NEET Chemistry

Amines

Question 1. Which of the following reactions will not give primary amine as the product?

Answer: 1

Question 2. Which of the following reactions is appropriate for converting acetamide to methanamine?

1. Hoffmann hypobromamide reaction
2. Stephens reaction
3. Gabriel phthalimide synthesis
4. Carbylamine reaction

Answer: 1. Hoffmann hypobromite reaction

Question 3. The method by which aniline cannot be prepared is

1. Degradation of benzamide with bromine in alkaline solution
2. Reduction of nitrobenzene with H₂/Pd in ethanol
3. Potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution
4. Hydrolysis of phenyhsocyamde with acidic solution

Answer: 3. Potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution

Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution reaction with potassium phthalimide under mild conditions.

Question 4. The electrolytic reduction of nitrobenzene in a strongly acidic medium produces

1. Azobenzene
2. Aniline
3. P-aminophenol
4. Azoxybenzene.

Answer: 3. P-aminophenol

Question 5. In a set of reactions m-bromobenzoic acid gave a product D. Identify the product D.

Answer: 3

Question 6. Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?

1. NaOH-Br₂
2. Sodalime
3. Hot conc.H₂SO₄
4. PCl₅

Answer: 1. NaOH-Br₂

⇒ \(\mathrm{CH}_3 \mathrm{CONH}_2+4 \mathrm{NaOH}+\mathrm{Br}_2 \rightarrow \mathrm{CH}_3 \mathrm{NH}_2+2 \mathrm{NaBr}+\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O}\)

This reaction is called the Hoffmann Bromamide degradation reaction.

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Question 7. Which one of the following on reduction with lithium aluminium hydride yields a secondary amine?

1. Methyl isocyanide
2. Acetamide
3. Methyl cyanide
4. Nitroethane

Answer: 1. Methyl isocyanide

Alkyl isocyanide on reduction with lithium aluminium hydride forms a secondary amine-containing methyl as one of the acyl groups.

Question 8. In a set of reactions propionic acid yielded a compound D. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\) \(\underrightarrow{\mathrm{SOCl}_2}\) B \(\underrightarrow{\mathrm{NH}_3}\) C  KOH/Br₂ D. The structure of D would be

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CONH}_2\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCH}_3\)

Answer: 1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2\)

Question 9. Electrolytic reduction of nitrobenzene in a weakly acidic medium gives

1. N-phenylhydroxylamine
2. Nitrosobenzene
3. Aniline
4. p-hydroxyproline

Answer: 3. Aniline

Electrolytic reduction of nitrobenzene in a weakly acidic medium gives aniline but in the strongly acidic medium, it gives aminophenol through the acid-catalysed rearrangement of the initially formed phenylhydroxylamine.

Question 10. Intermediates formed during the reaction of RCONH with Br₂ and KOH are

1. RCONHBr and RNCO
2. RNHCOBr and RNCO
3. RNH – Br and RCONHBr
4. RCONBr₂

Answer: 1. RCONHBr and RNCO

The reaction, RCONH₂ + Br₂ + KOH→ RNH₂ is known as Hoffmann bromamide degradation reaction. The mechanism of the reaction is

This reaction is used to descend the series, i.e., for preparing a lower homologue from a higher one.

Question 11. Amides may be converted into amines by a reaction named after

1. Hoffmann
2. Claisen
3. Perkin
4. Kekule.

Answer: 1. Hoffmann

This reaction is called the Hoffmannbromamide degradation reaction.

Question 12. Indicate which nitrogen compound amongst the following would undergo Hoffmann reaction (i.e. reaction with Br₂ and strong KOH) to furnish the primary amine (R-NH₂).

1. \(\mathrm{RCONHCH}_3\)
2. \(\mathrm{RCOONH}_4\)
3. \(\mathrm{RCONH}_2\)
4. \(R-\mathrm{CO}-\mathrm{NHOH}\)

Answer: 3. \(\mathrm{RCONH}_2\)

The amide (-CONH₂) group is converted into a primary amino group (-NH₂) by the Hoffman bromamide degradation reaction.

⇒ \(R \mathrm{CONH}_2+\mathrm{Br}_2+4 \mathrm{KOH}\) \(\underrightarrow{^{\Delta}}\) \(\underset{1^{\circ} \text { amine }}{R-\mathrm{NH}_2} +2 \mathrm{KBr}\) + \(\mathrm{K}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O}\)

Question 13. Given below are two statements:

• Statement-1: Primary aliphatic amines react with HNO₂ to give unstable diazonium salts.
• Statement-2: Primary aromatic amines react with HNO₂ to form diazonium salts which are stable even above 300 K.

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.

Primary aliphatic amines react with nitrous to form aliphatic diazonium salts which are unstable while aromatic amines react with nitrous acid at low temperatures (273-275 K) to form diazonium salts, a very important class of compounds used for the synthesis of a variety of aromatic compounds.

Question 14. Identify the compound that will react with Hinsberg’s reagent to give a solid which dissolves in alkali.

Answer: 4

Benzene sulphonyl chloride (C₆H₅SO₂Cl)₂ which is also known as Hinsberg’s reagent, reacts with primary amines to form sulphonamides. The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of a strong electron-withdrawing sulphonyl group. Hence, it is soluble in alkali.

Question 15. Which of the following amines will give the carbylamine test?

Answer: 1

Aliphatic and aromatic primary amines give an arylamine test. Secondary and tertiary amines do not show this reaction.

Question 16. The correct order of the basic strength of methyl-substituted amines in an aqueous solution is

1. \(\mathrm{CH}_3 \mathrm{NH}_2>\left(\mathrm{CH}_3\right)_2 \mathrm{NH}>\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\)
2. \(\left(\mathrm{CH}_3\right)_2 \mathrm{NH}>\mathrm{CH}_3 \mathrm{NH}_2>\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\)
3. \(\left(\mathrm{CH}_3\right)_3 \mathrm{~N}>\mathrm{CH}_3 \mathrm{NH}_2>\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\)
4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{~N}>\left(\mathrm{CH}_3\right)_2 \mathrm{NH}>\mathrm{CH}_3 \mathrm{NH}_2\)

Answer: 2. \(\left(\mathrm{CH}_3\right)_2 \mathrm{NH}>\mathrm{CH}_3 \mathrm{NH}_2>\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\)

The basicity of amines in an aqueous solution depends on the stability of the ammonium cation or conjugate acid formed by accepting a proton from water which in turn depends on the + I-effect of alkyl group, the extent of hydrogen bonding and the steric factor. AII these factors are favourable for 2° amines. Therefore, 2° amines are the strongest bases.

If the alkyl group is small i.e., CH₃ then there is no steric hindrance to H-bonding. Thus, the stability due to hydrogen bonding predominates over the stability due to the +I-effect of the -CH₃, group and hence primary amine is a stronger base than 3°amine. Hence, the overall decreasing basic strength for methylamines in an aqueous solution is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N

Quetsion 17. The amine that reacts with Hinsberg’s reagent to give an alkali-insoluble product is

Answer: 1

Secondary amines on reaction with Hinsberg’s reagent give N, N-divinylbenzene sulphonamide which does not contain any hydrogen atom attached to the N atom, it is not acidic and hence, insoluble in alkali. Tertiar,v amines do not react with Hinsberg’s reagent. Primary antine gives products which are soluble in alkali.

Question 18. Nitration of aniline in a strongly acidic medium also gives m-nitroaniline because

1. In spite of substituents nitro group always goes to only the m-position
2. In electrophilic substitution reactions amino group is meta-directive
3. In the absence of substituents nitro group always goes to the m-position
4. In an acidic (strong) medium aniline is present as an anilinium ion.

Answer: 4. In an acidic (strong) medium aniline is present as an anilinium ion.

The reason for the formation of an unexpected amount of m-nitroaniline is that under strongly acidic conditions of nitration, most of the aniline is converted into anilinium ion and since, -NH₃ is a m-directing group, therefore, a large amount of m-nitroaniline is also obtained.

Question 19. The correct increasing order of basic strength for the following compounds is

1. 3<1<2
2. 3<2<1
3. 2<1<3
4. 2<3<1

Answer: 3. 2<1<3

+I effect of the substituted group increases the basic strength while -I effect of the substituent decreases the basic strength of aniline.

Question 20. The correct statement regarding the basicity of arylamines is

1. Arylamines are generally more basic than alkylamines because of the aryl group
2. Arylamines are generally more basic than alkylamines because the nitrogen atom in arylamines is sp-hybridised
3. Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalised by interaction with the aromatic ring π electron system
4. Arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalised by interaction with the aromatic ring π-electron system.

Answer: 3. Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalised by interaction with the aromatic ring π-electron system

In arytrarnines, one pair of electrons on a nitrogen atom is delocalised over the benzene ring and, thus, not available for donation. So, arylamines are less basic than alkylamines

Question 21. On hydrolysis of a “compound”, two compounds are obtained. One of which on treatment with sodium nitrite and hydrochloric acid gives a product which does not respond to the iodoform test. The second one reduces Tollens’ reagent and Fehling’s solution. The “compound” is

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NC}\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{ON}=\mathrm{O}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CON}\left(\mathrm{CH}_3\right)_2\)

Answer: 1

HCOOH reduces Tollens’reagent and Fehling’s solution.

Question 22. Some reactions of amines are given. Which one is not correct?

Answer: 1

Aromatic tertiary amines undergo electrophilic substitution with nitrosonium ion at the p-position of the phenyl ring to form green-coloured nitrosamines.

Quetsion 23. An organic compound (C₃H₉N) (A), when treated with nitrous acid, gave an alcohol and N₂ gas was evolved. (A) on warming with CHCl₃ and caustic potash gave (C) which on reduction gave isopropylmethylamine. Predict the structure of (A).

Answer: 1

As A gives alcohol on treatment with nitrous acid thus, it should be primary amine. C₃H₉N has two possible structures with -NH₂ group.

As it gives isopropylmethylamine thus, it should be isopropyl amine, not n-propyl amine.

Question 24. Which of the following compounds is most basic?

Answer: 2

In benzylamine, the electron pair present on the nitrogen is not delocalised with the benzene ring.

Question 25. Which of the following statements about primary amines is false?

1. Alkyl amines are stronger bases than aryl amines.
2. Alkyl amines react with nitrous acid to produce alcohols.
3. Aryl amines react with nitrous acid to produce phenols.
4. Alkyl amines are stronger bases than ammonia.

Answer: 3. Aryl amines react with nitrous acid to produce phenols.

Aryl amines react with nitrous acid to produce diazonium salts.

Question 26. Match the compounds given in List 1 with their characteristic reactions given in List 2. Select the correct option.

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

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

Question 27. Predict the product.

Answer: 4

2° aliphatic and aromatic amines react with nitrous acid to form N-nitrosamine.

Question 28. Which of the following is more basic than aniline?

1. Benzylamine
2. Diphenylamine
3. Triphenylamine
4. p-Nitroaniline

Answer: 1. Benzylamine

Any group which when present on a benzene ring has electron-withdrawing nature (-NO₂,-CN,-SO₃H, -COOH, -CI, -C₆H₅ etc) decreases the basicity of aniline example, aniline is more basic than nitroaniline, Lone pair of electrons are more delocalised in diphenylamine and triphenylamine, thus these are less basic than aniline.

In benzylamine the electron pair present on nitrogen is not delocalised with the benzene ring hence, it is more basic than aniline.

Question 29. The final product C, obtained in this reaction, would be

Answer: 3

Question 30.

Answer: 2

‘C’ must be an isocyanide and it is obtained from a 1° amine by arylamine reaction (CHCI₃ + KOH). Further 1o amine can be obtained by the reduction of nitro compound so ‘A’ is nitrobenzene.

Question 31. Phenyl isocyanides are prepared by which of the following reaction?

1. Reimer-Tiemann reaction
2. Carbylamine reaction
3. Rosenmunds reaction
4. Wurtz reaction

Answer: 2. Carbylamine reaction

⇒ \(\mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{C}_6 \mathrm{H}_5-\mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}\)

The above reaction is called the arylamine reaction, which is a specific reaction of 1-amine.

Question 32. The compound obtained by heating a mixture of ethylamine and chloroform with ethanolic potassium hydroxide (KOH) is

1. An amide
2. An amide and nitro compound
3. An ethyl isocyanide
4. An alkyl halide.

Answer: 3. An ethyl isocyanide

Question 33. An aniline on nitration gives

Answer: 4

As NO⁺₂ electrophile can attack both ortho and para positions, therefore both (1) and (3) products will be obtained

Quetsion 34. The action of nitrous acid on an aliphatic primary amine gives

1. Secondary amine
2. Nitro alkane
3. Alcohol
4. Alkyl nitrite.

Answer: 3. Alcohol

R-NH₂ + HNO₂ → ROH + N₂ + H₂O

Question 35. Which one of the following orders is wrong, with respect to the property indicated?

1. Benzoic acid > phenol > cyclohexanol (add strength)
2. Aniline > cyclohexylamine > benzamide (basic strength)
3. Formic acid > acetic acid > propanoic acid (acid strength)
4. Fluoroacetic acid > chloroacetic acid > bromoacetic acid (acid strength)

Answer: 2. Aniline > cyclohexylamine > benzamide (basic strength)

Basic strength decreases as, cyclohexylamine > aniline > benzamide.

Lesser basicity in aniline and benzamide is due to the participation of a lone pair of electrons of -NH₂ group in resonance.

Question 36. For the arylamine reaction, we need hot alcoholic KOH and

1. Any primary amine and chloroform
2. Chloroform and silver powder
3. A primary amine and an alkyl halide
4. A monoalkylamine and trichloromethane.

Answer: 1. Any primary amine and chloroform

In arylamine reaction, primary amines on heating with chloroform in the presence of alcoholic KOH form isocyanides (or carbylamines). It is used to distinguish 1° amines from 2° and 3° amines.

⇒ \(R-\mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow R \mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}\)

Question 37. Which of the following will be the most stable diazonium salt RN⁺₂+X^–?

