CUET Chemistry

ALDEHYDE, KETONES AND CARBOXYLIC ACIDS 133 The above reaction is known as In the above sequence, Z is (a) —COOH (a) Backmann rearrangement (b) Benzoin condensation (c) Internal aldol condensation (d) Benzilic acid rearrangement. OH (b) O = =O 117. [(CH3)3 CO]3 Al ⎯⎯⎯⎯⎯→ X. The product X of COOH Acetone (c) COOH this reaction is OH O (a) (b) OH (d) HOOC— —COOH. O (c) O 122. In nucleophilic substitution reactions, the (d) reactivity of carbonyl compounds follows the order | OH OH (a) H2C = O > R2C = O > Ar2C = O > RCHO > ArCHO 118. Gas A is produced by dropping calcium carbide in water is bubbled through very dilute H2SO4 (b) H2C = O > RCHO > ArCHO > R2C = O containing mercury (II) sulphate to form B. > Ar2C = O B + X ⎯⎯→ Ethylidene dichloride. X can be (c) Ar2C = O > R2C = O > ArCHO > RCHO > H2C = O 119. (a) Cl2 123. (b) SOCl2/Pyridine (d) ArCHO > Ar2C = O > RCHO > R2C = O (c) SbCl5 > H2C = O. (d) CHCl3. Certain ketone A (capable of showing iodoform The enol form of acetone after treatment with test positive) on reduction gives B. B on acidic D2O gives dehydration at 443 K gives C. C on reductive ozonolysis gives only MeCHO. A and C in the OD O above sequence are respectively. ⏐⏐ ⏐ (b) CH3— C⎯CH2D (a) CH3COCH3 and CH3CH = CH2 (a) CH3 C=CH2 (b) CH3COC2H5, CH3CH2CH = CH2 (c) CH3COC2H5 and CH3CH = CHCH3 OH OD (d) CH3COC2H5, CH3CH2 = CH(OH)CH3. Cyclohexanone is treated with methylene ⏐ ⏐ (c) CH2 = C⎯CH2D (d) CD2 = C⎯CD3 . triphenyl phosphorane,(C6H5)3P = CH2. The 120. product formed is 124. Among ethanol (I), acetic acid (II), phenol (III) and benzoic acid (IV), the correct order of increasing acid strength is O (a) I < II < III < IV (b) I < III < II < IV (b) (c) I < III < IV < II (d) III < IV < I < II. OH (a) 125. The formula AcOAc represents CH3 CH2 (d) (a) Ester (b) Ether (c) (c) Carboxylic acids (d) Acid anhydride. 126. Which acid is produced when ethyl benzene is subjected to KMnO4 oxidation ? Vigorous NaOCl (a) Benzoic acid (b) Phenylacetic acid 121. — — ⎯⎯⎯⎯→ X ⎯⎯⎯→ Z (c) Ethenoic acid (d) Phthalic acid Oxidation 127. Which combination produces oxalyl chloride ? (a) (COOH)2 + Cl2 (b) (COOH)2 + KCl (c) (COOH)2 + SOCl2 (d) (COOH)2 + ZnCl2.

134 CHEMISTRY 128. Which reagent can convert α-chloroacetic acid 138. A concentrated solution of urea gives white to 2-chloroethanoyl chloride ? precipitate with (a) HCl (b) PCl5 (a) HNO3 (conc.) (b) HNO2 (c) Cl2 (d) MnCl2. (c) AgNO3 (d) NaOH. 129. Chloroacetic acid is allowed to react with moist 139. Chemical reaction of any amine with acid Ag2O (Ag2O/H2O), the main product is chloride is termed as (a) Silver lactate (a) Esterification (b) Condensation (b) α-hydroxyethanoic acid (c) Acylation (d) Saponification. (c) Silver pyruvate 140. NH2CONH2 is (a) monoacid base (d) The reaction does not occur. (b) monobasic acid 130. The mixture of formic acid and acetic acid (c) dibasic acid (d) diacid base. vapours are passed over heated manganous oxide at 575 K. The main product is 141. A compound ‘A’ with formula C4H10O3 on acylation with acetic anhydride give another (a) Ethyl ethanoate (b) Methyl formate compound with molecular mass 190. The number of hydroxyl group in the compound ‘A’ (c) Acetone (d) Acetaldehyde. are 131. Which of the following can provide distinction (a) two (b) one between formic acid and acetic acid ? (c) three (d) unpredictable. (a) Litmus solution 142. Which acid has lowest value of pKa ? (a) p-Methoxybenzoic acid (b) Alcoholic AgNO3 (c) Sodium bicarbonate (b) p-Chlorobenzoic acid (d) Diammine silver (I) hydroxide solution. (c) p-Aminobenzoic acid 132. Sulphuryl chloride and calcium propionate (d) p-Toluic acid. react to give 143. A halogen compound A on hydrolysis with dilute (a) Propyl hydrogen sulphate alkali followed by acidification gives acetic acid. The compound A is (b) Propyl acetate (c) Propionyl chloride (a) CH3CCl3 (b) CH3CHCl2 (d) Propanoic anhydride. (c) ClCH2CH2Cl (d) ClCH2CHCl2. 133. Acetone and Sodium hypochlorite react to give 144. N-Bromosuccimide is allowed to react with α- (a) Monochloroacetone (b) Dichloroacetone amino propionic acid. The main products are (c) Chloroform (d) Chloretone. (a) CH3CHO, CO2, NH3 134. A colourless gas evolved by the reaction of (b) CH3CH(Br)COOH propanamide with NaNO2/HCl is (c) CH3CH2CONH2 (a) Dinitrogen (b) Ammonia (d) CH3CH(NH)2CONH2. (c) Dioxygen (d) Propane. 145. Which of the following acid produces violet 135. Acetamide produces 1° amine with colour with FeCl3 solution ? (a) NaOH (b) HCl (a) Benzoic acid (b) Acetic acid (c) NaOH/Br2 (d) HgO. (c) Picric acid (d) Oxalic acid. 136. RCOCl reacts with sodium salt of hydrazoic acid 146. The calcium salt of which of the following acid on dry distillation produces 2, 4- to produce Dimethylpentane-3-one (a) Acylazide (b) 1° Amine (c) Urea (d) 2° Amine. (a) Propionic acid (b) Isobutyric acid 137. In esterification reaction, the role of conc. H2SO4 (c) Butyric acid (d) Adipic acid. is 147. Acetyl chloride and phenol react to produce (a) to act as dehydrating agent (a) Phenetole (b) to act as hydrolytic agent (b) Phenoxy chloride (c) to act as catalyst (c) Phenyl acetate (d) to act as catalyst as well as dehydrating (d) Methyl benzoate. agent.

ALDEHYDE, KETONES AND CARBOXYLIC ACIDS 135 148. The compound with a formula H2NCH2COOH 158. The main product formed when propionic acid behave as vapours are passed over heated manganese oxide at 575 K is (a) strong acid (b) strong base (a) propionic anhydride (b) propanal (c) amphoteric substance (c) 3-pentanone (d) propanone. (d) strong reducing agent. 159. Which of the following will be produced when ethanoic acid and diazomethane are allowed to 149. Phosgene and ammonia react to give react ? (a) Carbonic acid (b) Carbolic acid (c) Carbamic acid (d) Carbamide. (a) Ethyl acetate 150. Which of the following ester cannot undergo (b) Methyl methanoate Claisen self condensation ? (c) Methyl ethanoate (a) CH3(CH2)3COOC2H5 (b) C6H5COOCH2CH3 (d) Ethanoyl cyanide. (c) CH3CH2COOC6H5 (d) C6H5CH2COOC2H5. 160. Which compound will be formed when p-tolyl chloride and sodium p-toluate react to ethylacetate and excess of ethyl magnesium 151. bromide are allowed to react and product is produce hydrolysed ? (a) p-toluic anhydride (b) Toluic acid (a) 3-Ethyl pentan-3-ol (c) benzoic anhydride (d) p-toludene. (b) 3-Methyl pentan-3-ol 152. The chemical reaction between alcohol and (c) Hexan-3-ol esters in acidic medium (d) 2-Methyl-2-propanol. (a) involve exchange of alkyl groups 161. Which of the following reagent will distinguish 162. (b) are called transesterification between two functional isomers of formula (c) involve exchange of their alkyl parts C2H4O2 ? (d) are called self esterification. (a) NaHCO3 (b) CuSO4 153. Which carboxylic acid will show HVZ reaction? (c) C2H5Cl (d) All of the above. (a) Propionic acid HVZ reaction can be used to convert propionic (b) Trichloroethanoic acid acid to (c) Triphenyl acetic acid (a) Malonic acid (b) Stearic acid (d) 2, 2-dimethyl propanoic acid. (c) Lactic acid (d) Succinnic acid. 154. An ester on treatment with excess of ethyl 163. Claisen condensation involves the treatment of magnesium iodide and subsequent hydrolysis produces pentan-3-ol. The ester could be (a) Ethyl acetate and ethanol (b) Ethyl acetate and sodium ethoxide (a) ethyl formate (b) methylethanoate (c) Ethanol and peroxy acetic acid (c) ethyl acetate (d) ethyl propionate. (d) Ethyl acetate and Grignard’s reagent. 155. In the Hoffmann’s bromamide reaction, the 164. When ketene is passed through glacial acetic intermediate involved are acid. Which of the following product is formed? (a) RCONHBr (b) R—C—N = O (a) Ethyl acetate (c) R—NHBr (d) R—CONBr2. (b) Ethanoic anhydride 156. Reaction of ethyl formate with excess of MeMgI (c) Propanoic acid followed by hydrolysis gives (d) Pyruvic acid. (a) n-Propyl alcohol (b) Ethanal 165. Which of following aromatic acid has lowest value of Ka ? (c) Propanal (d) Isopropyl alcohol. (a) p-nitrobenzoic acid 157. The major product of nitration of benzoic acid (b) p-methoxybenzoic acid is (a) 3-Nitrobenzoic acid (c) p-chlorobenzoic acid (b) 4-Nitrobenzoic acid (d) Benzoic acid. (c) 2-Nitrobenzoic acid (d) 2, 4-Dinitrobenzoic acid.

