MCQs on Some Basic Concept of Chemistry for NEET

Some Basic Concepts Of Chemistry

Question 1. The dimensions of pressure are the same as that of

1. Force per unit volume
2. Energy per unit volume
3. Force
4. Energy

Answer: 2. Energy per unit volume

Pressure = \(\frac{\text { Force }}{\text { Area }}\)

Therefore, dimensions of pressure = \(\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

and dimensions of energy per unit volume = \(\frac{\text { Energy }}{\text { Volume }}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

Question 2. Given the numbers: 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for the three numbers is

1. 3, 3, and 4 respectively
2. 3, 4 and 4 respectively
3. 3,4 and 5 respectively
4. 3, 3, and 3 respectively.

Answer: 4. 3, 3, and 3 respectively.

Zeros placed left to the number are never significant, therefore the no. of significant figures for the numbers 161 cm, 0.161 cm, and 0.0161 cm are the same, i.e., 3.

Question 3. Equal masses of H₂, O₂, and methane have been taken in a container of volume V at a temperature of 27 °C in identical conditions. The ratio of the volumes of gases H₂, O₂ methane would be

1. 8:16:1
2. 16:8:1
3. 16:1:2
4. 8:1:2

Answer: 3. 16:1:2

According to Avogadrot’s hypothesis, the ratio of the volumes of gases will be equal to the ratio of their no. of moles.

So, number og molecules = \(\frac{\text{Mass}}{\text{Molecule mass}}\)

∴ \(n_{\mathrm{H}_2}=\frac{w}{2} ; n_{\mathrm{O}_2}=\frac{w}{32} ; n_{\mathrm{CH}_4}=\frac{w}{16}\)

So, the ratio is \(\frac{w}{2}: \frac{w}{32}: \frac{w}{16}\) or 16: 1: 2.

Read and Learn More NEET MCQs with Answers

Question 4. What volume of oxygen gas (O₂) measured at 0°C and 1 atm, is needed to burn completely 1 L of propane gas (C₃H₈) measured under the same conditions?

1. 5 L
2. 10 L
3. 7 L
4. 6 L

Answer: 1. 5 L

∴ \(\mathrm{C}_3\mathrm{H}_8+\underset{5\mathrm{vol}} 5\mathrm{O}_2 \rightarrow \underset{3\mathrm{vol}} 3\mathrm{CO}_2+\underset{4\mathrm{vol}} 4 \mathrm{H}_2\mathrm{O}\)

According to the above equation, 1 volume or 1 liter of propane requires 5 volume or 5 liters of O₂ to burn completely.

Question 5. 0. 24 g of a volatile gas, upon vaporization, gives 45 mL vapor at NTP. What will be the vapor density of the substance? (Density of H₂ = 0.089 g/L)

1. 95.93
2. 59.93
3. 95.39
4. 5.993

Answer: 2. 59.93

Weight of gas = 0.24 g,

Volume of gas = 45 mL = 0.045 litre and density of H₂ = 0.089 8/L

Weight of 45 mL of H₂ = density x volume = 0.089 x 0.045 = 4.005 x 10^-3 g

Therefore, Yapour density

= \(\frac{\text { Weight of certain volume of substance }}{\text { Weight of same volume of hydrogen }}=\frac{0.24}{4.005 \times 10^{-3}}=59.93\)

Question 6. The molecular weights of O₂ and SO₂ are 32 and 64 respectively. At 15°C and 150 mmHg pressure, one liter of O₂ contains ‘N’ molecules. The number of molecules in two liters of SO₂ under the same conditions of temperature and pressure will be

1. N/2
2. N
3. 2 N
4. 4 N

Answer: 3. 2 N

If 1L of one gas contains N molecules, 2L of any gas under the same conditions will contain 2N molecules.