1. \(\mathrm{CH}_3 \mathrm{~N}_2^{+} \mathrm{X}^{-}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} X^{-}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~N}_2^{+} \mathrm{X}^{-}\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{~N}_2^{+} \mathrm{X}\)

Answer: 2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} X^{-}\)

Aromatic diazonium salts are more stable due to the dispersal of the positive charge in the benzene ring

Question 38. Identify the product in the following reaction

Answer: 4

Question 39. The product formed from the following reaction sequence is

Answer: 4

Question 40. The reagent in ‘R’ in the given sequence of a chemical reaction is

1. CuCN/KCN
2. H₂O
3. CH₃CH₂OH
4. HI

Answer: 3. CH₃CH₂OH

Question 41. A given nitrogen-containing aromatic compound ‘A’ reacts with Sn/HCl, followed by HNO₂ to give an unstable compound ‘If. ‘If, on treatment with phenol, forms a beautiful coloured compound ‘C with the molecular formula C₁₂H₁₀N₂O. The structure of compound ‘A’ is

Answer: 2

Question 42. In the following reaction, the product (A) is

Answer: 4

Question 43. In the reaction A is

1. \(\mathrm{H}_3 \mathrm{PO}_2\) and \(\mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{H}^{+} \mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{HgSO}_4 \mathrm{H}_2 \mathrm{SO}_4\)
4. \(\mathrm{Cu}_2 \mathrm{Cl}_2\)

Answer: 1. \(\mathrm{H}_3 \mathrm{PO}_2\) and \(\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{H}_3 \mathrm{PO}_2 \text { and } \mathrm{H}_2 \mathrm{O} \text { reduces the }-\stackrel{+}{\mathrm{N}} 2 \mathrm{Cl}^{-} \text {to }-\mathrm{H} \text {. }\)

Question 44. Anline in a set of the following reactions yielded a coloured product Y.

Answer: 1

Quetsion 45. Aniline in a set of the following reactions yielded a coloured product D.

The structure of the product D would be

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHOH}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_2 \mathrm{CH}_3\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}\)

Answer: 4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}\)

Question 46. Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in the presence of dilute hydrochloric acid. The compound so formed is converted into a tetrafluoroborate which is subsequently heated to dry. The final product is

1. p-bromoaniline
2. p-bromofluorobenzene
3. 1, 3, 5-tribromobenzene
4. 2, 4, 6-tribromofluorobenzene

Answer: 4. 2, 4, 6-tribromofluorobenzene

Question 47. Which one of the following nitro compounds do not react with nitrous acid?

Answer: 3

Tertiary nitroalkanes do not react with nitrous acid as they do not contain cr-hydrogen atoms.

Question 48. Nitrobenzene on reaction with cone. HNO₃/H₂SO₄ at 80-100°C forms which one of the following products?

1. 1,4-Dinitrobenzene
2. 1,2,4-Trinitrobenzene
3. 1, 2-Dinitrobenzene
4. 1, 3-Dinitrobenzene

Answer: 4. 1, 3-Dinitrobenzene

Question 49. What is the product obtained in the following reaction?

Answer: 1

Question 50. Product ‘P in the above reaction is

Answer: 2

Question 51. Which product is formed, when acetonitrile is hydrolysed partially with cold concentrated HCl?

1. Methyl cyanide
2. Acetic anhydride
3. Acetic acid
4. Acetamide

Answer: 4. Acetamide

Biomolecules

Question 1. Sucrose on hydrolysis gives

1. β-D-glucose + α-D-fructose
2. α-D-glucose + β-D-glucose
3. α-D-glucose + β-D-fructose
4. α-D-fructose + β-D-fructose.

Answer: 3. α-D-glucose + β-D-fructose

In sucrose, two monosaccharides are held together by a glycosidic linkage between C-1 of α-D-glucose and C-2 of β-D-fructose.

Sucrose \(\underrightarrow{\text { Hydrolysis }}\) \(\alpha \text { – } D \text {-glucose }+\beta \text { – } D \text {-fructose }\)

Question 2. The difference between amylose and amylopectin is

1. Amylopectin have 1 → 4 α-linkage and 1 → 6 α-linkage
2. Amylose has 1 → 4 α-linkage and 1 → 6 β-linkage
3. Amylopectin have 1 →4 α-linkage and 1 → 6 β-linkage
4. Amylose is made up of glucose and galactose.

Answer: 1. Amylopectin have 1 → 4 α-linkage and 1 → 6 α-linkage

Amylose is a linear polymer of α-D-glucose held by C₁-C₄ glycosidic linkage whereas amylopectin is a branched chain polymer of α-D-glucose units in which the chain is held by C₁-C₄ glycosidic linkage while branching occurs by C₁-C₆ glycosidic linkage.

Question 3. The correct corresponding order of names of four aldoses with the configuration given below respectively, is

1. L-erythrose, L-threose, L-erythrose, D-threose
2. D-threose, D-erythrose, L-threose, L-erythrose
3. L-erythrose, L-threose, D-erythrose, D-threose
4. D-erythrose, D-threose, L-erythrose, L-threose.

Answer: 4. D-erythrose, D-threose, L-erythrose, L-threose.

Question 4. Which one given below is a non-reducing sugar?

1. Glucose
2. Sucrose
3. Maltose
4. Lactose

Answer: 2. Sucrose

All monosaccharides whether aldoses or ketoses are reducing sugars. Disaccharides such as sucrose in which the two monosaccharide units are linked through their reducing centres Le., aldehydic or ketonic groups are non-reducing.

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Quetsion 5. D(+)-glucose reacts with hydroxyl amine and yields an oxime. The structure of the oxime would be

Answer: 4

Question 6. Which one of the following sets of monosaccharides forms sucrose?

1. α-D-galactopyranose and α-D-glucopyranose
2. α-D-glucopyranose and β-D-fructofuranose
3. β-D-glucopyranose and α-D-fructofuranose
4. α-D-glucopyranose and β-D- fructopyranose

Answer: 2. α-D-glucopyranose and β-D fructofuranose

Sucrose is formed by the condensation of α-D- glucopyranose and β-D-fructofuranose.

Question 7. Which one of the following statements is not true regarding (+)-lactose?

1. On hydrolysis (+)-lactose gives equal amounts of D(+)-glucose and D(+)-galactose.
2. (+)-Lactose is a β-glucoside formed by the union of a molecule of D(+)-glucose and a molecule of D(+)-galactose.
3. (+)-Lactose is a reducing sugar and does not exhibit mutarotation.
4. (+)-Lactose, C₁₂H₂₂O₁₁ contains 8 -OH groups

Answer: 3. (+)-Lactose is a reducing sugar and does not exhibit mutarotation.

(+)-Lactose is a reducing sugar and all reducing sugars show mutarotation.

Question 8. Which one of the following does not exhibit the phenomenon of mutarotation?

1. (+)-Sucrose
2. (+)-Lactose
3. (+)-Maltose
4. (-)-Fructose

Answer: 1. (+)-Sucrose

Sucrose does not show mutarotation.

Mutarotation is the phenomenon of change in optical rotation shown by freshly prepared solutions of sugars. However, this property is not exhibited by all sugars.

Only those sugars which have a free aldehyde (-CHO) or ketone (C=O) group aldehyde are capable of ketone of the group showing and is therefore, incapable of showing mutarotation.

Question 9. Fructose reduces Tollens reagent due to

1. Asymmetric carbons
2. Primary alcoholic group
3. Secondary alcoholic group
4. Embolisation of fructose followed by conversion to aldehyde by base.

Answer: 4. Enolisation of fructose followed by conversion to aldehyde by base.

Under alkaline conditions of the reagent, fructose gets converted into a mixture of glucose and mannose (Lobry de Bruyn van Ekenstein rearrangement) both of which contain the -CHO group and hence, reduce Tollens’reagent to give silver mirror test

Question 10. The number of chiral carbons in β-D-(+) glucose is

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

Answer: 4. Four.

This structure of β-D-glucose has four asymmetric carbon atoms.

Question 11. Glycolysis is

1. Oxidation of glucose to glutamate
2. Conversion of pyruvate to citrate
3. Oxidation of glucose to pyruvate
4. Conversion of glucose to haem

Answer: 3. Oxidation of glucose to pyruvate

Glycolysis is the first stage in the oxidation of glucose. It is an anaerobic process and involves the degradation of glucose into two molecules of pyruvate with the generation of two molecules of ATP.

Question 12. Cellulose is a polymer of

1. Glucose
2. Fructose
3. Ribose
4. Sucrose.

Answer: 1. Glucose

Cellulose is a straight-chain polysaccharide composed of β-D-glucose units joined by β-glycosidic linkage between C₁ of one glucose unit and C₄ of the next glucose unit.

Question 13. Which of the following gives a positive Fehling solution test?

1. Sucrose
2. Glucose
3. Fats
4. Protein

Answer: 2. Glucose

Glucose reduces Fehling solution because glucose has a free -CHO group which is readily oxidised.

Question 14. α-D-glucose and β-D-glucose are

1. Epimers
2. Anomers
3. Enantiomers
4. Diastereomers.

Answer: 2. Anomers

Glucose forms a stable hemiacetal between the -CHO group and the -OH group on the 5th carbon. In this process, the 1st ‘C’ atom becomes asymmetric giving two isomers which differ in the configuration of the asymmetric carbon. These two isomers are called anomers.

Question 15. Which of the following is the sweetest sugar?

1. Fructose
2. Glucose
3. Sucrose
4. Maltose

Answer: 1. Fructose

Fructose is the sweetest among all the sugars and is highly soluble in water.

Question 16. Glucose molecule reacts with X number of molecules of phenylhydrazine to yield osazone. The value of X is (8)

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

Answer: 4. Five

Glucose first reacts with phenyl hydrazine giving phenylhydrazine. Then the adjacent -CHOH group is oxidized by a 2d phenyl hydrazine molecule and itself is reduced to aniline. The resulting carbonyl group reacts with 3’d phenyl hydrazine molecule giving osazone.

Question 17. The oxidation of glucose is one of the most important reactions in a living cell. What is the number of ATP molecules generated in cells from one molecule of glucose?

1. 28
2. 38
3. 12
4. 18

Answer: 2. 38

⇒ \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}+38 \mathrm{ATP}\)

Question 18. The α-D-glucose and β-D-glucose differ from each other due to differences in carbon atoms with respect to its

1. Number of OH groups
2. Size of hemiacetal ring
3. Conformation
4. Configuration.

Answer: 4. Configuration.

These isomers differ only in the orientation (or configuration) of the Cl atom.

Question 19. Chemically considering digestion is basically

1. Anabolism
2. Hydrogenation
3. Hydrolysis
4. Dehydrogenation.

Answer: 3. Hydrolysis

Question 20. On hydrolysis of starch, we finally get

1. Glucose
2. Fructose
3. Both (1) and (2)
4. Sucrose.

Answer: 1. Glucose

Glucose is produced commercially by the hydrolysis of starch by boiling it with dil. H₂SO₄ at 393 K under pressure of 2-3 bar.

Question 21. Which of the following is a basic amino acid?

1. Serine
2. Alanine
3. Tyrosine
4. Lysine

Answer: 4. Lysine

Lysine is a basic amino acid

Question 22. The non-essential amino acids among the following is

1. Lysine
2. Valine
3. Leucine
4. Alanine.

Answer: 4. Alanine

Question 23. Which structure(s) of proteins remains (s) intact during the denaturation process?

1. Both secondary and tertiary structures
2. Primary structure only
3. Secondary structure only
4. Tertiary structure only

Answer: 2. Primary structure only

During the denaturation of Proteins, 2° and 3° structures are destroyed but 1o structure remains intact.

Question 24. Which of the following compounds can form a zwitter ion?

1. Aniline
2. Acetanilide
3. Benzoic acid
4. Glycine

Answer: 4. Glycine

⇒ \(\mathrm{HOOC}-\underset{\text { Glycine }}{\mathrm{CH}_2-} \mathrm{NH}_2 \rightleftharpoons-\mathrm{OOC} \underset{\text { Zwitter ion }}{\mathrm{CH}_2}-\stackrel{+}{\mathrm{N}} \mathrm{H}_3\)

Question 25. In a protein molecule, various amino acids are linked together by

1. Peptide bond
2. Dative bond
3. α-glycosidic bond
4. β-Glycosidic bond.

Answer: 1. Peptide bond

Question 26. Which of the statements about “Denaturation” given below are correct?

1. Denaturation of proteins causes loss of secondary and tertiary structures of the protein.
2. Denaturation leads to the conversion of a double strand of DNA into a single strand.
3. Denaturation affects the primary structure which gets distorted.

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

Anbswer: 3. (1) and (2)

Denaturation does not change the primary structure of the protein.

Question 27. Which functional group participates in disulphide bond formation in proteins?

1. Thioester
2. Thioether
3. Thiol
4. Thiolactone

Answer: 3. Thiol

Disulphide bond may be reduced to thiol by means of reagents i.e., NaBH₄, which shows the presence of thiol group in disulphide bond formation.

Question 28. Which of the following structures represents the peptide chain?

Answer: 3

In peptide linkage i.e., – CONH – group, the carboxyl group of one amino acid molecule forms an amide by combination with the amino group of the next amino acid molecule with the liberation of a water molecule.

Question 29. The correct statement in respect to protein haemoglobin is that it

1. Functions as a catalyst for biological reactions
2. Maintains blood sugar level
3. Acts as an oxygen carrier in the blood
4. Forms antibodies and offers resistance to diseases.

Answer: 3. Acts as an oxygen carrier in the blood

Four Fe²⁺ ions of each haemoglobin can bind with four molecules of O₂ and it is carried as oxyhaemoglobin.

Question 30. The helical structure of protein is stabilised by

1. Dipeptide Bonds
2. Hydrogen Bonds
3. Ether Bonds
4. Peptide Bonds.

Answer: 2. Hydrogen Bonds

α-Hellt structure is formed when the chain of α-amino acids coil as a right-handed screw because of the formation of hydrogen bonds between amide groups of the same peptide chain, i.e., NH group in one unit is linked to carbonyl oxygen of the fourth unit by hydrogen bonding. This H-trending is responsible for holding the helix in a stable position.

Question 31. Which is not a true statement?

1. α-Carbon of α-amino acid is asymmetric.
2. All proteins are found in the L-form.
3. The human body can synthesise all the proteins it needs.
4. At pH = 7, both ammo and carboxylic small groups exist in ionised form.

Answer: 2. All proteins are found in L-form

Some proteins are also found in the D-form.

Quetsion 32. (peptide bond)

Which statement is incorrect about peptide bonds?

1. The C – N bond length in proteins is longer than the usual bond length of the N – C bond.
2. Spectroscopic analysis shows the planar structure of the   group.
3. The C – N bond length in proteins is smaller than the usual bond length of the C – N bond.
4. None of the above.

Answer: 1. The C – N bond length in proteins is longer than the usual bond length of the N – C bond.

A Peptide bond is formed by the reaction of the -COOH group of one amino acid with the -NH₃ group of another amino acid and is represented as

As some double bond character is found between C-N bonds, the bond length of C-N in protein should be smaller than the usual C-N bond.

Question 33. Which is the correct statement?

1. Starch is a polymer of α-glucose.
2. Amylose is a component of cellulose.
3. Proteins are composed of only one type of amino acid.
4. In the cyclic structure of fructose, there are four carbons and one oxygen atom.

Answer: 1. Starch is a polymer of α-glucose.

Starch is also known as amylum which occurs in all green plants. A molecule of starch (C₆H₁₀O₅)_n is built of a large number of α-glucose rings joined through oxygen atoms.