136 CHEMISTRY 166. Which of the following is present in Vinegar ? 178. Reduction of esters with sodium and alcohol is referred to as (a) Acetic acid (b) Citric acid (c) Tartaric acid (d) Lactic acid. (a) Bouvealt-Blanc reaction 167. Which of the following is strongest reducing (b) Mendius reaction agent ? (c) Clemensen’s reduction (a) benzoic acid (b) acetic acid (d) MPV reduction. (c) isobutyric acid (d) methanoic acid. 179. Which of the following will convert carboxylic acids to alkane with same number of carbon 168. The percentage of acetic acid in pyroligneous atoms ? acid is approximately (a) 2% (b) 10% (a) H2/Adkin’s catalyst (b) HI (c) 20% (d) 15%. (c) NaOH/CaO (d) LiAlH4. 169. Ammonium acetate on heating gives 180. Which acid is formed when sucrose is heated (a) N2 gas (b) ammonia gas with conc. HNO3 ? (c) acetic acid vapours (d) ethanamide. (a) gluconic acid (b) ascorbic acid 170. Alkali metal salt of palmitic acid is known as (c) oxalic acid (d) citric acid. (a) an epoxide (b) a soap 181. Bees wax is (c) an alkaloid (d) a lipid. (a) Myricyl palmitate 171. Acetamide and nitrous acid react to form (b) Myricyl stearate (a) CH3NH2 and N2 (b) CH3OH and NH3 (c) Mixture of higher hydrocarbon (c) CH3COOH and N2 (d) CO2CH3NH2. (d) Myricyloleate. 172. Which among the following reagents will justify 182. Which of the following is known as oil of winter- green ? the acidic nature of amides ? (a) Methyl salicylate (a) HgO (b) HNO2 (b) Phenyl salicylate (c) Br2/KOH (d) HCl. (c) Ethyl salicylate 173. Which acid derivative react with hydroxyl (d) Ethyl, p-hydroxybenzoate. amine to produce hydroxamic acid ? (a) Esters (b) Acid chloride 183. Chemically banana oil is (c) both (a) and (b) (d) None of these. (a) mixture of hydrocarbons obtained from banana leaves 174. Which of the following is not a hydtroxy acid ? (b) mixture of compounds having 3-centre bonds (a) Lactic acid (b) Tartaric acid in their molecules (c) Citric acid (d) Succinic acid. (c) isoamyl acetate 175. Distinction between oxalic acid, malonic acid (d) isooctyl butyrate. and succinic acid can be achieved by 184. What is formed when diethyl ether reacts with (a) heating (b) acidified KMnO4 CO at about 423 K of 500 atm pressure in the presence of boron trifluoride ? (c) Br2 water (d) PCl5. 176. Which one of the following is expected to be (a) Ethyl acetate (b) (C2H5)2CO3 (c) Et OH (d) Ethylpropionate. highly ionised in aqueous solution ? (a) CH2ClCH2CH2COOH 185. Which dicarboxylic acid gives AcOH on strong (b) CH3CHClCH2COOH heating ? (c) CH3CH2CHClCOOH (d) CH3CH2CCl2COOH. (a) Ethanedioic acid Amides contain C = O group, yet they do not (b) Malonic acid give characteristic reactions of C = O group 177. (c) Glutaric acid because (d) Butane-1, 4-dioic acid. (a) they dimerise 186. Silver salt of which of following acid on treatment with bromine will produce (b) of resonance 2-bromopropane (c) they possess cyclic structures (a) propanoic acid (b) butyric acid (d) of attached alkyl group. (c) isobutyric acid (d) glutaric acid.

ALDEHYDE, KETONES AND CARBOXYLIC ACIDS 137 187. Which of the following acid undergoes In this sequence, the gaseous product Y is decarboxylation most easily ? (a) Ethyne (b) Ethane (a) C6H5COCH2COOH (b) C6H5CO COOH (c) Ethene (d) Propane. 195. The IUPAC name of (c) C6H5 CH COOH (d) C6H5 CH – COOH O ⏐ ⏐ ⏐⏐ OH NH2 CH3O—C — 188. The hydrolysis of which of the following takes —COOH is the longest time ? (a) CH3COCl (b) CH3COOC2H5 (a) 4-Methoxybenzoic acid (c) CH3COOCOCH3 (d) CH3CONH2. (b) 4-Carboxy- methylbenzoate PCl5 C2H6 (c) 4-Methoxycarbonyl-benzoic acid 189. Propionic acid ⎯⎯⎯⎯→ X ⎯⎯⎯⎯→ Y (d) None of these. AlCl3 CH3COOH (excess) Distill In the above sequence Y is 196. C2H2 A ⎯⎯⎯⎯→ B + C. ⎯⎯⎯⎯⎯⎯⎯⎯→ (a) Phenyl ethyl ketone (b) Phenyl acetate HgSO4 (c) Ethyl benzene (d) Ethyl benzoate. In the above sequence B and C are 190. —CH3 X —COOH (a) Ethanal, Ethanol Y (b) Ethanal, Ethanoic anhydride (c) Acetic anhydride, ethanol In the above sequence of reaction X and Y are (d) Acetic acid. Which of following acid can show optical respectively 197. isomerism ? (a) H2/Pt, Br2 (b) KMnO4, H2/Pt (a) 2, 2-Dimethylpropanoic acid (b) 2-Methylpropanoic acid (c) KMnO4 (aq), HI/P (c) 2-Methylbutanoic acid (d) Ethanoic acid. (d) NH2 – NH2/KOH, HI/P. Mg CH3COCl 191. CH3Cl ⎯⎯⎯⎯→ X ⎯⎯⎯⎯→ ...... Ether Heat The end product in the above reaction will be 198. X ←⎯⎯⎯⎯⎯⎯ Malonic acid (a) Acetone P2O5 (b) Ethanol Heat (c) Tertiary butyl alcohol ⎯⎯⎯⎯⎯⎯→ Y. (d) Isopropyl alcohol. 423 K (i) CO2 X and Y in the above sequence are respectively 192. C2H5MgBr ⎯⎯⎯⎯→ CnH2n+1COOH. (a) C3O2, CH3COOH (b) CO2, CH3COOH (ii) H2O (c) CO2, CO2 (d) CH3CHO, CH3COOH. The value of n is A mixture of colourless organic compound and (a) 1 (b) 2 NaNO2 produces brisk effervescence when dilute HCl is added to it. The organic compound (c) 3 (d) 0. could be HCN 199. 193. Acetaldehyde ⎯⎯→ X ⎯⎯⎯⎯→ Y Heat Z. ⎯⎯⎯→ H2O / H + In the above sequence, the end product Z is (a) Glucose (b) Urea (a) Prop-2-enoic acid (b) But-2-enoic acid (c) Benzoic acid (d) Ethanol. (c) Lactic acid (d) Tartaric acid. 200. Which of following anion is a strongest base ? NaOH / I2 (a) C6H5COO– (b) HCOO– 194. CH3COC2H5 ⎯⎯⎯⎯→ (c) CH3COO– (d) (CH3)2CHCOO–. X Yellow ppt. Ag ⎯⎯⎯→ Y(g).

138 CHEMISTRY NaOH H3O+ H ⎯⎯⎯→ 201. X ⎯⎯⎯⎯→ Y + CHCl3, Y ethanoic OOO acid. 208. H H + Conc. H2SO4 ⎯→ ? O O (Fuming) Here, X is (a) Cl3CCOCH3 (b) Cl3CCOC2H5 COOH (c) Cl2CHCOCH3 (d) Cl2CHCOC2H5. 202. Sodium phthalamate The main product in the above reaction is (a) Acetone Br2 / NaOH H3O+ (b) Acetone dicarboxylic acid ⎯⎯⎯⎯⎯→ X ⎯⎯⎯⎯→ Y. Here, Y is (c) Dihydroxy acetone (a) Anthranilic acid (d) Citranoic acid. (b) Phthalic acid 209. Anhyd. FeCl3 (c) terephthalic acid CH3COOH + Cl2 ⎯⎯⎯⎯⎯⎯→ ...... In this (d) o-phenylene diamine. reaction acetic acid gets converted into 203. The refluxing of (CH3)2 NCOCH3 with acid gives (a) CH3COCl (b) CHCl3 204. (a) 2CH3NH2 + CH3COOH (c) Cl3CH(OH)2 (d) Cl3CCOOH. (b) 2CH3OH + CH3CONH2 (c) (CH3)2NH + CH3COOH 210. What is formed as a product in the following (d) (CH3)2NCOOH + CH4. The principal product formed in the reaction reaction ? Rh Methanol + CO ⎯⎯⎯→ ...... (a) Ethanol (b) Ethanoic acid Peroxide (c) Methylmethanoate (d) Methanal. CH2 = CH(CH2)8COOH + HBr ⎯⎯⎯⎯→ is Heat (a) CH3CH(Br)(CH2)8COOH 211. H2NCONH2 ⎯⎯⎯⎯→ X. In the given reaction ‘X’ is (b) CH2 = CH(CH2)8COBr (a) NH3 + CO2 (b) urotropine (c) CH2Br(CH2)9COOH (d) CH2 = CH(CH2)7CH(Br)COOH. OO ⏐⏐ ⏐⏐ NH3 Y (c) H2N— C —NH—C —NH2 C6H5COCl ⎯⎯⎯→ X ⎯⎯⎯→ C6H5CN. In 205. (d) urethane. this sequence Y is (a) KCN (b) Ba(CN)2 212. Which of the following acid does not contain carboxylic group ? (c) P2O5 (d) P4. 206. Which of the following will liberate ethane as (a) Styphinic acid (b) Oxalic acid one of the product ? (c) Pivalic acid (d) Salicylic acid. (a) Soda lime decarboxylation of sodium 213. What is formed when ethanoic acid and methanoate hydrazoic acid are allowed to react in the (b) Electrolysis of aqueous solution of sodium presence of H2SO4 ? ethanoate (a) Methanamine (b) Ethanamide (c) Reduction of ethanoic acid with HI/P (c) Ethanamine (d) Oxamide. (d) Treatment of ethanoic acid with H2/Pd. X 207. The increasing order of acid strength is correct 214. Lactic acid ⎯⎯⎯→ Lactile chloride. in The reagent X can be (a) Phenol, Acetic acid, Ethanol, Chloroacetic (a) PCl5 (b) Cl2 acid (c) NaCl (d) ICl. (b) Acetic acid, Ethanol, Phenol, Chloroacetic 215. CS2 acid Silver acetate + I2 ⎯⎯⎯→ ...... The main (c) Ethanol, Phenol, Acetic acid, Chloroacetic product formed in the reaction is acid (a) CH3I (b) CHI3 (d) Chloroacetic acid, Acetic acid, Ethanol, (c) CH3COI (d) CH3COOCH3. Phenol.

