mathrm{NO}_{2(g)}+1 / 2 \mathrm{O}_{2(g)}\)

For the given reaction the rate is written as \(\frac{-d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{2 d\left[\mathrm{O}_2\right]}{d t}\)

Given that \(\frac{-d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

∴ \(\frac{d\left[\mathrm{NO}_2\right]}{d t}=2 \times 6.25 \times 10^{-3}=1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

and \(\frac{d\left[\mathrm{O}_2\right]}{d t}=\frac{6.25 \times 10^{-3}}{2}=3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Question 4. For the reaction, \(\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3\) if \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)  the value of \(\frac{-d\left[\mathrm{H}_2\right]}{d t}\) would be

1. \(4 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
2. \(6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
3. \(1 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
4. \(3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Answer: 4. \(3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

For reaction, \(\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3\)

Rate = \(\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}=-\frac{d\left[\mathrm{~N}_2\right]}{d t}\)

Given, \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

∴ \(-\frac{d\left[\mathrm{H}_2\right]}{d t}=\frac{3}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}=\frac{3}{2} \times 2 \times 10^{-4}\)

⇒ \(-\frac{d\left[\mathrm{H}_2\right]}{d t}=3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Question 5. In the reaction: \(\mathrm{BrO}_{3(\text { aq) }}^{-}+5 \mathrm{Br}_{(\text {aq) }}+6 \mathrm{H}_{(\text {aq) }}^{+} \rightarrow 3 \mathrm{Br}_{2(t)}+3 \mathrm{H}_2 \mathrm{O}_{(l)}\). The rate of disappearance of bromide ions as

1. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{5}{3} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)
2. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=\frac{5}{3} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)
3. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)
4. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)

Answer: 4. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)

For the given reaction, \(\mathrm{BrO}_3^{-}{ }_{(a q)}^{-}+5 \mathrm{Br}_{(a q)}^{-}+6 \mathrm{H}^{+}{ }_{(a q)} \rightarrow 3 \mathrm{Br}_{2(l)}+3 \mathrm{H}_2 \mathrm{O}_{(l)}\)

Rate of reaction in terms of \(\mathrm{Br}_2\) and \(\mathrm{Br}^{-}\) is,

Rate = \(\frac{1}{3} \frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{1}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t} \)

∴ \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)

Question 6. Consider the reaction: \(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)}\) The equality relationship between, \(\frac{d\left[\mathrm{NH}_3\right]}{d t}\) and \(-\frac{d\left[\mathrm{H}_2\right]}{d t}\) is

1. \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{d\left[\mathrm{H}_2\right]}{d t}\)
2. \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)
3. \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)
4. \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{3}{2} \frac{d\left[\mathrm{H}_2\right]}{d t}\)

Answer: 3. \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)

⇒ \(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)}\)

Rate = \(\frac{-d\left[\mathrm{~N}_2\right]}{d t}=-\frac{d\left[\mathrm{H}_2\right]}{3 d t}=+\frac{d\left[\mathrm{NH}_3\right]}{2 d t}\)

Hence, \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)

Question 7. For the reaction, 2A + B → 3C + D,  which of the following does not express the reaction rate?

1. \(-\frac{d[A]}{2 d t}\)
2. \(-\frac{d[C]}{3 d t}\)
3. \(-\frac{d[B]}{d t}\)
4. \(\frac{d[D]}{d t}\)

Answer: 2. \(-\frac{d[C]}{3 d t}\)

2 A+B → 3 C+D

rate = \(\frac{-d[A]}{2 d t}=-\frac{d[B]}{d t}=\frac{d[C]}{3 d t}=\frac{d[D]}{d t}\)

A negative sign shows a decrease in concentration.

Question 8. 3A → 2B, rate of reaction \(\frac{+d[B]}{d t}\) is equal to

1. \(-\frac{3}{2} \frac{d[A]}{d t}\)
2. \(-\frac{2}{3} \frac{d[A]}{d t}\)
3. \(-\frac{1}{3} \frac{d[A]}{d t}\)
4. \(+2 \frac{d[A]}{d t}\)

Answer: 2. \(-\frac{2}{3} \frac{d[A]}{d t}\)

3 A → 2 B

Rate of the reaction = \(\frac{1}{2} \frac{d[B]}{d t}=-\frac{1}{3} \frac{d[A]}{d t}\)

⇒ \(\frac{d[B]}{d t}=-\frac{2}{3} \frac{d[A]}{d t}\)

Question 9. For the reaction, \(\mathrm{H}^{+}+\mathrm{BrO}_3^{-}+3 \mathrm{Br}^{-} \longrightarrow 5 \mathrm{Br}_2+\mathrm{H}_2 \mathrm{O}\) which of the following relations correctly represents the consumption and formation of products?

1. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)
2. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)
3. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{5}{3} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)
4. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=\frac{5}{3} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

Answer: 1. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

Rate of reaction = \(-\frac{1}{3} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}=+\frac{1}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

∴ \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

Question 10. For the reaction, \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}\) the rate of reaction is expressed as

1. \(\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=-\frac{\Delta[\mathrm{HI}]}{\Delta t}\)
2. \(-\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{HI}]}{\Delta t}\)
3. \(\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{\Delta[\mathrm{HI}]}{2 \Delta t}\)
4. None of these.

Answer: 2. \(-\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{HI}]}{\Delta t}\)

For \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(g)} \rightarrow 2 \mathrm{HI}_{(g)}\), the rate of reaction is

⇒ \(-\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{HI}]}{\Delta t}\)

The negative sign shows the disappearance of the reactant and the positive sign shows the appearance product

Question 11. For a certain reaction, the rate = k[A]²[B], when the initial concentration of A is tripled keeping the concentration of B constant, the initial rate would

1. Increase by a factor of nine
2. Increase by a factor of three
3. Decrease by a factor of nine
4. Increase by a factor of six.

Answer: 1. Increase by a factor of nine

⇒ \(r_1=k[A]^2[B]\)

Keeping concentration of B constant and tripling conc. of A, new rate would be \(r_2=k[3 A]^2[B]=9 k[A]^2[B]\)

∴ \(r_2=9 \times r_1\)

Question 12. Mechanism of a hypothetical reaction, \(X_2+Y_2 \longrightarrow 2 X Y\), is given below:

1. \(X_2 \rightarrow X+X\) (fast)
2. \(X+Y_2 \rightleftharpoons X Y+Y\) (slow)
3. \(X+Y \longrightarrow X Y\) (fast)

The overall order of the reaction will be

1. 2
2. 0
3. 1.5
4. 1

Answer: 3. 1.5

Correct the reactions given in question as \(X_2 \rightleftharpoons{ } X+X\) fast; \(X+Y_2 \longrightarrow X Y+Y\) (slow)

The slow step is the rate-determining step.

Rate \(=k[X]\left[Y_2\right]\)…(1)

Equilibrium constant for fast step, \(K=\frac{[X]^2}{\left[X_2\right]}\)

[X] = \(\sqrt{K\left[X_2\right]}\)

By substituting [X] in equation (1), we get

Rate = \(k \sqrt{K\left[X_2\right]}\left[Y_2\right]=k\left[X_2\right]^{1 / 2}\left[Y_2\right]\)

∴ Order of reaction = \(\frac{1}{2}+1=\frac{3}{2}=1.5\)

Question 13. The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. It is because the

1. The rate is proportional to the surface coverage
2. Rate is inversely proportional to the surface coverage
3. Rate is independent of the surface coverage
4. The rate of decomposition is very slow.

Answer: 1. Rate is proportional to the surface coverage

At low pressure, the rate is proportional to the surface coverage and is of first order while at high pressure, it follows zero-order kinetics due to the complete coverage of the surface area.

Question 14. The rate constant of the reaction A → B is 0. 6 x 10^-3 mol L^-1 s^-1. If the concentration of A is 5 M, then the concentration of B after 20 minutes is

1. 3.60 M
2. 0.36 M
3. 0.72 M
4. 1.08 M

Answer: 3. 0.72 M

The reaction is of zero order as the unit of rate constant is mol L^-1 s^-1.

Concentration of B =k x t = 0.6 x 10^-3 x 20 x 60 =0.72M

Question 15. For a reaction between A and B, the order with respect to A is 2 and the order with respect to B is 3. The concentrations of both A and B are doubled, the rate will increase by a factor of

1. 12
2. 16
3. 32
4. 10

Answer: 3. 32

Rate \(_1=k[A]^2[B]^3\)

Rate \(_2=k[2 A]^2[2 B]^3\)

Rate \(_2=32 k[A]^2[B]^3\)

∴ Rate \(_2=32\left(\text { Rate }_1\right)\)

Question 16. In a reaction, A + B → product, the rate is doubled when the concentration of B is doubled, and the rate increases by a factor of 8 when the concentration of both the reactants (A and B) are doubled, the rate law for the reaction can be written as

1. Rate = k[A][B]²
2. Rate = k[A]²[B]²
3. Rate = k[A][B]
4. Rate = k[A]²[B]

Answer: 4. Rate = k[A]²[B]

⇒ \(\begin{array}{ccc}
{[A]} & {[B]} & \text { Rate } \\
x & y & R …….(1)\\
x & 2 y & 2 R ……(2)\\
2 x & 2 y & 8 R….(3)
\end{array}\)

Let the rate law; rate \(=k[A]^a[B]^b\)

From data given, \((x)^a(y)^b=R\)….(4)

⇒ \((x)^a(2 y)^b=2 R\)…….(5)

Dividing eqn. (5) by (4), \(\frac{(2 y)^b}{(y)^b}=\frac{2 R}{R} \Rightarrow(2)^b=2=2^1\)

Thus, b=1

From data of (3) experiment, \((2 x)^a(2 y)^b=8 R\)….(6)

From equation (5) and (6), \(\frac{(2 x)^a}{(x)^a}=\frac{8 R}{2 R} \Rightarrow(2)^a=4=2^2\)

Thus, a=2. By replacing the values of a and b in rate law; rate \(=k[A]^2[B]\)

Question 17. Which one of the following statements for the order of a reaction is incorrect?

1. Order can be determined only experimentally.
2. Order is not influenced by the stoichiometric coefficient of the reactants.
3. The order of a reaction is the sum of power to the concentration terms of reactants to express the rate of reaction.
4. The order of reaction is always a whole number.

Answer: 4. Order of reaction is always a whole number.

The order of a reaction is not always a whole number. It can be zero, or fractional also.

Question 18. The unit of rate constant for a zero-order reaction is

1. mol L^-1 s^-1
2. L mol^-1 s^-1
3. L² mol^-2 s^-1
4. s^-1

Answer: 1. mol L^-1 s^-1

Rate = k[A]⁰

mol L^-1 s^-1 = k

Thus, the unit of rate constant is mol L^-1 s^-1

Question 19. During the kinetic study of the reaction, 2A + B → C + D, following results were obtained:

Based on the above data which one of the following is correct?

1. Rate = k[A]²[B]
2. Rate = k[A][B]
3. Rate = k[A]²[B]²
4. Rate = k[A][B]²

Answer: 4. Rate = k[A][B]²

Let the rate of reaction be given by rate = \(k[A]^a[B]^b\)

Now consider 2 and 3 where [A] is constant, \(\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}=\frac{[0.3]^a[0.2]^b}{[0.3]^a[0.4]^b}\)

⇒ \(\frac{1}{4}=\left(\frac{1}{2}\right)^b \Rightarrow b=2\)

Now consider 1 and 4, \(\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}=\frac{[0.1]^a[0.1]^b}{[0.4]^a[0.1]^b}\)

⇒ \(\frac{1}{4}=\left(\frac{1}{4}\right)^a \Rightarrow a=1\)

Rate = \(k[A][B]^2\)

Question 20. For the reaction, A + B → products, it is observed that

1. On doubling the initial concentration of an only, the rate of reaction is also doubled and
2. On doubling the initial concentration of both a and b, there is a change by a factor of 8 in the rate of the reaction.

The rate of this reaction is given by

1. Rate = k[A] [B]²
2. Rate = k[A]²[B]²
3. Rate = k[A][B]
4. Rate = k[A]²[B]

Answer: 1. Rate = k[A][B]²

R = \(k[A]^m[B]^n\)….(1)

2R = \(k[2 A]^m[B]^n\)…..(2)

from (1), (2) and (3), m=1, n=2….(3)

So, rate \(=k[A][B]^2\)

Question 21. The bromination of acetone that occurs in an acid solution is represented by this equation. \(\mathrm{CH}_3 \mathrm{COCH}_{3(a q)}+\mathrm{Br}_{2(a q)} \longrightarrow \mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{Br}_{(a q)}\) + \(\mathrm{H}_{(a q)}^{+}+\mathrm{Br}_{(a q)}^{-}\)

These kinetic data were obtained for given reaction concentrations. Initial concentrations, M

Based on these data, the rate equation is

1. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\left[\mathrm{H}^{+}\right]^2\)
2. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\left[\mathrm{H}^{+}\right]\)
3. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{H}^{+}\right]\)
4. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\)

Answer: 3. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{H}^{+}\right]\)

From the first two experiments, it is clear that when the concentration of Br₂ is doubled, the initial rate of disappearance of Br₂ remains unaltered. So, the order of reaction with respect to Br₂ is zero. Thus, the probable rate law for the reaction will be k[CH₃COCH₃][H⁺]

Question 22. The reaction of hydrogen and iodine monochloride is given as: \(\mathrm{H}_{2(g)}+2 \mathrm{ICl}_{(g)} \rightarrow 2 \mathrm{HCl}_{(g)}+\mathrm{I}_{2(g)}\) This reaction is of first order with respect to H_2(g) and ICl_(g), following mechanisms were proposed.