Question 34. Haemoglobin is

1. A Vitamin
2. A Carbohydrate
3. An Enzyme
4. A Globular Protein.

Answer: 4. A Globular Protein.

Haemoglobin is a globular protein of four subunits, each subunit having a heme moiety and a polypeptide chain (Two α and two β chains).

Question 35. The secondary structure of a protein refers to

1. Regular folding patterns of continuous portions of the polypeptide chain
2. Three-dimensional structure, especially the bond between amino acid residues that are distant from each other in the polypeptide chain
3. Mainly denatured proteins and structures of prosthetic groups
4. Linear sequence of amino acid residues in the polypeptide chain.

Answer: 1. Regular folding patterns of continuous portions of the polypeptide chain

Question 36. During the process of digestion, the proteins present in food materials are hydrolysed to amino acids. The two enzymes involved in the process proteins \(\underrightarrow{\text { enzyme }(A)}\) polypeptides \(underrightarrow{\text { enzyme }(B)}\) amino acids, are respectively.

1. Invertase and zymase
2. Amylase and maltase
3. Diastase and lipase
4. Pepsin and trypsin

Answer: 4. Pepsin and trypsin

Question 37. Enzymes are made up of

1. Edible proteins
2. Proteins with specific structure
3. Nitrogen-containing carbohydrates
4. Carbohydrates.

Answer: 2. Proteins with specific structure

Question 38. Which of the following is correct?

1. Cycloheptane is an aromatic compound.
2. Diastase is an enzyme.
3. Acetophenone is an ether.
4. All of these.

Answer: 2. Diastase is an enzyme.

Diastase is an enzyme that hydrolyses starch into maltose

Question 39. The function of enzymes in the living system is to

1. Catalyse biochemical reactions
2. Provide energy
3. Transport oxygen
4. Provide immunity.

Answer: 1. Catalyse biochemical reactions

Question 40. Which of the following statements about enzymes is true?

1. Enzymes catalyse chemical reactions by increasing the activation energy.
2. Enzymes are highly specific both in binding chiral substrates and in catalysing their reactions.
3. Enzymes lack in nucleophilic groups.
4. Pepsin is a proteolytic enzyme.

Answer: 4. Pepsin is a proteolytic enzyme.

Question 41. Enzymes take part in a reaction and

1. Decrease the rate of a chemical reaction
2. Increase the rate of a chemical reaction
3. Both (1) and (2)
4. None of these.

Answer; 2. Increase the rate of a chemical reaction

Enzyrnes being a biocatalyst increases the rate of a chemical reaction by providing alternative lower activation energy pathways.

Question 42. RBC deficiency is a deficiency disease of

1. Vitamin B₂
2. Vitamin B₁₂
3. Vitamin B₆
4. Vitamin B₁

Answer: 2. Vitamin B₁₂

Anaemia is a disease in which the body does not have enough red blood cells (RBC) due to a lack of vitamin B-12. This vitamin is needed to make red blood cells.

Question 43. A deficiency of vitamin B₁ causes the disease

1. Convulsions
2. Beriberi
3. Cheilosis
4. Sterility.

Answer: 2. Beri-beri

Question 44. Which of the following is not a fat-soluble vitamin?

1. Vitamin B complex
2. Vitamin D
3. Vitamin E
4. Vitamin A

Answer: 1. Vitamin B complex

Vitamin B complex is not a fat-soluble vitamin. It is a water-soluble vitamin.

Question 45. Which of the following vitamins is water soluble?

1. Vitamin E
2. Vitamin K
3. Vitamin A
4. Vitamin B

Answer: 4. Vitamin B

Vitamins B and C are water soluble whereas vitamins A, D, E and K are fat soluble

Question 46. The human body does not produce

1. Enzymes
2. DNA
3. Vitamins
4. Hormones.

Answer: 3. Vitamins

Certain organic substances required for regulating some of the body processes and preventing certain diseases are called vitamins, which cannot be synthesised by the human body.

Question 47. Vitamin B₁₂ contains

1. Fe (2)
2. Co (3)
3. Zn (2)
4. Ca (2)

Answer: 2. Co (3)

Vitamin B₁₂ is chemically named a cyanocobalamin having molecular formula C₆₃H₈₈O₁₄N₁₄PCo.

Question 48. Given below are two statements.

• Statement 1: A unit formed by the attachment of a base to 1′-position of sugar is known as a nucleoside.
• Statement 2: When nucleoside is linked to phosphorous acid at the 5′-position of the sugar moiety, we get nucleotide.

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

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

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

When nucleoside is linked to phosphoric acid (not phosphorous acid) at the 5′-position of the sugar moiety, we get nucleotide.

Question 49. The central dogma of molecular genetics states that genetic information flows from

1. Amino acids → Proteins → DNA
2. DNA → Carbohydrates → Proteins
3. DNA → RNA → Proteins
4. DNA → RNA → Carbohydrates

Answer: 3. DNA → RNA → Proteins

Genetic information flows from

DNA \(\underrightarrow{\text { Transcription }}\) RNA \(\underrightarrow{\text { Translation }}\) Proteins

Quetsion 50. The correct statement regarding RNA and DNA, respectively is

1. The sugar component in RNA is arabinose and the sugar component in DNA is ribose
2. The sugar component in RNA is 2′-deoxyribose and the sugar component in DNA is arabinose
3. The sugar component in RNA is arabinose and the sugar component in DNA is 2′-deoxyribose
4. The sugar component in RNA is ribose and the sugar component in DNA is 2′-deoxyribose.

Answer: 4. The sugar component in RNA is ribose and the sugar component in DNA is 2′-deoxyribose.

Question 51. In DNA, the linkages between different nitrogenous bases are

1. Phosphate linkage
2. H-bonding
3. Glycosidic linkage
4. Peptide linkage.

Answer: 2. H-bonding

Nitrogenous bases are linked together by hydrogen bonds.

Question 52. The segment of DNA which acts as the instrumental manual for the synthesis of the protein is

1. Ribose
2. Gene
3. Nucleoside
4. Nucleotide.

Answer: 2. Gene

Genes are responsible for protein synthesis.

Question 53. In DNA, the complementary bases are

1. Adenine and guanine; thymine and cytosine
2. Uracil and adenine; cytosine and guanine
3. Adenine and thymine; guanine and cytosine
4. Adenine and thymine; guanine and uracil

Answer: 3. Adenine and thymine; guanine and cytosine

DNA contains two types of nitrogenous bases

Purine → Adenine (A) and guanine (G)

Pyrimidine → Cytosine (C) and thymine (T)

The purine and pyrimidine bases pair only in certain combinations. Adenine pairs with thymine (A:T) by two hydrogen bonds and guanine with cytosine (G: C) by three hydrogen bonds.

Question 54. RNA and DNA are chiral molecules, their chirality is due to

1. Chiral bases
2. Chiral phosphate ester units
3. D-sugar component
4. L-sugar component.

Answer: 3. D-sugar component

The constituents of nucleic acids are nitrogenous bases, sugar and phosphoric acid. The sugar present in DNA is D(-)-2-deoxyribose and the sugar present in RNA is D(-)- ribose. Due to these D(-)-sugar components, DNA and RNA molecules are chiral molecules.

Question 55. A sequence of how many nucleotides in messenger RNA makes a codon for an amino acid?

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

Answer: 1. Three

The four bases in mRNA: adenine, cytosine, guanine and uracil have been shown to act in the form of triplets; each triplet behaving as a code for the synthesis of a particular amino acid

Question 56. Chargaff’s rule states that in an organism

1. The amount of adenine (A) is equal to that of thymine (T) and the amount of guanine (G) is equal to that of cytosine (C)
2. The amount of adenine (A) is equal to that of guanine (G) and the amount of thymine (T) is equal to that of cytosine (C)
3. The amount of adenine (A) is equal to that of cytosine (C) and the amount of thymine(T) is equal to that of guanine (G)
4. Amounts of all bases are equal.

Answer: 1. Amount of adenine (A) is equal to that of thymine (T) and the amount of guanine (G) is equal to that of cytosine (C)

Amount of A = T and that of G = C

Question 57. Which of the following is correct about H-bonding in nucleotide?

1. A – T, G – C
2. A – G, T – C
3. G – T, A – C
4. A – A, T – T

Answer: 1. A – T, G – C

Question 58. An example of a biopolymer is

1. Teflon
2. Neoprene
3. Nylon-6,6
4. DNA.

Answer: 4. DNA.

DNA is an example of a biopolymer

Question 59. The couplings between base units of DNA is through 

1. Hydrogen bonding
2. Electrostatic bonding
3. Covalent bonding
4. Van der Waal’s forces

Answer: 1. Hydrogen bonding

Question 60. Which of the following statements is not correct?

1. Ovalbumin is a simple food reserve in egg whites.
2. Blood proteins thrombin and fibrinogen are involved in blood clotting.
3. Denaturation makes the proteins more active.
4. Insulin maintains sugar levels in the blood of the human body.

Answer: 3. Denaturation makes the proteins more active.

Question 61. Which of the following hormones is produced under the conditions of stress which stimulate glycogenolysis in the liver of human beings?

1. Thyroxin
2. Insulin
3. Adrenaline
4. Estradiol

Answer: 3. Adrenaline

Adrenaline hormone helps to release fatty acids from fat and glucose from liver glycogen under the condition of stress. Hence, it is also called flight or fight hormone.

Question 62. Which of the following hormones contains iodine?

1. Testosterone
2. Adrenaline
3. Thyroxine
4. Insulin

Answer: 3. Thyroxine

Question 63. Which of the following is an amine hormone?

1. Insulin
2. Progesterone
3. Thyroxine
4. Oxypurin

Answer: 3. Thyroxine

Thyroxine is an amine hormone and water-soluble hormone containing an amino group.

Question 64. Which one of the following is a peptide hormone?

1. Adrenaline
2. Glucagon
3. Testosterone
4. Thyroxine

Answer: 2. Glucagon

Glucagon is a peptide hormone, synthesised by the a-cells of the pancreas.

Question 65. The hormone that helps in the conversion of glucose to glycogen is

1. Cortisone
2. Bile acids
3. Adrenaline
4. Insulin.

Answer: 4. Insulin

Insulin is a hormone secreted by the pancreas that lowers blood glucose levels by promoting the uptake of glucose by cells and the conversion of glucose to glycogen by the liver and skeletal muscle

Question 66. Which one is responsible for the production of energy in biochemical reactions?

1. Thyroxine
2. Adrenaline
3. Oestrogen
4. Progesterone

Answer: 1. Thyroxine

It is a hormone secreted from the thyroid gland. It controls various biochemical reactions involving the burning of proteins, carbohydrates, and fats to release energy.

Question 67. The cell membranes are mainly composed of

1. Fats
2. Proteins
3. Phospholipids
4. Carbohydrates.

Answer: 3. Phospholipids

Cell membranes are mainly composed of phospholipids.

Question 68. Phospholipids are esters of glycerol with

1. Three carboxylic acid residues
2. Two carboxylic acid residues and one phosphate group
3. One carboxylic acid residue and two phosphate groups
4. Three phosphate groups

Answer: 2. Two carboxylic acid residues and one phosphate group

Phospholipids may be regarded as derivatives of glycerol in which two of the hydroxy groups are esterified with fatty acids while the third is esterified with some derivatives of phosphoric acid.

Question 69. The number of molecules of ATP produced in the lipid metabolism of a molecule of palmitic acid is

1. 56
2. 36
3. 130
4. 86

Answer: 3. 130

In lipid metabolism, a molecule of palmitic acid (C₁₅H₃₁ – COOH) produces 130 adenosine triphosphate molecules (ATP).

Polymers

Question 1. Which statement regarding polymers is not correct?

1. Elastomers have polymer chains held together by weak intermolecular forces.
2. Fibres possess high tensile strength.
3. Thermoplastic polymers are capable of repeatedly softening and hardening m on heating and cooling respectively.
4. Thermosetting polymers are reusable.

Answer: 4. Thermosetting polymers are reusable.

Thermosetting polymers on heating undergo extensive cross-linking and become infusible. Hence, these cannot be reused.

Question 2. Which of the following molecules on polymerization produces neoprene?

Answer: 4

Neoprene or polychloroprene is formed by the free radical polymerisation of chloroprene.

Question 3. Which of the following polymers is prepared by addition polymerisation?

1. Dacron
2. Teflon
3. Nylon-6,6
4. Novolac

Answer: 2. Nylon-6,6

Among the given polymers tellon polymer is prepared by addition polymerisation.

Question 4. Which of the following is a natural polymer?

1. cis-1, 4-polyisoprene
2. poly (Butadiene-styrene)
3. polybutadiene
4. poly (Butadiene-acrylonitrile)

Answer: 1. cis-1, 4-polyisoprene

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Question 5. The polymer that is used as a substitute for wool in making commercial fibres is

1. Melamine
2. Nylon-6, 6
3. Polyacrylonitrile
4. Buna-N.

Answer: 3. Polyacrylonitrile

Question 6. Regarding cross-linked or network polymers, which of the following statements is incorrect?

1. They contain covalent bonds between various linear polymer chains.
2. They are formed from bi- and tri-functional monomers.
3. Examples are bakelite and melamine.
4. They contain strong covalent bonds in their polymer chains.

Answer: 4. They contain strong covalent bonds in their polymer chains.

Question 7. Which one of the following structures represents nylon 6, 6 polymers?

Answer: 4

Nylon-6,6 is obtained by condensing adipic acid (HOOC(CH₂)₄COOH) with hexamethylenediamine (H₂N(CH₂)₆NH₃).

Quetsion 8. Natural rubber has

1. Alternate cis- and trans-configuration
2. Random cis- and fraus-configuration
3. All cis-configuration
4. All fraus-configuration.

Answer: 3. All cis-configuration

Natural rubber is cls-polylsoprene.

Question 9. Caprolactam is used for the manufacture of

1. Teflon
2. Terylene
3. Nylon 6, 6
4. Nylon 6.

Answer: 4. Nylon 6.

Question 10. Which one of the following is an example of a thermosetting polymer?

Answer: 4

1. Neoprene rubber (elastomer)
2. PVC (thermoplastic polymer)
3. Nylon-6,6 (fibre)
4. Novolac which further undergoes cross-linking to produce bakelite (thermosetting polymer)

Question 11. Which of the following organic compounds polymerizes to form the polyester dacron?

1. Propylene and para HO—(C₆H₄)—OH
2. Benzoic acid and ethanol
3. Terephthalic acid and ethylene glycol
4. Benzoic acid and para HO—(C₆H₄)—OH

Answer: 3. Terephthalic acid and ethylene glycol

Question 12. Nylon is an example of

1. Polyamide
2. Polythene
3. Polyester
4. Polysaccharide.

Answer: 1. Polyamide

Question 13. Which is the monomer of neoprene in the following?

Answer: 1

Question 14. Which one of the following is not a condensation polymer?