Question 7. What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

1. 2.8 kg
2. 6.4 kg
3. 9.6 kg
4. 96 kg

Answer: 3. 9.6 kg

∴ \({\mathrm{C}_2 \mathrm{H}_4}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

For complete combustion, \(2.8 \mathrm{~kg} \text { of } \mathrm{C}_2 \mathrm{H}_4\) requires = \(\frac{96}{28} \times 2.8 \times 10^3 \mathrm{~g}\)

= \(9.6 \times 10^3 \mathrm{~g}=9.6 \mathrm{~kg}\) of \(\mathrm{O}_2\)

Question 8. An element, X has the following isotopic composition:

1. ²⁰⁰X: 90%
2. ¹⁹⁹X: 8.0%
3. ²⁰²X: 2.0%

The weighted average atomic mass of the naturally occurring element X is closest to

1. 201 amu
2. 202 amu
3. 199 amu
4. 200amu

Answer: 4. 200amu

Average isotopic mass of X = \(\frac{200 \times 90+199 \times 8+202 \times 2}{90+8+2}\)

= \(\frac{18000+1592+404}{100}=199.96 \mathrm{amu}=200 \mathrm{amu}\)

Question 9. Boron has two stable isotopes, ¹⁰B(19%) and ¹¹B(81%). Calculate the average weight of boron in the periodic table.

1. 10.8
2. 10.2
3. 11.2
4. 10.0

Answer: 1. 10.8

Average atomic mass = \(\frac{19 \times 10+81 \times 11}{100}=10.81\)

Question 10. Which one of the following has a maximum number of atoms?

1. 1 g of Ag_(s) [Atomic mass of Ag = 108]
2. 1 g of Mg_(s) [Atomic mass of Mg = 24]
3. 1 g of O_(2)(g) [Atomic mass of O = 16]
4. 1 g of Li_(s) [Atomic mass of Li = 7]

Answer: 4. 1 g of Li_(s) [Atomic mass of Li = 7]

1 mole of substance = \(N_A\) atoms

108 g of \(\mathrm{Ag}=N_A\) atoms \(\Rightarrow 1 \mathrm{~g}\) of \(\mathrm{Ag}=\frac{N_A}{108}\) atoms

24 g of \(\mathrm{Mg}=N_A\) atoms \(\Rightarrow 1 \mathrm{~g}\) of \(\mathrm{Mg}=\frac{N_A}{24}\) atoms

32 g of \(\mathrm{O}_2=N_A\) molecules \(=2 N_A\) atoms

⇒ \(1 \mathrm{~g}\) of \(\mathrm{O}_2=\frac{N_A}{16}\) atoms

7 g of \(\mathrm{Li}=N_A\) atoms \(\Rightarrow 1 \mathrm{~g}\) of \(\mathrm{Li}=\frac{N_A}{7}\) atoms

Therefore, 1 g of Li_(s) has a maximum number of atoms.

Question 11. In which case is a number of molecules of water maximum?

1. 18 mL of water
2. 0.18 g of water
3. 0.00224 L of water vapors at 1 atm and 273 K
4. 1^-3 mol of water

Answer: 1. 18 mL of water

1. Mass of water= V x d = 18 x 1 = 18g

Molecules of water = mole x \(N_A\) = \(\frac{18}{18} N_A=N_A\)

2. Molecules of water = mole x \(\mathrm{N}_{\mathrm{A}}=\frac{0.18}{18} \mathrm{~N}_{\mathrm{A}}\)

3. Moles of water = \(\frac{0.00224}{22.4}=10^{-4}\)

Molecules of water = mole x N_A = 10^-4 N_A

4. Molecules of water = mole x N_A = 10^-3 N_A

Question 12. Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂– When 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂ weighs 9 g, the atomic weights of X and Y are

1. 40,30
2. 60,40
3. 20, 30
4. 30, 20

Answer: 1. 40,30

Let the atomic weight of element X is x and that of element Y is y.

For \(X Y_2, n=\frac{w}{\text { Molecule weight}}\)

0.1 = \(\frac{10}{x+2 y} \Rightarrow x+2 y=\frac{10}{0.1}\) =100……(1)

For \(X_3 Y_2\), n= \(\frac{w}{\text { Molecule weight}}\)

0.05 = \(\frac{9}{3 x+2 y} \Rightarrow 3 x+2 y=\frac{9}{0.05}=180\)….(2)

On solving equations (1) and (2), we get x = 40

40 + 2y = 100

⇒ = 2y = 60

∴ y = 30

Question 13. If Avogadro number N_A, is changed from 6.022 x 10²³ mol^-1 to 6.022 x 10²⁰ mol^-1, this would change

1. The mass of one mole of carbon
2. The ratio of chemical species to each other in a balanced equation
3. The ratio of elements to each other in a compound
4. The definition of mass in units of grams.