ALDEHYDE, KETONES AND CARBOXYLIC ACIDS 139 216. In general, the order of reactivity towards aryl 223. —OH group in ethanol is almost neutral but 217. substitution among acid derivatives is the same is acidic in AcOH because (a) RCOCl > RCONH2 > RCOOR′ > (RCO)2O (a) in ethanol —OH is linked to electron (b) RCOCl > (RCO)2O > RCOOR′ > RCONH2 releasing C2H5 —group (c) (RCO)2O > RCOCl > RCONH2 > RCOOR′ (d) RCOOR′ > RCOCl > (RCO)2O > RCONH2. (b) in AcOH, C = O group causes electron Given three acids withdrawing effect CH3— —COOH (X), —COOH (Y) and (c) Resonance cause the development of partial positive charge on O atom of O—H group in AcOH. (d) All the above reasons are correct. 224. Which of the following is p-anisic acid ? CH3 CH3 (a) CH3O— —COOH H3C— —COOH (Z). CH3 (b) CH3— —COOH The correct order of ease of acid catalysed (c) NH2— —COOH esterification is (a) X > Z > Y (b) X > Y > Z (d) O2N— —COOH. (c) Z > X > Y (d) Y > Z > X. 218. Among α-nitro acetic acid (1), α-fluoroacetic 225. X ¾A¾lk¾ali¾ne® Y ¾H¾Cl¾(d¾il.)® Z acid (2) α-bromoacetic acid (3), α-cyanoacetic acid (4), the correct order of increasing acid Hydrolysis strength is Acetonitrile H3O+ (a) 3 < 2 < 4 < 1 (b) 1 < 2 < 3 < 4 In the above sequence X is (c) 2 < 1 < 3 < 4 (d) 4 < 1 < 2 < 3. (a) ClCH2CH2Cl (b) CH3CHCl2 219. Certain ester ‘A’ is boiled with KOH. The (c) CH2Cl . CHCl2 (d) CH3CCl3. product is cooled and acidified with conc. HCl to get white crystalline precipitate. The ester 226. H2 /Pt is likely to be Propanamide ⎯⎯⎯⎯⎯→ X (a) Ethylacetate (b) Methylbenzoate ⎯⎯⎯⎯⎯⎯⎯⎯→ Y (c) Isopropyl acetate (d) Methyl formate. Br2 /KOH In the above sequence of reactions 220. What is not true about cinnamic acid ? (a) X and Y are isomers (a) It produces brisk effervescence with (b) X is 1° amine and Y is 2° amine NaHCO3 (c) X and Y are homologous (b) It contains —COOH and —OH groups (d) Y is bromo derivative of X. (c) It decolourises bromine solution 227. What is Z in the following sequence of reactions? (d) It decolourises alkaline KMnO4. Zn CH3Cl Alkaline 221. Rate of esterification is highest for the acid Phenol ⎯⎯⎯⎯→ X ⎯⎯⎯⎯→ Y ⎯⎯⎯⎯→ Z dust AlCl3 KMnO4 (a) CH3(CH2)2COOH (b) (CH3)2CHCOOH (a) Toluene (b) Benzene (c) CH3COOH (d) Cl(CH2)4COOH. (c) Benzaldehyde (d) Benzoic acid. 222. 0.122 g of which of the following acid will 228. Which of the following order is incorrect ? require 20 ml of M/20 KOH solution for (a) Formic acid > Acetic acid > Propionic acid neutralisation. (a) C6H5COOH (b) CH3COOH (ACID STRENGTH) (c) ClCH2COOH (d) (CH3)3CHCOOH. (b) Cyclohexanol > Phenol > Benzoic acid (ACID STRENGTH)

140 CHEMISTRY (c) Benzamide < Aniline < Cyclohexylamine (BASIC STRENGTH) (a) = O (b) O (d) FCH2COOH > ClCH2COOH > BrCH2COOH O (ACID STRENGTH) CH2—CH2COOH 229. Identify Z in the following sequence of reactions (d) O 230. (c) Δ P2O5 H2O / H + CH2—CH2COOH ⎯⎯⎯→ ⎯⎯⎯→ ⎯⎯⎯→ CH3COONH4 X Y Z O (a) CH3CH2CONH2 (b) CH3CN 235. The end product C in the following reaction is (c) (CH3CO)2O (d) CH3COOH. PCl5 An ester (A) with molecular formula C9H10O2 CH3OH ⎯⎯⎯⎯→ A was treated with excess of CH3MgBr and the compound so formed was treated with conc. KCN H3O+ ⎯⎯⎯⎯→ B ⎯⎯⎯⎯→ C. H2SO4 to form olefin (B). Ozonolysis of B gave (a) Methanoic acid (b) Ethanoic acid ketone with formula C8H8O which shows iodoform test positive. The structure of A is (c) Methanamide (d) Ethanamide. 236. Formic acid reduces Tollen’s reagent because (a) C6H5COOC2H5 (a) It is strongest carboxylic acid (b) CH3OCH2COC6H5 (b) It does carry CHO group in its molecule (c) CH3CO—C6H5—COCH3 (c) Oxidation number of C in it is + 2 (d) C6H5COOC6H5. (d) It is weakest carboxylic acid. 231. Alc. NaCN 237. CH3COCl X Y C2H5OH Acetal. BrCH2CH2CH2Br ⎯⎯⎯⎯⎯→ X ⎯⎯→ ⎯⎯⎯⎯→ H3O+ H2SO4 ⎯⎯⎯⎯→ Y. In this sequence X is The product Y in the above sequence is (a) H2/Pt (b) H2/Pd – BaSO4 (c) Cr2O72–/H+ (d) Zn-Hg/HCl. (a) Glutaric acid COOCH3 (b) 1, 3-propanedioic acid 238. COOH COOH (c) Succinic acid H——OH H——OH H——OH H——OH H——OH H——OH (d) Melonic acid. 232. In order to convert C6H5CONH2 to COOH COOCH3 COOCH3 C6H5—CH3, which of the following sequence of steps is correct ? (A) (B) (C) What is correct about A, B, C ? (a) Dil. NaOH ; CaO/NaOH ; CH3Cl/AlCl3 (a) A and B are diastereomers (b) NaNO3/HCl ; NH3 ; NaOH (b) A and B are identical (c) KOH/Br2 ; K2Cr2O7/H+ ; CH3Cl/AlCl3 (c) A and B are enantiomers (d) CH3Cl/AlCl3 ; CaO/NaOH ; Dil. NaOH. (d) A and C are enantiomers. X2 /OH− 233. 239. CH3COCH2CH2COOC2H5 NaBH4 X. R—COCH3 ⎯⎯⎯⎯→ CHX3 + Carboxylate ⎯⎯⎯⎯→ H+ The product X is ion ⎯⎯⎯⎯→ Carboxylic Acid (a) CH3CH2CH2CH2COOC2H5 In the above sequence, the carboxylic acid (b) CH3COCH2CH2CH2OH + C2H5OH obtained is (c) CH3 CH – CH2CH2 CH2OH + C2H5OH (a) CH3COOH (b) HCOOH ⏐ OH (c) RCOOH (d) RCH2COOH. (d) CH3 CH CH2 CH2 COOC2H5AN 234. When adipic acid is heated strongly what is ⏐ formed ? OH

ALDEHYDE, KETONES AND CARBOXYLIC ACIDS 141 240. Which of the following compounds does not 249. Which of the following gives aldol condensation? react with sodium bisulphite? (a) Propionaldehyde (b) Formaldehyde (a) Benzene carbaldehyde (c) Benzaldehyde (d) None of these (b) Hypnone 250. Acetaldehyde and acetone can be distinguished by (c) Acetone (d) Acetaldehyde (a) Bromoform test (b) Solubility in water 241. Aldehyde and ketone can be distinguished by (c) Tollen’s test (d) Molisch test (a) Benedicts solution (b) H2SO4 solution 251. Which of the following does not give brick red (c) NH3 precipitate with Fehling’s solution? KMnO4/OH– (d) 242. Grignard reagent does not react with which of (a) Acetaldehyde (b) Formalin the followings (c) D glucose (d) Acetone (a) Aldehyde (b) Ester 252. Cannizzaro’s reaction is not given by (c) Ether (d) Alcohol (a) Trimethyl acetaldehyde 243. Formaldehyde, on heating with potassium (b) Benzaldehyde hydroxides, gives (a) Methyl alcohol (b) Ethyl formate (c) Acetaldehyde (c) Methane (d) Acetylene (d) Formaldehyde 244. Reduction of Benzoyl chloride with Pd and 253. In Etard’s reaction toluene is oxidised to BaSO4 produces benzaldehyde using (a) Benzoyl cyanide (b) Benzaldehyde (a) H2O2 (b) Cl2 (c) Benzoic acid (d) None of these (c) Chromyl chloride (d) KMnO4 245. Aldehydes react with alcohols in the presence 254. Bimolecular reduction of acetone gives of dry HCl gas to form acetals. This reaction is an example of (a) Diacetone amine (b) Pinacol (a) nucleophilic substitution (c) Chloretone (d) Propane (b) nucleophilic addition 255. Which of the following cannot reduce Fehling solution? (c) electrophilic substitution (a) Formic acid (b) Acetic acid (d) electrophilic addition (c) Formaldehyde (d) Acetaldehyde 246. Pentan-2-one and pentan-3-one can be distinguished with the help of 256. Which will not give acetamide on reaction with ammonia? (a) HCN (b) I2 + NaOH (a) Acetic acid (b) Acetyl chloride (c) NaHSO3 (d) Both (b) and (c) 247. The reaction of phenyl acetate with AlCl3 as (c) Acetic anhydride (d) Methyl propionate catalyst to form o-and p-hydroxy acetophenone 257. Acid present in tomatoes is is called (a) Lactic acid (b) Oxalic acid (a) Beckmann’s rearrangement (c) Citric acid (d) Tartaric acid (b) Fries rearrangement 258. The key step in Cannizzaro’s reaction is the intermolecular shift of (c) Friedel’s Craft reaction (d) Reimer-Tiemann reaction (a) Proton (b) Hydride ion 248. Pyruvic acid is obtained by (c) Hydronium ion (d) Hydrogen bond (a) Dehydration of Lactic acid 259. Benzophenone does not react with (b) Acidic hydrolysis of acetaldehyde (a) MeMgBr (b) NaHSO3 cyanohydrin (c) NaOH (d) Na2CO3 (c) Acidic hydrolysis of formaldehyde cyanohydrin (d) Reaction of HCN with CH3CHO followed by treatment with NaOH/I2