1. Melamine
2. Glyptal
3. Dacron
4. Neoprene

Answer: 4. Neoprene

Neoprene is an additional polymer.

Question 15. Which of the following statements is false?

1. Artificial silk is derived from cellulose.
2. Nylon-6,6 is an example of an elastomer.
3. The repeat unit in natural rubber is isoprene.
4. Both starch and cellulose are polymers of glucose.

Answer: 2. Nylon-6,6 is an example of an elastomer.

Nylon-6,6 is an example of fibre.

Question 16. Of the following which one is classified as polyester polymer?

1. Terylene
2. Bakelite
3. Melamine
4. Nylon-6,6

Answer: 1. Terylene

Terylene (Dacron) is a polyester polymer because it is made by monomer units of ethylene glycol and terephthalic acid.

Question 17. Which of the following structures represents neoprene polymer?

Answer: 1

It is a polymer of chloroprene.

Question 18. Structures of some common polymers are given. Which one is not correctly presented

Answer: 1

Neoprene is a polymer of chloroprene.

The rest of the polymers are correctly represented.

Question 19. Which one of the following statements is not true?

1. Buna-S is a copolymer of butadiene and styrene.
2. Natural rubber is a 1,4-polymer of isoprene.
3. In vulcanization, the formation of sulphur bridges between different chains makes rubber harder and stronger.
4. Natural rubber has the trans-configuration at every double bond.

Answer: 4. Natural rubber has the trans-configuration at every double bond.

Natural rubber is cis-1,4-polyisoprene and has only cis-configuration about the double bond as shown below

whereas in Gutta-percha, only trans configuration exists about the double bond.

Question 20. Which one of the following polymers is prepared by condensation polymerisation?

1. Teflon
2. Natural rubber
3. Styrene
4. Nylon-6,6

Answer: 4. Nylon-6,6

Nylon-6,6 is a condensation polymer of adipic acid and hexamethylenediamine.

n HOOC-(CH₂)₄-COOH + n H₂N – (CN₂)₆ – NH₂

Question 21.

1. Homopolymer
2. Copolymer
3. Addition polymer
4. Thermosetting polymer.

Answer: 2. Copolymer

Formed by the condensation of adipic acid and hexamethylenediamine. It is a copolymer (a polymer made from more than one type of monomer molecule is referred to as a copolymer).

Question 22. The monomer of the polymer

Answer: 1

Question 23. Which one of the following is a chain-growth polymer?

1. Starch
2. Nucleic acid
3. Polystyrene
4. Protein

Answer: 3. Polystyrene

Chain-growth polymers involve a series of reactions each of which consumes a reactive particle and produces another similar one. The reactive particles may be free radicals or ions (cation or anion) to which monomers get added by a chain reaction. It is an important reaction of alkenes and conjugated dienes or indeed of all kinds of compounds that contain C – C double bonds.

Question 24. Acrilan is a hard, horny and a high melting material. Which one of the following represents its structure?

Answer: 1

Acrilan is an additional polymer of acrylonitrile.

Question 25. Monomer of

1. 2-Methylpropene
2. Styrene
3. Propylene
4. Ethene.

Answer: 1. 2-Methylpropene

Quetsion 26. Which of the following is not correctly matched?

Answer: 3

Terylene is an example of condensation po1ymer and the condensation of terephthalic and ethylene glycol.

Quetsion 27. CF₂ = CF₂ is the monomer of

1. Teflon
2. Orlon
3. Polythene
4. Nylon-6.

Answer: 1. Teflon

Question 28. Which compound forms linear polymer due to H-bond?

1. H₂O
2. NH₃
3. HF
4. HCl

Answer: 3. HF

H-F—H-F—H-F—H-F

Dotted lines represent hydrogen bonds between HF molecules and hence, it is a linear polymer. Due to the high electronegativity value of ‘F’ atom, it forms effective hydrogen bonding.

Question 29. Natural rubber is a polymer of

1. Styrene
2. Ethyne
3. Butadiene
4. Isoprene.

Answer: 4. Isoprene.

Polyisoprene is the natural rubber, which is the polymer of isoprene.

Question 30. Terylene is a condensation polymer of ethylene glycol and

1. Salicylic acid
2. Phthalic acid
3. Benzoic acid
4. Terephthalic acid

Answer: 4. Terephthalic acid

Terylene is the condensation polymer of ethylene glycol and terephthalic acid

Question 31. Which one of the following is used to make ‘non¬stick’ cookware?

1. Polyethylene terephthalate
2. Polytetrafluoroethylene
3. PVC
4. Polystyrene

Answer: 2. Polytetrafluoroethylene

Polytetrafluoroethylene or Teflon is a tough material, resistant to heat and a bad conductor of electricity. It is used for coating the cookware to make them non-sticky

Question 32. The bakelite is prepared by the reaction between

1. Phenol and formaldehyde
2. Tetramethylene glycol
3. Urea and formaldehyde
4. Ethylene glycol.

Answer: 1. Phenol and formaldehyde

Phenol and formaldehyde undergo condensation polymerisation under two different conditions to give a cross-linked polymer called bakelite

Question 33. The biodegradable polymer is

1. Buna-5
2. Nylon-6,6
3. Nylon-2-nylon 6
4. Nylon-6.

Answer: 3. Nylon-2-nylon 6

Nylon-2-nylon-6 is a biodegradable polymer.

Question 34. Which one of the following sets forms the biodegradable polymer?

Answer: 2

Nylon-2-nylon-6 is an alternating polyamide copolymer of glycine (H₂NCH₂COOH) and aminocaproic acid (H₂N(CH₂)₂COOH) and is biodegradable

 

Chemistry In Everyday Life

Question 1. Some tranquilizers are listed below. Which one of the following belongs to barbiturates?

1. Valium
2. Veronal
3. Chlordiazepoxide
4. Meprobamate

Answer: 2. Veronal

Veronal tranquilizer belongs to barbiturates

Quetsion 2. Match List-1 with List-2

correct answer from the options given below:

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

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

Question 3. Given below are two statements:

Statement 1: Aspirin and Paracetamol belong to the class of narcotic analgesics.

Statement 2: Morphine and Heroin are non-narcotic analgesics.

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: 3. Both statement 1 and statement 2 are false.

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Aspirin and paracetamol belong to the class of non-narcotic analgesics, while morphine and many of its homologs like heroin belong to the class of narcotic analgesics.

Question 4. Among the following, the narrow-spectrum antibiotic is

1. Chloramphenicol
2. Penicillin G
3. Ampicillin
4. Amoxycillin.

Answer: 2. Penicillin G

Penicillin G has a narrow spectrum. Chloramphenicol is a broad-spectrum antibiotic. Ampicillin and amoxicillin are synthetic modifications of penicillins. These have a broad spectrum.

Question 5. A mixture of chloroxylenol and terpineol acts as

1. Antiseptic
2. Antipyretic
3. Antibiotic
4. Analgesic.

Answer: 1. Antiseptic

Dettol which is a well-known antiseptic is a mixture of chloroxylenol and α-terpineol in a suitable solvent

Question 6. Which of the following is an analgesic?

1. Streptomycin
2. Chloromycetin
3. Novalgin
4. Penicillin

Answer: 3. Novalgin

Streptomycin, chloromycetin, and penicillin are antibiotics while novalgin is an analgesic.

Question 7. Bithional is generally added to the soaps as an additive to function as an

1. Buffering agent
2. Antiseptic
3. Softener
4. Dryer.

Answer: 2. Antiseptic

Question 8. Antiseptics and disinfectants either kill or prevent the growth of microorganisms. Identify which of the following statements is not true.

1. Dilute solutions of boric acid and hydrogen peroxide are strong antiseptics.
2. Disinfectants harm the living tissues.
3. A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant.
4. Chlorine and iodine are used as strong disinfectants.

Answer: 1. Dilute solutions of boric acid and hydrogen peroxide are strong antiseptics.

Dilute solutions of boric acid and hydrogen peroxide are weak antiseptics.

Question 9. Dettol is a mixture of

1. Chloroxylenol and bithionol
2. Chloroxylenol and terpineol
3. Phenol and iodine
4. Terpineol and bithionol.

Answer: 2. Chloroxylenol and terpineol

Dettol is a mixture of chloroxylenol and α-terpineol.

Question 10. Chloramphenicol is an

1. Antifertility drug
2. Antihistamine
3. Antiseptic and disinfectant
4. Antibiotic-broad spectrum.

Answer: 4. Antibiotic-broad spectrum.

Question 11. Which one of the following is employed as Antihistamine?

1. Chloramphenicol
2. Diphenhydramine
3. Norethindrone
4. Omeprazole

Answer: 2. Diphenhydramine

Diphenylhydramine is employed as an antihistamine drug.

Question 12. Which one of the following is employed as a tranquilizer drug?

1. Promethazine
2. Valium
3. Naproxen
4. Mifepristone

Answer: 2. Valium

Vaiium is a tranquilizer.

Question 13. Which one of the following is employed as a tranquilizer?

1. Naproxen
2. Tetracycline
3. Chlorpheniramine
4. Equanil

Answer: 4. Equanil

Equanil is used for the treatment of stress, and mild and severe mental diseases i.e., as a tranquilizer.

Question 14. Chloropicrin is obtained by the reaction of

1. Steam on carbon tetrachloride
2. Nitric acid on chlorobenzene
3. Chlorine on picric acid
4. Nitric acid on chloroform.

Asnwer: 4. Nitric acid on chloroform.

When chloroform is treated with concentrated nitric acid, its hydrogen is replaced by nitro group.

⇒ \(\mathrm{CHCl}_3+\mathrm{HONO}_2 \rightarrow \underset{\text { Chloropicrin }}{\mathrm{CNO}_2 \mathrm{Cl}_3}+\mathrm{H}_2 \mathrm{O}\)

Question 15. Aspirin is an acetylation product of

1. m-hyd hydroxybenzoic acid
2. o-dihydroxybenzene
3. o-hydroxybenzoic acid
4. p-dihydroxybenzene.

Answer: 3. o-hydroxybenzoic acid

Aspirin is acetylsalicylic acid, which is formed by the acetylation of o-hydroxybenzoic acid.

Question 16. Which of the following can possibly be used as an analgesic without causing addiction and mood modification?

1. Diazepam
2. Tetrahydrocatinol
3. Morphine
4. N-Acetyl-para-aminophenol.

Answer: 4. N-Acetyl-para-aminophenol.

N-acetyl-para-aminophenol (or paracetamol) is an antipyretic that can also be used as an analgesic to relieve pain without addiction and mood modification.

Question 17. Which one of the following statements is not true?

1. Ampicillin is a natural antibiotic.
2. Aspirin is both analgesic and antipyretic.
3. Sulphadiazine is a synthetic antibacterial drug.
4. Some disinfectants can be used as antiseptics.

Answer: 1. Ampicillin is a natural antibiotic.

Ampicillin is a modification of penicillin and thus is a synthetic antibiotic.

Question 18. Diazo coupling is useful for preparing some

1. Pesticides
2. Dyes
3. Proteins
4. Vitamins.

Answer: 2. Dyes

Azo dyes are derived by coupling of a phenol adsorbed on the surface of a fabric with a diazonium salt. Dyes can be prepared by diazo coupling. For example,

Question 19. The artificial sweetener is stable at cooking temperature and does not provide calories is

1. Saccharin
2. Aspartame
3. Sucralose
4. Alitame.

Answer: 3. Sucralose

Sucralose is a trichloro derivative of sucrose. Its appearance and taste is like sugar. It is stable at cooking temperature and it does not provide calories

Question 20. Artificial sweeteners which are stable under cold conditions only is

1. Saccharine
2. Sucralose
3. Aspartame
4. Alitame.

Answer: 3. Aspartame

Aspartame is stable under cold conditions and unstable at cooking temperatures.

Question 21. Which of the following is a cationic detergent?

1. Sodium lauryl sulphate
2. Sodium stearate
3. Cetyl trimethyl ammonium bromide
4. Sodium dodecylbenzene sulphonate

Asnwer: 3. Cetyltrimethyl ammonium bromide

Cetyltrimethyl ammonium bromide is a. cationic detergent.

Question 22. Which of the following forms cationic micelles above a certain concentration?

1. Sodium dodecyl sulphate
2. Sodium acetate
3. Urea
4. Cetyltrimethylammonium bromide

Answer: 4. Cetyltrimethylammonium bromide

Cefyltrimethylammonium bromide is a popular cationic detergent.

Aldehydes Ketones And Carboxylic Acids

Question 1. The intermediate compound X in the following chemical reaction is

Answer: 2

Question 2. Identify compound X in the following sequence of reactions:

Answer: 3

Question 3. Reaction by which benzaldehyde cannot be prepared

Answer: 2

Clemmensen reduction in the presence of Zn-Hg and con. HCl reduces aldehydes and ketones to -CH₂ group but the carboxylic acid group remains unaffected.

Question 4. Consider the following reaction.

The product A is

Answer: 1

It is Rosenmund’s reduction.

Question 5. Which one of the following can be oxidised to the corresponding carbonyl compound?

1. 2-Hydroxypropane
2. Ortho-Nitrophenol
3. Phenol
4. 2-Methyl-2-hydroxy-propane

Answer: 1. 2-Hydroxypropane

Read and Learn More NEET MCQs with Answers

Question 6. In the following reaction, product P is

1. RCH₂OH
2. RCOOH
3. RCHO
4. RCH₃

Answer: 3. RCHO

This is Rosenmund reduction

Question 7. Which alkene on ozonolysis gives CH₃CH₂CHO and CH₃COCH₃?

Answer: 1

Question 8.

1. Acetaldehyde
2. Ethanolamine
3. Acetone
4. Dimethylamine.

Answer: 1. Acetaldehyde

Question 9. Ketones [RCOR₁] where R = R₁ = alkyl group. It can be obtained in one step by

1. Oxidation of tertiary alcohol
2. Reaction of acid halide with alcohols
3. Hydrolysis of esters
4. Oxidation of primary alcohol.

Answer: 1. Oxidation of tertiary alcohol

A tertiary alcohol is difficult to oxidise. But when it is treated with an acidic oxidising agent under some conditions, it is oxidised to ketone and then to acids. Both the ketone and acid contain a lesser number of carbon atoms than the starting alcohol.

Question 10. The oxidation of toluene to benzaldehyde by chromyl chloride is called

1. Etard reaction
2. Riemer-Tiemann reaction
3. Wurtz reaction
4. Cannizzaro’s reaction.

Answer: 1. Etard reaction

The oxidation of toluene (C₆H₅CH₃) with chromyl chloride (CrO₂Cl₂) in CCl₄ or CS₂ to give benzaldehyde is called the Etard reaction. In this reaction, the chromyl chortle first forms a brown complex, which is separated and then decomposed with H₂O to give benzaldehyde (C₆H₅CHO).