Answer: 1. The mass of one mole of carbon

Mass of 1 mo1 (6.022x 10²³ atoms)of carbon =12g

If Avogadro number is changed to 6.022 x 10²⁰ atoms then the mass of 1 mol of carbon

= \(\frac{12 \times 6.022 \times 10^{20}}{6.022 \times 10^{23}}=12 \times 10^{-3} \mathrm{~g}\)

Question 14. The number of water molecules is maximum in

1. 1.8 grams of water
2. 18 grams of water
3. 18 moles of water
4. 18 molecules of water.

Answer: 3. 18 molecules of water.

1.8 gram of water = \(\frac{6.023 \times 10^{23}}{18} \times 1.8\)

= \(6.023 \times 10^{27}\) molecules

= 6.023 x 10²² molecules

18 gram of water = 6.023 x 10²³ molecules

18 moles of water = 18 x 6.023 x 10²³ molecules

Question 15. A mixture of gases contains H₂ and O₂ gases in the ratio of 1: 4 (w/w). What is the molar ratio of the two gases in the mixture?

1. 16:1
2. 2: 1
3. 1: 4
4. 4: 1

Answer: 4. 4: 1

Number of moles of \(\mathrm{H}_2=1 / 2\)

Number of moles of \(\mathrm{O}_2=\frac{4}{32}\)

Hence, molar ratio = \(\frac{1}{2}: \frac{4}{32}=4: 1\)

Question 16. Which has the maximum number of molecules among the following?

1. 44 g CO₂
2. 48 g O₃
3. 8 g H₂
4. 64 g SO₂

Answer: 3. 64 g SO₂

8 g H₂ has 4 moles while the others have 1 mole each

Question 17. The number of atoms in 0.1 mol of a triatomic gas is (N_A = 6.02 x 10²³ mol^-1)

1. 6.026 x 10²²
2. 1.806 x 10²³
3. 3.600 x 10²³
4. 1.800 x10²²

Answer: 2. 1.806 x 10²³

No. of atoms = N_A x No. of moles x 3 = 6.023 x 10²³ x 0.1 x 3 = 1.806 x 10²³

Question 18. The maximum number of molecules is present in

1. 15 L of H₂ gas at STP
2. 5 L of N₂ gas at STP
3. 0.5g of H₂ gas
4. 10g of O₂ gas.

Answer: 1. 15 L of H₂ gas at STP

AT STP, 22.4L=6.023x 10²³ molecules

15 L \(\mathrm{H}_2=\frac{6.023 \times 10^{23} \times 15}{22.4}=4.033 \times 10^{23}\) molecules

5 L \(\mathrm{~N}_2=\frac{6.023 \times 10^{23} \times 5}{22.4}=1.344 \times 10^{23}\) molecules

2 g \(\mathrm{H}_2=6.023 \times 10^{23}\) molecules

0.5 g \(\mathrm{H}_2=\frac{6.023 \times 10^{23} \times 0.5}{2}=1.505 \times 10^{23}\) molecules

32 g \(\mathrm{O}_2=6.023 \times 10^{23}\)

10 g of \(\mathrm{O}_2=\frac{6.023 \times 10^{23} \times 10}{32}=1.882 \times 10^{23}\) molecules

Question 19. Which has the maximum molecules?

1. 7g N₂
2. 2g H₂
3. 16g NO₂
4. 16g O₂

Answer: 2. 2g H₂

number of molecules = moles \(\times N_A\)

Molecules of \(\mathrm{N}_2=\frac{7}{14} N_A=0.5 N_A\), Molecules of \(\mathrm{H}_2=N_A\)

Molecules of \(\mathrm{NO}_2=\frac{16}{46} N_A=0.35 N_A\)

Molecules of \(\mathrm{O}_2=\frac{16}{32} N_A=0.5 N_A\)

∴ \(2 \mathrm{~g} \mathrm{H}_2\left(1\right. mole \left.\mathrm{H}_2\right)\) contains maximum molecules.

Question 20. The specific volume of cylindrical virus particle is 6.2 x 10^-2 cc/g whose radius and length are 7 A and 10 A respectively. If N_A = 6.02 x 10²², find the molecular weight of the virus.