Mechanism A: \(\mathrm{H}_{2(g)}+2 \mathrm{ICl}_{(g)} \rightarrow 2 \mathrm{HCl}_{(g)}+\mathrm{I}_{2(g)}\)

Mechanism B: \(\mathrm{H}_{2(g)}+\mathrm{ICl}_{(g)} \rightarrow \mathrm{HCl}_{(g)}+\mathrm{HI}_{(g)} \text {; slow } \); \(\mathrm{HI}_{(g)}+\mathrm{ICl}_{(g)} \rightarrow \mathrm{HCl}_{(g)}+\mathrm{I}_{2(g)} \text {; fast }\)

Which of the above mechanism(s) can be consistent with the given information about the reaction?

1. A and B both
2. Neither A nor B
3. A only
4. B only

Answer: 4. B only

The slow step is the rate-determining step and it involves 1 molecule of H_2(g) and 1 molecule of ICI_(g). Hence, the rate will be, r = k[H_2(g)] [ICl_(g)]

i.e., the reaction is 1st order with respect to H_2(g) and ICl_(g).

Question 23. The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is

1. 2
2. -2
3. 1
4. -1

Answer: 2. -2

Rate of reaction \(=k[A] \alpha[B] \beta\)

α → order of reaction with respect to A

β → order of reaction with respect to B

⇒ \(r_1=k[A]^\alpha[B]^\beta\)

⇒ \(r_2=r_1 / 4=k[A]^\alpha[2 B]^\beta\)

⇒ \(\frac{r_1}{r_2}=\frac{k[A]^\alpha[B]^\beta}{k[A]^\alpha[2 B]^\beta} \Rightarrow 4=\left(\frac{1}{2}\right)^\beta \Rightarrow \beta=-2\)

Question 24. If the rate of the reaction is equal to the rate constant, the order of the reaction is

1. 0
2. 1
3. 2
4. 3

Answer: 1. 0

A → products

If \(-\frac{d x}{d t}=k\), it means \(-\frac{d x}{d t}=k[A]^0=k\)

Hence, the order of reaction must be zero.

Question 25. 2A → B + C, It would be a zero-order reaction when

1. The rate of reaction is proportional to the square of the concentration of A
2. The rate of reaction remains the same at any concentration of A
3. The rate remains unchanged at any concentration of B and C
4. The rate of reaction doubles if the concentration of B is increased to double.

Answer: 2. The rate of reaction remains the same at any concentration of A

2A → B+C

The rate equation of this reaction may be expressed as r = k[A]⁰, Order= 0, r=k

∴ The rate is independent of the concentration of the reactant A

Question 26. For the reaction; \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\) rate and rate constant are 1.02 x 10⁴ and 3.4 x 10^-5 sec^-1 respectively, then concentration of N₂O₅ at that time will be

1. 1.732
2. 3
3. 1.02 x 10^-4
4. 3.4 x 10⁵

Answer: 2. 3

⇒ \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\)

This is a first-order reaction.

∴ rate = \(k\left[\mathrm{~N}_2 \mathrm{O}_5\right]\)

⇒ \([\mathrm{N}_2 \mathrm{O}_5]=\text { rate } / k=\frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}=3\)

Question 27. The experimental data for the reaction, 2A + B₂ →2AB is

⇒ \(\begin{array}{cccl}
\text { Experiment } & {[\boldsymbol{A}]} & {\left[\boldsymbol{B}_2\right]} & \text { Rate }\left(\text { mole s }^{-1}\right) \\
1 & 0.50 & 0.50 & 1.6 \times 10^{-4} \\
2 & 0.50 & 1.00 & 3.2 \times 10^{-4} \\
3 & 1.00 & 1.00 & 3.2 \times 10^{-4}
\end{array}\)

The rate equation for the above data is

1. Rate = k [A]²[B]²
2. Rate = k [A]²[B]
3. Rate = k [B₂]
4. Rate = k [B₂]²

Answer: 3. Rate = k [B₂]

For the reaction, \(2 A+B_2 \rightleftharpoons 2 A B\),

Rate \(\propto[A]^x\left[B_2\right]^y\).

From experiment 1, \(1.6 \times 10^{-4} \propto[0.50]^x[0.50]^y\)….(1)

From experiment 2, \(3.2 \times 10^{-4} \propto[0.50]^x[1.00]^y\)…(2)

From experiment 3, \(3.2 \times 10^{-4} \propto[1.00]^x[1.00]^y\)….(3)

On dividing equation (3) by (2), we get, 1 = \(\left[\frac{1.00}{0.50}\right]^x \Rightarrow 1=2^x \Rightarrow 2^0=2^x \Rightarrow x=0\)

Now, divide equation (2) by equation (1) we get, 2 = \(\left[\frac{1.00}{0.50}\right]^y \Rightarrow 2=2^y \Rightarrow y=1\)

Thus, rate equation is : Rate \(=k[A]^0\left[B_2\right]^1=k\left[B_2\right]\)

Question 28. The given reaction, \(2 \mathrm{FeCl}_3+\mathrm{SnCl}_2 \rightarrow 2 \mathrm{FeCl}_2+\mathrm{SnCl}_4 \) is an example of

1. Third order reaction
2. First order reaction
3. Second order reaction
4. None of these.

Answer: 1. Third-order reaction

For a general reaction, xA + yB + zC → product, the order of the reaction is x + y + z.

Since three molecules undergo a change in concentration, therefore it is a third-order reaction

Question 29. The data for the reaction A + B → C, is

⇒ \(\begin{array}{lccc}
\text { Exp. } & {[A]_0} & {[B]_0} & \text { Initial rate } \\
1 & 0.012 & 0.035 & 0.10 \\
2 & 0.024 & 0.070 & 0.80 \\
3 & 0.024 & 0.035 & 0.10 \\
4 & 0.012 & 0.070 & 0.80
\end{array}\)

The rate law corresponds to the above data is

1. Rate = k[A][B]³
2. Rate = k[A]²[B]²
3. Rate = k[B]³
4. Rate = k[B]⁴.

Answer: 3. Rate = k[B]³

A+B → C

Let rate = k[A]^x [B]^y

where order ofreaction is (x + y).

Putting the values of exp. 1, 2, and 3, we get the following equations.

0.10 = k [0.012]^x [0.035]^y ….(1)

0.80=k [0.024]^x [0.070]^y ……(2)

0.10 = k [0.024]^x [0.035]^y …..(3)

Dividing (2) by (3), we get \(\frac{0.80}{0.10}=\left(\frac{0.070}{0.035}\right)^y \Rightarrow 2^y=8 \Rightarrow y=3\)

Keeping [A] constant, [B] is doubled, and the rate becomes B times. Dividing eq. (3) by eq. (1), we get \(\frac{0.10}{0.10}=\left(\frac{0.024}{0.012}\right)^x \Rightarrow 2^x=1 \Rightarrow x=0\)

Keeping [B] constant, [A] is doubled, and the rate remains unaffected. Hence, the rate is independent of [A].

rate ∝ [B]³.

Question 30. For a first-order reaction A → Products, the initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. The rate constant for the reaction in min^-1 is

1. 1.3818
2. 0.9212
3. 0.4606
4. 0.2303

Answer: 2. 0.9212

For a first-order reaction

k = \(\frac{2.303}{t} \log \frac{a}{a-x}\)

k = \(\frac{2.303}{5} \log \frac{0.1}{0.001}\)

k = \(\frac{2.303}{5} \log 10^2=\frac{2.303 \times 2}{5}=0.9212 \mathrm{~min}^{-1}\)

Question 31. The given graph is a representation of the kinetics of a reaction.

The y and x axes for zero and first-order reactions, respectively are

1. Zero order (y = concentration and x = time), first order (y = t_1/2 and x = concentration)
2. Zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
3. Zero order (y = rate and x = concentration), first order (y = t_1/2 and x = concentration)
4. Zero order (y = rate and x = concentration), first order (y = rate and x = t_1/2)

Answer: 3. Zero order (y = rate and x = concentration), first order (y = t_1/2 and x = concentration)

Question 32. The rate constant for a first-order reaction is 4.606 x 10^-3 s^-1 The time required to reduce 2.0 g of the reactant to 0.2 g is

1. 100 s
2. 200s
3. 500 s
4. 1000 s

Answer: 3. 500 s

For a first-order reaction,

k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)

t = \(\frac{2.303}{4.606 \times 10^{-3} \mathrm{~s}^{-1}} \log \left(\frac{2}{0.2}\right)=\frac{2.303 \times 10^3}{4.606}=500 \mathrm{~s}\)

Question 33. If the rate constant for a first-order reaction is k, the time (t) required for the completion of 99% of the reaction is given by

1. t = 2303/k
2. t = 0.693/k
3. t = 6.909/k
4. t = 4.606/k

Answer: 4. t = 4.606/k

For 1 st order reaction,

t = \(\frac{2.303}{k} \log \frac{a}{a-x}=\frac{2.303}{k} \log \frac{100}{100-99}\)

= \(\frac{2.303}{k} \log 10^2=\frac{2.303}{k} \times 2 \times \log 10=\frac{4.606}{k}\)

Question 34. A first-order reaction has a rate constant of 2.303 x 10^-3 s^-1. The time required for 40 g of this reactant to reduce to 10 g will be [Given that log₁₀ 2 = 0.3010]

1. 230.3 s
2. 301 s
3. 2000 s
4. 602 s

Answer: 4. 602 s

For a first-order reaction, k = \(\frac{2.303}{t} \log \frac{[A]_0}{[A]_t}\)

⇒ \(2.303 \times 10^{-3}=\frac{2.303}{t} \log \frac{40}{10}\)

t = \(\frac{1}{10^{-3}} \log 2^2=\frac{2}{10^{-3}} \log 2=\frac{2}{10^{-3}} \times 0.3010=602 \mathrm{~s}\)

Question 35. The correct difference between first and second-order reactions is that

1. The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
2. The half-life of a first-order reaction does not depend on [A]0₂; the half-life of a second-order reaction does depend on [A]₀
3. A first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
4. The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations.

Answer: 2. The half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀

For the first order reaction, \(t_{1 / 2}=\frac{0.693}{k}\) independent of initial concentration [A]₀

For second order reaction, \(t_{1 / 2}=\frac{1}{k[A]_0}\)

Half-life depends on the initial concentration of the reactant

Question 36. When the initial concentration of the reactant is doubled, the half-life period of a zero-order reaction

1. Is halved
2. Is doubled
3. Is tripled
4. Remains unchanged.

Answer: 2. Is doubled

⇒ \(\left(t_{1 / 2}\right)_{\text {zero }}=\frac{[A]_0}{2 k}\)

The half-life of a zero-order reaction is directly proportional to the initial concentration.

∴ If [A]₂ = doubled then, t_1/2 = doubled

Question 37. A first-order reaction has a specific reaction rate of 10^-2 sec^-1. How much time will it take for 20 g of the reactant to reduce to 5 g?

1. 138.6 sec
2. 346.5 sec
3. 693.0 sec
4. 238.6 sec

Answer: 1. 138.6 sec

For a first-order reaction, k = \(\frac{2.303}{t} \log \frac{[A]_0}{[A]_t}\) or \(10^{-2}=\frac{2.303}{t} \log \frac{20}{5}\)

⇒ \(10^{-2}=\frac{2.303 \times 0.6020}{t}\) or \(t=138.6 \mathrm{sec}\)

Question 38. The rate of the first-order reaction is 0.04 mol L^-1 s^-1 at 10 seconds and 0.03 mol L^-1 s^-1 at 20 seconds after the initiation of the reaction. The half-life period of the reaction is

1. 44.1 s
2. 54.1s
3. 24.1 s
4. 34.1 s

Answer: 3. 24.1 s

For a first-order reaction, A → products and for the concentration of the reactant at two different times,

k = \(\frac{2.303}{t_2-t_1} \log \frac{[A]_1}{[A]_2}\)

k = \(\frac{2.303}{t_2-t_1} \log \frac{(\text { rate })_1}{(\text { rate })_2}\) (because rate ∝ [A])

k = \(\frac{2.303}{(20-10)} \log \left(\frac{0.04}{0.03}\right)=0.0287 \mathrm{sec}^{-1}\)

∴ \(t_{1 / 2}= \frac{0.693}{k}=\frac{0.693}{0.0287 \mathrm{sec}^{-1}}=24.14 \mathrm{sec}\)

Question 39. When the initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is

1. Second
2. More than zero but less than the first
3. Zero
4. First.

Answer: 4. First

The half-life period of a first-order reaction is independent of initial concentration, \(t_{1 / 2}=\frac{0.693}{k}\)

Question 40. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of the reaction is

1. 1
2. 2
3. 3
4. 0

Answer: 1. 1

As t_75% = 2 x t_50%, the order of the reaction is one.

Question 41. The half-life of a substance in a certain enzyme-catalyzed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L^-1 to 0.04 mg L^-1 is

1. 414 s
2. 552 s
3. 690 s
4. 276 s

Answer: 3. 690 s

A fall of concentration from 1.28 mg L^-1 to 0.04 mg L^-1 requires 5 half-lives.