Question 11. Given below are two statements:

• Statement-1: The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole-dipole interactions.
• Statement 2: The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. 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: 1. Both statement 1 and statement 2 are correct.

Both the given statements are correct.

Question 12. Identify the product (A) in the following reaction

Answer: 3

This is the Ciemmensen reduction reaction.

Question 13. Identify the product [D] obtained in the following sequence of reactions.

Answer: 3

Question 14. Identify the major product obtained in the following reaction.

Answer: 1

Question 15. Match the List 1 with List 2.

Choose the correct answer from the options given below:

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

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

• Aldehydes react with HCN to give cyanohydrin.
• Aldehydes react with alcohol to form acetal.
• Aldehydes react with amine to give Schifft base.
• Aldehydes react with NH₂OH to give oxime.

Question 16. Which one of the following is not formed when acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating?

Answer: 2

When acetone reacts with 2-pentanone in the presence of dii. NaOH, the following products are formed

Question 17. What is the IUPAC name of the organic compound formed in the following chemical reaction?

1. 2-Methylbutan-2-ol
2. 2-Methylpropan-2-ol
3. Pentan-2-ol
4. Pentan-3-ol

Answer: 1. 2-Methylbutan-2-ol

Question 18. The reaction between benzaldehyde and acetophenone in the presence of dilute NaOH is known as

1. Aldol condensation
2. Cannizzaros reaction
3. Cross Cannizzaros reaction
4. Cross Aldol condensation.

Answer: 4. Cross Aldol condensation.

Cross aldol condensation

Question 19. Consider the reactions,

Identify A, X, Y and Z.

1. A-Methoxymethane, X-Ethanol, Y-Ethanoic acid, Z-Semicarbazide.
2. A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone.
3. A-Ethanol, X-Acetaldehyde, Y-Butanone, Z- Hydrazone.
4. A-Methoxymethane, X-Ethanoic acid, Y-Acetate ion, Z-Hydrazine.

Answer: 2. A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone.

Since A gives a silver mirror test, it must be an aldehyde and aldehydes are formed by oxidation of 1° alcohols’ Thus, ‘X’ is a 1° alcohol, i.e., CH₃CH₂OH.

Question 20. Of the following, which is the product formed when cyclohexanone undergoes aldol condensation followed by heating?

Answer: 1

Question 21. The correct structure of the product ‘A’ formed in the reaction

Answer: 2

C = C bond is reduced faster than C = O bond with H₂(Pd-C).

Question 22. Which of the following reagents would distinguish ds-cyclopenta-1,2-diol from the trans-isomer?

1. MnO₂
2. Aluminium isopropoxide
3. Acetone
4. Ozone

Answer: 3. Acetone

Trans-isomer does not react with acetone

Question 23. The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon is

1. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation
2. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism
3. A carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol
4. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration.

Answer: 2. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism

Keto-enol tautomerism

Question 24. The product formed by the reaction of an aldehyde with a primary amine is

1. Carboxylic acid
2. Aromatic acid
3. Schiff’sbase
4. Ketone.

Answer: 3. Schiff’sbase

Question 25. Reaction of a carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is

1. Hydrazine in the presence of a feebly acidic solution
2. Hydrocyanic acid
3. Sodium hydrogen sulphite
4. A Grignard reagent.

Answer: 1. Hydrazine in the presence of a feebly acidic solution

Carbonyl compounds react with ammonia derivatives in the weakly acidic medium as follows

Question 26. Which one is most reactive towards nucleophilic addition reaction?

Answer: 4

Aromatic aldehydes are more reactive than alkyl aryl ketones. The electron withdrawing group (-NO₂) increases the reactivity towards nucleophilic addition reactions whereas the electron donating group (-CH₃) decreases the reactivity towards nucleophilic addition reactions.

Therefore, the order is

Question 27. The order of stability of the following tautomeric compounds is

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

Answer: 4. 3>2>1

Question 28. Predict the products in the given reaction.

Answer: 3

Aldehyde having no α,-hydrogen atoms on heating with concentrated alkali solution (50%) undergoes Cannizzaro’s reaction.

Question 29. Acetone is treated with excess ethanol in the presence of hydrochloric acid. Hie product obtained is

Answer: 4

Question 30. CH₃CHO and C₆H₅CH₂CHO can be distinguished chemically by

1. Benedict’s test
2. Iodoform test
3. Tollens’ reagent test
4. Fehling’s solution test.

Answer: 2. Iodoform test

Acetaldehyde acetone and methyl ketones having CH₃CO- group undergo a haloform reaction. Thus CH₃CHO will give a yellow precipitate with I₂ and NaOH but C₆H₅CH₂CHO will not.

Question 31. Consider the reaction  RCHO + NH₂NH₂ → RCH=N — NH₂ What sort of reaction is it?

1. Electrophilic addition-elimination reaction
2. Free radical addition-elimination reaction
3. Electrophilic substitution-elimination reaction
4. Nucleophilic addition-elimination reaction

Answer: 4. Nucleophilic addition-elimination reaction

Question 32. Which of the following compounds will give a yellow precipitate with iodine and alkali?

1. Acetophenone
2. Methyl acetate
3. Acetamide
4. 2-Hydroxypropane

Answer: 1. Acetophenone and 4. 2-Hydroxypropane

This example shows the iodoform reaction.

The compound with the group will give a yellow precipitate of iodoform (CH₃) when reacting with iodine and alkali.

Question 33. Clemmensen reduction of a ketone is carried out in the presence of which of the following?

1. Glycol with KOH
2. Zn-Hg with HCl
3. LiAIH₄
4. H₂ and Pt as catalyst

Answer: 2. Zn-Hg with HCl

The Carbonyl group is reduced to -CH₂ group when treated with amalgamated zinc and conc. HCI. This process is called Clemmensent reduction.

Question 34. The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds :

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

Answer: 4. 1>2>3

The greater the number of alkyl/phenyl groups attached to the carbonyl groups lower its reactivity 1 > 2 > 3. +R-effect is stronger than +I-effect

Question 35. Which of the following reactions will not result in the formation of carbon-carbon bonds?

1. Reimer-Tiemann reaction
2. Cannizzaro reaction
3. Wurtz reaction
4. Friedel-Crafts acylation

Answer: 2. Cannizzaro reaction

From the above examples, it is evident that C-C bond formation does not take place in the Cannizzaro reaction.

Question 36. Which one of the following compounds will be most readily dehydrated?

Answer: 3

The ease of dehydration of the given compounds can be explained on the basis of the stability of the carbocation formed. In the case of options (1), (2) and (4), a secondary carbocation is formed but the presence of an electron-withdrawing group adjacent to the positively charged carbon, intensifies the charge and hence, destabilises the species.

However, in the case of option (3), a secondary carbocation is formed, but the electron-withdrawing group is present farther away, as a result, the effect of this group is diminished and hence, the carbocation is relatively more stable

Question 37. The following compounds are given,

Which of the above compound(s), on being warmed with iodine solution and NaOH, will give iodoform?

1. 1, 3 and 4
2. Only 2
3. 1, 2 and 3
4. 1 and 2

Answer: 3. 1, 2 and 3

Compounds with  give positive iodoform hence, (1), (2) and (3) will give positive iodoform not (4).

Question 38. Acetophenone when reacted with a base, C₂H₅ONa, yields a stable compound which has the structure

Answer: 3

The first step is a simple condensation reaction. The last step is an example of the E1cB mechanism and the leaving group is hydroxide, which is unusual. Still, this step manages to take place owing to the stability incorporated therein the product, which is a conjugated carbonyl compound.

Question 39. A strong base can abstract an α-hydrogen from

1. Ketone
2. Alkane
3. Alkene
4. Amine.

Answer: 1. Ketone

The base (OH^–) ion removes one of the α-hydrogen atoms (which is somewhat acidic) from aldehydes and ketones to form a carbanion or the enolate ion. The acidity of a-hydrogen is due to resonance stabilization of enolate anion.

Question 40. Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and cone. HCl is called

1. Cope reduction
2. Dow reduction
3. Wolff-Kishner reduction
4. Clemmensen reduction.

Answer: 4. Clemmensen reduction.

Aldehydes and ketones are converted to alkanes when treated with zinc amalgam and conc. HCl. This is known as Clemmensen reduction. Here group is reduced to the group.

Question 41. Which one of the following on treatment with 50% aqueous sodium hydroxide yields the corresponding alcohol and acid?

Answer: 1

Aldehydes which do not have α-H atoms, in the presence of 50% NaOH or 50% KOH undergo a disproportionation reaction to produce alcohol and sodium salt of the acid. This reaction is known as Cantizzaro reaction. C₆H₅CHO containing no α-H atom undergoes the Canrtizzaro reaction to produce benzyl alcohol and sodium benzoate.

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\) \(\underrightarrow{50 \% \mathrm{NaOH}}\) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}\)

Question 42. The product formed in aldol condensation is

1. A beta-hydroxy aldehyde or a beta-hydroxy ketone
2. An alpha-hydroxy aldehyde or ketone
3. An alpha, beta-unsaturated ester
4. A beta-hydroxy acid.

Answer: 1. A beta-hydroxy aldehyde or a beta-hydroxy ketone

The aldehydes or ketones containing α-H atom in the presence of dilute alkali undergo a self-condensation reaction to form β-hydroxyaldehyde or β-hydroxyketone. This reaction is known as aldol condensation.

Question 43. Nucleophilic addition reaction will be most favoured in

Answer: 1

Question 44. A carbonyl compound reacts with hydrogen cyanide to form cyanohydrin which on hydrolysis forms a racemic mixture of a-hydroxy acid. The carbonyl compound is

1. Formaldehyde
2. Acetaldehyde
3. Acetone
4. Diethyl ketone.

Answer: 2

Question 45. The major organic product formed from the following reaction:

Answer: 2

Question 46. In this reaction, an asymmetric centre is generated. The acid obtained would be

⇒ \(\mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCN}\rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CN}\) \(\underrightarrow{\mathrm{HOH}}\) \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\)

An asymmetric centre is generated. The acid obtained would be

1. D-isomer
2. L-isomer
3. 50% D + 50% L-isomer
4. 20% D + 80% L-isomer.

Answer: 3. 50% D + 50% L-isomer

Lactic acid (CH₃CH(OH)COOH) is an optically active compound due to the presence of an asymmetric carbon atom. It exists, in D- and L-form, the ratio of which is found to be (1: 1), i.e., a racemic mixture is obtained.

Question 47. When m-chlorobenzaldehyde is treated with 50% KOH solution, the product(s) obtained is (are)

Answer: 2

The above reaction is known as Cannizzaro’s reaction.

Question 48. A and B in the following reactions are 

Answer: 4

Question 49. are

1. Resonating structures
2. Tautomers
3. Geometrical isomers
4. Optical isomers.

Answer: 1. Resonating structures

They are resonating forms because the position of the atomic nuclei remains the same and only electron redistribution has occurred.

Question 50. Which of the following is incorrect?

1. FeCl₃ is used in the detection of phenol.
2. Fehling solution is used in the detection of glucose.
3. Tollens reagent is used in the detection of unsaturation.
4. NaHSO₃ is used in the detection of carbonyl compounds.

Answer: 3. Tollens reagent is used in the detection of unsaturation.

Tollens reagent is a solution of ammoniacal silver nitrate and is used for the detection of -CHO group. Aldehydes reduce Tollens reagent and get oxidised to convert Ag⁺ ions to Ag powder which forms the silver-coloured mirror in the test tube. So, this test is also known as the silver mirror test.

⇒ \(R-\mathrm{CHO}+\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} \rightarrow R-\mathrm{COO}^{-}+\mathrm{Ag}\)

Question 51. Polarisation in acrolein can be described as

Answer: 4

O-atom is more electronegative than C-atom, therefore O-atom bears a partial -ve charge and C-atom to which it is attached bears a partial +ve charge.

Question 52. The first product of the reaction between PCHO and NH₂NH₂ is

1. \(\mathrm{RCH}=\mathrm{NNH}_2\)
2. \(\mathrm{RCH}=\mathrm{NH}\)
3. \(\mathrm{RCH}_2 \mathrm{NH}_2\)
4. \(\mathrm{RCON}_3\)

Answer: 1. \(\mathrm{RCH}=\mathrm{NNH}_2\)

It is a simple condensation reaction which proceeds with the elimination of water.

Quetsion 53. An ester (A) with the molecular formula, C₉H₁₀O₂ was treated with an excess of CH₃MgBr and the complex so formed, was treated with H₂SO₄ to give an olefin(B). Ozonolysis of (B) gave a ketone with molecular formula C₈H₈O which shows +ve iodoform test. The structure of (A) is

1. \(\mathrm{H}_3 \mathrm{CCH}_2 \mathrm{COC}_6 \mathrm{H}_5\)
2. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOC}_6 \mathrm{H}_5\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)
4. \(p-\mathrm{H}_3 \mathrm{CO}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{COCH}_3\)

Answer: 3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)

Question 54. The iodoform test is not given by

1. Ethanal
2. Ethanol
3. 2-Pentanone
4. 3-Pentanone.

Answer: 4. 3-Pentanone.

Compounds containing  group show iodoform test.

So iodoform test is not given by 3-pentanone.

Question 55. Phenylmethanol can be prepared by reducing the benzaldehyde with

1. CH₃Br and Na
2. CH₃I and Mg
3. CH₃Br
4. Zn and HCl

Answer: 4. Zn and HCl

Question 56. The oxidation of toluene with CrO₃ in the presence of (CH₃CO)₂O gives a product A, which on treatment with aqueous NaOH produces

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}\)
2. 2,4-diacetyl toluene
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\)
4. \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}\)

Question 57. When aniline reacts with the oil of bitter almonds (C₆H₅CHO) condensation takes place and a benzal derivative is formed. This is known as

1. Schiffs base
2. Benedicts reagent
3. Millon’sbase
4. Schiff’s reagent

Answer: 1. Schiff base

Benzaldehyde reacts with primary aromatic amines to form Schiff’s base (Benzylidene aniline)

Question 58. Compound A has a molecular formula C₂Cl₃OH. It reduces Fehling’s solution and on oxidation, it gives a monocarboxylic acid B. If A is obtained by the action of chlorine on ethyl alcohol, then compound A is

1. Methyl chloride
2. Monochloroacetic acid
3. Chloral
4. Chloroform

Answer: 3. Chloral

The compound ‘A’ reduces Fehling’s solution thus, it must have a free -CHO group.

Thus, the compound A is chloral.

Question 59. Which of the following compounds will undergo self-aldol condensation in the presence of cold dilute alkali?

1. \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{CHO}\)
2. \(\mathrm{CH}_2=\mathrm{CHCHO}\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

Answer: 4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

Since CH₃CH₂CHO has an α-hydrogen atom, therefore it will undergo aldol condensation in the presence of cold dilute alkali.