142 CHEMISTRY CASE BASED QUESTIONS Case Study 1 (c) will not discharge the colour of Baeyer’s reagent Read the passage given below and answer the following questions: (d) is an unsaturated ketone. 5. Mark the correct statement 1. The starting material (A) in the reaction is (a) (a) compound A and B are geometrical isomers (b) (b) oxidation of C gives benzoic acid (c) (c) ozonolysis of compound A followed by (d) hydrolysis in the presence of zinc will produce 3-hydroxy propanal as one of the 2. The product (D) on warming with amm. AgNO3 products will produce (d) compound A is optically active in nature. (a) oxalic acid (b) malonic acid Case Study 2 Read the passage given below and answer the following questions: An organic monobasic acid (A) (C = 60.87%, H = 4.35%) is found to produce violet colour with neutral FeCl3 solution. When the organic acid is heated with a mix- ture of NaOH and CaO it produces another product (B) which again also gives violet colour with FeCl3. B when heated with zinc dust produces product (C) whose ozonolysis gives a product (D) which on warm- ing with conc. NaOH produces glycollic acid. 1. The monobasic acid A is (a) CH3CH2CH2CH2COOH COOH (b) OH (c) CO2H OH (c) glycollic acid (d) acetic acid. 3. The structure of (C) in the reaction (d) CH3 COOH (a) (b) CO2H CHO CH3 2 The product B in the reaction is (a) benzoic acid (b) phenol CH2CHO CH2COOH (c) acetic acid (d) catechol (c) (d) 3. The conversion of product D to glycollic acid is known as 4. The compound B given in the reaction (a) Aldol condensation (a) will produce white fume with PCl5 (b) Oppaenaner oxidation (b) will reduce amm. AgNO3 solution (c) M.P.V. reduction (d) Internal Cannizzaro’s reaction.

ALDEHYDE, KETONES AND CARBOXYLIC ACIDS 143 4. Which of the following isomers of the reactant 2. The compound R is A will produced red colour with neutral ferric (a) chloride? (b) (a) ortho isomer (b) meta isomer (c) para isomer (d) none of these. (c) (d) 5. The ozonolysis product of C will be CHO (b) CH2OH (a) CHO CHO COOH COOH (c) (d) CHO COOH Case Study 3 Read the passage given below and answer the following questions: Two aliphatic aldehydes P and Q react in the pres- ence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below. 3. The compound S is 1. The compounds P and Q respectively are (a) (a) (b) (b) (c) (c) (d) (d)

15. POLYMERS MULTIPLE CHOICE QUESTIONS Choose the correct answer from the alternatives 8. Which pair of polymers have similar properties? given in each of the following questions : (a) PAN, PTFE 1. Which among the following is a crossed linked (b) PCTFE, PTFE polymer ? (c) Nylon, PVC (a) Amylopectin (d) Bakelite, Alkyd resin. (b) Melamine formaldehyde resin 9. Which of the following is a natural polymer ? (c) Glycogens (a) Polythene (b) Polysaccharides (d) Polyesters. (c) Nylon (d) Terylene. 2. Which of the following is an elastomer ? 10. Which among the following is a synthetic polymer ? (a) Vulcanized rubber (b) Dacron (c) Polystyrene (d) Malamine. (a) Proteins 3. Bakelite is an example of (b) Polysaccharides (a) Elastomer (c) Natural rubber (b) Fibre (d) Phenol-formaldehyde resin. (c) Thermoplastic 11. Which among the following is a linear polymer? (d) Thermosetting polymer. (a) High density polythene 4. Dacron is an example of (b) Low density polythene (a) Elastomer (c) Bakelite (b) Fibre (d) Vulcanized rubber. (c) Thermoplastic 12. Which among the following is a branched chain polymer ? (d) Thermosetting polymer. 5. Nylon-66 is obtained by condensation polymeri- (a) Nylons (b) Polyesters zation of (c) Glycogens (d) Bakelite. (a) Adipic acid and hexamethylene diamine (b) Phenol and formaldehyde 13. Which of the following type of forces are present in Nylon-66 ? (c) Terephthalic acid and ethylene glycol (a) Van der Waal’s forces of attraction (d) Sebacic acid and hexamethylene diamine. (b) Hydrogen bonding 6. The monomer of PMMA is (c) Three dimensional network of bonds (a) Methyl methacrylate (d) None of these. (b) Ethyl acrylate 14. Which among the following is a chain-growth (c) Acrylonitrile polymer ? (d) Methyl acrylate. (a) Nylon (b) Bakelite 7. The monomer of Acrilan is (c) Terylene (d) Teflon. (a) Methyl methacrylate 15. Which among the following is a step-growth (b) Ethyl acrylate polymer ? (c) Acrylonitrile (d) Methyl acrylate. (a) PTFE (b) PVC (c) Polyester (d) Polythene. 169

170 CHEMISTRY 16. Polyacrylates contain the linkages of 28. Caprolactam is obtained from (a) Amides (b) Esters (a) Cyclohexane (c) Alcohols (d) Nitriles. (b) Hexane 17. The synthetic polymer which resemble natural (c) Adipic acid rubber is (d) Adipic acid and hexamethylene diamine. (a) Neoprene (b) Buna-S 29. Which of the following is not a polyamide polymer ? (c) Nylon (d) Rayon. (a) Nylon-66 (b) Nylon-6 18. Terylene is produced by reaction of ethylene glycol and (c) Nylon-610 (d) Glyptal. (a) Phthalic acid (b) Terephthalic acid 30. Which of the following is not an addition polymer ? (c) Metaphthalic acid (d) All of the above. (a) Polyethylene (b) Polypropylene 19. Glyptal is a polymer of ethylene glycol and (c) Polystyrene (d) Dacron. (a) Phthalic acid (b) Terephthalic acid 31. The monomer of neoprene is (c) Metaphthalic acid (d) Adipic acid. (a) Butadiene 20. Which of the following is thermoplastic ? (b) Chloroprene (a) Dacron (b) Nylon (c) Isoprene (c) Polythene (d) All of the above. (d) 2-methyl 1, 3-butadiene. 21. The process of vulcanization of rubber makes it 32. The monomer of natural rubber is (a) Hard (b) Soft (a) Butadiene (c) Less elastic (d) None of these. (b) Chloroprene 22. Polystyrene is an example of (c) Isoprene (a) Elastomer (d) Butadiene and Styrene. (b) Fibre (c) Thermoplastic 33. For natural polymers PDi is generally (d) Thermosetting polymer. 23. Monomer of PTFE is (a) 0 (b) 1 (c) 100 (d) 1000. 34. Buna-S is (a) Ethylene (a) Natural rubber (b) Sulphur rubber (b) Propylene (c) Butadiene (c) Synthetic rubber (d) None of these. (d) Tetra fluoroethylene. 24. Which of the following type of forces are present 35. Which is not a polyacrylate ? in vulcanized rubber ? (a) PMMA (b) Acrilan (c) Poly acrylonitrile (d) PCTFE 36. Which of the following is a polyester ? (a) Weakest intermolecular forces (a) Nylon-66 (b) Hydrogen bonding (c) Three dimensional network of bonds (b) Nylon-610 (d) None of these. 25. Which of the following is polycarbonate ? (c) Alkyd resin (d) Phenol-formaldehyde resin. 37. The alternative name of Glyptal is (a) Acrilan (b) Lexan (a) Alkyd resin (c) PAN (d) Buna-S. (b) Phenol-formaldehyde resin 26. The starting materials for polyurethanes are (c) Melamine-formaldehyde resin toluene-2, 4 dicyanate and (d) Melmac. (a) Ethylene glycol (b) Phenol 38. The alternative name of PMMA is (c) Cresol (d) None of these. (a) Alkyd resin 27. The monomer of polyacrylonitrile is (b) Phenol-formaldehyde resin (a) Vinyl cyanide (b) Vinyl chloride (c) Plexiglass (c) Vinyl alcohol (d) None of these. (d) None of these.