∴ Time required = 5 x t_1/2= 5 x 138 = 690 s

Question 42. The half-life period of a first-order reaction is 1386 seconds. The specific rate constant of the reaction is

1. \(0.5 \times 10^{-2} \mathrm{~s}^{-1}\)
2. \(0.5 \times 10^{-3} \mathrm{~s}^{-1}\)
3. \(5.0 \times 10^{-2} \mathrm{~s}^{-1}\)
4. \(5.0 \times 10^{-3} \mathrm{~s}^{-1}\)

Answer: 2. \(0.5 \times 10^{-3} \mathrm{~s}^{-1}\)

Given, \(t_{1 / 2}=1386 \mathrm{~s}\)

For a first-order reaction, \(t_{1 / 2}=\frac{0.693}{k}\) (k = rate constant)

1386 = \(\frac{0.693}{k} \Rightarrow k=5 \times 10^{-4} \mathrm{~s}^{-1}=0.5 \times 10^{-3} \mathrm{~s}^{-1}\)

Question 43. If 60% of a first-order reaction was completed in 60 minutes, 50% of the same reaction would  be completed in approximately (log 4 = 0.60, log 5 = 0.69)

1. 45 minutes
2. 60 minutes
3. 40 minutes
4. 50 minutes

Answer: 1. 45 minutes

For a first-order reaction, \(k=\frac{2.303}{t} \log \frac{a}{a-x}\)

k = \(\frac{2.303}{60} \log \frac{100}{40}=\frac{2.303}{60} \times \log 2.5=0.0153\)

Again, \(t_{1 / 2}=\frac{2.303}{k} \log \frac{100}{50}=\frac{2.303}{0.0153} \times \log 2=45.31 \mathrm{~min}\)

Question 44. In a first-order reaction, A → B, if k is rate constant and the initial concentration of the reactant A is 0.5 M, then the half-life is

1. \(\frac{\log 2}{k}\)
2. \(\frac{\log 2}{k \sqrt{0.5}}\)
3. \(\frac{\ln 2}{k}\)
4. \(\frac{0.693}{0.5 k}\)

Answer: 3. \(\frac{\ln 2}{k}\)

For a 1st t order reaction, \(k=\frac{2.303}{t} \log _{10} \frac{a}{a-x}\)

At \(t_{1 / 2}, k=\frac{2.303}{t_{1 / 2}} \log _{10} \frac{a}{a-\frac{a}{2}}\) or \(t_{1 / 2}=\frac{2.303}{k} \log _{10} 2=\frac{\ln 2}{k}\)

Question 45. For a first-order reaction A → B the reaction rate at a reactant concentration of 0.01 M is found to be 2.0 x 10^-5 mol L^-1 s^-1. The half-life period of the reaction is

1. 30 s
2. 220 s
3. 300 s
4. 347 s

Answer: 4. 347 s

A → B

Rate of reaction = 2 x 10^-5 mol L^-1 s^-1

⇒ order of reaction is n = 1, rate = k [A]ⁿ = k[A]

⇒ k is the rate constant.

[A] = 0.01M

⇒ k = \(\frac{2 \times 10^{-5}}{0.01}=2 \times 10^{-3}, k=\frac{0.693}{t_{1 / 2}}\)

\(t_{1 / 2}=\frac{0.693}{2 \times 10^{-3}}=346.5 \approx 347 \mathrm{~s}\)

Question 46. The rate of a first-order reaction is 1.5 x 10^-2 mol L^-1 min^-1 at 0.5 M concentration of the reactant. The half-life of the reaction is

1. 0.383 min
2. 23.1 min
3. 8.73 min
4. 7.53 min

Answer: 2. 23.1 min

Rate \(\left(\frac{d x}{d t}\right)=k C\)

i.e., \(1.5 \times 10^{-2}=k \times 0.5$ or, $k=\frac{1.5 \times 10^{-2}}{0.5}\)

For first order reaction, \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693 \times 0.5}{1.5 \times 10^{-2}}=23.1 \mathrm{~min} \)

Question 47. The reaction A → B follows first-order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0. 9 mole of A to produce 0.675 mole of B?

1. 1 hour
2. 0.5 hour
3. 0.25 hour
4. 2 hours

Answer: 1. 1 hour

The time taken for the completion of the same fraction of change is independent of initial concentration.

Question 48. For a first-order reaction, the half-life period is independent of

1. The first power of final concentration
2. Cube root of initial concentration
3. Initial concentration
4. The square root of the final concentration.

Answer: 3. Initial concentration

For the first order reaction, the rate constant is given by, \(k_1=\frac{1}{t} \ln \frac{a}{a-x}\)

a = initial concentration,(a-x)= concentration at t time

At t = \(t_{1 / 2}, x=a /2\)

⇒ \(k_1=\frac{1}{t_{1 / 2}} \ln \frac{a}{a-a / 2} \Rightarrow k_1=\frac{1}{t_{1 / 2}} \ln 2 \Rightarrow k_1=\frac{0.693}{t_{1 / 2}}\)

Therefore, t_1/2 is independent of initial concentration.

Question 49. Given below are two statements: One is labeled as Assertion A and the other is labeled as Reason R:

Assertion A: A reaction can have zero activation energy.

Reason R: The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy.

In light of the above statements, choose the correct answer from the options given below.

1. A is true but R is false.
2. A is false but R is true.
3. Both A and R are true and R is the correct explanation of A.
4. Both A and R are true and R is not the correct

Answer: 2. A is false but R is true.

If \(E_a=0\), then according to Arrhenius equation, \(k=A e^{-E_0 / R T} \Rightarrow k=A e^{0 / R T}=A\)

This implies that every collision results in a chemical reaction which cannot be true. So, a reaction cannot have zero activation energy.

Question 50. For reaction A → B, the enthalpy of the reaction is -4.2 kJ mol^-1 and the enthalpy of activation is 9.6 kJ mol^-1. The correct potential energy profile for the reaction is shown in the option.

Answer: 3

As the enthalpy of the reaction is negative, hence it is an exothermic reaction.

Question 51. The slope of Arrhenoius (In k vs 1/T) of a first-order reaction is -5 x 10³ K. The value of E_a of the reaction is [Given: R = 8.314 J K^-1 mol^-1)

1. \(-83 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(41.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(83.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(166 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 2. \(41.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

ln k = \(-\frac{E_a}{R T}+\ln A\)

–\(\frac{E_a}{R}=-5 \times 10^3=-5000\)

⇒ \(E_a=5000 \times 8.314=41570 \mathrm{~J} \mathrm{~mol}^{-1}\)

= \(41.57 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 52. For a reaction, activation energy E_a = 0, and the rate constant at 200 K is 1.6 x 10⁶ s^-1. The rate constant at 400 K will be [Given that gas constant R = 8.314 J K^-1 mol^-1]

1. \(3.2 \times 10^4 \mathrm{~s}^{-1}\)
2. \(1.6 \times 10^6 \mathrm{~s}^{-1}\)
3. \(1.6 \times 10^3 \mathrm{~s}^{-1}\)
4. \(3.2 \times 10^6 \mathrm{~s}^{-1}\)

Answer: 2. \(1.6 \times 10^6 \mathrm{~s}^{-1}\)

According to Arrhenius equation, \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_2 T_1}\right]\)

⇒ \(\log \frac{k_2}{1.6 \times 10^6}=0 ; \frac{k_2}{1.6 \times 10^6}=1\)

⇒ \(k_2=1.6 \times 10^6 \mathrm{~s}^{-1}\)

Question 53. The addition of a catalyst during a chemical reaction alters which of the following quantities?

1. Enthalpy
2. Activation energy
3. Entropy
4. Internal energy

Answer: 2. Activation energy

A catalyst provides an alternate path to the reaction which has lower activation energy.

Question 54. The activation energy of a reaction can be determined from the slope of which of the following graphs?

1. \(\ln k v s \frac{1}{T}\)
2. \(\frac{T}{\ln k} v s \frac{1}{T}\)
3. \(\ln k v s T\)
4. \(\frac{\ln k}{T} v s T\)

Answer: 1. \(\ln k v s \frac{1}{T}\)

According to Arrhenius equation, \(k=A e^{-E_a / R T}\)

ln k = ln A – \(\frac{E_a}{R T}\)

Hence, if ln k is plotted

against 1/T, slope of the line will be \(-\frac{E_a}{R}\)

Question 55. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20 °C to 35 °C? (R = 8.314 J mol^-1 K^-1)

1. \(34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(15.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(342 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(269 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 1. \(34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)

⇒  \(k_2=2 k_1, T_1=20+273=293 \mathrm{~K}\) or \(T_2=35+273=308 \mathrm{~K}\)

R = \(8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

⇒  \(\log 2=\frac{E_a}{2.303 \times 8.314}\left(\frac{1}{293}-\frac{1}{308}\right)\)

0.3010 = \(\frac{E_a}{19.147} \times \frac{15}{293 \times 308}\)

⇒  \(E_a=34673 \mathrm{~J} \mathrm{~mol}^{-1}=34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 56. In a zero-order reaction, for every 10 °C rise in temperature, the rate is doubled. If the temperature is increased from 10 °C to 100 °C, the rate of the reaction will become

1. 256 times
2. 512 times
3. 64 times
4. 128 times.

Answer: 2. 512 times

At 10°C rise, the rate increases by 2

⇒ \(\frac{r_{100^{\circ} \mathrm{C}}}{r_{10^{\circ} \mathrm{C}}}=2^{\left(\frac{100-10}{10}\right)}=2^9=512 \text { times }\)

Question 57. The activation energy (E_a) and rate constants (k₁ and k₂) of a chemical reaction at two different temperatures (T₁) and T₂) are related by

1. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)
2. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\)
3. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}+\frac{1}{T_1}\right)\)
4. \(\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)

Answer: 2. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) and 4. \(\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)

⇒ \(k_1=A e^{-E_a / R T_1}, k_2=A e^{-E_a / R T_2}\)

ln \(k_1=\ln A-E_a / R T_1\)…(1)

ln \(k_2=\ln A-E_a / R T_2\)…(2)

From eq.(1) and (2), we have ln \(k_2-\ln k_1=\ln A-\frac{E_a}{R T_2}-\ln A+\frac{E_a}{R T_1}\)

ln \(\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \Rightarrow \ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\)

Question 58. The rate of the reaction, 2NO + Cl₂ → 2NOCl is given by the rate equation, rate = k[NO²[Cl₂]. The value of the rate constant can be increased by

1. Increasing the temperature
2. Increasing the concentration of NO
3. Increasing the concentration of the Cl₂
4. Doing all of these.

Answer: 1. Increasing the temperature

The rate constant is independent of the initial concentration of the reactants. It has a constant value at a fixed temperature. According to the Arrhenius equation, the value of the rate constant can be increased by increasing the temperature.

Question 59. The rate constants k₁ and k₂ for two different reactions are \(10^{16} \cdot e^{-2000 / T}\) and \(10^{15^2} \cdot e^{-1000 / T}\), respectively. Tire temperature at which k₁ = k₂ is

1. 2000 K
2. \(\frac{1000}{2.303}\) K
3. 1000 K
4. \(\frac{2000}{2.303}\) K

Answer: 2. \(\frac{1000}{2.303}\) K

⇒ \(k_1=10^{16} e^{-2000 / T}, k_2=10^{15} e^{-1000 / T}\)

When, \(k_1=k_2, 10^{16} e^{-2000 / T}=10^{15} e^{-1000 / T}\)

or \(10 e^{-2000 / T}=e^{-1000 / T}\)

Taking the natural logarithm of both sides, we get ln \(10-\frac{2000}{T}=\frac{-1000}{T}\) or \(2.303-\frac{2000}{T}=\frac{-1000}{T}\)

or, \(\frac{1000}{T}=2.303 \text { or } T=\frac{1000}{2.303} \mathrm{~K}\)

Question 60. The temperature dependence of the rate constant (k) of a chemical reaction is written in terms of the Arrhenius equation, k = A x e^-E*/RT. The activation energy (E*) of the reaction can be calculated by plotting

1. \(k v s T\)
2. \(k v s \frac{1}{\log T}\)
3. \(\log k v s \frac{1}{T}\)
4. \(\log k v s \frac{1}{\log T}\)

Answer: 3. \(\log k v s \frac{1}{T}\)

On plotting log k vs l/T, we get a straight line, the slope indicates the value of activation energy.

Question 61. The activation energy for a simple chemical reaction A \(\rightleftharpoons\) B is E_a in the forward direction. The activation energy for the reverse reaction

1. Is negative of E_a
2. Is always less than E_a
3. Can be less than or more than E_a
4. Is always double of E_a

Answer: 3. Can be less than or more than E_a

Activation energy is the minimum amount of energy required to convert reactant into product. The activation energy for reverse reaction can be less than or more than E_a depending on whether the reaction is exothermic or endothermic.

Question 62. When a biochemical reaction is carried out in the laboratory, outside the human body in the absence of an enzyme, then the rate of reaction obtained is 10^-6 times, and the activation energy of the reaction in the presence of an enzyme is

1. 6/RT
2. P is required
3. Different from E_a obtained in the laboratory
4. Can’t say anything.

Answer: 3. Different from Ea obtained in the laboratory

According to k = \(A e^{-E a / R T}\) (Arrhenius equation), the activation energy of a reaction in the presence of an enzyme is different from E_a obtained in the laboratory.

Question 63. How do enzymes increase the rate of reactions?

1. By lowering the activation energy
2. By increasing activation energy
3. By changing the equilibrium constant
4. By forming enzyme-substrate complex

Answer: 1. By lowering activation energy

Enzymes act like catalysts in biochemical reactions. The presence of an enzyme increases the rate of reaction by lowering the activation energy of the reactant.