Question 60. Which of the following compounds will give a positive test with Tollens reagent?

1. Acetic acid
2. Acetone
3. Acetamide
4. Acetaldehyde

Answer: 4. Acetaldehyde

Acetaldehyde reduces Toiiens’ reagent to the silver mirror.

Question 61. (CH₃)₂C = CHCOCH₃ can be oxidised to (CH₃)₂C = CHCOOH by

1. Chromic acid
2. NaOI
3. Cu at 300°C
4. KMnO₄

Answer: 2. NaOI

⇒ \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CHCOCH}_3\) \(\underrightarrow{\mathrm{NaOI}}\) \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CHCOOH}+\mathrm{CHI}_3\)

(NaOH + I₂)/NaOI is the most suitable reagent for the given reaction.

Question 62. In which of the following, the number of carbon atoms does not remain the same when carboxylic acid is obtained by oxidation?

1. \(\mathrm{CH}_3 \mathrm{COCH}_3\)
2. \(\mathrm{CCl}_3 \mathrm{CH}_2 \mathrm{CHO}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

Answer: 1. \(\mathrm{CH}_3 \mathrm{COCH}_3\)

Ketones on oxidation give carboxylic acids with lesser number of carbon atoms i.e \(\mathrm{CH}_3 \mathrm{COCH}_3\) \(\underrightarrow{[\mathrm{O}]}\) \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 63. Acetaldehyde reacts with

1. Electrophiles only
2. Nucleophiles only
3. Free radicals only
4. Both electrophiles and nucleophiles.

Answer: 2. Nucleophiles only

Question 64. The reagent which can be used to distinguish acetophenone from benzophenone is

1. 2,4-dinitrophenylhydrazine
2. Aqueous solution of NaHSO₃
3. Benedict reagent
4. I₂ and NaoH

Answer: 4. I₂ and NaoH

Acetophenone reacts with NaOH and I₂ to give yellow ppt. of CHI₃ but benzophenone (C₆H₅COC₆H₅) does not. Hence, it can be used to distinguish between them.

Question 65. The above-shown polymer is obtained when a carbonyl compound is allowed to stand. It is a white solid. The polymer is

1. Trioxane
2. Formose
3. Paraformaldehyde
4. Metaldehyde.

Answer: 1. Trioxane

Question 66. The given compound describes a condensation polymer which can be obtained in two ways: either by treating 3 molecules of acetone (CH₃COCH₃) with cone. H₂SO₄ or passing propyne (CH₃C ≡ CH) through a red hot tube. The polymer is

1. Phorone
2. Mesityl oxide
3. Deacetonyl alcohol
4. Mesitylene.

Answer: 4. Mesitylene.

Acetone forms mesitylene ( 1, 3, 5- trimethylbenzene) on distillation with conc. H₂SO₄.

Question 67. This polymer (B) is obtained when acetone is saturated with hydrogen chloride gas, B can be

1. Phorone
2. Formose
3. Diacetone alcohol
4. Mesityl oxide.

Answer: 1. Phorone

Question 68. If formaldehyde and KOH are heated, then we get

1. Methane
2. Methyl alcohol
3. Ethyl formate
4. Acetylene.

Answer: 2. Methyl alcohol

⇒ \(\mathrm{HCHO}+\mathrm{KOH}\) \(\underrightarrow{50 \% \mathrm{KOH}}\) \(\mathrm{HCOOK}+\mathrm{CH}_3 \mathrm{OH}\)

The above reaction is called as Canrizzards reaction.

Question 69. Formalin is an aqueous solution of

1. Fluorescein
2. Formic acid
3. Formaldehyde
4. Furfuraldehyde.

Answer: 3. Formaldehyde

The formula is an aqueous solution of 40% HCHO.

Question 70. Complete the following reaction:

Answer: 2

Question 71. What is Y in the above reaction?

1. \(\mathrm{RCOO}^{-} \mathrm{Mg}^{+} \mathrm{X}\)
2. \(R_3 \mathrm{CO}^{-} \mathrm{Mg}^{+} X\)
3. \(\mathrm{RCOO}^{-} \mathrm{X}^{+}\)
4. \((\mathrm{RCOO})_2 \mathrm{Mg}\)

Answer: 1. \(\mathrm{RCOO}^{-} \mathrm{Mg}^{+} \mathrm{X}\)

Question 72. The reaction that does not give benzoic acid as the major product is

Answer: 3

PCC (Pyridium chlorochromate) stops oxidation at the aldehyde stage, thereby preventing the further oxidation of aldehydes to carboxylic acids.

Question 73. Which one of the following esters gets hydrolysed most easily under alkaline conditions?

Answer: 4

Electron-withdrawing groups increase the reactivity towards nucleophilic substitution reaction and -NO₂ is a strong electron-withdrawing group.

Question 74. Consider the following compounds

The correct decreasing order of their reactivity towards hydrolysis is

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

Answer: 3. 2>4<>1>3

Question 75.

In the above reaction product P is

Answer: 2

The product (P) is benzoic acid.

Question 76. Which of the following compounds gives benzoic acid on hydrolysis?

1. Chlorobenzene
2. Benzoyl chloride
3. Chlorophenol
4. Chlorotoluene

Answer: 2. Benzoyl chloride

⇒ \(\begin{aligned}
\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{HCl} \\
\text { Benzoyl chloride } \quad \text { Benzoic acid }
\end{aligned}\)

Question 77. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their

1. Formation of intramolecular H-bonding
2. Formation of carboxylate ion
3. More extensive association of carboxylic acid via van der Waals’ forces of attraction
4. Formation of intermolecular H-bonding.

Answer: 4. Formation of intermolecular H-bonding.

Question 78. The weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in the presence of calcium oxide is

1. 30
2. 18
3. 16
4. 32

Answer: 4. 32

This is the Kolbe electrolysis reaction. The weight of two moles of methane is 32 g

Question 79. Match List 1 with List 2

Choose the correct answer from the options given below.

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

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

Question 80. The major product of the following reaction is

Answer: 3

Question 81. The correct order of strengths of the carboxylic acids

Answer: 2

Acidic strength ∝ – I effect

As oxygen is more electron withdrawing (2) and (3) show greater – I effect than (1). Thus, (1) is the least acidic. Out of (2) and (3), (2) is more acid than (3) as the distance of O increases from the -COOH group and acidic strength decreases

Quetsion 82. The correct order of decreasing acid strength of trichloroacetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is

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

Answer: 1. B>A>D>C

As the -I effect increases, the -COOH group becomes more electron deficient and the tendency to lose H⁺ ions increases ie., acid strength increases. As the +I effect increases acid strength decreases.

Thus, the correct order of acid strength is

Question 83. Which one of the following is most reactive towards electrophilic reagents?

Answer: 4

Question 84. An organic compound A on treatment with NH₃ gives B, which on heating gives C. C when treated with Br₂ in the presence of KOH produces ethyl amine. Compound A is

Answer: 4

The compound will be CH₃CH₂COOH

Question 85. Propionic acid with Br₂/P yields a dibromo product. Its structure would be

Answer: 3

This is Hell-Volhard-Zelinksy reavtion. In this reaction, acids containing α-H react with X₂/red P giving a product in which the hydrogens are substituted by X.

Question 86. Which of the following represents the correct order of the acidity in the given compounds?

1. \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)> \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\)
2. \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\) > \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)
3. \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\) > \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)
4. \(\mathrm{CH}_3 \mathrm{COOH}>\mathrm{BrCH}_2 \mathrm{COOH}\)>\(\mathrm{ClCH}_2 \mathrm{COOH}>\mathrm{FCH}_2 \mathrm{COOH}\)

Answer: 3. \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\) > \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)

Question 87. In a set of reactions acetic add yielded a product D.

The structure of D would be COOH

Answer: 4

Quetsion 88. The – OH group of an alcohol or the – COOH group of a carboxylic acid can be replaced by – Cl using

1. Phosphorus pentachloride
2. Hypochlorous acid
3. Chlorine
4. Hydrochloric acid.

Answer: 1. Phosphorus pentachloride

⇒ \(\mathrm{ROH}+\mathrm{PCl}_5 \rightarrow \mathrm{RCl}+\mathrm{POCl}_3+\mathrm{HCl}\)

RCOOH+\(\mathrm{PCl}_5 \rightarrow R \mathrm{ROCl}+\mathrm{POCl}_3+\mathrm{HCl}\)

Question 89. Which one of the following orders of acid strength is correct?

1. RCOOH > ROH > HOH > HC ≡ CH
2. RCOOH > HOH > ROH > HC ≡ CH
3. RCOOH > HOH > HC ≡ CH > ROH
4. RCOOH > HC ≡ CH > HOH > ROH

Answer: 2. RCOOH > HOH > ROH > HC ≡ CH

Carboxylic acid is much stronger than water and alcohol. Since the carboxylate ion after the removal of the proton is stabilised by resonating structures. The -OH in alcohols is almost neutral. Acetylene is also the weakest acid among the given examples.

Question 90. In a set of the given reactions, acetic acid yielded a product C.

Answer: 4

Question 91. Ethyl benzoate can be prepared from benzoic acid by using

1. Ethyl alcohol
2. Ethyl alcohol and dry HCl
3. Ethyl chloride
4. Sodium ethoxide.

Answer: 2. Ethyl alcohol and dry HCl

Ethyl benzoate can be prepared by heating benzoic acid with alcohol in the presence of dry HCI or conc H₂SO₄. The reaction is called as esterification reaction.

Question 92. Reduction by LiAlH₄ of the hydrolysed product of an ester gives

1. Two alcohols
2. Two aldehydes
3. One acid and one alcohol
4. Two acids.

Answer: 1. Two alcohols

The reduction of the hydrolysed product of ester by LiAlH₄ produces two alcohols.

Question 93. Which one of the following compounds will react with the NaHCO₃ solution to give sodium salt and carbon dioxide?

1. Acetic acid
2. n-Hexanol
3. Phenol
4. Both (2) and (3)

Answer: 1. Acetic acid

NaHCO₃ is weakly basic so it can the acid CH₃COOH. While phenol is weakly acidic and r-hexanol is neutral, they do not react with NaHCO₃.

⇒ \(\mathrm{CH}_2 \mathrm{COOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 94. Which one of the following products is formed when adipic acid is heated?

Answer: 1

Question 95. An acyl halide is formed when PCl₅ reacts with an

1. Amide
2. Ester
3. Acid
4. Alcohol.

Answer: 3. Acid

⇒\(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{PCl}_5 \rightarrow \mathrm{CH}_3 \mathrm{COCl}+\mathrm{POCl}_3+\mathrm{HCl}\)

Question 96. Benzoic acid gives benzene on being heated with X and phenol gives benzene on being heated with Y. Therefore, X and Y are respectively

1. Soda-lime and copper
2. Zn dust and NaOH
3. Zn dust and soda-lime
4. Soda-lime and zinc dust.

Answer: 4. Soda-lime and zinc dust.

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) \(\underrightarrow{\text { Soda-lime }(X)}\) \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{Na}_2 \mathrm{CO}_3\)

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\) \(\underrightarrow{\mathrm{Zn} \text { dust }(Y)}\) \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{ZnO}\)

Question 97. A is a lighter phenol and B is an aromatic carboxylic acid. Separation of a mixture of A and B can be carried out easily by using a solution of

1. Sodium hydroxide
2. Sodium sulphate
3. Calcium chloride
4. Sodium bicarbonate.

Answer: 4. Sodium bicarbonate.

Carboxylic acids dissolve in NaHCO₃ but phenols do not.

Question 98. The compound formed when malonic acid is heated with urea is

1. Cinnamic acid
2. Butyric acid
3. Barbituric acid
4. Crotonic acid.

Answer: 3. Barbituric acid

Question 99. Among the following the strongest acid is

1. \(\mathrm{CH}_3 \mathrm{COOH}\)
2. \(\mathrm{CH}_2 \mathrm{ClCH}_2 \mathrm{COOH}\)
3. \(\mathrm{CH}_2 \mathrm{ClCOOH}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\)

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

The strongest acid is CH₂ClCOOH as -I effect of CI atom decreases with the increase in distance.

Question 100. Which of the following is the correct decreasing order of acidic strength of

1. Methanoic acid
2. Ethanoic acid
3. Propanoic acid
4. Butanoic acid

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

Answer: 1. 1>2>3>4

+I effect of the alkyl group increases from CH₃ to CH₃CH₂ to CH₃CH₂CH₂ to CH₃CH₂CH₂CH₂ resulting in a decrease of acid character. Therefore, the order is (1) > (2) > (3) > (4).

Question 101. The product formed in the following chemical reaction is

Answer: 1

NaBH₄ is a less powerful reducing agent than LiAlH₄. It is only powerful enough to reduce aldehydes, ketones and acid chlorides to alcohols. Ester amides acids and nitrites are not reduced.

Question 102. Match the compounds given in List 1 with List 2 and select the suitable option using the codes given below.

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

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

Quetsion 103. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is

1. \(\mathrm{CH}_3 \mathrm{COOCH}_3\)
2. \(\mathrm{CH}_3 \mathrm{CONH}_2\)
3. \(\mathrm{CH}_3 \mathrm{COOCOCH}_3\)
4. \(\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 4. \(\mathrm{CH}_3 \mathrm{COCl}\)

CH₃COCI is most susceptible to nucleophilic attack. The susceptibility of a substrate towards nucleophilic attack depends on how well a leaving group is attached to it. Cl^– is a weak base and therefore, a good leaving group

Quetsion 104. The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of

1. Acid anhydride > amide > ester > acyl chloride
2. Acyl chloride > ester > acid anhydride > amide
3. Acyl chloride > acid anhydride > ester > amide
4. Ester > acyl chloride > amide > acid anhydride.

Answer: 3. Acyl chloride > acid anhydride > ester > amide

Question 105. Self-condensation of two moles of ethyl acetate in the presence of sodium ethoxide yields

1. Ethyl propionate
2. Ethyl butyrate
3. Acetoacetic ester
4. Methyl acetoacetate.

Answer: 3. Acetoacetic ester

Ethyl acetate undergoes Claisen condensation in the presence of sodium ethoxide involving an α-hydrogen atom in which two molecules of ethyl acetate combine together to form an acetoacetic ester.

Question 106. Which one of the following esters cannot undergo Claisen self-condensation?

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5\)
4. \(\mathrm{C}_6 \mathrm{H}_{11} \mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5\)

Answer: 2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)

The esters having active methylene group (-CH₂-), show Claisen condensation reaction. As C₆H₅ – COOC₂H₅ has no α-hydrogen atom or active methylene group, it cannot undergo Claisen condensation reaction.