POLYMERS 171 39. Which of the following is intermediate in (c) Sub-macromolecules preparation of Nylon-6 ? (d) Sub-micromolecules. (a) Cyclohexane 48. Which of the following has an ester linkage ? (b) Cyclohexanone oxime (a) Nylon-66 (b) Decron (c) Caprolactum (c) Nylon-610 (d) Bakelite. (d) Adipic acid and hexamethylene diamine. 49. Isoprene is a monomer of 40. The monomer of PVC is (a) Natural rubber (b) Synthetic rubber (a) Ethene (c) Salol (d) Saran. (b) Chloroethene 50. The natural occurring polymer among the following is (c) Dichloroethene (d) Tetra chloroethene. (a) Rayon (b) Nylon 41. The S in Buna-S refers to (c) Decron (d) Proteins. (a) Sulphur (b) Styrene 51. Which of the following polymers, has formaldehyde as one of the starting material ? (c) Sodium (d) None of these. (a) Melmac (b) Alkyd resin 42. The repeating units of Acrilan are (c) PAN (d) Acrilon H 52 Which of the following have 1, 2-ethanediol as starting material ? ⏐ (a) H2C== C —COOCH3 (a) Nylon 6 (b) PMMA H (c) PTFE (d) Dacron. ⏐ 53. Which names are associated with PMMA ? (b) H2C== C —COOC2H5 (a) Mdmac (b) Acrilan H (c) Saran (d) Acrilite. ⏐ (c) H2C== C —CN 54. Which of the following is/are thermoplastic polymers ? CH3 (a) Nylon 6, 6 (b) Bakelite ⏐ (d) H2C== C—COOCH3 . (c) Polystyrene (d) Terylene. 43. Which of the following is a polymer containing 55. Which of the following is used in the nitrogen ? preparation of nylon ? (a) Adipic acid (b) Butadiene (a) Nylon (b) Polyester (c) Isoprene (d) Ethylene. (c) Polydiene (d) Polyolefin. 56. The catalyst used in the manufacture of polyethene by zeigler method is 44. Alkyd resins are polymers obtained by reaction of glycol or glycerol with (a) Lithium tetrachloride and triphenyl aluminium (a) Malonic acid (b) Maleic acid (c) Phthalic acid (d) Acetic acid. (b) Titanium tetrachloride and trimethyl aluminium 45. Teflon is an example of polymer which is a/an (a) polyamide (c) Titanium oxide (b) addition polymer (d) Titanium isoperoxide. (c) polyester 57. Synthetic rubber is (d) formaldehyde resin. (a) Polyester (b) Polyamide 46. Perlon is another name of (c) Polysaccharide (d) Polyhalodiene. (a) Nylon-66 (b) Nylon-6 58. Which of the following is a polyamide ? (c) Nylon-610 (d) Nylon-44. 47. Polymers can also be called (a) Nylon (b) Orlon (a) Macromolecules (c) Teflon (d) Terelene. (b) Micromolecules

172 CHEMISTRY 59. Synthetic polymer prepared from caprolactum 71. Cellulose is a natural polymer of is called (a) α-D glucose (b) β-D glucopyranose (a) Nylon 6, 10 (b) Teflon (c) α-D fructofuranose (d) Maltose. (c) Terelene (d) Nylon 6. 72. Which of the following element is present in natural silk but not in artificial silk ? 60. Hair wigs are made from a copolymer of vinyl chloride and acrylonitrite, and is called (a) Sulphur (b) Chlorine (a) PVC (b) Dynel (c) Nitrogen (d) Hydrogen. (c) Polyacrylonitrile (d) Celldilose. 73. Which of following rubber is not a polydiene ? 61. Which of the following is not an example of (a) Polyisoprene (b) Polychloroprene addition polymer ? (c) Thiokol rubber (d) Nitrile rubber. (a) Polyethene (b) Polystyrene 74. Which of the following is not a cellulose product? (c) Neoprene (d) Terylene. (a) Gun cotton (b) Rayon 62. Terylene is a condensation polymer of ethylene (c) Dacron (d) Celluloid. glycol and 75. Which of the following polymer has a monomer (a) Benzoic acid (b) Phthalic acid unit named prop-2-ene-1-nitrile. (c) Salicylic acid (d) Terephthalic acid. (a) Orlon (b) Saran 63. Teflon, styron and neoprene are all (c) Dacron (d) Tetron. (a) Co-polymers 76. Which of the following is not a biopolymer ? (b) Condensation polymer (a) Starch (b) Proteins (c) Homopolymers (c) Nucleic acids (d) Sucrose. (d) Monomers. 77. Which of the following is a branched chain polymer ? 64. Soft drink bottles and baby feeders are generally made of (a) Polyester (b) Nylon (a) Polyesters (b) Polyurethane (c) PVC (d) LDP. (c) Polyurea (d) Polystyrene. 78. Which of the following is non-biodegradable ? 65. Polymer used in bullet proof glass is (a) Soap (b) Paper (a) PMMA (b) Lexane (c) Polyethene (d) Wood. (c) Nomex (d) Kevlar. 79. Nomex is 66. Which one among the following is not a correct (a) Polyester (b) Polyenes match ? (c) Polydiene (d) Polyamide. (a) Silk (Polyamide) (b) Lipase (Ester) 80. alyptal is formed by reacting glycerol with (c) Indigo (Azo dye) (d) Karatin (Protein). (a) Malonic acid (b) Pthalic acid 67. Which of the following can absorb over 90% of (c) Maleic acid (d) Acetic acid. its own mass of water and does not stick to wound ? 81. Butyl rubber is a polymer of (a) Rayon (b) Gun cotton O OH (c) Cotton (d) Thiokol. (a) HO 68. A co-polymer of isobutylene and isoprene is O called and (a) Buna-S (b) Butyl rubber NH2 O (c) Thiokol (d) None of these. H2N O 69. A co-polymer of CH2==CHCl and CH2==CH(OCOCH3) is called (b) HO OH and HO OH (a) Saran (b) Orlon (c) Dynal (d) Vinyon. (c) and CN 70. Which one of the following is a protein fibre ? O CH3 (d) and O (a) Polyester (b) Cotton (c) Silk (d) Rayon.

POLYMERS 173 82. The weakest interparticle forces are present in O (a) thermosetting polymers (b) thermoplastic polymers (b) NH2 and H2N NH2 (c) fibres (d) elastomers O 83. Which of the following is an example of (c) OH and H H co-polymer? (a) Buna-N (b) PAN (c) Polythene (d) PTFE Cl 84. Poly acrylonitrile is an example of H (a) additional polymer (d) OH and Cl Cl (b) condensation polymer 91. Which is an example of thermosetting polymer? (c) natural polymer (a) Polythene (b) Neoprene (d) biodegradable polymer (c) PVC (d) Bakelite 85. Which of the following polymer of glucose is 92. One of these polymers consists of coil-like stored by animals? polymer chains: (a) Cellulose (b) Amylose (a) Thermoplasts (b) Thermosets (c) Amylopectin (d) Glycogen (c) Elastomers (d) All polymers 86. Cellulose is a condensation polymer of 93. Which of the following statements is not true about low density polythene? (a) Maltose (b) β-Glucose (c) α-Glucose (d) β-fructose (a) It is obtained through free radical addition 87. Natural silk is a (b) It is hard and less flexible (a) Polypeptide (b) Polyacrylate (c) It is poor conductor of electricity (c) Polyester (d) Polysaccharide (d) It has a highly branched structure 88. Which of the following is not a semisynthetic 94. Following is the unique to polymeric materials: polymer? (a) Elasticity (a) cis-polyisoprene (b) Viscoelasticity (b) Cellulose nitrate (c) Cellulose acetate (c) Plasticity (d) Vulcanized rubber (d) None 95. Elastic deformation in polymers is due to ...... . 89. One of characteristic properties of polymer (a) Slight adjust of molecular chains material is (b) Slippage of molecular chains (a) High temperature stability (c) Straightening of molecular chains (b) High mechanical strength (d) Severe of covalent bonds (c) High elongation 96. Which of the following polymer can be formed by using the following monomer unit (d) Low hardness 90. Bakelite is a product of the reaction between O O H2N H? (a) H OH (a) Nylon 6, 6 O H and HO (b) Nylon 2, 6 (c) Nylon 6, 10 (d) Nylon

174 CHEMISTRY CASE BASED QUESTIONS Case Study 1 2. The monomeric unit of the polymer is Read the passage given below and answer the following questions: (a) N (b) N N A polymer is a compound of high molecular mass N formed by the combination of large number of monomeric units. Polymers are classified into different (c) (d) types. The process of polymerisation occurs through either addition polymerisation or condensation 3. The commercial name of polymer is polymerisation. Daily life without the varied applications would not be easier and colourful. (a) Plexiglass (b) Orlon A certain monomeric unit of C, H and N (c) Lucite (d) Vinyon polymerises to produce a polymer. 4. On heating the monomeric unit of the above NNN polymer with buta-1,3-diene it results in the formation of (a) Saran (b) SAN (c) PEA (d) ABS n 5. The polydispersity index (PDI) of the given polymer is 1. The above polymer is a (a) zero (b) equal to 1 (a) homopolymer (b) copolymer (c) > 1 (d) < 1 (c) natural polymer (d) condensation polymer.

ANSWERS 1. THE SOLID STATE MULTIPLE CHOICE QUESTIONS 1. (b) 2. (b) 3. (a) 4. (d) 5. (d) 6. (d) 7. (a) 8. (b) 12. (b) 13. (a) 14. (c) 15. (d) 16. (c) 9. (a) 10. (b) 11. (d) 20. (b) 21. (b) 22. (b) 23. (b) 24. (a) 28. (a) 29. (c) 30. (c) 31. (d) 32. (b) 17. (b) 18. (b) 19. (c) 36. (b) 37. (a) 38. (d) 39. (c) 40. (c) 44. (d) 45. (c) 46. (b) 47. (d) 48. (b) 25. (b) 26. (a) 27. (d) 52. (d) 53. (c) 54. (b) 55. (c) 56. (c) 60. (d) 61. (c) 62. (b) 63. (b) 64. (a) 33. (c) 34. (a) 35. (a) 68. (d) 69. (d) 70. (a) 71. (c) 72. (a) 76. (b) 77. (a) 78. (c) 79. (d) 80. (c) 41. (b) 42. (a) 43. (d) 84. (b) 85. (d) 86. (c) 87. (a) 88. (b) 92. (c) 93. (c) 94. (d) 95. (a) 96. (d) 49. (b) 50. (b) 51. (a) 100. (c) 101. (d) 102. (a) 103. (c) 104. (a) 108. (d) 109. (d) 110. (b) 111. (b) 112. (b) 57. (a) 58. (b) 59. (c) 116. (b) 117. (c) 118. (b) 119. (b) 120. (b) 124. (d) 125. (c) 126. (c) 127. (a) 128. (b) 65. (a) 66. (c) 67. (a) 132. (b) 133. (b) 134. (b) 135. (c) 136. (d) 140. (c) 141. (a) 142. (a) 143. (a) 144. (a) 73. (c) 74. (b) 75. (c) 148. (d) 149. (b) 150. (a) 151. (b) 152. (b) 156. (a) 157. (c) 158. (b) 159. (d) 160. (c) 81. (c) 82. (a) 83. (c) 164. (b) 165. (c) 166. (b) 167. (c) 168. (b) 89. (b) 90. (d) 91. (b) 97. (c) 98. (c) 99. (b) 105. (a) 106. (b) 107. (d) 113. (b) 114. (a) 115. (b) 121. (b) 122. (b) 123. (d) 129. (d) 130. (a) 131. (d) 137. (d) 138. (d) 139. (d) 145. (b) 146. (d) 147. (d) 153. (a) 154. (c) 155. (a) 161. (d) 162. (b) 163. (a) 169. (c) CASE BASED QUESTIONS Case Study 1 1. (c) 2. (b) 3. (a) 4. (c) 5. (d) 4. (a) 5. (a) Case Study 2 1. (d) 2. (c) 3. (d) HINTS/SOLUTIONS 30. Icc is a molecular solid. 34. Glass being a pseudo solid shows isotropy. 31. In primitive cube R = a/2. 35. The diagram given is a CsCl type unit cell. B+ ions are present at the corners of cube. Hence 32. Electron trapped in place of anion vacancy con- it is simple cubic cell. stitutes F-centre. 36. The anion A– is surrounded by 8 cations. Thus 33. The defect caused by displacement of cation to the site is cubical void. some other interstial space is called Frenkel defect. 180