Question 64. The activation energy of a chemical reaction can be determined by

1. Evaluating rate constants at two different temperatures
2. Evaluating velocities of reaction at two different temperatures
3. Evaluating rate constant at standard temperature
4. Changing concentration of reactants.

Answer: 1. Evaluating rate constants at two different temperatures

According to Arrhenius equation: \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_2 T_1}\right]\)

where E_a = activation energy

R = gas constant = 8.314 J K^-1 mol^-1

k₁ and k₂ are rate constants of the reaction at two different temperatures T₁ and T₂ respectively.

Question 65. By the action of enzymes, the rate of biochemical reaction

1. Does not change
2. Increases
3. Decreases
4. Either (1) or (3).

Answer: 2. Increases

Since the enzymes are regarded as biological catalysts, therefore their action increases the rate of biological reaction.

Question 66. An increase in the concentration of the reactants of a reaction leads to a change in

1. Activation energy
2. Heat of reaction
3. Threshold energy
4. Collision frequency.

Answer: 4. Collision frequency.

Collision frequency ∝ number of reacting molecules or atoms

The higher the concentration of reactant molecules, the higher is the probability of collision and so the collision frequency.

 

Surface Chemistry

Question 1. The correct option representing a Freundlich adsorption isotherm is

1. \(\frac{x}{m}=k p^{0.3}\)
2. \(\frac{x}{m}=k p^{2.5}\)
3. \(\frac{x}{m}=k p^{-0.5}\)
4. \(\frac{x}{m}=k p^{-1}\)

Answer: 1. \(\frac{x}{m}=k p^{0.3}\)

Freundlich adsorption isotherm equation is \(\frac{x}{m}=k p^{\frac{1}{n}}\left(1 \geq \frac{1}{n} \geq 0\right)\)

Question 2. Which one of the following characteristics is associated with adsorption?

1. ΔG and ΔH are negative but ΔS is positive.
2. ΔG and ΔS are negative but ΔH is positive.
3. ΔG is negative but ΔH and ΔS are positive.
4. ΔG, ΔH and ΔS all are negative.

Answer: 4. ΔG, ΔH and ΔS all are negative.

As the molecules of the adsorbate are held on the surface of the solid adsorbent, entropy decreases i.e., ΔS = -ve.

As ΔG=ΔH- TΔS

For the adsorption to occur, ΔG = -ve and it is possible only if ΔH= -ve

Question 3. In Freundlich adsorption isotherm, the value of 1/n is

1. Between 0 and 1 in all cases
2. Between 2 and 4 in all cases
3. 1 In case of physical absorption
4. 1 In case of chemisorption.

Answer: 1. Between 0 and 1 in all cases

Freundlich adsorption isotherm: \(\frac{x}{m}=k \cdot p^{1 / n} ; \quad 0 \leq \frac{1}{n} \leq 1\)

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Question 4. If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to the adsorption process?

1. x/m =f(p) at constant T
2. x/m =f(T) at constant p
3. p =f(T) at constant (x/m)
4. \(\frac{x}{m}\) = p x T

Answer: 4. \(\frac{x}{m}\) = p x T

⇒ \(\frac{x}{m}=p \times T\) is the incorrect relation.

The correct relation is amount of absorption \(\frac{x}{m} \propto \frac{p}{T}\)

Question 5. The Langmuir adsorption isotherm is deduced using the assumption

1. The adsorption sites are equivalent in their ability to adsorb the particles
2. The heat of adsorption varies with coverage
3. The adsorbed molecules interact with each other
4. The adsorption takes place in multilayers.

Answer: 1. The adsorption sites are equivalent in their ability to adsorb the particles

Langmuir adsorption isotherm is based on the assumption that every adsorption site is equivalent and that the ability of a particle to bind there is independent of whether nearby sites are occupied or not.

Question 6. A plot of log (x/m) versus log P for the adsorption of a gas on a solid gives a straight line with a slope equal to

1. log k
2. -log k
3. n
4. 1/n

Answer: 4. 1/n

This is according to Freundlich adsorption isotherm

Question 7. Which is not correct regarding the adsorption of a gas on the surface of a solid?

1. On increasing temperature, adsorption increases continuously.
2. Enthalpy and entropy change is negative.
3. Adsorption is more for some specific substance.
4. It is a reversible reaction.

Answer: 1. On increasing temperature adsorption increases continuously.

Adsorption is the ability of a substance to concentrate or hold gases, liquids or dissolved substances upon its surface. Solids absorb greater amounts of substances at lower temperatures. In general, adsorption decreases with an increase in temperature.

Question 8. Which one is an example of heterogeneous catalysis?

1. Decomposition of ozone in the presence of nitrogen monoxide,
2. Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron.
3. Oxidation of sulphur dioxide into sulphur trioxide in the presence of oxides of nitrogen.
4. Hydrolysis of sugar catalysed by H⁺ ions.

Answer: 2. Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron.

Question 9. The incorrect statement regarding enzymes is

1. Enzymes are biocatalysts
2. Like chemical catalysts enzymes reduce the activation energy of bioprocesses
3. Enzymes are polysaccharides
4. Enzymes are very specific for a particular reaction and substrate.

Answer: 3. Enzymes are polysaccharides

Enzymes are protein molecules of high molecular mass and form colloidal solutions in water.

Question 10. Which one of the following statements is not correct?

1. The value of the equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium.
2. Enzymes catalyse mainly biochemical reactions.
3. Coenzymes increase the catalytic activity of enzymes.
4. The catalyst does not initiate any reaction.

Answer: 1. The value of the equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium.

Catalysts do not change the value of equilibrium constant as they affect forward as well as backward reactions equa1ly.

Question 11. Which one of the following statements is incorrect about enzyme catalysis?

1. Enzymes are mostly proteinous in nature.
2. Enzyme action is specific.
3. Enzymes are denatured by ultraviolet rays and at high temperatures.
4. Enzymes are least reactive at optimum temperature.

Answer: 4. Enzymes are least reactive at optimum temperature.

The enzyme activity rises rapidly with temperature and becomes maximum at a definite temperature, called optimum temperature.

Question 12. The enzyme which hydrolyses triglycerides to fatty acids and glycerol is called

1. Maltase
2. Lipase
3. Zymase
4. Pepsin.

Answer: 2. Lipase

Question 13. According to the adsorption theory of catalysis, the speed of the reaction increases because

1. The concentration of reactant molecules at the active centres of the catalyst becomes high due to adsorption
2. In the process of adsorption, the activation energy of the molecules becomes large
3. Adsorption produces heat which increases the speed of the reaction
4. Adsorption lowers the activation energy of the reaction.

Answer: 4. Adsorption lowers the activation energy of the reaction.

Adsorption causes a decrease in surface energy which appears as heat. thus, adsorption is an exothermic process and hence lowers the activation energy of the reaction.

Question 14. A colloidal system has particles of which of the following sizes?

1. 10^-9 m to 10^-12 m
2. 10^-6 m to 10^-9 m
3. 10^-4 m to 10^-10 m
4. 10^-5 m to 10^-7 m

Answer: 2. 10^-6 m to 10^-9 m

The particle size of colloids lies in the range of 10^-6 m to 10^-9 m. Particles themselves are invisible even under the most powerful microscope.

Question 15. Pumice stone is an example of

1. Solid sol
2. Foam
3. Sol
4. Gel.

Answer: 1. Solid sol

Pumice stone is an example of a solid sol in which the dispersed phase is gas and the dispersion medium is solid.

Question 16. Given below are two statements:

Statement-1: In the coagulation of a negative sol, the flocculating power of the three given ions is in the order: \(\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^{+}\)

Statement-2: In the coagulation of a positive sol, the flocculating power of the three given salts is in the order: \(\mathrm{NaCl}>\mathrm{Na}_2 \mathrm{SO}_4>\mathrm{Na}_3 \mathrm{PO}_4\)

In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both statement-1 and statement-2 are correct.
2. Both statement-1 and statement-2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but statement 2 is correct.

Answer: 3. Statement 1 is correct but statement 2 is incorrect.

In the coagulation of a positive sol, the flocculating power is in the order: \(\mathrm{PO}_4^{3-}>\mathrm{SO}_4^{2-}>\mathrm{Cl}^{-}\) or, \(\mathrm{NaCl}<\mathrm{Na}_2 \mathrm{SO}_4<\mathrm{Na}_3 \mathrm{PO}_4\)

Question 17. The right option for the statement “Tyndall effect is exhibited by” is

1. Urea solution
2. NaCl solution
3. Glucose solution
4. Starch solution.

Answer: 4. Starch solution.

As the starch solution is a colloidal solution hence it exhibits the Tyndall effect. NaCl, urea and glucose form a true solution.

Question 18. Measuring zeta potential is useful in determining which property of colloidal solution?

1. Viscosity
2. Solubility
3. Stability of the colloidal particles
4. Size of the colloidal particles

Answer: 3. Stability of the colloidal particles

Measuring, zeta potential is useful in determining the stability of the colloidal particles.

Question 19. Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI]I^-1 sol?

1. 50 mL of 0.1 M AgNO₃ + 50 mL of 0.1 M KI
2. 50 mL of 1 M AgNO₃ + 50 mL of 1.5 M KI
3. 50 mL of 1 M AgNO₃ + 50 mL of 2 M KI
4. 50 mL of 2 M AgNO₃ + 50 mL of 1.5 M KI

Answer: 2. 50 mL of 1 M AgNO₃ + 50 mL of 1.5 M KI

If the colloidal sol of AgI is prepared by adding KI solution to AgNO₃ till KI is in slight excess, iodide ion (I^–) will be adsorbed on the surface of AgI thereby, giving a negative charge to the sol.

⇒ \(\mathrm{AgI}+\underset{(\text { From } \mathrm{KI})}{\mathrm{I}^{-}} \longrightarrow \underset{\text { Negative sol }}{\mathrm{AgI}: \mathrm{I}^{-}}\)

Question 20. On which of the following properties does the coagulating power of an ion depend?

1. The magnitude of the charge on the ion alone
2. The size of the ion alone
3. Both the magnitude and sign of the charge on the ion
4. The sign of charge on the ion alone

Answer: 3. Both the magnitude and sign of the charge on the ion

According to the Hardy-Schulze rule, the coagulating power of an electrolyte depends on both the magnitude and sign of the charge of the effective ion or electrolyte.

Question 21. The coagulation values in millimoles per litre of the electrolytes used for the coagulation of As₂S₃ are given below:

1. (NaCl) = 52,
2. (BaCl₂) = 0.69,
3. (MgSO₄) = 0.22

The correct order of their coagulating power is

1. 1 > 2 > 3
2. 2 > 1 > 3
3. 3 > 2 > 1
4. 3 > 1 > 2

Answer: 3. 3 > 2 > 1

Coagulating power \(\propto \frac{1}{\text { Coagulation value }}\)

The lower the coagulation value, the higher is the coagulating power so, the correct order is: \(\underset{3} {\mathrm{MgSO}_4}>\underset{2} {\mathrm{BaCl}_2}>\underset{1}{\mathrm{NaCl}}\)

Question 22. Fog is a colloidal solution of

1. Solid in gas
2. Gas in gas
3. Liquid in gas
4. Gas in liquid.

Answer: 3. Liquid in gas

Fog is an example of aerosol in which the dispersed phase is liquid and the dispersion medium is gas.

Question 23. Which property of colloidal solution is independent of charge on the colloidal particles?

1. Electroosmosis
2. Tyndall effect
3. Coagulation
4. Electrophoresis

Answer: 2. Coagulation

Tyndall effect is the scattering of light by colloidal particles which is independent of the charge on them.

Question 24. The protecting power of lyophilic colloidal sol is expressed in terms of

1. Coagulation Value
2. Gold Number
3. Critical Micelle Concentration
4. Oxidation Number.

Answer: 2. Gold Number

Question 25. Which one of the following forms micelles in an aqueous solution above a certain concentration?

1. Dodecyl trimethyl ammonium chloride
2. Glucose
3. Urea
4. Pyridinium chloride

Answer: 1. Dodecyl trimethyl ammonium chloride

Question 26. Position of non-polar and polar parts in micelle

1. Polar at the outer surface but non-polar at the inner surface
2. Polar at the inner surface non-polar at the outer surface
3. Distributed over all the surface
4. Are present in the surface only.

Answer: 1. Polar at the outer surface but non-polar at the inner surface

Micelles are the clusters or aggregates formed in solution by association of colloids. Usually, such molecules have a lyophobic group and a lyophilic group. The long hydrocarbon is the lyophobic portion which tries to recede away from the solvent water and the ionisable lyophitic group which tends to go into water resulting in ions.

As the concentration is increased the lyophobic parts receding away from the solvent approach each other and form a cluster. Thus, the lyophobic ends are in the interior and lyophilic groups projecting outward in contact with the solvent.

Question 27. Which one of the following methods is commonly used method for the destruction of colloids?

1. Dialysis
2. Condensation
3. Filtration by animal membrane
4. By adding electrolyte

Answer: 4. By adding electrolyte

By adding electrolytes the colloidal particles are precipitated. The electrolytes neutralise the charge of colloids leading to their coagulation and thus, destroying the colloid.