1. 15.4 kg/mol
2. 1.54 x 10⁴ kg/mol
3. 3.08 x 10⁴ kg/mol
4. 3.08 x 10³ kg/mol

Answer: 1. 15.4 kg/mol

Specific volume (volume of 1 g) of cylindrical virus

particle = 6.02 x 10^-2 cc/g

Radius of virus, r = 7Å = 7x 10^-8 cm

Volume of virus = πr²l

= \(\frac{22}{7} \times\left(7 \times 10^{-8}\right)^2 \times 10 \times 10^{-8}=154 \times 10^{-23} c c\)

weight of one virus particle = \(\frac{\text { Volume }(\mathrm{cc})}{\text { Specific volume }(\mathrm{cc} / \mathrm{g})}\)

= \(\frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \mathrm{~g}\)

∴ Molecular weight of virus = weight of  \(N_A\) particles

= \(\frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \times 6.02 \times 10^{23} \mathrm{~g} / \mathrm{mol}\)

= \(15400 \mathrm{~g} / \mathrm{mol}=15.4 \mathrm{~kg} / \mathrm{mol}\)

Question 21. The number of atoms in 4.25 g of NH₃ is approximately

1. 4 x 10²³
2. 2 x 10²³
3. 1 x 10²³
4. 6 x 10²³

Answer: 4. 6 x 10²³

17 g of \(\mathrm{NH}_3=4 N_A\) atoms

4.25 g of \(\mathrm{NH}_3=\frac{4 N_A}{17} \times 4.25\) atoms \(=N_A\) atoms \(=6 \times 10^{23}\) atoms

Question 22. Haemoglobin contains 0.334% of iron by weight. The molecular weight of hemoglobin is approximately 67200. The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of hemoglobin is

1. 4
2. 6
3. 3
4. 2

Answer: 1. 4

Quantity of iron in one molecule = \(\frac{67200}{100} \times 0.334=224.45 \mathrm{amu}\)

No. of iron atoms in one molecule of haemoglobin = \(\frac{224.45}{56}\) = 1

Question 23. The number of moles of oxygen in one liter of air containing 21% oxygen by volume, under standard conditions, is

1. 0.0093 mol
2. 2.10 mol
3. 0.186 mol
4. 0.21 mol

Answer: 1. 0.0093 mol

Volume of oxygen in one litre of air = \(\frac{21}{100}\) x 1000 = 210 mL.

Therefore, no. of moles = \(\frac{210}{224000}\) = 0.0093 mol

Question 24. The total number of valence electrons in 4.2 g of N₃^– ion is (NA is the Avogadro’s number)

1. 2.1 N_A
2. 4.2 N_A
3. 1.6 N_A
4. 3.2 N_A

Answer: 3. 1.6 N_A

Each nitrogen atom has 5 valence electrons, therefore the total number of valence electrons in N₃^– ion is 16.

Since the molecular mass of N₃^– is 42, therefore the total number of valence electrons in 4.2 g of \(\mathrm{N}_3^{-} \text {ion }=\frac{4.2}{42} \times 16 \times N_A=1.6 N_A\)

Question 25. The number of gram molecules of oxygen in 6.2 x 10²⁴ CO molecules is

1. 10 g molecules
2. 5 g molecules
3. lg molecule
4. 0.5 g molecules.

Answer: 2. 5 g molecules

Avogadro’s number, N_A = 6.02 x 10²³ molecules = 1 mole

∴ 6.02 x 10²⁴ CO molecules = 10 moles CO

= 10 g atoms of O = 5 g molecules of O₂

Question 26. The ratio of C_p and C_v of a gas ‘X’ is 1.4. The number of atoms of the gas ‘X’ present in 11.2 liters of it at NTP will be

1. 6.02 x 10²³
2. 1.2 x 10²³
3. 3.01 x 10²³
4. 2.01 x 10²³

Answer: 1. 6.02 x 10²³

Here, C_p/C_v = 1.4, which shows that the gas is diatomic.