ANSWERS 181 37. Coordination number of A– as well as B+ is 8 59. In antifluorite structure oxide ions adopt ccp shown in diagram. arrangement. Thus, C.N. of each is 8. 38. Distance between A– and A– is same as that 60. Some anionic sites are occupied by an electron bet-ween B+ and B+ i.e. a. which constitute F-centres. 39. Shortest distance between A– and B+ is half of 61. In bcc unit cell, the number of atoms per unit the body diagonal and is equal to 3a cell is 8 + 1 = 2. 0.866 a. or 8 2 62. Among the given substances, MnO2 is a para- magnetic compound. 40. In cubic closed packed arrangement, there are one octahedral void and two tetrahedral voids 63. Statement (i) and (iv) are correct about the per sphere. magnetic properties. 41. Electron cannot be present at the lattice site 64. The number of anions = 8 = 1. in a pure crystal. 8 42. The electrical conductivity of semiconductors 65. SrF2 adopts fluorite type structure. is of the order of 10– 9 – 102 ohm– 1cm– 1. 66. MgFe2O4 is ferrite. 67. Spinel is MgAl2O4. 43. TiO2 is diamagnetic and not paramagnetic as 68. Fe3O4 is FeII Fe2III O4 and it adopts inverse it has no unpaired electrons. spinel structure. 44. fcc unit cell adopts hexagonal close packing. 69. Statements I and IV donot pertain to amor- Thus, a given lattice point has 12 nearest neighbours. phous solids. 70. In zinc blende structure, S2– ions adopt ccp zM 45. Mass per unit cell = NA ; Volume of unit arrangement but Zn2+ occupy 50% of tetrahedral holes. In Na2O, O2– ions adopt ccp cell = a3 ; Density = zM . structure while Na+ ions occupy all tetrahedral a3N A holes. 46. The shaded plane joins two diagonally opposite 71. Statements II and IV do not pertain to solid edges. Hence it is known as diagonal plane of state. symmetry. 72. Both the statements, I and IV are correct about 47. There are twelve edges. Hence, there are six CsCl crystal structure. diagonal planes. 73. Match Box has dimensions a ≠ b ≠ c ; α = β = γ 48. Line ac is body diagonal as well as three fold = 90 which refers to orthorhombic crystal axis of symmetry. system. 49. Since there are eight corners. Hence, there are 74. A pentavalent impurity in the crystal lattice four body diagonals. of germanium makes it n-type semiconductor. 50. It has only one centre of symmetry. 75. SO3, I2 and KCl are solid at 10°C. 76. Site Y is octahedral void. 51. Crystalline substance do not exhibit isotropy. 77. There are four X atoms and four Y atoms per 52. Among the given species Cr2O3 is unit cell. Hence, the formula units are Four. ferromagnetic substance. 78. Y being octahedral void has co-ordination 53. TiO is a diamagnetic substance. number = 6. 54. A binary compound AB with radius ratio 0.52 79. NaCl adopts the given type of structure. is likely to adopt rock salt structure. 8 55. The co-ordination number of each sphere in ccp 80. No. of B atoms in the unit cell = 8 = 1 arrangement is 12. 2 56. For a cation to occupy a cubic void the radius No. of A atoms in the unit cell = 2 = 1. ratio should be ≥ 0.732. 81. For cation to occupy cubical void. The radius ratio should fall in the range 0.732 – 1.00. 57. In BCC, only 68% of the available space is occupied as compared to 74% in HCP and CCP. 58. In HCP, with each sphere there is one octahedral void.

182 CHEMISTRY 82. The specific cleavage pattern shown by crystal 98. The no. of O2– ions = 8+6 =4 is because of the constituent particles are ar- 82 ranged in planes. Octahedral voids = 4 83. In a closed packed arrangement of atom the 2 8 ratio r(oct. void) = 0.414. Octahedral voids occupied by M = 4 × 3 = 3 r(atom) ∴ Formula is M8/3 O4 or M2/3 O1 or M2O3. 99. In rock salt structure the octahedral voids oc- 100. 84. In a rock-salt structure the CN of cation and 101. cupied are 100%. anion are respectively 6,6. 102. In antifluorite structure all the tetrahedral 85. In antifluorite structure, the number of for- 103. voids are occupied by cations. mula units per unit cell = 4. Hence statement 104. There are 12 edges of a cube and Na+ ions are d is false. present at the each edge centre. 105. In zinc blende structure S2– ions adopt ccp ar- 86. For N atoms having closed packed rangement and zinc ions occupy 50% of the tet- arrangement, there is 2N tetrahedral voids. rahedral voids. Thus, the correct answer is 1 : 2. ∴ Total no. of ZnS units per unit cell = 4. 87. For hcp structure, the CN is 12. In antifluorite structure tetrahedral voids are 88. It refers to polymorphism. occupied and not the octahedral voids. 89. The given dimensions refer to triclinic system. NaCl structure is fcc both with respect to Na+ and Cl– ions. Hence, interchanging the position 90. The occupied space in ccp arrangement is 74%. of these ions does not bring any structural change. 91. 12-16 compounds are binary compounds of the elements of group 12 and 16 respectively. 92. Substances with given characteristics are In diamond all the C atoms are tetrahedrally piezo-electrics. arranged because of sp3 hybrid state. Hence, diamond has ccp arrangement in which corners 93. Substance prepared by doping of Si with gal- and face centres of the unit cell are occupied lium is p-type semiconductor whereas that by C atoms. In addition to these, four of the doped with As is n-type semiconductor. eight tetrahedral voids are also occupied by C atoms. 94. In rock salt structure anions are at the corners of the cube whereas cations are at the edge cen- 1 tres. The edge length of such a unit cell is equal 106. Number of gold atoms = 8 × 8 = 1 to twice the sum of radius of cation and radius of anion. 1 Number of copper atoms = 6 × 2 = 3 Edge length = 2(r+ + r–) = 2(rNa+ + rCl–) Hence the empirical formula is AuCu3. = 2(95 + 181) = 552 pm. In face centred cubic unit cell the distance of 95. Since the radius ratio is in the range 107. closest approach between two particles is equal 0.414 – 0.732, the cation A+ would prefer to be in octahedral void and hence the compound AB to half the face diagonal. is likely to have rock salt structure. 1 Face diagonal = 2 . Edge length = 1.414a. 96. Number of atoms of X = 8 × 8 + 1 × 1 = 2 ∴ Distance of closest approach 1 Number of atoms of Y = 2 × 6 = 3 108. 1.414a Thus, the formula is X2Y3. = 2 = 0.707a. 97. If a is the edge length of a cube, then fraction In antifluorite structure fcc lattice sites are of space occupied by the spheres is given by occupied by the anions (Pt Cl6)2–. (Volume of sphere) × No. of sphere per unit cell 109. The number of anions = 8+6 = 4. Volume of unit cell 82 4 π(a/2)3 × 1 ∴ Octahedral voids = 4 ; = 3 a3 tetrahedral void = 8; = 0.5238 or 52.4%. Total voids = 12 ;

ANSWERS 183 Voids occupied by the ions = 3 121. Edge length = 2r+ + 2r– or 508 = r+ + r– 2 ∴ Fraction of voids occupied = 3 = 1 . 12 4 ∴ r– = 254 – 110 = 144 pm. 110. nλ = 2d sin θ or d = nλ = 1 × 0.134 122. Anions in unit cell = 8 + 6 = 4 ; 2 sin θ 2 × 0.1822 123. 8 2 GF JICations = 12 + 1 + 4 = 8. = 0.368 nm. H K4 Hence the ratio of cations : anions = 2 : 1. 111. In normal spinel structure A2+ ions occupy 1 th The fraction of total volume occupied in simple 8 Volume of particles 4/3 π(a / 2)3 π cube = Volume of cube = a3 = 6. of tetrahedral voids and B3+ ions occupy 1 i.e., 112. 2 124. 0.95 125. The radius ratio (r+/r–) = 1.81 = 0.54.Hence 50% of octahedral voids. coordination number of cation should be six. Thus, substance is likely to adopt rock salt d represents distance between consective structure. planes. Face diagonal of a cube 113. For fcc unit cell, r = a or a = 2 2 ×r 22 = 2 × 1.414 × 500 = 2 × a = 2 × 400 = 566 pm. = 1414 pm. λ = 2d sin θ = 2 × 2 × 0.2 × sin 8.4 126. Body diagonal = a 3 n1 114. or a = Body diagonal = 866 = 2 × 0.2 × 0.146 = 0.0584 nm. 33 115. Both NaCl and CsCl are likely to show Schottky 116. type defect. 127. = 499.98 ≈ 500 pm. 117. Statements I, II are correct about NaCl. 128. It belongs to a primitive unit cell. 118. AgBr and ZnS are likely to show Frenkel type defect. In fluorite structure Ca2+ ions adopt ccp ar- Applying the formula rangement while F– ions are in all tetrahedral holes. N = z × A × 100 = 4× 100 = 5 × 1024. 129. Number of P atoms per unit cell = 8+ 6 = 4. d × a3 × (2 × 10−8 8 2 10 ) Number of Q atoms 119. Density = Molar mass × 2 = 4(octahedral holes) NA × a3 + 4(half of tetrahedral holes) 100 × 2 The formula is P4Q8 or PQ2. = 6.02 × 1023 × (400 × 10−10 )3 130. In diamond all C atoms are sp3 hybridised. Dia- 200 mond has ccp arrangement in which corners = 38.528 and face centres are occupied by C atoms. In = 5.188 g cm– 3. addition to this, four of the eight tetrahedral voids are also occupied by C atoms. 8 131. The statement a, b, c are correct for 13-15 com- 120. W atoms per unit cell = 8 = 1 ; Na atoms per pounds but the statement d-does not refer to such compounds. unit cell = 1. 132. For ferromagnetic substances there is spon- 12 taneous alignment of magnetic momomenta in O atoms per unit cell = 4 = 3 same direction. Hence formula = NaWO3. 133. The correct matching among them is BaTiO3 which is ferroelectric.