Question 28. At the critical micelle concentration (CMC) the surfactant molecules

1. Associate
2. Dissociate
3. Decompose
4. Become Completely Soluble.

Answer: 1. Associate

The soap concentration at which micelles (spherical colloid molecules) first appear is called critical micelle concentration (CMC). At this condition, the surfactant molecules associate with each other

Question 29. The ability of an anion, to bring about coagulation of a given colloid, depends upon

1. The magnitude of the charge
2. Both magnitude and charge
3. Its charge only
4. Sign of the charge alone.

Answer: 2. Both magnitude and charge

Both magnitudes of charge and the nature of charge affect the coagulation of a given colloid. The greater the magnitude of the charge, the quicker will be the coagulation.

Question 30. When a few typical solutes are separated by a particular selective membrane such as protein particles, or blood corpuscles, this process is called

1. Transpiration
2. Endosmosis
3. Dialysis
4. Diffusion.

Answer: 3. Dialysis

Dialysis is the process of separating the particles of colloids from the particles of crystalloids by means of diffusion through a selective membrane placed in water.

General Principles And Processes Of Isolation Of Elements

Question 1. Match List 1 and List 2

Choose the correct answer from the options given below:

1. (1) -(A), (2) -(B), (3) -(C), (4) -(D)
2. (1) -(C), (2) -(A), (3) -(B), (4) -(D)
3. (1) -(C), (2) -(A), (3) -(D), (4) -(B)
4. (1) -(A), (2) -(C), (3) -(h), (4) -(D)

Answer: 2. (1) -(C), (2) -(A), (3) -(B), (4) -(D)

Haematite \(-\mathrm{Fe}_2 \mathrm{O}_3\)

Magnetite –\(\mathrm{Fe}_3 \mathrm{O}_4\)

Calamine –\(\mathrm{ZnCO}_3\)

Kaolinite –\(\mathrm{Al}_2 \mathrm{O}_3 \cdot 2 \mathrm{SiO}_2 \cdot 2 \mathrm{H}_2 \mathrm{O}\)

Question 2. Which one is malachite from the following?

1. \(\mathrm{CuCO}_3 \cdot \mathrm{Cu}(\mathrm{OH})_2\)
2. \(\mathrm{CuFeS}_2\)
3. \(\mathrm{Cu}(\mathrm{OH})_2\)
4. \(\mathrm{Fe}_3 \mathrm{O}_4\)

Answer: 1. \(\mathrm{CuCO}_3 \cdot \mathrm{Cu}(\mathrm{OH})_2\)

Malachite: CuCO₃ · Cu(OH)₂

Question 3. Identify the incorrect statement.

1. The scientific and technological process used for isolation of the metal from its ore is known as metallurgy.
2. Minerals are naturally occurring chemical substances in the earth’s crust.
3. Ores are minerals that may contain a metal.
4. Gangue is an ore contaminated with undesired materials.

Answer: 4. Gangue is an ore contaminated with undesired materials.

An ore rarely contains only a desired substance. it is usually contaminated with earthly or undesired materials known as gangue.

Question 4. “Metals are usually not found as nitrates in their ores.” Out of the following two (1 and 2) reasons which is/are true for the above observation?

1. Metal nitrates are highly unstable.
2. Metal nitrates are highly soluble in water.

1. 1 is false but 2 is true.
2. 1 is true but 2 is false.
3. 1 and 2 are true.
4. 1 and 2 are false

Answer: 1. 1 is false but 2 is true.

All nitrates are soluble in water and are quite stable as they do not decompose easily on heating.

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Question 5. Which one of the following is a mineral of iron?

1. Malachite
2. Cassiterite
3. Pyrolusite
4. Magnetite

Answer: 4. Magnetite

Magnetite is Fe₃O₄ and contains up to 70% of iron metal.

Question 6. Cassiterite is an ore of

1. Sb
2. Ni
3. Mn
4. Sn

Answer: 4. Sn

Cassiterite is also known as tin stone (SnO₂), an ore of tin (Sn).

Question 7. Sulphide ores of metals are usually concentrated by the froth floatation process. Which one of the following sulphide ores offers an exception and is concentrated by chemical leaching?

1. Galena
2. Copper pyrite
3. Sphalerite
4. Argentite

Answer: 4. Argentite

The leaching process involves the treatment of the ore with a suitable reagent as to make it soluble while impurities remain insoluble. The ore is recovered from the solution by a suitable chemical method.

Argentite or silver glance, Ag₂S is an ore of silver. Silver is extracted from argentite by the MacArthur and Forest cyanide process (leaching process).

⇒ \(\mathrm{Ag}_2 \mathrm{~S}+4 \mathrm{NaCN} \longrightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{Na}_2 \mathrm{~S}\)

⇒ \(2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{Zn} \longrightarrow \mathrm{Na}_2\left[\mathrm{Zn}(\mathrm{CN})_4\right]+2 \mathrm{Ag}\)

Question 8. The roasting of sulphides gives the gas X as a byproduct. This is a colourless gas with a choking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic, acts as a reducing agent and its acid has never been isolated. The gas X is

1. \(\mathrm{CO}_2\)
2. \(\mathrm{SO}_3\)
3. \(\mathrm{H}_2 \mathrm{~S}\)
4. \(\mathrm{SO}_2\)

Answer: 4. \(\mathrm{SO}_2\)

Question 9. The reaction that does not take place in a blast furnace between 900 K to 1500 K temperature range during extraction of iron is

1. \(\mathrm{C}+\mathrm{CO}_2 \rightarrow 2 \mathrm{CO}\)
2. \(\mathrm{CaO}+\mathrm{SiO}_2 \rightarrow \mathrm{CaSiO}_3\)
3. \(\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{FeO}+\mathrm{CO}_2\)
4. \(\mathrm{FeO}+\mathrm{CO} \rightarrow \mathrm{Fe}+\mathrm{CO}_2\)

Answer: 3. \(\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{FeO}+\mathrm{CO}_2\)

Question 10. The maximum temperature that can be achieved in a blast furnace is

1. Upto 5000 K
2. Upto 1200 K
3. Upto 2200 K
4. Upto 1900 K.

Answer: 3. Upto 2200 K

It can withstand up to approximately 2000°C or 2200K.

Question 11. Considering the Ellingham diagram, which of the following metals can be used to reduce alumina?

1. Fe
2. Zn
3. Mg
4. Cu

Answer: 3. Mg

Any metal oxide with lower value of ΔG° is more stable than a metal oxide with higher ΔG°. This implies that the metal oxide placed higher in the Eliingham diagram can be reduced by the metal involved in the formation of the oxide placed lower in the diagram.

The relative tendency of the various metals to act as reducing agents is: Ca>Mg>AI>Zn>Fe>Cu

Thus, Mg being more reducing in nature, can reduce aluminium oxide (alumina) to aluminium.

Question 12. In the extraction of copper from its sulphide ore, the metal is finally obtained by the reduction of cuprous oxide with

1. Carbon monoxide
2. Copper (1) sulphide
3. Sulphur dioxide
4. Iron (2) sulphide.

Answer: 2. Copper (1) sulphide

It is an example of auto reduction. \(2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO}_2\)

Question 13. The metal oxide which cannot be reduced to metal by carbon is

1. Al₂O₃
2. PbO
3. ZnO
4. Fe₂O₃

Answer: 1. Al₂O₃

Oxides of less reactive metals (like PbO, ZnO, Fe₂O₃) can be reduced by carbon, while oxides of very reactive metals (like Al₂O₃) can be reduced only by the electrolytic method.

Question 14. Which of the following elements is present as the impurity to the maximum extent in the pig iron?

1. Manganese
2. Carbon
3. Silicon
4. Phosphorus

Answer: 2. Carbon

Pig iron contains about 470 carbon and many impurities such as S, Mn, P, Si, etc’ in smaller amounts.

Question 15. The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.

1. \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(b)}+3 \mathrm{CO}_{2(g)}\)
2. \(\mathrm{CaCO}_{3(s)} \rightarrow \mathrm{CaO}_{(s)}+\mathrm{CO}_{2(g)}\)
3. \(\mathrm{CaO}_{(s)}+\mathrm{SiO}_{2(s)} \rightarrow \mathrm{CaSiO}_{3(s)}\)
4. \(2 \mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{(g)}\)

Answer: 3. \(\mathrm{CaO}_{(s)}+\mathrm{SiO}_{2(s)} \rightarrow \mathrm{CaSiO}_{3(s)}\)

Slag is formed by the reaction \(\mathrm{CaO}+\mathrm{SiO}_2 \rightarrow \mathrm{CaSiO}_3\)

Question 16. Which of the following statements, about the advantage of roasting of sulphide ore before reduction is not true?

1. The ΔG_f° of the sulphide is greater than those for CS₂ and H₂S.
2. The ΔG_f° is negative for roasting of sulphide ore to oxide.
3. Roasting of the sulphide to the oxide is thermodynamically feasible.
4. Carbon and hydrogen are suitable reducing agents for metal sulphides.

Answer: 4. Carbon and hydrogen are suitable reducing agents for metal sulphides.

The standard free energies of formation (ΔG_f°) of most of the sulphides are greater than those of CS₂ and H₂S. Hence, neither carbon nor hydrogen can reduce metal sulphides to metal. The standard free energies of the formation of oxides are much lower than those of SO₂ Therefore, oxidation of metal sulphides to metal oxides is thermodynamically favourable. Hence, sulphide ore is roasted to the oxide before reduction.

Question 17. Nitriding is the process of surface hardening of steel by treating it in an atmosphere of

1. NH₃
2. O₃
3. N₂
4. H₂S

Answer: 1. NH₃

When steel is heated in the Presence of NH₃, iron nitride on the surface of the steel is formed which imparts a hard coating. This process is called nitriding.

Question 18. Aluminium is extracted from alumina (Al₂O₃) by electrolysis of a molten mixture of

1. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{HF}+\mathrm{NaAlF}_4\)
2. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{CaF}_2+\mathrm{NaAlF}_4\)
3. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{Na}_3 \mathrm{AlF}_6+\mathrm{CaF}_2\)
4. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{KF}+\mathrm{Na}_3 \mathrm{AlF}_6\)

Answer: 3. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{Na}_3 \mathrm{AlF}_6+\mathrm{CaF}_2\)

The electrolytic mixture contains alumina(Al₂O₃), cryolite(Na₃AlF₆) and fluorspar(CaF₂) in the ratio of 20:40:20. Due to the presence of these, the conductivity of alumina increases and fusion temperature decreases from 2000°C to 900°c

Question 19. The purification of aluminium, by electrolytic refining, is known as

1. Hoopes process
2. Baeyers process
3. Hall’s process
4. Serpecks process.

Answer: 1. Hoopes process

Aluminium metal obtained from the Hoopet electrolytic refining process is about 99.9% pure. The cell used for this process consists of three layers. The upper layer is pure Al, which acts as the cathode, and the middle layer is a mixture of fluorides of Al and Ba, which acts as an electrolyte. The lowest layer is impure Ali which acts as anode. On electrolysis, pure Al is transferred from the bottom to the top layer, through the middle layer.

Question 20. Calcium is obtained by

1. Reduction of calcium chloride with carbon
2. Electrolysis of molten anhydrous calcium chloride
3. Roasting of limestone
4. Electrolysis of solution of calcium chloride in H₂O

Answer: 2. Electrolysis of molten anhydrous calcium chloride

Calcium is obtained by the electrolysis of a fused mixture of anhydrous CaCl₂ and CaF₂ in a graphite-lined tank which serves as anode. The cathode is a hollow movable iron rod which is kept cool. During electrolysis, calcium is deposited at the cathode while Cl₂ is liberated at the anode.

Question 21. Extraction of gold and silver involves leaching with CN^– ion. Silver is later recovered by

1. Distillation
2. Zone refining
3. Displacement with Zn
4. Liquation.

Answer: 3. Displacement with Zn

Extraction of gold and silver involves leaching the metal with CN^– and the metals silver and gold are later recovered by displacement method.

⇒ \(4 M_{(s)}+8 \mathrm{CN}_{(a q)}^{-}+2 \mathrm{H}_2 \mathrm{O}_{(a q)}+\mathrm{O}_{2(g)} \rightarrow 4\left[M(\mathrm{CN})_2\right]_{(a q)}^{-}+4 \mathrm{OH}_{(a q)}^{-}\)

⇒ \(2\left[\mathrm{M}(\mathrm{CN})_2\right]_{(\mathrm{miq})}^{-}+\mathrm{Zn}_{(s)} \rightarrow 2 \mathrm{M}_{(x)}+\left[\mathrm{Zn}(\mathrm{CN})_4\right]^{2-}{ }_{(\omega q)}\)

Question 22. Which one of the following methods can be used to obtain highly pure metal which is liquid at room temperature?

1. Zone refining
2. Electrolysis
3. Chromatography
4. Distillation

Answer: 4. Distillation

Question 23. Identify the correct statement from the following

1. Wrought iron is impure iron with 4% carbon.
2. Blister copper has a blistered appearance due to the evolution of CO₂.
3. Vapour phase refining is carried out for nickel by the van Arkel method.
4. Pig iron can be moulded into a variety of shapes.

Answer: 4. Wrought iron is impure iron with 4% carbon.

1. Pig iron is impure iron with 4% carbon
2. Blister copper has a blistered appearance due to the evolution of SO₂.
3. Vapour phase refining is carried out for the nickel bv Mondt process.
4. Pig iron can be moulded into a variety of shapes

Question 24. Match items of Column 1 with the items of Column 2 and assign the correct code:

Answer: 3

Question 25. Which of the following pairs of metals is purified by the van Arkel method?

1. Ga and In
2. Zr and Ti
3. Ag and Au
4. Ni and Fe

Answer: 2. Zr and Ti

van Arkel method is used for the purification of Zr and Ti.