22.4L at NTP = 6.02 x 10²³ molecules

∴ 11.2 L at NTP = 3.01 x 10²³ molecules

Since gas is diatomic

∴ 11.2 L at NTP = 2 x 3.01 x 10²³ atoms =6.02 x 10²³ atoms

Question 27. The number of oxygen atoms in 4.4 g of CO₂ is

1. 1.2 x 10²³
2. 6 x 10²²
3. 6 x 10²³
4. 12 x 10²³

Answer: 1. 1.2 x 10²³

1 mol of CO₂ = 44gof CO₂

∴ 4.4 g CO₂ = 0.1 mol CO₂= 6 x 10²² molecules

[Since, 1 mole CO₂ = 6 x 10²³ molecules]

=2 x 6 x 10²² atoms of O₂ = 1.2 x 10²³ atoms of O

Question 28. 1 cc N₂O at NTP contains

1. \(\frac{1.8}{224} \times 10^{22}\) atoms
2. \(\frac{6.02}{22400} \times 10^{23}\) molecules
3. \(\frac{1.32}{224} \times 10^{23}\) electrons
4. All of the above.

Answer: 4. All of the above.

22400 cc of \(\mathrm{N}_2{O}\) contain \(6.02 \times 10^{23}\) molecules

∴ 1 cc of \(\mathrm{N}_2 \mathrm{O}\) contain \(\frac{6.02 \times 10^{23}}{22400}\) molecules

Since in \(\mathrm{N}_2 \mathrm{O}\) molecule there are 3 atoms.

∴ \(1 \mathrm{cc} \mathrm{N}_2 \mathrm{O}=\frac{3 \times 6.02 \times 10^{23}}{22400}\) atoms \(=\frac{1.8 \times 10^{22}}{224}\) atoms

No. of electrons in a molecule of \(\mathrm{N}_2 \mathrm{O}=7+7+8=22\)

Hence, no. of electrons in \(1 \mathrm{cc}\) of \(\mathrm{N}_2 \mathrm{O}\)

= \(\frac{6.02 \times 10^{23}}{22400} \times 22\) electrons \(=\frac{1.32}{224} \times 10^{23}\) electrons

Question 29. An organic compound contains 78% (by wt) carbon and the remaining percentage of hydrogen. The right option for the empirical formula of this compound is [Atomic weight of C is 12, H is 1]

1. CH₄
2. CH
3. CH₂
4. CH₃

Answer: 4. CH₃

Given the percentage of carbon – 78% and hence the percentage of hydrogen – 22%

∴ Empirical formula = CH₃

Question 30. An organic compound contains carbon, hydrogen, and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the compound would be

1. CHO
2. CH₄O
3. CH₃O
4. CH₂O

Answer: 3. CH₃O

Hence, the empirical formula of the compound would be CH₃O

Question 31. The percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (atomic weight = 78.4) then the minimum molecular weight of peroxidase anhydrous enzyme is

1. 1.568 x 10⁴
2. 1.568 x 10³
3. 15.68
4. 2.136 x 10⁴

Answer: 1. 1.568 x 10⁴

In peroxidase anhydrous enzyme, 0.5% Se is present means, 0.5 g Se is present in 100 g of the enzyme.

In a molecule of enzyme one Se atom must be present.

Hence, 78.4 g Se will be present in \(\frac{100}{0.5}\) x 78.4 =1.568 x 10⁴

∴ The minimum molecular weight of the enzyme is 1.568 x 10⁴

Question 32. Which of the following fertilizers has the highest nitrogen percentage?

1. Ammonium sulphate
2. Calcium cyanamide
3. Urea
4. Ammonium nitrate

Answer: 3. Urea

Urea(NH₂CONH₂), %of N = \(\frac{28}{60}\) x 100= 46.66%

Similarly, % of N in other compounds are:

(NH₄)₂SO₄ = 21.2%; CaCN₂ = 35.0% and NH₄NO₃ = 35.0%

Question 33. The right option for the mass of CO₂ produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40) \(\mathrm{CaCO}_3\) \(\underrightarrow{1200 \mathrm{~K}}\) CaO + CO₂

1. 2.64 g
2. 1.32 g
3. 1.12 g
4. 1.76 g

Answer: 4. 1.76 g

∴ \(\underset{40+12+16×3=100 \mathrm{~g}}{\mathrm{CaCO_3}}\) \(\underrightarrow{1200 K}\) \(\underset{40+16=56 \mathrm{~g}}{\mathrm{CaO}}+\underset{44 \mathrm{~g}}{\mathrm{CO}_2}\)