Question 107. Sodium formate on heating yields

1. Oxalic acid and H₂
2. Sodium oxalate and H₂
3. CO₂ and NaOH
4. Sodium oxalate

Answer: 2. Sodium oxalate and H₂

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

 

 

Organic Chemistry Some Basic Principles And Techniques

Question 1. The number of sigma (σ) and pi (π) bonds in pent-2-en-4-yne is

1. 13 σ bonds and no π bond
2. 10 σ bonds and 3π bonds
3. 8 σ bonds and 5π bonds
4. 11 σ bonds and 2π bonds.

Answer: 2. 10 σ bonds and 3π bonds

Question 2. Which of the following molecules represents the order of hybridisation sp², sp², sp, sp from left to right atoms?

1. HC ≡ C — C ≡ CH
2. CH₂ = CH — C ≡ CH
3. CH₂ = CH — CH = CH₂
4. CH₃ — CH = CH — CH₃

Answer: 2. CH₂ = CH — C = CH

⇒ \(\stackrel{s p^2}{\mathrm{CH}_2}=\stackrel{s p^2}{\mathrm{C}} \mathrm{H}-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}} \mathrm{H}\)

Question 3. The total number of π-bond electrons in the following structure is

1. 12
2. 16
3. 4
4. 8

Answer: 4. 8

There are four double bonds. Hence, no. of E-electrons =2×4=8.

Question 4. The state of hybridisation of C₂, C₃, C₅ and C₆ of the hydrocarbon,

is in the following sequence

1. sp³, sp², sp² and sp
2. sp, sp², sp² and sp³
3. sp, sp², sp³ and sp²
4. sp, sp³, sp² and sp³

Answer: 4. sp, sp³, sp² and sp³

∴ \(\mathrm{C}_2-s p, \mathrm{C}_3-s p^3, \mathrm{C}_5-s p^2 \text { and } \mathrm{C}_6-s p^3\)

Question 5. In the hydrocarbon, \(\underset{6}{\mathrm{CH}_3}-\underset{5}{\mathrm{CH}}=\underset{4}{\mathrm{CH}}-\underset{3}{\mathrm{CH}_2}-\underset{2}{\mathrm{C}} \equiv \underset{1}{\mathrm{C}} \mathrm{H}\) the state of hybridisation of carbons 1, 3 and 5 are in the following sequence

1. sp, sp², sp³
2. sp³, sp², sp
3. sp², sp, sp³
4. sp, sp³, sp²

Answer: 4. sp, sp³, sp²

The state of hybridisation of carbon in the 1,3 and 5 positions are sp, sp³ and sp².

Question 6. In which of the following compounds there is more than one kind of hybridisation (sp, sp², sp³) for carbon?

1. CH₂ = CH-CH = CH₂
2. H – C ≡ C – H
3. CH₃CH₂CH₂CH₃
4. CH₃-CH = CH-CH₃

Answer: 4. CH₃-CH = CH-CH₃

Question 7. A straight-chain hydrocarbon has the molecular formula C₈H₁₀. The hybridisation of the carbon atoms from one end of the chain to the other is respectively sp³, sp², sp², sp³, sp², sp², sp and sp. The structural formula of the hydrocarbon would be

1. CH₃C ≡ CCH₂-CH = CHCH = CH₂
2. CH₃CH2 – CH = CHCH CHC ≡ CH
3. CH₃CH = CHCH₂-C ≡ CCH = CH₂
4. CH₃CH = CHCH₂ – CH = CHC ≡ CH

Answer: 4. CH₃CH = CHCH₂ – CH = CHC ≡ CH

⇒ \(\stackrel{s p^3}{\mathrm{CH}_3} \stackrel{s p^2}{\mathrm{C H}}=\stackrel{s p^2}{\mathrm{C H}}-\stackrel{s p^3}{\mathrm{C H}}_2-\stackrel{s p^2}{\mathrm{C H}}=\stackrel{s p^2}{\mathrm{C H}} -\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C H}}\)

Question 8. Which of the following possesses a sp-carbon in its structure?

1. CH₂=CCl-CH = CH₂
2. CCl₂ = CCl₂
3. CH₂ = C= CH₂
4. CH₂ = CH-CH=CH₂

Answer: 3. CH₂ = C= CH₂

⇒ \(\stackrel{s p^2}{\mathrm{C H}}_2=\stackrel{s p}{\mathrm{C}}=\stackrel{s p^2}{\mathrm{C H}_2}\)

Question 9. The Cl-C-Cl angle in 1, 1, 2, 2-tetrachloro-ethene and tetrachloromethane respectively will be about

1. 120° and 109.5°
2. 90° and 109.5°
3. 109.5° and 90°
4. 109.5° and 120°

Answer: 1. 120° and 109.5°

Tetrachloroethene being an alkene has sp²-hybridised C-atoms and hence the angle CI-C-CI is 120° while in tetrachioromethane is, carbon is sp³ hybridised, therefore the angle Cl-C-Cl is109°28′.

Question 10. The number of σ bonds, π bonds and lone pair of electrons in pyridine, respectively are

1. 11,3, 1
2. 12,2,1
3. 11,2,0
4. 12,3,0

Answer: 1. 11,3, 1

σ bonds:11

π bonds : 3

Lone pair: 1

Question 11. An organic compound X(molecular formula C₆H₇O₂N) has six carbon atoms in a ring system, two double bonds and a nitro group as a substituent, X is

1. Homocyclic but not aromatic
2. Aromatic but not homocyclic
3. Homocyclic and aromatic
4. Heterocyclic and aromatic.

Answer: 1. Homocyclic but not aromatic

Hence, it is homocyclic (as the ring system is made of one thee of atoms, i.e., carbon) but not aromatic.

Question 12. The correct IUPAC name of the following compound is

1. 1-bromo-5-chloro-4-methylhexan-3-ol
2. 6-bromo-2-chloro-4-methylhexan-4-ol
3. 1-bromo-4-methyl-5-chlorotoxin-3-ol
4. 6-bromo-4-methyl-2-chlorohexan-4-ol.

Answer: 1. 1-bromo-5-chloro-4-methylhexan-3-ol

Question 13. The correct Structure of 2, 6-dimethyl-dec-4-ene is

Answer: 2

Question 14. The IUPAC name of the compound

1. 5-formylhex-2-en-3-one
2. 5-methyl-4-oxohex-2-en-5-al
3. 3-keto-2-methylhex-5-enal
4. 3-keto-2-methylhex-4-enal

Answer: 4. 3-keto-2-methylhex-4-enal

Question 15. The structure of the isobutyl group in an organic compound is

Answer: 3

Question 16. The structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-yonic acid is

Answer: 4

IUPAC name of the compound is 3 -ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid.

Question 17. Which nomenclature is not according to the IUPAC system?

Answer: 1

Question 18. The correct IUPAC name for the compound

1. 4-ethyl-3-propylhex-1-ene
2. 3-ethyl-4-ethenyl heptane
3. 3-ethyl-4-propylhex-5-ene
4. 3-(1-ethylpropyl)hex-1-ene.

Answer: 1. 4-ethyl-3-propyl hex-1-ene

Question 19. The IUPAC name of the following compound is

1. Trans-2-chloro-3-iodo-2-pentene
2. cis-3-iodo-4-chloro-3-pentane
3. Trans-3-iodo-4-chloro-3-pentene
4. cis-2-chloro-3-iodo-2-pentene.

Answer: 1. Trans-2-chloro-3-iodo-2-pentene

Question 20. The IUPAC name of the compound CH₃CH=CHC≡CH is

1. Pent-4-yn-2-ene
2. Pent-3-en-1-yne
3. Pent-2-en-4-yne
4. Pent-1-yn-3-ene.

Answer: 2. Pent-3-en-1-yne

If a molecule contains both carbon-carbon double or triple bonds, the two are treated as seeking the lowest number combination. However, if the sum of numbers turns out to be the same starting from either of the carbon chains, then the lowest number is given to the C=C double bond.

Question 21. The IUPAC name of the compound having the formula CH≡C—CH=CH₂ is

1. 1-butyne-3-ene
2. but-1-yne-3-ene
3. 1-butene-3-yne
4. 3-butene-1-yne.

Answer: 3. 1-butene-3-yne

Since the sum of numbers starting from either side of the carbon chain turns out to be the same, so lowest number is given to the C=C end.

Question 22. The IUPAC name is

1. 1-chloro-1-oxo-2,3-dimethyl pentane
2. 2-ethyl-3-methyl butanol chloride
3. 2,3-dimethyl pentanol chloride
4. 3,4-dimethylpentanoyl chloride.

Answer: 3. 2,3-dimethyl pentanol chloride

It is 2,3-dimethylpentanoyl chloride.

Question 23. Names of some compounds are given. Which one is not in the IUPAC system?

Answer: 1

Question 24. The name of the compound given below is

1. 4-ethyl-3-methyl octane
2. 3-methyl-4-ethylacetate
3. 2,3-diethyl heptane
4. 5-ethyl-6-methyloctane.

Answer: 1. 4-ethyl-3-methyl octane

Question 25. IUPAC name of the following is CH₂ = CH – CH₂ – CH₂ – C ≡ CH

1. 1,5-hexene
2. 1-hexene-5-yne
3. 1-hexyne-5-ene
4. 1,5-hexynene.

Answer: 2. 1-hexene-5-yne

⇒ \(\stackrel{1}{\mathrm{C}} \mathrm{H}_2=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{3}{\mathrm{C}} \mathrm{H}_2-\stackrel{4}{\mathrm{C}} \mathrm{H}_2-\stackrel{5}{\mathrm{C}} \equiv \stackrel{6}{\mathrm{C}} \mathrm{H}\)

The double bond gets priority over the triple bond. therefore, the correct IUPAC name is 1-hexene-5-yne.

Question 26. The incorrect IUPAC name is

Answer: 1

The    group should get priority over methyl group.;

Correct IUPAC name is

Question 27. The IUPAC name of (CH₃)₂CH – CH₂ – CH₂Br is

1. 1-bromo-3-methyl butane
2. 2-methyl-3-bromopropane
3. 1-bromopentane
4. 2-methyl-4-bromobutane.

Answer: 1. 1-bromo-3-methyl butane

Question 28. The IUPAC name is

1. 3-amino-5-heptenoic acid
2. β-amino-ε-heptenoic acid
3. 5-amino-2-heptenoic acid
4. 5-amino-hex-2-enecarboxylic acid.

Answer: 1. 3-amino-5-heptanoic acid

As -COOH group is the highest priority group, it is numbered one. So, the IUPAC name is 3-amino-5-heptenoic acid.

Question 29. 2-Methyl-2-butene will be represented as

Answer: 2

Question 30. The IUPAC name is

1. 4-hydroxy-1-methylpentanal
2. 4-hydroxy-2-methyl pent-2-en-1-al
3. 2-hydroxy-4-methyl pent-3-en-5-al
4. 2-hydroxy-3-methylpent-2-en-5-al.

Answer: 2. 4-hydroxy-2-methyl pent-2-en-1-al

Question 31. The compound which shows metamerism is

1. C₄H₁₀O
2. C₅H₁₂
3. C₃H₈O
4. C₃H₆O

Answer: 1. C₄H₁₀O

Metamerism occurs when the compound has a different number of carbon atoms on either side of the functional group.

Question 32. Which among the given molecules can exhibit tautomerism?

1. 3 only
2. Both 1 and 2
3. Both 1 and 3
4. Both 2 and 3

Which of the given compounds can exhibit tautomerism?

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

Answer: 1. 3 only

α-Hydrogen at bridge carbon never participates in tautomerism. Thus, only (3) exhibit tautomerism.

Question 33. Given:

Which of the given compounds can exhibit tautomerism?

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

Answer: 2. 1, 2 and 3

In keto-enol tautomerism,

Question 34. The enolic form of ethyl acetoacetate as shown below has

1. 9 sigma bonds and 2 pi-bonds
2. 9 sigma bonds and 1 pi-bond
3. 18 sigma bonds and 2 pi-bonds
4. 16 sigma bonds and 1 pi-bond.

Answer: 3. 18 sigma bonds and 2 pi-bonds

The enolic form of ethyl acetoacetate has 18 σ-bonds and 2π-bonds.

Question 35. Which one of the following pairs represents stereoisomerism?

1. Structural isomerism and geometrical isomerism
2. Optical isomerism and geometrical isomerism
3. Chain isomerism and rotational isomerism
4. Linkage isomerism and geometrical isomerism

Answer: 2. Optical isomerism and geometrical isomerism

Question 36. The molecular formula of diphenylmethane,

How many structural isomers are possible when one of the hydrogen is replaced by a chlorine atom?

1. 6
2. 4
3. 8
4. 7

Answer: 2. 4

Only four structural isomers are possible for mono-chlorinated diphenylmethane.

Question 37. Tautomerism is exhibited by

1. R₃CNO₂
2. RCH₂NO₂
3. (CH₃)₃CNO
4. (CH₃)₂NH

Answer: 2. RCH₂NO₂

It is a special type of functional isomerism, in which both the isomers are represented by one and the same substance and are always present in equilibrium. It is exhibited by nitroalkane (RCH₂NO₂) and isonitroalkane.

Question 38. The number of isomers in C₄H₁₀O will be

1. 7
2. 8
3. 5
4. 6

Answer: 1. 7

There are 7 isomers in C₄H₁₀O. Out of these, 4 are alcohols and 3 are ethers.

Question 39. Isomers of a substance must have the same

1. Structural formula
2. Physical properties
3. Chemical properties
4. Molecular formula.

Answer: 4. Molecular formula.

Isomers must have the same molecular formula but a different structural formula.

Question 40. How many chain isomers could be obtained from the alkane C₆H₁₄?

1. Four
2. Five
3. Six
4. Seven

Answer: 2. Five

5-chain isomers are obtained from alkane C₆H₁₄

Question 41. A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?

1. -I effect of – CH₃ groups
2. +R effect of – CH₃ groups
3. -R effect of – CH₃ groups
4. Hyperconjugation

Answer: 4. Hyperconjugation

tert-Bttylcarbocation, (CH₃)₂C⁺ is more stable than sec-butyl carbocation (CH₃)₂C⁺H due to hyperconjugation.

⇒ \(\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{C}}\) has nine C – H bonds while \(\left(\mathrm{CH}_3\right)_2 \stackrel{+}{\mathrm{C}}\) has six C – H bonds. Thus, there are more hyperconjugative structures in tert-butyl carbocation.

Question 42. The most stable carbocation, among the following, is

1. (CH₃)₃ C—CH—CH₃
2. CH₃—CH₂—CH—CH₂—CH₃
3. CH₃—CH—CH₂—CH₂—CH₃
4. CH₃—CH₂—CH₂

Answer: 3. CH₃—CH—CH₂—CH₂—CH₃

Among the given carbocations, \(\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3\) is most stable carbocation.

As is consists of a maximum number of u-hydrogens and is stabilised by hyperconjugation.