Question 26. The method of zone refining of metals is based on the principle of

1. Greater mobility of the pure metal than that of the impurity
2. Higher melting point of the impurity than that of the pure metal
3. The greater noble character of the solid metal than that of the impurity
4. Greater solubility of the impurity in the molten state than in the solid.

Answer: 4. Greater solubility of the impurity in the molten state than in the solid.

Elements which are used as semiconductors such as Si, Ge, Ga, etc. are refined by this method, which is based on the difference in solubility of impurities in the molten and solid state of the metal.

 

p Block Elements

Question 1. In which of the following compounds, nitrogen exhibits the highest oxidation state?

1. \(\mathrm{N}_2 \mathrm{H}_4\)
2. \(\mathrm{NH}_3\)
3. \(\mathrm{N}_3 \mathrm{H}\)
4. \(\mathrm{NH}_2 \mathrm{OH}\)

Answer: 3. \(\mathrm{N}_3 \mathrm{H}\)

N₂H₂ ⇒ 2x + 4(+1) = 0

⇒ 2x +4=0 ⇒ x = -2

N₃H ⇒ x + 3(+1) = 0 ⇒ x =-3

N₃H ⇒ 3x + 1(+1) = 0

⇒ 3x + 1 = 0 ⇒ x = -1/3

NH₂OH ⇒ x + 2 + 1(-2) + 1 = 0

⇒ x + 1 = 0 ⇒ x = -1

Thus, the highest oxidation state is -1/3

Question 2. Nitrogen forms N₂, but phosphorus does not form P₂, however, it converts P₄, reason is

1. Triple bond present between phosphorus atom
2. pπ – pπ bonding is weak
3. pπ – pπ bonding is strong
4. Multiple bonds form easily.

Answer: 2. pπ – pπ bonding is weak

For strong n-bonding, pπ – pπ bonding should be strong. In the case of P, due to its larger size as compared to N-atom, pπ- Pπ bonding is not so strong.

Question 3. Which of the following oxides is most acidic?

1. \(\mathrm{As}_2 \mathrm{O}_5\)
2. \(\mathrm{P}_2 \mathrm{O}_5\)
3. \(\mathrm{N}_2 \mathrm{O}_3\)
4. \(\mathrm{Sb}_2 \mathrm{O}_5\)

Answer: 3. \(\mathrm{N}_2 \mathrm{O}_3\)

Among N, P, As and Sb, the former has the highest electronegativity (EN) so its oxide is the most acidic.

As the electronegativity value of the element increases, the acidic character of the oxide also increases.

Read and Learn More NEET MCQs with Answers

Question 4. Which of the following has the highest dipole moment?

1. \(\mathrm{SbH}_3\)
2. \(\mathrm{AsH}_3\)
3. \(\mathrm{NH}_3\)
4. \(\mathrm{PH}_3\)

Answer: 3. \(\mathrm{NH}_3\)

Due to the greater electronegativity of nitrogen, the dipole moment for NH₃ is greater.

Question 5. The basic character of hydrides of the V group elements decreases in the order

1. \(\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3\)
2. \(\mathrm{SbH}_3>\mathrm{AsH}_3>\mathrm{PH}_3>\mathrm{NH}_3\)
3. \(\mathrm{SbH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{NH}_3\)
4. \(\mathrm{NH}_3>\mathrm{SbH}_3>\mathrm{PH}_3>\mathrm{AsH}_3\)

Answer: 1. \(\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3\)

AI1 the hydrides of group V elements have one lone pair of electrons on their central atom. Therefore, they can act as Lewis bases. The basic character of these hydrides decreases down the group.

Question 6. Among the following oxides, the lowest acidic is

1. \(\mathrm{As}_4 \mathrm{O}_6\)
2. \(\mathrm{As}_4 \mathrm{O}_{10}\)
3. \(\mathrm{P}_4 \mathrm{O}_6\)
4. \(\mathrm{P}_4 \mathrm{O}_{10}\)

Answer: 1. \(\mathrm{As}_4 \mathrm{O}_6\)

The acidic character of the oxides decreases with the decrease in the oxidation state and also decreases down the group

Question 7. Which of the following fluorides does not exist?

1. NF₅
2. PF₅
3. ASF₅
4. SbF₅

Answer: 1. NF₅

Nitrogen cannot form pentahalides because it cannot expand its octet due to non-availability of d-orbitals.

Question 8. Which one has the lowest boiling point?

1. \(\mathrm{NH}_3\)
2. \(\mathrm{PH}_3\)
3. \(\mathrm{AsH}_3\)
4. \(\mathrm{SbH}_3\)

Answer: 2. \(\mathrm{PH}_3\)

The boiling point of hydrides increases with an increase in atomic number but ammonia has an exceptionally high boiling point due to hydrogen bonding. Thus, the correct order of boiling point is, BiH₃ > SbH₃ > NH₃ > AsH₃ > PH₃

Question 9. The number of electrons shared in the formation of nitrogen molecules is

1. 6
2. 10
3. 2
4. 8

Answer: 1. 6

Nitrogen molecule is diatomic containing a triple bond between two N atoms, \(\ddot{\mathrm{N}} \equiv \ddot{\mathrm{N}}\) therefore, nitrogen molecule is formed by sharing six electrons

Question 10. Nitrogen is a relatively inactive element because

1. Its atom has a stable electronic configuration
2. It has a low atomic radius
3. Its electronegativity is fairly high
4. The dissociation energy of its molecule is fairly high.

Answer: 4. Dissociation energy of its molecule is fairly high.

N₂ molecule contains a triple bond between N atoms having very high dissociation energy (946 kJ mol^-1) due to which it is relatively inactive.

Question 11. Pure nitrogen is prepared in the laboratory by heating a mixture of

1. \(\mathrm{NH}_4 \mathrm{OH}+\mathrm{NaCl}\)
2. \(\mathrm{NH}_4 \mathrm{NO}_3+\mathrm{NaCl}\)
3. \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NaOH}\)
4. \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NaNO}_2\)

Answer: 4. \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NaNO}_2\)

Question 12. Which of the following statements is not correct for nitrogen?

1. Its electronegativity is very high.
2. d-orbitals are available for bonding.
3. It is a typical non-metal.
4. Its molecular size is small.

Answer: 2. d-orbitals are available for bonding.

In the case of nitrogen, d-orbitals are not available for bonding. N: 1s² 2s² 2p³

Question 13. Urea reacts with water to form A which will decompose to form B. B When passed through Cu_(aq)²⁺, deep blue colour solution C is formed. What is the formula of C from the following?

1. \(\mathrm{CuSO}_4\)
2. \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}\)
3. \(\mathrm{Cu}(\mathrm{OH})_2\)
4. \(\mathrm{CuCO}_3 \cdot \mathrm{Cu}(\mathrm{OH})_2\)

Answer: 2. \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}\)

Question 14. Aqueous solution of ammonia consists of

1. \(\mathrm{H}^{+}\)
2. \(\mathrm{OH}^{-}\)
3. \(\mathrm{NH}_4^{+}\)
4. \(\mathrm{NH}_4^{+}\) and \(\mathrm{OH}^{-}\)

Answer: 4. \(\mathrm{NH}_4^{+}\) and \(\mathrm{OH}^{-}\)

Aqueous solution of ammonia contains \(\mathrm{NH}_4^{+}\) and OH- ions.

Question 15. Which of the following oxides of nitrogen is paramagnetic?

1. \(\mathrm{NO}_2\)
2. \(\mathrm{N}_2 \mathrm{O}_3\)
3. \(\mathrm{N}_2 \mathrm{O}\)
4. \(\mathrm{N}_2 \mathrm{O}_5\)

Answer: 1. \(\mathrm{NO}_2\)

NO₂ is paramagnetic due to the presence of one unpaired electron.

Question 16. Which of the following is a nitric acid anhydride?

1. \(\mathrm{NO}\)
2. \(\mathrm{NO}_2\)
3. \(\mathrm{N}_2 \mathrm{O}_5\)
4. \(\mathrm{N}_2 \mathrm{O}_3\)

Answer: 3. \(\mathrm{N}_2 \mathrm{O}_5\)

When two molecules of nitric acid undergo heating, lose a water molecule to form an anhydride.

Thus, N₂O₅ is nitric acid anhydride.

Question 17. When copper is heated with a cone. HNO₃ it produces

1. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2,\mathrm{NO}\) and \(\mathrm{NO}_2\)
2. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{N}_2 \mathrm{O}\)
3. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{NO}_2\)
4. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{NO}\)

Answer: 3. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{NO}_2\)

Question 18. Zn gives H₂ gas with H₂SO₄ and HCl but not with HNO₃ because

1. Zn acts as an oxidising agent when reacting with HNO₃
2. HNO₃ is a weaker acid than H₂SO₄ and HCl
3. In the electrochemical series, Zn is above the hydrogen
4. NO₃ is reduced in preference to hydronium ion.

Answer: 4. NO₃ is reduced in preference to hydronium ion.

Zinc is in the top position of hydrogen in the electrochemical series. So, Zn displaces H₂ from dilute H₂SO₄ and HCI with liberation of H₂.

⇒ \(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{ZnSO}_4+\mathrm{H}_2\)

On the other hand, HNO₂ is one oxidising agent. Hydrogen obtained in the reaction is converted into H₂O.

⇒ \(\mathrm{Zn}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+2 \mathrm{H}\)

⇒ \(2 \mathrm{HNO}_3 \rightarrow \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}_2+\mathrm{O}\)

⇒ \(2 \mathrm{H}+\mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{O}\)

Question 19. Sugarcane in reaction with nitric acid gives

1. CO₂ and SO₂
2. (COOH)₂
3. 2HCOOH(two moles)
4. No reaction.

Answer: 2. (COOH)₂

⇒ \(\underset{\text { Cane sugar }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\underset{\text { From } \mathrm{HNO}_3}{18[\mathrm{O}]} \longrightarrow \underset{\text { Oxalic acid }}{6(\mathrm{COOH})_2}+5 \mathrm{H}_2 \mathrm{O}\)

Question 20. Which of the following phosphorus is the most reactive?

1. Scarlet phosphorus
2. White phosphorus
3. Red phosphorus
4. Violet phosphorus

Answer: 2. White phosphorus

White phosphorus has low ignition temperature so it is the most reactive among all the allotropes.

Question 21. Each of the following is true for white and red phosphorus except that they

1. Are both soluble in CS₂
2. Can be oxidised by heating in air
3. Consist of the same kind of atoms
4. Can be converted into one another.

Answer: 1. Are both soluble in CS₂

Red phosphorus is insoluble in CS₂ and only white P is soluble in C₂.

Question 22. A compound X upon reaction with H₂O produces a colourless gas ‘Y with rotten fish smell. Gas Y is absorbed in a solution of CuSO₄ to give Cu₃P₂ as one of the products. Predict the compound ‘X.

1. \(\mathrm{Ca}_3 \mathrm{P}_2\)
2. \(\mathrm{NH}_4 \mathrm{Cl}\)
3. \(\mathrm{As}_2 \mathrm{O}_3\)
4. \(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2\)

Answer: 1. \(\mathrm{Ca}_3 \mathrm{P}_2\)

⇒ \(\underset{\mathrm{Y}}{\mathrm{Ca}_3 \mathrm{P}_2}+6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_2+ \underset{\mathrm{Y}}{2 \mathrm{PH}_{3(g)}}\)

⇒ \(3 \mathrm{CuSO}_4+ \underset{\mathrm {Y}}{2 \mathrm{PH}_3} \rightarrow \underset{\text {Copper phosphide}}{\mathrm{Cu}_3 \mathrm{P}_2}+3 \mathrm{H}_2 \mathrm{SO}_4\)

Question 23. PH₄I + NaOH forms

1. \(\mathrm{PH}_3\)
2. \(\mathrm{NH}_3\)
3. \(\mathrm{P}_4 \mathrm{O}_6\)
4. \(\mathrm{P}_4 \mathrm{O}_{10}\)

Answer: 1. \(\mathrm{PH}_3\)

⇒ \(\mathrm{PH}_4 \mathrm{I}+\mathrm{NaOH} \rightarrow \mathrm{NaI}+\mathrm{PH}_3+\mathrm{H}_2 \mathrm{O}\)

Question 24. Identify the incorrect statement related to PCl₅ from the following:

1. PCl₅ molecule is non-reactive.
2. Three equatorial P – Cl bonds make an angle of 120° with each other.
3. Two axial P – Cl bonds make an angle of 180° with each other.
4. Axial P – Cl bonds are longer than equatorial P – Cl bonds.

Answer: 1. PCl₂ molecule is non-reactive.

It is a reactive gas as it easily provides Cl₂ gas.

Question 25. PCl₃ reacts with water to form

1. \(\mathrm{PH}_3\)
2. \(\mathrm{H}_3 \mathrm{PO}_3, \mathrm{HCl}\)
3. \(\mathrm{POCl}_3\)
4. \(\mathrm{H}_3 \mathrm{PO}_4\)

Answer: 2. \(\mathrm{H}_3 \mathrm{PO}_3, \mathrm{HCl}\)

⇒ \(\mathrm{PCl}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \mathrm{HCl}\)

Question 26. Which of the following oxoacids of phosphorus has the strongest reducing property?

1. \(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7\)
2. \(\mathrm{H}_3 \mathrm{PO}_3\)
3. \(\mathrm{H}_3 \mathrm{PO}_2\)
4. \(\mathrm{H}_3 \mathrm{PO}_4\)

Answer: 3. \(\mathrm{H}_3 \mathrm{PO}_2\)

Acids which contain P – H bonds have strong reducing properties. Among the given compounds, H₃PO₂ is the strongest reducing agent as it contains two P – H bonds.