184 CHEMISTRY 134. The radius ratio r+ = 146 = 0.675 141. In CCP arrangement, there would be 4 oxide r− 216 142. ions per unit cell. With these 4 oxide ions 8 143. tetrahedral voids and 4 octahedral voids would As it lies in the range 0.414 – 0.732. It exhibits be associated. One tetrahedral void (one out rock salt type structure. of eight) would be occupied by A while two octahedral voids (half of four) would be 135. 148 occupied by B. So, formula of the compound is 136. The ratio r+ /r– = 195 = 0.759. Since this value AB2O4. 137. is more than 0.732 it implies that Rb+ ion oc- In CCP, with each oxide ion there would be 2 cupies a cubical hole in the crystal lattice. In tetrahedral voids and one octahedral void. One- other words, the co-ordination number of Rb+ third octahedral voids are occupied by B while is 8. one-sixth tetrahedral voids are occupied by A. Thus formula of the compound may be derived Barium titanate is BaTiO3. Since Ti4+ is at the as: body centre, it represents octahedral hole. To- tal octahedral holes are four out of which only A1/3B2/6O or A2B2O6 or ABO3. one is occupied by Ti4+. This makes it 25% of The structure of AB2 refers to fluorite type the holes. structure. No. of atoms of B per cube 11 144. ABAB ...... pattern of close packing gives hex- = 8 × 8 (at corners) + 2 × 6 (at face centre) = 4 agonal unit cell. The number of particles per With each atom of B there would be 2 tetrahe- dral voids in the close packing and all of these unit cell is equal to 12 + 2 + 3 = 6. are occupied by A. 6 21 No. of atoms of A per cube = 4 × 2 = 8 145. In zinc blende structure, the anions adopt ccp arrangement the nearest anions touch along Thus, formula of the compound is A8B4 or A2B. face diagonal. Thus distance of closest 138. Edge length in rock salt structure approach is equal to 1 of face diagonal. = 2(rA+ + rB–) 2 520 = 2(80 + rB–) rB– = 520 − 160 = 180 pm. zM 4 × 74.5 2 NAa3 6.02 × 1023 × (745 × 10−10 )3 146. d= = zM = 1.197 g cm– 3. 139. Density = a3NA = 2 × 23 147. d= 119 × z or z = 2.75 × NA × a3 (424 × 10−10 )3 × 6.02 × 1023 NA × a3 119 = 1.00 g cm– 3. 140. Line passing through the centre of cube and = 2.75 × 6.02 × 1023 × (654 × 10−10 )3 centres of two opposite faces is four fold axis. 119 Since there are six faces. Hence, a cube has three four fold axis. Body diagonal is three fold = 3.9 ≈ 4. axis. Therefore the number of three fold axis in a cube is equal to four. 2 × 39 × 10−3 6.02 × 1023 × 4 × 4.52 × 10−3 FHG JKI148.d= 3 23 = 2 × 39 × 100 = 900 kg m3. 6.02 × 1.44 149. No. of atoms per unit cell = 2 ∴ No. of atoms in 12.08 × 1023 unit cells = 24.16 × 1023. Four fold axis Three fold axis

ANSWERS 185 150. Let the number of Fe atoms present in + 3 state 156. Distance between K+ and F– in be x. KF = r(K+) + r(F–) = 133 + 136 = 269 pm. ∴ Number of atoms present in + 2 state = (0.93 – x). 88 157. r+/r– = 200 = 0.44. Now total magnitude of + ve charge must be equal to total magnitude of – ve charge Since radius ratio lies between 0.414 – 0.732. Thus, x × 3 + (0.93 – x)2 = 2 This implies that X+ must occupy octahedral hole. Hence co-ordination number of X+ is six. or 3x – 2x + 1.86 = 2 or x = 0.14 158. Since Cu : O ratio is less. This implies that some Cu+ ions have been replaced by cu2+ ions. To % of Fe3+ = 0.14 × 100 = 15.05%. maintain electrical neutrality each cu2+ ion 0.93 takes the place for two cu+ ions, thereby creat- 6 ing a hole, which is responsible for its p-type 151. Number of A atoms per unit cell = 8 semiconductor nature. 6 159. d = zM/NAa3 Number of B atoms per unit cell = 2 = 3 Here, HFG KJIz = 4 × 1 − 0.2 The formula is A6/8B3 or A6B24 or AB4. 100 = 3.992 152. Nearest neighbours in bcc touch along body 3.992 × 40 diagonal, while next nearest neighbours are ∴ d = 6.02 × 1023 × (5.7 × 10−8)3 at the corners or edge or the unit cell. = 1.432 g cm–3. Now, body diagonal = 2 × 433 Length of the edge = Body diagonal = 2 × 433 160. 33 = 500 pm. 153. According to Bragg’s equation nλ = 2d sin θ Here, λ = d ; n = 1 ∴ 1 × λ = 2λ sin θ. or sin θ = 1 or θ = 30°. 161. n-type semiconductor is formed by doping of 2 element of group 14 with those of group 15. Here, dopant provides excess electron to conduction 154. Volume of the unit cell = 1.3 × 10–22 cm band hence choice ‘d’ is correct. Let number of H2O molecules = x 162. In sphalerite structure anions occupy the fcc Mass of unit cell = x + 18 g latticed points and cations occupy 50% of tetrahedral voids. Octahedral voids are not NA occupied at all. Density of crystal = M/V x + 18 163. No. of O atoms = NA × 1.3 × 10−22 per unit cell = 4 18x or NA × 1.3 × 10−22 = 0.92 No. of octahedral voids = 4 or x = 0.92 × 6.02 × 1023 × 1.3 × 10−22 = 3.99 ≈ 4. No. of octahedral voids occupied = 4 × 66% 18 155. The number of oxide ions in hexagonal unit = 4 × 2 ≈ 8 cell = 6 3 3 The formula is = Al8/3O4 ∴ Octahedral voids = 6 or Al2O3. 2 164. Number of (AlF6)–3 per unit cell Octahedral voids occupied by M = 6 × = 4 Octahedral voids = 4 3 Thus formula is M4O6 or M2O3. Tetrahedral voids = 8

186 CHEMISTRY Total Na+ ions = 12 168. There are fourteen faces in the truncated Formula is Na12 (AlF6)4 or Na3AlF6. octahedron, (8 hexagonal and 6 square faces). 165. The statements I and IV are based on facts. Total number of vertices are 24, and and 36 166. Refer to the crystal description of NaCl and ZnS. edges. 167. As shown in the construction of truncated octahedron, it has 8 hexagonal faces. 169. There is no charge on the semiconductor as such. It is neutral. 2. SOLUTIONS MULTIPLE CHOICE QUESTIONS 1. (a) 2. (b) 3. (c) 4. (a) 5. (d) 6. (b) 7. (a) 8. (c) 12. (b) 13. (d) 14. (b) 15. (b) 16. (c) 9. (c) 10. (a) 11. (b) 20. (b) 21. (b) 22. (c) 23. (b) 24. (a) 28. (b) 29. (c) 30. (c) 31. (c) 32. (a) 17. (a) 18. (b) 19. (d) 36. (c) 37. (c) 38. (a) 39. (c) 40. (b) 44. (a) 45. (b) 46. (a) 47. (c) 48. (c) 25. (a) 26. (b) 27. (b) 52. (a) 53. (a) 54. (c) 55. (c) 56. (a) 60. (a). 61. (c) 62. (b) 63. (b) 64. (b) 33. (a) 34. (b) 35. (c) 68. (b) 69. (b) 70. (b) 71. (d) 72. (c) 76. (d) 77. (a) 78. (a) 79. (d) 80. (d) 41. (c) 42. (d) 43. (a) 84. (b) 85. (d) 86. (c) 87. (c) 88. (a) 92. (a) 93. (c) 94. (b) 95. (d) 96. (a) 49. (c) 50. (b) 51. (a) 100. (b) 101. (a) 102. (d) 103. (d) 104. (b) 108. (c) 109. (c) 110. (d) 111. (b) 112. (c) 57. (d) 58. (a) 59. (a) 116. (b) 117. (b) 118. (b) 119. (a) 120. (c) 124. (b) 125. (c) 126. (c) 127. (a) 128. (b) 65. (d) 66. (d) 67. (b) 132. (b) 133. (d) 134. (c) 135. (c) 136. (b) 140. (a) 141. (a) 142. (d) 143. (a) 144. (d) 73. (a) 74. (b) 75. (a) 148. (a) 149. (c) 150. (b) 151. (d) 152. (c) 156. (b) 157. (c) 158. (c) 159. (a) 160. (b) 81. (b) 82. (a) 83. (b) 164. (b) 165. (a) 166. (d) 167. (c) 168. (a) 172. (d) 173. (b) 174. (c) 175. (b) 176. (b) 89. (b) 90. (d) 91. (a) 180. (a) 181. (b) 182. (c) 97. (a) 98. (a) 99. (d) 5. (d) 105. (b) 106. (b) 107. (a) 5. (c) 113. (c) 114. (a) 115. (b) 121. (c) 122. (d) 123. (d) 129. (b) 130. (a) 131. (c) 137. (d) 138. (c) 139. (c) 145. (b) 146. (a) 147. (b) 153. (a) 154. (c) 155. (b) 161. (b) 162. (d) 163. (b) 169. (a) 170. (b) 171. (d) 177. (c) 178. (c) 179. (b) CASE BASED QUESTIONS Case Study 1 1. (c) 2. (b) 3. (a) 4. (b) 4. (d) Case Study 2 1. (a) 2. (b) 3. (b)