Question 43. Which of the following is correct with respect to- I effect of the substituents? (R = alkyl)

1. – NH₂ < – OR < – F
2. – NR₂ < – OR < – F
3. – NH₂ > – OR > – F
4. – NR₂ > – OR > – F

Answer: 1. – NH₂ < – OR < – F and 2. – NR₂ < – OR < – F

I-effect increases on increasing the electro negativity of the atom.

∴ -NH₂<-OR<-F (-I effect)

Also, – NR₂ < – OR < – F (-I effect)

Question 44. Which of the following carbocations is expected to be most stable?

Answer: 3

-NO₂ group is meta-directing, thus will stabilize an electrophile at m-position.

Question 45. The correct statement regarding electrophiles is

1. Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile
2. Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
3. Electrophiles can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
4. Electrophiles is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile.

Answer: 3. Electrophiles can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile

Question 46. Which of the following statements is not correct for a nucleophile?

1. Ammonia is a nucleophile.
2. Nucleophiles attack low-density sites.
3. Nucleophiles are not electron-seeking.
4. A nucleophile is a Lewis acid.

Answer: 4. Nucleophile is a Lewis acid.

Nucleophiles are electron-rich species hence, they are Lewis bases.

Question 47. Treatment of cyclopentanone with methyl lithium gives which of the following species?

1. Cyclopentanonyl radical
2. Cyclopentanonyl biradical
3. Cyclopentanonyl anion
4. Cyclopentanonyl cation

Answer: 3. Cyclopentanonyl anion

Question 48. Which of the following is the most correct electron displacement for a nucleophilic reaction to take place?

Answer: 1. Nucleophiles will attack a stable carbocation (S_N1 reaction).

Question 49. Consider the following compounds:

Hyperconjugation occurs in

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

Answer: 1. 3 only

Hyperconjugation can occur only in compound 3 as it has an α-hydrogen atom.

Question 50. In which of the following compounds, does the C—Cl bond ionisation give the most stable carbonium ion?

Answer: 4

is most stable due to hyperconjugation.

Question 51. The radical is aromatic because it has

1. 7 p-orbitals and 7 unpaired electrons
2. 6 p-orbitals and 7 unpaired electrons
3. 6 p-orbitals and 6 unpaired electrons
4. 7 p-orbitals and 6 unpaired electrons.

Answer: 3. 6 p-orbitals and 6 unpaired electrons

Question 52. Arrange the following in increasing order of stability.

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

Answer: 1. 4<3<1<2

The greater the number of electron-donating alkyl groups (+1 effect), the greater the stability of carbocations.

+I effect is in the order:

The more the number of hyperconjugation structures of carbocations, the more is the stability.

Hence, the order of stability of carbocations is 5<4<3<1<2.

Question 53. What is the hybridization state of benzyl carbonium ion

1. sp²
2. spd²
3. sp²d
4. sp³

Answer: 1. sp²

Question 54. Homolytic fission of the following alkanes forms free radicals CH₃ — CH₃, CH₃ — CH₂ — CH₃, (CH₃)₂ CH — CH₃, CH₃ — CH₂ — CH(CH₃)₂. Increasing the order of stability of the radicals is

1. \(\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}}-\mathrm{CH}_2-\mathrm{CH}_3<\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3\)\(<\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}_2<\left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}}\)
2. \(\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}_2<\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3\)\(<\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}}-\mathrm{CH}_2-\mathrm{CH}_3<\left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}}\)
3. \(\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}_2<\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3\)\(<\left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}}<\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}}-\mathrm{CH}_2-\mathrm{CH}_3\)
4. \(\left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}}<\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}}-\mathrm{CH}_2-\mathrm{CH}_3\)\(<\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3<\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}_2\)

Answer: 2. \(\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}_2<\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3\)\(<\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}}-\mathrm{CH}_2-\mathrm{CH}_3<\left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}}\)

The more the number of hyperconjugation structures, the greater is the stability.

Question 55. Which one is a nucleophilic substitution reaction among the following?

Answer: 3

A nucleophilic substitution reaction involves the displacement of a nucleophile by another.

Question 56. Which one of the following is most reactive towards electrophilic reagents?

Answer: 2

+R effect of the -OH group is greater than that of the -OCH₃ group.

Question 57. Which of the following species is not electrophilic in nature?

1. \(\stackrel{+}{\mathrm{Cl}}\)
2. BH₃
3. H₃O⁺

Answer: 3. H₃O⁺

Question 58. The stability of carbanions in the following:

is in the order of

1. (4) > (2) > (3) > (1)
2. (1) > (3) > (2) > (4)
3. (1) > (2) > (3) > (4)
4. (2) > (3) > (4) > (1)

Answer: 3. (1) > (2) > (3) > (4)

The higher the no. of electron-releasing groups lower be stability of carbanion, and vice-versa. So, the order of stability of carbanions is

Question 59. For (1) I^–, (2) Cl^–, and (3) Br^–, the increasing order of nucleophilicity would be

1. CI^– < Br^– < I^–
2. I^– < Cl^– < Br^–
3. Br^– < Cl^– < I^–
4. I^– < Br^– < Cl^–

Answer: 1. Cl^– < Br^– < I^–

In the case of different nucleophiles, but present in the same group in the periodic table, the larger is the atomic mass, the higher is the nucleophilicity. Hence, the increasing order of nucleophilicity of the halide ions is F^–<Cl^–<Br^–<I^–.

Question 60. Which amongst the following is the most stable carbocation?

Answer: 4

3° carbon is more stable due to the stabilization of the charge by three methyl groups (or +I effect). It can also be explained on the basis of hyperconjugation. The greater the number of hyper conjugative α-H atoms, the more the hyperconjugation structures and the more will be stability.

Question 61. Which of the following is the most stable carbocation (carbonium ion)?

1. \(\mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)
2. \(\left(\mathrm{CH}_3\right)_2 \stackrel{+}{\mathrm{CH}}\)
3. \(\left(\mathrm{CH}_3\right)_3 \mathrm{C}\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)

Answer: 3. \(\left(\mathrm{CH}_3\right)_3 \mathrm{C}\)

3° > 2° > 1° The more the delocalisation of the positive charge, the more is its stability.

Question 62. Cyclic hydrocarbon ‘A’ has all the carbon and hydrogen atoms in the plane. All the carbon-carbon bonds have the same length, less than 1.54 Å, but more than 1.34 Å. The bond angle will be

1. 109°28′
2. 100°
3. 180°
4. 120°

Answer: 4. 120°

All the properties mentioned in the question suggest that it is a benzene molecule. Since in benzene, all carbons are sp²-hybridised, therefore, C – C – C angle is 120°

Question 63. Paper chromatography is an example of

1. Adsorption chromatography
2. Partition chromatography
3. Thin layer chromatography
4. Column chromatography.

Answer: 2.  Partition chromatography

Paper chromatography is a type of partition chromatography.

Question 64. The most suitable method of separation of 1:1 mixture of ortho and para-nitrophenols is

1. Chromatography
2. Crystallisation
3. Steam distillation
4. Sublimation.

Answer: 3. Steam distillation

The o- and p-nitrophenols are separated by steam distillation since the o-isomer is steam volatile due to intramolecular H-bonding while-isomer is not steam volatile due to the association of molecules by intermolecular H-bonding.

Question 65. The best method for the separation of naphthalene and benzoic acid from their mixture is

1. Distillation
2. Sublimation
3. Chromatography
4. Crystallisation.

Answer: 4. Crystallisation.

Both naphthalene and benzoic acid are sublimable. Hence they cannot be separated by sublimation method. They can be separated by crystallisation with hot water. Benzoic acid dissolves in hot water whereas naphthalene does not.

Question 66. In steam distillation of toluene, the pressure of toluene in vapour is

1. Equal to the pressure of a barometer
2. Less than the pressure of a barometer
3. Equal to vapour pressure of toluene in simple distillation
4. More than the vapour pressure of toluene in simple distillation.

Answer: 2. Less than the pressure of the barometer

In steam distillation of toluene, the pressure of toluene in the vapour is less than the pressure of a barometer, because it is carried out when a solid or liquid is insoluble in water and is volatile with steam but the impurities are non-volatile.

Question 67. Which of the following techniques is most suitable for the purification of cyclohexanone from a mixture containing benzoic acid, isoamyl alcohol, cyclohexane and cyclohexanone?

1. Sublimation
2. Evaporation
3. Crystallisation
4. IR spectroscopy

Answer: 4. IR spectroscopy

In IR spectroscopy, each functional group appears at a certain peak (in cm^-1). So, cyclohexanone can be identified by carbonyl peak.

Question 68. In Lassaignes extract of an organic compound, both nitrogen and sulphur are present, which gives blood red colour with Fe due to the formation of

1. \(\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}\)
2. \([\mathrm{Fe}(\mathrm{SCN})]^{2+}\)
3. \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{H}_2 \mathrm{O}\)
4. \(\mathrm{NaSCN}\)

Answer: 2. \([\mathrm{Fe}(\mathrm{SCN})]^{2+}\)

In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives a blood-red colour and no Prussian blue since there is no free cyanide.

⇒ \(\mathrm{Na}+\mathrm{C}+\mathrm{N}+\mathrm{S} \longrightarrow \mathrm{NaSCN}\)

⇒ \(\mathrm{Fe}^{3+}+\mathrm{SCN}^{-} \longrightarrow \underset{\text { Blood red }}{[\mathrm{Fe}(\mathrm{SCN})]^{2+}}\)

Question 69. Nitrogen detection in an organic compound is carried out by the Lassaignes test. The blue colour formed corresponds to which of the following formulae?

1. \(\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2\)
2. \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\)
3. \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2\)
4. \(\mathrm{Fe}_5\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\)

Answer: 2. \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\)

⇒ \(\underset{\text { Sodium ferrocyanide }}{3 \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]}+2 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \rightarrow  \underset{ \text { Ferric ferrocyanide
(Prussian blue) }}{\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3}+6 \mathrm{Na}_2 \mathrm{SO}_4\)

Question 70. The Lassaignes extract is boiled with a cone. HNO₃ while testing for halogens. By doing so it

1. Decomposes Na₂S and NaCN formed
2. Helps in the precipitation of AgCl
3. Increases the solubility product of AgCl
4. Increases the concentration of NO^–₃ ions.

Answer: 1. Decomposes Na₂S and NaCN, formed

In the case of Lassaigne’s test of halogens, it is necessary to remove sodium cyanide and sodium sulphide from the sodium extract if nitrogen and sulphur are present. This is done by boiling the sodium extract with conc. HNO₃.

⇒ \(\mathrm{NaCN}+\mathrm{HNO}_3 \rightarrow \mathrm{NaNO}_3+\mathrm{HCN} \uparrow \)

⇒ \(\mathrm{Na}_2 \mathrm{~S}+2 \mathrm{HNO}_3 \rightarrow 2 \mathrm{NaNO}_3+\mathrm{H}_2 \mathrm{~S} \uparrow\)

Question 71. In the sodium fusion test of organic compounds, the nitrogen of the organic compound is converted into

1. Sodamide
2. Sodium cyanide
3. Sodium nitrite
4. Sodium nitrate.

Answer: 2. Sodium cyanide

Sodium cyanide (Na + C + N → NaCN)

Question 72. Lassaignes test is used in qualitative analysis to detect

1. Nitrogen
2. Sulphur
3. Chlorine
4. All of these.

Answer: 4. All of these.

Question 73. A blue colouration is not obtained when

1. Ammonium hydroxide dissolves in copper sulphate
2. Copper sulphate solution reacts with K₄[Fe(CN)₆]
3. Ferric chloride reacts with sod. ferrocyanide
4. Anhydrous CuSO₄ is dissolved in water.

Answer: 2. Copper sulphate solution reacts with K₄[Fe(CN)₆]

⇒ \(2 \mathrm{CuSO}_4+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow \underset{\text { chocolate ppt. }}{\mathrm{Cu}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]+2 \mathrm{~K}_2 \mathrm{SO}_4}\)

Question 74. Kjeldahl’s method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds?

Answer: 3

Kjeldahtr’s method is not applicable to compounds containing nitrogen in the nitro group, azo groups and nitrogen present in the ring (for example, pyridine).

Question 75. In Dumas’s method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is

1. 16.76
2. 15.76
3. 17.36
4. 18.20

Answer: 1. 16.76

Mass of organic compound = 0.25 g

Experimental values, At STP

⇒ \(V_1=40 \mathrm{~mL}, V_2=?, T_1=300 \mathrm{~K}, T_2=273 \mathrm{~K}\)

⇒ \(P_1=725-25=700 \mathrm{~mm}, P_2=760 \mathrm{~mm}\)

⇒  \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \Rightarrow V_2=\frac{P_1 V_1 T_2}{T_1 P_2}=\frac{700 \times 40 \times 273}{300 \times 760}=33.52 \mathrm{~mL}\)

22400 mL of \(\mathrm{N}_2\) at STP weighs = 28 g

∴ 33.52 mL of \(\mathrm{N}_2\) at STP weighs = \(\frac{28 \times 33.52}{22400}=0.0419 \mathrm{~g}\)

% of \(\mathrm{N}=\frac{\text { Mass of nitrogen at STP }}{\text { Mass of organic compound taken }} \times 100\)

= \(\frac{0.0419}{0.25} \times 100=16.76 \%\)

Question 76. In Kjeldahl’s method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1 M H₂SO₄. The percentage of nitrogen in the soil is

1. 37.33
2. 45.33
3. 35.33
4. 43.33

Answer: 1. 37.33

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{NH}_3 \rightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\)

10 mL of 1 M H₂SO₄ = 10 m mol

[M x V_(mL) = mmol]

The acid used for the absorption of ammonia = 10 mL of 2 N (or 1 M) H₂SO₄

% of N = \(\frac{1.4 \times N \times V}{W}=\frac{1.4 \times 2 \times 10}{0.75}=37.33 \%\)

Question 77. In Dumas’ method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at 300 K = 15 mm)

1. 15.45
2. 16.45
3. 17.45
4. 14.45

Answer: 2. 16.45

Given: V₁ = 55mL, V₂ = ?

P₁ = 715 – 15 = 700 mm, P₂ = 760 mm

T₁ = 300K, T₂ = 273K

General sas equation, \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

Volume of nitrogen at STP, \(V_2=\frac{P_1 V_1 T_2}{P_2 T_1}=\frac{700 \times 55 \times 273}{760 \times 300}=46.099 \mathrm{~mL}\)

% of nitrogen = Vz , where, W= the mass of organic compound.

% of N = \(=\frac{46.099}{8 \times 0.35}\) = 16.46

Question 78. Kjeldahl’s method is used in the estimation of

1. Nitrogen
2. Halogens
3. Sulphur
4. Oxygen.

Answer: 1. Nitrogen

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

Read and Learn More NEET MCQs with Answers

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.