100 g CaCO₂ produced CO₂ gas = 44 g

20 gCaCO₃ will produce CO₂ gas = \(\frac{44}{100}\) = 8.8 g

If sample is 100% pure, CO₂ produced = 8.8 g

If sample is 20% pure, CO₂ produced = \(\frac{8.8}{100}\) x 20 = 17.6 g

Question 34. What mass of 95% pure CaCO₃ will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction \(\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_{(a q)} \rightarrow \mathrm{CaCl}_{2(a q)}+\mathrm{CO}_{2(g)}+2 \mathrm{H}_2 \mathrm{O}_{(l)}\)

1. 1.25 g
2. 1.32 g
3. 3.65 g
4. 9.50 g

Answer: 2. 1.32 g

Volume of HCl = \(50 \mathrm{~mL}=0.05 \mathrm{~L}\)

Molarity of HCl= 0.5M

∴ Moles of HCl = \(0.05 \times 0.5=0.025\) moles

∴ \(\mathrm{CaCO}_3+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

∴ For 2 moles of \(\mathrm{HCl}, \mathrm{CaCO}_3\) required =1 mole

For 0.025 moles of \(\mathrm{HCl}, \mathrm{CaCO}_3\) required = \(\frac{0.025}{2}\) moles

Mass of \(\mathrm{CaCO}_3\) required = \(100 \times \frac{0.025}{2}=1.25 \mathrm{~g}\)

For 95% pure \(\mathrm{CaCO}_3\), mass of \(\mathrm{CaCO}_3\) required \(=\frac{1.25}{95} \times 100 \mathrm{~g}\)

= \(1.315 \mathrm{~g} \approx 1.32 \mathrm{~g}\)

Question 35. The number of moles of hydrogen molecules required to produce 20 moles of ammonia through the Habers process is

1. 40
2. 10
3. 20
4. 30

Answer: 4. 30

Habert process, N₂ + 3H₂ → 2NH₃

2 moles of NH₃ are formed by 3 moles of H₂

∴ 20 moles of NH₃ will be formed by 30 moles of H₂

Question 36. The density of 2 M aqueous solution of NaOH is 1. 28 g/cm³. The molality of the solution is [Given that molecular mass of NaOH = 40 g mol^-1]

1. 1.20 m
2. 1.56 m
3. 1.67 m
4. 1.32 m

Answer: 3. 1.67 m

Density = 1.28 g/cc Concentration of solution = 2 M

Molar mass of NaOH = 40 g mol^-1

Volume of solution = 1 L = 1000 mL

Massof solution=d x V=1.28 x 1000= 1280g

Mass of solute = n x Molar mass = 2 x 40 = 80 g

Mass of solvenl = (1280 – 80) C = 1200 g

Number of moles of solute = \(\frac{80}{40}\) = 2

∴ Molarity = \(\frac{2 \times 1000}{1200}\) = 1.67 m

Question 37. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with a cone. H₂SO₄. The evolved gaseous mixture is passed through KOH pellets. The weight (in g) of the remaining product at STP will be

1. 1.4
2. 3.0
3. 2.8
4. 4.4

Answer: 3. 2.8

H₂O gets absorbed by conc. H₂SO₄ Gaseous mixture (containing CO and CO₂) when passed through KOH pellets, CO₂ gets absorbed.

Moles of CO left (unabsorbed)= \(\frac{1}{20}\) + \(\frac{1}{20}\) = \(\frac{1}{10}\)

Mass of CO = moles x molar mass = \(\frac{1}{10}\) x 28 = 2.8 g

Question 38. What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO₃ is mixed with 50 mL of 5.8% NaCl solution? (Ag = 107.8, N = 14,0 = 16, Na = 23, Cl = 35.5)

1. 3.5 g
2. 7 g
3. 14 g
4. 28 g

Answer: 2. 7 g

16.9% solution of AgNO₃ means 16.9 g of AgNO₃ in 100 mL of solution.

= 8.45 g of AgNO₃ in 50 mL solution.

Similarly, 5.8 g of NaCI in 100 mL solution

= 2.9 g of NaCl in 50 mL solution.