Question 27. Which is the correct statement for the given acids?

1. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.
2. Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid.
3. Both are diprotic acids.
4. Both are triprotic acids.

Answer: 1. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.

Question 28. Strong reducing behaviour of H₃PO₂ is due to

1. High electron gain enthalpy of phosphorus
2. High oxidation state of phosphorus
3. Presence of two —OH groups and one P—H bond
4. Presence of one —OH group and two P—H bonds.

Answer: 4. Presence of one —OH group and two P—H bonds.

All oxyacids of phosphorus which have P-H bonds act as strong reducing agents. H₃PO₂ has two P-H bonds hence, it acts as a strong reducing agent.

Question 29. Which of the following statements is not valid for oxoacids of phosphorus?

1. Orthophosphoric acid is used in the manufacture of triple superphosphate.
2. Hypophosphorous acid is a diprotic acid.
3. All oxoacids contain tetrahedral four-coordinated phosphorus.
4. All oxoacids contain at least one P = O unit and one P—OH group.

Answer: 2. Hypophosphorous acid is a diprotic acid.

Hypophosphorous acid is a monoprotic acid

Question 30. Oxidation states of P in \(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5, \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6, \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7\) are respectively

1. +3, +5, +4
2. +5, +3, +4
3. +5, +4, +3
4. +3, +4, +5

Answer: 4. +3, +4, +5

The oxidation state can be calculated as:

H₄P₂O₅ : +4 + 2x+ 5(- 2) = 0 = 2x – 6 = 0 = x = +3

H₄P₂O₆ : +4 + 2x + 6(- 2) = 0 = 2x – 8 = 0 t x= +4

H₄P₂O₇ : +4 + 2x + 7(- 2) = 0 + 2x – 10 = 0 = x = +5

Question 31. How many bridging oxygen atoms are present in P₄O₁₀?

1. 6
2. 4
3. 2
4. 5

Answer: 1. 6

Question 32. The structural formula of hypophosphorous acid is

Answer: 3

The formula of hypophosphorous acid is H₃PO₂ as shown in (3). It is a monobasic acid

Question 33. H₃PO₂ is the molecular formula of an acid of phosphorus. Its name and basicity respectively are

1. Phosphorous acid and two
2. Hypophosphorous acid and two
3. Hypophosphorous acid and one
4. Hypophosphoric acid and two.

Answer: 3. Hypophosphorous acid and one

H₃PO₂ is named as hypophosphorous acid. As it contains only one P-OH group, its basicity is one.

Question 34. Which one of the following substances is used in the laboratory for fast drying of neutral gases?

1. Phosphorus pentoxide
2. Active charcoal
3. Anhydrous calcium chloride
4. Na₃PO₄

Answer: 1. Phosphorus pentoxide

P₂O₅ absorbs moisture much more readily than anhydrous CaCl₂

Question 35. P₂O₅ is heated with water to give

1. Hypophosphorous acid
2. Phosphorous acid
3. Hypophosphoric acid
4. Orthophosphoric acid.

Answer: 4. Orthophosphoric acid.

⇒ \(\mathrm{P}_2 \mathrm{O}_5+3 \mathrm{H}_2 \mathrm{O}\) \(\underrightarrow{^{\Delta}}\) \(2 \mathrm{H}_3 \mathrm{PO}_4\)

Question 36. The basicity of orthophosphoric acid is

1. 2
2. 3
3. 4
4. 5

Answer: 2. 3

Orthophosphoric acid, H₃PO₄ contains three P-OH groups and is, therefore, tribasic

Question 37. When orthophosphoric acid is heated to 600°C, the product formed is

1. \(\mathrm{PH}_3\)
2. \(\mathrm{P}_2 \mathrm{O}_5\)
3. \(\mathrm{H}_3 \mathrm{PO}_3\)
4. \(\mathrm{HPO}_3\)

Answer: 4. \(\mathrm{HPO}_3\)

On heating, it gives pyrophosphoric acid at 525 K and metaphosphoric acid at 875 K

Question 38. Given below are two statements:

Statement-1: The boiling points of the following hydrides of group 16 elements increase in the order: H₂O < H₂S < H₂Se < H₂Te.

Statement 2: The boiling points of these hydrides increase with the increase in molecular mass.

In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both statement-1 and statement-2 are correct.
2. Both statement-1 and statement-2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but statement 2 is correct.

Answer: 2. Both statement-1 and statement-2 are incorrect.

Question 39. Which is the correct thermal stability order for H₂E (E = O, S, Se, Te and Po)?

1. \(\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}<\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S}\)
2. \(\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}\)
3. \(\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}\)
4. \(\mathrm{H}_2 \mathrm{Po}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}\)

Answer: 4. \(\mathrm{H}_2 \mathrm{Po}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}\)

The thermal stability of hydrides decreases from H₂O to H₂Po. This is because as the size of atom E it H₂E increases, the bond H-E becomes weaker and thus, breaks on heating. Therefore, the correct order of thermal stability is H₂Po < H₂Te < H₂Se < H₂S < H₂O.

Question 40. The acidity of diprotic acids in aqueous solutions increases in the order

1. \(\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\)
2. \(\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Te}\)
3. \(\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}\)
4. \(\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{~S}\)

Answer: 1. \(\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\)

As the atomic size increases down the group, the bond length increases the bond strength decreases and the cleavage of the E-H bond becomes easier thus, more will be acidity. Thus, the correct order is: H₂S < H₂Se < H₂Te.

Question 41. Which of the following bonds has the highest energy?

1. S-S
2. O-O
3. Se-Se
4. Te-Te

Answer: 1. S-S

The bond energy of S – S is exceptionally high due to its catenation tendency.

Question 42. Which of the following does not give oxygen on heating?

1. \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\)
2. \(\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7\)
3. \(\mathrm{KClO}_3\)
4. \(\mathrm{Zn}\left(\mathrm{ClO}_3\right)_2\)

Answer:

⇒ \(\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{N}_2+\mathrm{Cr}_2 \mathrm{O}_3+4 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Zn}\left(\mathrm{ClO}_3\right)_2\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{ZnCl}_2+3 \mathrm{O}_2\)

⇒ \(\mathrm{KClO}_3\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{KCl}+3 / 2 \mathrm{O}_2\)

⇒ \(2 \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) \(\underrightarrow{^{\Delta}}\) \(2 \mathrm{~K}_2 \mathrm{CrO}_4+\mathrm{Cr}_2 \mathrm{O}_3+3 / 2 \mathrm{O}_2\)

Question 43. Which would quickly absorb oxygen?

1. Alkaline solution of pyrogallol
2. Cone. H₂SO₄
3. Lime water
4. Alkaline solution of CuSO₄

Answer: 1. Alkaline solution of pyrogallol

The alkaline solution of pyrogallol absorbs oxygen quickly.

Question 44. Oxygen will directly react with each of the following elements except

1. P
2. Cl
3. Na
4. S

Answer: 2. Cl

Chlorine does not react directly with oxygen.

Question 45. It is possible to obtain oxygen from air by fractional distillation because

1. Oxygen is in a different group of the periodic table from nitrogen
2. Oxygen is more reactive than nitrogen
3. Oxygen has higher b.pt. Than nitrogen
4. Oxygen has a lower density than nitrogen.

Answer: 3. Oxygen has higher b.pt. Than nitrogen

Air is liquefied by making use of the |oule Thomson effect (cooling by expansion of the gas). Water vapour and CO₂ are removed by solidification.

The remaining constituents of liquid air i.e., liquid oxygen and liquid nitrogen are separated by means of fractional distillation as fractional distillation is a process of separation of mixture based on the difference in their boiling points. (b.pt. of O₂ = – 183°C : b.pt. of N₂ = -195.8°C)

Question 46. Match the following: 

Which of the following is the correct option?

1. 1-A, 2-B, 3-C, 4-D
2. 1-B, 2-A, 3-D, 4-C
3. 1-C, 2-D, 3-A, 4-B
4. 1-D, 2-C, 3-B, 4-A

Answer: 2. 1-B, 2-A, 3-D, 4-C

CO – neutral, BaO – basic,

Al₂O₃ – amphoteric and Cl₂O₇ – acidic.

Question 47. The angular shape of the ozone molecule (O₃) consists of

1. 1σ and 1π bond
2. 2σ and 1π bond
3. 1σ and 2π bonds
4. 2σ and 2π bonds

Answer: 2. 2σ and 1π bond

The angular shape of the ozone molecule (O₃)

Question 48. The gases respectively absorbed by alkaline pyrogallol and oil of cinnamon are

1. \(\mathrm{O}_3, \mathrm{CH}_4\)
2. \(\mathrm{O}_2, \mathrm{O}_3\)
3. \(\mathrm{SO}_2, \mathrm{CH}_4\)
4. \(\mathrm{N}_2 \mathrm{O}, \mathrm{O}_3\)

Answer: 2. \(\mathrm{O}_2, \mathrm{O}_3\)

Alkaline pyrogallol absorbs O₂ and oil of cinnamon absorbs O₃

Question 49. Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other?

1. Is soluble in water.
2. Is used as a food preservative.
3. Forms acid-rain.
4. Is a reducing agent.

Answer: 2. Is used as a food preservative.

NO₂ is not used as a food preservative.

Question 50. Sulphur trioxide can be obtained by which of the following reactions?

1. \(\mathrm{CaSO}_4+\mathrm{C}\) \(\underrightarrow{^{\Delta}}\)
2. \(\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \underrightarrow{^{\Delta}}\)
3. \(\mathrm{S}+\mathrm{H}_2 \mathrm{SO}_4 \underrightarrow{^{\Delta}}\)
4. \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{PCl}_5 \underrightarrow{^{\Delta}}\)

Answer: 2. \(\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \underrightarrow{^{\Delta}}\)

⇒ \(\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{SO}_3\)

Question 51. Match List 1 with List 2

Choose the correct answer from the options given below:

1. 1-A, 2-C, 3- D, 4-B
2. 1-C, 2-D, 3-B, 4-A
3. 1-A, 2-C, 3-B, 4-D
4. 1-C, 2-D, 3-A, 4-B

Answer: 4. 1-C, 2-D, 3-A, 4-B

Question 52. Which of the following oxoacids of sulphur has — O — O — linkage

1. H₂SO₃, sulphurous acid
2. H₂SO₄, sulphuric acid
3. H₂S₂O₈, peroxodisulphuric acid
4. H₂S₂O₇, pyrosulphuric acid

Answer: 3. H2S₂O₂, peroxodisulphuric acid

Peroxodisulphuric acid, H₂S₂O₈ has – O – O linkage.

Question 53. Identify the correct following:

1. \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\)
2. \(\mathrm{H}_2 \mathrm{SO}_3\)
3. \(\mathrm{H}_2 \mathrm{SO}_4\)
4. \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8\)

Answer: 1. \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\)

Question 54. In which pair of ions do both species contain an S — S bond?

1. \(\mathrm{S}_4 \mathrm{O}_6^{2-}, \mathrm{S}_2 \mathrm{O}_3^{2-}\)
2. \(\mathrm{S}_2 \mathrm{O}_7^{2-}, \mathrm{S}_2 \mathrm{O}_8^{2-}\)
3. \(\mathrm{S}_4 \mathrm{O}_6^{2-}, \mathrm{S}_2 \mathrm{O}_7^{2-}\)
4. \(\mathrm{S}_2 \mathrm{O}_7^{2-}, \mathrm{S}_2 \mathrm{O}_3^{2-}\)

Answer: 1. \(\mathrm{S}_4 \mathrm{O}_6^{2-}, \mathrm{S}_2 \mathrm{O}_3^{2-}\)

Question 55. Oleum is

1. Castor oil
2. Oil of vitriol
3. Fuming H₂SO₄
4. None of these.

Answer: 3. Fuming H₂SO₄

Pyrosulphuric acid or oleum (+6) is H₂S₂O₇ which is obtained by dissolving SO₃ and is called fuming sulphuric acid.

Question 56. Match List 1 (substances) with List 2 (processes) employed in the manufacture of the substances and select the correct option.

1. 1-A, 2-D, 3-B, 4-C
2. 1-A, 2-B, 3-C, 4-D
3. 1-D, 2-C, 3-B, 4-A
4. 1-D, 2-B, 3-C, 4-A

Answer: 4. 1-D, 2-B, 3-C, 4-A

Question 57. Statement 1: Acid strength increases in the order given as HF « HCl« HBr « HI.

Statement 2: As the size of the elements F, Cl, Br, and I increases down the group, the bond strength of HF, HC1, HBr, and HI decreases and so the acid strength increases.

In light of the above statements, choose the correct answer from the options given below.

1. Statement 1 is incorrect but statement 2 is true.
2. Both statement 1 and statement 2 are true.
3. Both statement 1 and statement 2 are false.
4. Statement 1 is correct but statement 2 is false.

Answer: 2. Both statement 1 and statement 2 are true.

Question 58. In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?

1. \(\mathrm{CO}_2<\mathrm{SiO}_2 <\mathrm{SnO}_2<\mathrm{PbO}_2\): Increasing oxidizing power
2. \(\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}\) : Increasing acidic strength
3. \(\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S} <\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\) : Increasing pK values
4. \(\mathrm{NH}_3<\mathrm{PH}_3<\mathrm{AsH}_3<\mathrm{SbH}_3\) : Increasing acidic character

Answer: 3. \(\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S} <\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\) : Increasing pK values

pK_a

H₂O: 14

H₂S: 7

H₂Se: 3.872

H₂Te: 2.6

Question 59. Which of the following statements is not true for halogens?