The reaction can be represented as

∴ Mass of AgCl precipitated = 0.049 x 143.3 = 7.02 ≈7 g

Question 39. 20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g of magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (Atomic weight of Mg = 24)

1. 96
2. 60
3. 84
4. 75

Answer: 3. 84

∴ \(\underset{84 \mathrm{~g}}{\mathrm{MgCO}_{3(s)}}\) \(\underrightarrow{\Delta}\) \(\underset{\mathrm{40 g}}{\mathrm{MgO}}{(s)}+\mathrm{CO}_{2(g)}\)

84 g of \(\mathrm{MgCO}_3 \equiv 40 \mathrm{~g}\) of \(\mathrm{MgO}\)

∴ 20 g of \(\mathrm{MgCO}_3 \equiv \frac{40}{84} \times 20=9.52 \mathrm{~g}\) of MgO

Actual yield =8 g of MgO

∴ % purity = \(\frac{8}{9.52} \times 100=84 \%\)

Question 40. When 22.4 litres of H_2(g) is mixed with 11.2 litres of Cl_2(g), each at STP, the moles of HCl_(g) formed is equal to

1. l mol of HCl_(g)
2. 2 mol of HCl_(g)
3. 0.5 mol of HCl_(g)
4. 1.5 mol of HCl_(g)

Answer: 1. l mol of HCl_(g)

∴ \(n_{\mathrm{H}_2}=\frac{22.4}{22.4}=1 \mathrm{~mol} ; n_{\mathrm{Cl}_2}=\frac{11.2}{22.4}=0.5 \mathrm{~mol}\)

The reaction is as,

Here, Cl₂ is the limiting reagent. So, 1 mole of HCl_(g) is formed

Question 41. 1.0 g of magnesium is burnt with 0.56 g O₂ in a closed vessel. Which reactant is left in excess and how much? (Atomic weight Mg = 24, O = 16)

1. Mg, 0.16 g
2. 02, 0.16 g
3. Mg, 0.44 g
4. 02, 0.28 g

Answer: 1. Mg, 0.16 g

∴ \(n_{\mathrm{Mg}}=\frac{1}{24}=0.0416\) moles

∴ \(n_{\mathrm{O}_2}=\frac{0.56}{32}=0.0175\) mole

The balanced equation is

⇔ \(\begin{array}{lccc}
& 2 \mathrm{Mg}+ & \mathrm{O}_2 \longrightarrow & 2 \mathrm{MgO} \\
\text { Initial } & 0.0416 \text { mole } & 0.0175 \text { mole } & 0 \\
\text { Final } & (0.0416-2 \times 0.0175) & 0 & 2 \times 0.0175 \\
& =0.0066 \text { mole } & &
\end{array}\)

Here, O₂ is limiting the reagent

∴ Mass of Mg left in excess = 0.0066 x 24 = 0.16 g

Question 42. 6.02 x 10²⁰ molecules of urea are present in 100 mL of its solution. The concentration of the solution is

1. 0.001 M
2. 0.1 M
3. 0.02 M
4. 0.01 M

Answer: 4. 0.01 M

Moles of urea = \(\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}}=0.001\)

Concentration of solution = \(\frac{0.001}{100} \times 1000=0.01 \mathrm{M}\)

Question 43. An experiment, it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO₃, which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine)?

1. X₂Cl₂
2. XCl₂
3. XCl₄
4. X₂Cl

Answer: 2. XCl₂

Miltimoles of solution of chloride = 0’05 x 10 = 0.5

Millimoles of AgNO₃ solution = 10 x 0.1 = 1

So, the millimoles of AgNO₃ are double that of the chloride solution

∴ XCl₂ + 2AgNO₃ →  2AgCl + X(NO₃)₂

Question 44. 25.3 g of sodium carbonate, Na₂CO₃ is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, the molar concentration of sodium ion, Na⁺ and carbonate ions, CO^–₂ are respectively (Molar mass of Na₂CO₃ = 106 g mo^-1)

1. 0.955 M and 1.910 M
2. 1.910 M and 0.955 M
3. 1.90 M and 1.910 M
4. 0.477 M and 0.477 M

Answer: 2. 1.910 M and 0.955 M

Given that the molar mass of Na₂CO₃= 106 g

Molarity of solution = \(\frac{25.3 \times 1000}{106 \times 250}=0.955 \mathrm{M}\)

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{Na}^{+}+\mathrm{CO}_3^{2-}\)

∴ \({\left[\mathrm{Na}^{+}\right]=2\left[\mathrm{Na}_2 \mathrm{CO}_3\right]=2 \times 0.955=1.910 \mathrm{M}}\)

∴ \({\left[\mathrm{CO}_3^{2-}\right]=\left[\mathrm{Na}_2 \mathrm{CO}_3\right]=0.955 \mathrm{M}}\)

Question 45. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be

1. 3 mol
2. 4 mol
3. 1 mol
4. 2 mol

Answer: 2. 4 mol

10 g of H₂ = 5 mol and 64 g of O₂= 2 mol

∴ In this reaction, oxygen is the limiting reagent so the amount of H₂O produced depends on the amount of O₂.

Since 0.5 mol of O₂ gives I mol of H₂O

∴ 2 mol of O₂ will give 4 mol of H₂O

Question 46. How many moles of lead(2) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g HCl?

1. 0.011
2. 0.029
3. 0.044
4. 0.333

Answer: 2. 0.029

The formation of moles of lead(2) chloride depends upon the no. of moles of PbO which acts as a limiting reagent here. So, a number of moles of PbCl₂ formed will be equal to the no. of moles of PbO i.e.,0.029.

Question 47. The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminum metal from bauxite by the Hall process is

1. 270 kg
2. 540 kg
3. 90 kg
4. 180 kg

(Atomic mass: AI = 27)

Answer: 3. 90 kg

∴ \(\underset{\text { (From bauxite) }}{3 \mathrm{C}+2 \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow 4 \mathrm{Al}+3 \mathrm{CO}_2}\)

4 moles of AI is produced by 3 moles of C.

1 mole of Al is produced by \(\frac{3}{4}\) mole of C

∴ \(\frac{270 \times 1000}{27}=10^4\) moles of Al is produced by \(\frac{3}{4} \times 10^4\) moles of C

Amount of carbon used = \(\frac{3}{4} \times 10^4 \times 12 \mathrm{~g}\)

= \(\frac{3}{4} \times 10 \times 12 \mathrm{~kg}=90 \mathrm{~kg}\)

Question 48. The molarity of liquid HCl, if the density of the solution is 1.17 g/cc is

1. 36.5
2. 18.25
3. 32.05
4. 42.10

Answer: 3. 32.05

Density= 1.17 g/cc.

⇒ 1 cc. the solution contains 1.17 g of HCl

∴ Molarity = \(\frac{1.17 \times 1000}{36.5 \times 1}=32.05\)

Question 49. The volume of CO₂ obtained by the complete decomposition of 9.85 g of BaCO₃ is

1. 2.24 L
2. 1.12 L
3. 0.84 L
4. 0.56 L

Answer: 2. 1.12 L

9.85 g of BaCO₃ will produce 1.118 L of CO₂ at N.T.p, on the complete decomposition.

Question 50. In the reaction  \(4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_2 \mathrm{O}_{(l)}\), when 1 mole of ammonia and 1 mole of O₂ are made to react to completion

1. All the oxygen will be consumed
2. 1.0 mole of NO will be produced
3. 1.0 mole of H₂O is produced
4. All the ammonia will be consumed.

Answer: 1. All the oxygen will be consumed

∴ \(\underset{4 moles}4 \mathrm{NH}_{3(g)}+\underset{5 moles}5 \mathrm{O}_{2(g)} \rightarrow \underset{4 moles}4 \mathrm{NO}_{(g)} + \underset{6 moles} 6 \mathrm{H}_2 \mathrm{O}_{(l)}\)

⇒ 1 mole of NH₃ requires = 514 = 1.25 mole of oxygen while

I mole of O₂ requires =4/5 = 0.8 mole of NH₃

Therefore, all oxygen will be consumed.

Question 51. The amount of zinc required to produce 224 mL of H₂ at STP on treatment with dilute H₂SO₄ will be

1. 65 g
2. 0.065 g
3. 0.65 g
4. 6.5 g

Answer: 3. 0.65 g

Since 65 g of zinc reacts to liberate 22400 mL of H₂ at STP, therefore the amount of zinc needed to produce 224 mL of H, at STP = \(\frac{65}{224000}\) x 224 = 0.65 g

Question 52. At STP the density of CCl₄ vapor in g/L will be nearest to

1. 6.87
2. 3.42
3. 10.26
4. 4.57

Answer: 1. 6.87

Weight of 1 mol of CCl₄ vapour = 12 + 4 x 35.5 = 154 g

∴ Density of CCl₄ vapour = \(\frac{154}{22.4}\) gL^-1 = 6.875 g L^-1