1. All form monobasic oxyacids.
2. All are oxidizing agents.
3. All but fluorine show positive oxidation states.
4. Chlorine has the highest electron-gain enthalpy.

Answer: 3. All but fluorine show positive oxidation states.

All halogens show both positive and negative oxidation states while fluorine shows only negative oxidation states except +1 in HOF.

Question 60. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?

1. \(\mathrm{Br}_2>\mathrm{I}_2>\mathrm{F}_2>\mathrm{Cl}_2\)
2. \(\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2\)
3. \(\mathrm{I}_2>\mathrm{Br}_2>\mathrm{Cl}_2>\mathrm{F}_2\)
4. \(\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2\)

Answer: 4. \(\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2\)

The order of bond dissociation enthalpy is

A reason for this anomaly is the relatively large electron-electron repulsion among the lone pairs in F₂ a molecule where they are much closer to each other than in the case of Cl₂

Question 61. The variation of the boiling points of the hydrogen halides is in the order HF > HI > HBr > HCl. What explains the higher boiling point of hydrogen fluoride?

1. There is strong hydrogen bonding between HF molecules.
2. The bond energy of HF molecules is greater than in other hydrogen halides.
3. The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule.
4. The electronegativity of fluorine is much higher than for other elements in the group.

Answer: 1. There is strong hydrogen bonding between HF molecules.

HF forms strong intermolecular H-bonding due to the high electronegativity of F. Hence, the boiling point of HF is abnormally high. Boiling points of other hydrogen halides gradually increase from HCl to HI due to an increase in the size of halogen atoms from Cl to I which further increases the magnitude of van der Waals forces.

Question 62. Among the following which is the strongest oxidising agent?

1. Br₂
2. I₂
3. Cl₂
4. F₂

Answer: 4. F₂

Standard reduction potentials of halogens are positive and decrease from fluorine to iodine. So, F₂ is the strongest oxidising agent.

Question 63. Which one of the following arrangements does not give the correct picture of the trends indicated against it?

1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy
2. F₂ > Cl₂ > Br₂ > I₂: Electronegativity
3. F₂ > Cl₂ > Br₂ > I₂: Oxidizing power
4. F₂ > Cl₂ > Br₂ > I₂: Electron gain enthalpy

Answer: 1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy and 4. F₂ > Cl₂ > Br₂ > I₂: Electron gain enthalpy

In the case of diatomic molecules (X₂) of halogens, the bond dissociation energy decreases in the order: Cl₂ > Br₂ > F₂> I₂.

This is due to the relatively large electron-electron repulsion among the lone pairs is F, then in the case of Cl₂.

The oxidising power, electronegativity and reactivity decrease in the order: F₂ > C₂ > Br₂ > I₂

Electron gain enthalpy of halogens follows the given order: Cl₂>F₂>Br₂>I₂

The low value of electron gain enthalpy of fluorine is probably due to the small size of the fluorine atom.

Question 64. Which one of the following orders is not in accordance with the property stated against it?

1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy
2. F2 > Cl2 > Br2 > I₂: Oxidising power
3. HI > HBr > HCl > HF: Acidic property in water
4. F₂ > Cl₂ > Br₂ > I₂: Electronegativity

Answer: 1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy

The lower value of bond dissociation energy of fluorine is due to the high inter-electronic repulsions between non-bonding electrons in the 2p-orbitals of fluorine. As a result, the F – F bond is weaker in comparison to the CI – Cl and Br – Br bonds.

Question 65. Which statement is wrong?

1. Bond energy of F₂ > Cl₂
2. Electronegativity of F > Cl
3. F is more oxidising than Cl
4. Electron affinity of Cl > F

Answer: 1. Bond energy of F₂ > Cl₂

Due to more repulsion in between non-bonding electron pairs (2p) of two fluorines (due to the small size of F-atom) in comparison to non-bonding electron pairs (3p) in chlorine, the bond energy of F, is less than Cl₂.

B.E(F₂) = 158.8 kJ/mole and

B.E. (Cl₂) =242.6 kJ/ mole

Question 66. Which of the following has the greatest electron affinity?

1. I
2. Br
3. F
4. Cl

Answer: 4. Cl

In general, the electron affinity decreases from top to bottom in a group. But in group 17, fluorine has lower electron affinity as compared to chlorine due to the very small size of the fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2s orbitals of fluorine and thus, the incoming electron does not experience much attraction

Question 67. Which of the following displaces Br₂ from an aqueous solution containing bromide ions?

1. \(\mathrm{I}_2\)
2. \(\mathrm{I}_3^{-}\)
3. \(\mathrm{Cl}_2\)
4. \(\mathrm{Cl}^{-}\)

Answer: 3. \(\mathrm{Cl}_2\)

Since chlorine is a stronger oxidising agent than bromine, therefore it will displace bromine from an aqueous solution containing bromide ions. \(\mathrm{Cl}_2+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_2\)

Question 68. Which of the following species has four lone pairs of electrons?

1. I
2. O
3. Cl^–
4. He

Answer: 3. Cl^–

Outer electronic configuration of \(\mathrm{Cl}=3 s^2 3 p_x^2 3 p_y^2 3 p_z^1\)

Outer electronic configuration of \(\mathrm{Cl}^{-}=3 s^2 3 p_x^2 3 p_y^2 3 p_z^2\),

i.e., 4 lone pair of electrons

Question 69. Match the following:

Which of the following is the correct option?

1. 1-D, 2-C, 3-B, 4-A
2. 1-A, 2-B, 3-C, 4-D
3. 1-B, 2-D, 3-A, 4-C
4. 1-C, 2-D, 3-B, 4-A

Answer: 1. 1-D, 2-C, 3-B, 4-A

Question 70. When Cl₂ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

1. Zero to +1 and zero to -5
2. Zero to -1 and zero to +5
3. Zero to -1 and zero to +3
4. Zero to +1 and zero to -3

Answer: 2. Zero to -1 and zero to +5

⇒ \(\stackrel{0}{\mathrm{C}} \mathrm{l}_2+\underset{\text{(hot and conc.)}}{6 \mathrm{NaOH}} \rightarrow 5 \stackrel{-1}{\mathrm{NaCl}}+\mathrm{NaClO}_3+3 \mathrm{H}_2 \mathrm{O}\)

This is an example of a disproportionation reaction and oxidation state of chlorine changes from 0 to -1 and +5

Question 71. Which of the following is used in the preparation of chlorine?

1. Both MnO₂ and KMnO₄
2. Only KMnO₄
3. Only MnO₂
4. Either MnO₂ or KMnO₄

Answer: 1. Both MnO₂ and KMnO₄

⇒ \(\mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \uparrow\) \(2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{MnCl}_2+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{Cl}_2 \uparrow\)

Question 72. Which of the following elements is extracted commercially by the electrolysis of an aqueous solution of its compound?

1. Cl
2. Br
3. Al
4. Na

Answer: 1. Cl

Chlorine is obtained by the electrolysis of brine (concentrated NaCl solution). Chlorine is liberated at the anode.

Question 73. When chlorine is passed over dry slaked lime at room temperature, the main reaction product is

1. Ca(ClO₂)₂
2. CaCl₂
3. CaOCl₂
4. Ca(OCl)₂

Answer: 3. CaOCl₂

Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 74. In the manufacture of bromine from seawater, the mother liquor containing bromides is treated with

1. Carbon dioxide
2. Chlorine
3. Iodine
4. Sulphur dioxide.

Answer: 2. Chlorine

Bromide in the mother liquor (containing MgBr₂) is oxidised to Br₂ by passing CI₂ which is a stronger oxidising agent.

⇒ \(2 \mathrm{Br}^{-}+\mathrm{Cl}_2 \rightarrow \mathrm{Br}_2+2 \mathrm{Cl}^{-}\)

Question 75. The bleaching action of chlorine is due to

1. Reduction
2. Hydrogenation
3. Chlorination
4. Oxidation.

Answer: 4. Oxidation.

The bleaching action of chlorine is due to oxidation in the presence of moisture. The bleaching effect is permanent.

⇒ \(\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}+[\mathrm{O}]\)

Colouring matter + [O] → Colourtress matter

Question 76. Bleaching powder reacts with a few drops of cone. HCl to give

1. Chlorine
2. Hypochlorous Acid
3. Calcium Oxide
4. Oxygen.

Answer: 1. Chlorine

⇒ \(\mathrm{CaOCl}_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2\)

Question 77. Among the following, the correct order of acidity is

1. \(\mathrm{HClO}_2<\mathrm{HClO}<\mathrm{HClO}_3<\mathrm{HClO}_4\)
2. \(\mathrm{HClO}_4<\mathrm{HClO}_2<\mathrm{HClO}<\mathrm{HClO}_3\)
3. \(\mathrm{HClO}_3<\mathrm{HClO}_4<\mathrm{HClO}_2<\mathrm{HClO}\)
4. \(\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4\)

Answer: 4. \(\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4\)

The acidic character of the oxoacids increases with an increase in the oxidation number of the halogen atom i.e., \(\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4\).

This can be explained on the basis of the relative stability of the anions Ieft after the removal of a proton. Since the stability of the anion decreases in the order : \(\mathrm{ClO}_4^{-}>\mathrm{ClO}_3^{-}>\mathrm{ClO}_2^{-}>\mathrm{ClO}^{-}\) acid strength also decreases in the same order.

Question 78. Which of the statements given below is incorrect?

1. O₃ molecule is bent.
2. ONF is isoelectronic with O₂N^–.
3. OF₂ is an oxide of fluorine.
4. Cl₂O₇ is an anhydride of perchloric acid.

Answer: 3. OF₂ is an oxide of fluorine.

OF₂ (oxygen difluoride) is a fluoride of oxygen because fluorine is more electronegative than oxygen.

Question 79. The correct order of increasing bond angles in the following species is

1. \(\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2<\mathrm{ClO}_2^{-}\)
2. \(\mathrm{ClO}_2<\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2^{-}\)
3. \(\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2^{-}<\mathrm{ClO}_2\)
4. \(\mathrm{ClO}_2^{-}<\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2\)

Answer: 4. \(\mathrm{ClO}_2^{-}<\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2\)

Question 80. Which one of the following oxides is expected to exhibit paramagnetic behaviour?

1. CO₂
2. SiO₂
3. SO₂
4. ClO₂

Answer: 4. ClO₂

Question 81. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): ICi is more reactive than I₂.

Reason (R): The I — Cl bond is weaker than the I — I bond. In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both (A) and (R) are correct and (R) is the correct explanation of (A).
2. Both (A) and (R) are correct but (R) is not the correct explanation of (A).
3. (A) is correct but (R) is not correct.
4. (A) is not correct but (R) is correct.

Answer: 1. Both (A) and (R) are correct and (R) is the correct explanation of (A).

In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because the X-X’ (ICl) bond in interhalogens is weaker than the X-X (I-I) bond in halogens except for the F-F bond.

Question 82. Match the interhalogen compounds of column A with the geometry in column B and assign the correct code.

1. 1-C, 2-A, 3-D, 4-B
2. 1-E, 2-D, 3-C, 4-B
3. 1-D, 2-C, 3-B, 4-A
4. 1-C, 2-D, 3-A, 4-B

Answer: 1. 1-C, 2-A, 3-D, 4-B

Question 83. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

Assertion A: Helium is used to dilute oxygen in the diving apparatus.

Reason R: Helium has a high solubility in O₂.

In light of the above statements, choose the correct answer from the options given below.

1. A is true but R is false.
2. A is false but R is true.
3. Both A and R are true and R is the correct explanation of A.
4. Both A and R are true and R is not the correct explanation of A.

Answer: 1. A is true but R is false.

Helium is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

Question 84. Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.

1. Noble gases have large positive values of electron gain enthalpy.
2. Noble gases are sparingly soluble in water.
3. Noble gases have very high melting m and boiling points.
4. Noble gases have weak dispersion forces.

Answer: 3. Noble gases have very high melting m and boiling points.

Noble gases have very low melting and boiling points because the only type of interatomic interaction in these elements is weak dispersion forces.

Question 85. Match the Xenon compounds in Column A with their structure in Column B and assign the correct code.

1. 1-C, 2-D, 3-A, 4-B
2. 1-A, 2-B, 3-C, 4-D
3. 1-B, 2-C, 3-D, 4-A
4. 1-B, 2-C, 3-A, 4-D

Answer: 3. 1-B, 2-C, 3-D, 4-A

Question 86. Identify the incorrect statement, regarding the molecule Xe04.

1. XeO₄ molecule is square planar.
2. There are four pπ – dπ bonds.
3. There are four sp³ – p, σ bonds.
4. XeO₄ molecule is tetrahedral.

Answer: 1. XeO₄ molecule is square planar.

Question 87. Which compound has a planar structure?

• \(\mathrm{XeF}_4\)
• \(\mathrm{XeOF}_2\)
• \(\mathrm{XeO}_2 \mathrm{~F}_2\)
• \(\mathrm{XeO}_4\)

Answer: 1. \(\mathrm{XeF}_4\)

In XeF₄ the ‘Xe’ atom is sp³d² hybridised, which contains tu,o lone pair orbitals and four bond pair orbitals. Therefore, the shape of XeF₄ molecule is square plar.rar, with one lone pair orbital over and the other below the plane.