ANSWERS 187 HINTS/SOLUTIONS 1. Moles = V(L) × molarity = 2.5 × 0.5 = 1.25. 21. Freezing point will be highest if ΔTf will be 2. A solution showing positive deviation has higher lowest. ΔTf will be lowest for C6H12O6 because other substances will undergo dissociation in vapour pressure and lower boiling point. aqueous solutions causing increase in the conc. of solute particles. 3. If vapour pressure is lowered, the freezing point will also get lowered. 22. Azeotropic solutions boil at constant temperature without undergoing any change in 4. Solubility of NaCl increases by rise in the composition. temperature. 23. π = CRT or π ∝ C. 5. Mass of solvent is not known, which is required to calculate mole fraction of solute/solvent. 24. The solutions with same molality will also have same mole fraction. However, their molarities, 6. Due to trimerisation the number of solute normalities, etc. may differ because of difference particle will be reduced to one-third. in densities. 7. Dissociation increases the number of solute 25. ΔTf = Kf m or Kf = ΔTf /m. particles. Hence, observed molecular mass 26. C6H5COOH, (benzoic acid dimerises in benzene). decreases and van’t Hoff factor becomes > 1. Hence it exhibits molecular mass corresponding 8. The conc. of urea and glucose will be same, KCl to (C6H5COOH)2 i.e., 244. and CaCl2 being electrolytes will cause a change 27. Elevation of boiling point in case of AlCl3 is in the concentration of solute particles. relatively larger than that of CaCl2. 28. both I and II curves represent plot of pressure 9. Concentration being same, the osmotic pressure vs composition for ideal binary solution. of two solutions will also be equal. 29. Pressure expected from Raoult’s law 10. Copper ferrocyanide can act as semipermeable membrane. = 80 × 0.4 + 120 × 0.6 = 32 + 72 = 104 mm 11. Concentration of Mg2+ Pressure observed = 100 mm = 0.01 × 106 = 10 ppm. Since observed pressure is less than the expected 1000 value from Raoult’s law, therefore, such a solution will show negative deviations from 12. 4.9 × 10−3 × 106 = 4.9 ppm. Raoult’s law. 103 30. Mixture of benzene and toluene forms ideal 13. 0.1 Ca(NO3)2 and 0.1 M Na2SO4 would produce solution. For such solution ΔmixG < 0. same conc. of ions, on dissociation, in solution. 31. Azeotropic solutions keeps on distilling at 14. ΔSmix. is always > 0. constant composition and cannot be separated 15. Non-electrolyte does not dissociate, but it can into constituents liquids by simple distillation. undergo association depending upon its nature. 32. Assuming complete dissociation of particles the solute particles concentration is as 16. Among the given properties, osmotic pressure is the only one which is colligative property. 0.001 M MgCl2 = 0.003 M 0.001 M NaCl = 0.002 M 17. Osmosis of water will occur into the egg shell increasing its size. 0.001 M urea = 0.001 M 18. π = CRT, for same value of C, π ∝ T. 0.01 M NaCl = 0.01 M 19. Deliquescence is a phenomenon by virtue of Thus correct increasing order of boiling points which a crystalline substance on standing in air is 0.001 M urea < 0.001 NaCl < 0.001 MgCl2 < forms a solution by absorbing water vapours 0.01 M NaCl. from the atmosphere. For deliquescence to occur the substance should be very soluble in water 33. This is as per the statement of Raoult’s law. and its saturated solution should have low vapour pressure as compared to the pressure of 34. Such a solution may exhibit positive or negative water vapours in air at particular temperature. deviations depending upon the nature of A and B. 20. The molar conc. of 10% aq. solution of glucose is lower than that of 10% aqueous solution of urea. Hence π1 < π2.

188 CHEMISTRY 35. In cyclohexanol-ethanol pair, the solute-solvent 54. Ca(NO3)2 ⎯→ Ca2+ + 2NO3–. interactions will be weaker than solute-solute 55. Depression in freezing point (ΔTf) would be and solvent-solvent interactions. maximum in case of 1% KCl as it contains 36. 0.1 M Al2(SO4)3 would produce maximum conc. maximum number of particles in a given volume of ions in solution. of solution. 37. The shrinking is due to osmosis of fluid from 56. ΔTf = Kf × m the egg to the concentrated sodium chloride 1.1 = 1.86 × 1.25 × 1000 solution. M × 20 38. On cooling the dilute solution, the solvent will M = 1.86 × 1.25 × 1000 = 105.7. start crystallising. Hence, ice will separate first. 1.1 × 20 39. ΔTb = i Kb × m = 2 × 0.53 × 1 = 1.06 57. Mole fraction of solute = x2 Freezing point = 0 – 1.06 = 1.06°C. (x1 + x2 ) 40. Solution B having greater vapour pressure p = p° × mole fraction of solute. would have lower concentration of solute. 58. ΔSmix is always positive. 41. 4.5 = x or x = 1.5%. 59. KH increases with increase in temperature. 180 60 60. Kb is independent of the nature of solute. 61. pA° xA + pB° xB = 550 mm 42. Molarity and normality depend upon volume of the solution. Volume of the solution changes 400  1 1 x  + 600  x x  = 550 or x = 3. with change in temperature.  +   1 +     43. Mango shrivels because it gains water by osmosis. 62. NaCl ionises as (NaCl ⎯→ Na+ + Cl–). Hence elevation in boiling point of solution is almost 44. Sodium chloride and sodium sulphate are ionic double the elevation for glucose solution. compounds. They dissociate in solution producing more number of particles. 63. Concentration of F 45. Solutions having same vapour pressure are = 0.2 × 106 = 4 × 102 ppm. called isopeistic solutions. 500 46. Solubility of X decreases with rise in 64. 0.8 mole are present in = 1000 ml temperature. This can happen only if ΔsolH < 0. 0.1 mole would be present in 47. 1% solution of NaCl has maximum concentration 1000 of particles in the solution. = × 0.1 = 125 ml. 48. Osmotic pressure, being a colligative property 0.8 increases with increase in number of solute 65. Suppose 1000 cm3 of solution molecules. 1000 × 15 49. Acetic acid dimerises in benzene. Mass of solute = 100 = 150 g Mass of solution = 1000 × 1.1 = 1100 g 50. Molarity depends on volume of the solution and Mass of solvent = 950 g volume of solution increases with increase in temperature and hence molarity decreases. Molality = 150 = 1000 = 1.6 mol kg–1. 950 98 51. Among the given solutions 0.1 M CaCl2 would contain maximum number of particles. CaCl2 66. In aqueous solution of KBr ion-dipole are the gives three mole ions per mole. major interactions between solute and solvent particles. 52. Since Kf is solvent dependent. Thus, ΔTf for solution of same solute in different solvents have 67. Total mole of KOH in 6 L differents values of ΔTf. = (4 × 0.2 + 2 × 0.5) = 1.8 53. Conc. of urea = 12 = 0.2 M 1.8 mol 60 ∴ molarity = 6L = 0.3 M. Conc. of sucrose = 68.4 = 0.2 M. 342

ANSWERS 189 68. 10 = 5 ∴ M = 10 × 180 = 360 g mol–1. WA = MA . pA = 18 × 5 = 190 or 1 M 180 5 WB MB . pB 137 × 1 137 1.52 . 69. xBenzene = 2 , xliquid = 3 80. Ba(NO3)2 dissociates into three particles and 5 5 has i = 3 while i = 1 for sugar. Hence their pBenzene = 2 × 266 = 106.4, pliquid = 3 × 236 = 141.6 colligative properties will be in the ratio of 3 : 1. 5 5 81. 3 = 2 p = 141.6 + 106.4 = 248 mm. 180 M 70. Calculate moles of each in 100 g of solution. M = 180 × 2 = 120. 3 66 6 Glucose = 180 ; Urea = 60 ; Sucrose = 342 Δp 4 1 Molar concentration of urea is maximum. Hence 82. pA = 4 + 55.5 = 59.5 = 0.067. urea solution will have highest value of ΔTf . 83. (NH4)2SO4 + Ba(OH)2 71. Mass of solution = 100 g (say) ⎯→ 2NH3(aq) + 2H2O + BaSO4. Volume of solution = 100 ml 25 × 1 + 75 × 1 0.6 2 5 = 0.275 M. 84. M(resulting) = 100 Mass of solute = 35 g ∴ Moles of solute = 35 = 1 85. Molarity of mixture of 6 M and 3 M HCl 35 = 350 × 6 + 650 × 3 = 4.05 mol L–1 Molarity = 1 × 1000 = 6 mol L–1. 1000 100 / 0.6 Now, apply dilution formula 1 × 10 M1V1 = M2V2 or V2 = 4.05 × 1000 = 1350 ml 72. 1 × 10 = V × 2.5 or V = 2.5 = 4 L 3 Water required = 4 – 1 = 3 L. Volume of water to be added 100 = 1350 – 1000 = 350 ml. 73. Volume of 100 g of solution = d ml 86. α= i −1 = i −1 ∴ M = 20 × d × 1000 1 −1 1 −1 100 × 98 m2 2.55 × 100 × 98 α or i= 1− α . or d = 20 × 1000 = 1.249 ≈ 1.25. =1–i 22 74. m = 34.2 × 103 = 0.55 m. 87. The relative lowering of vapour pressure will be 181.8 × 342 least for glucose. The other solutes being ionic 75. Mass = m × 100 × 58.5 = 0.1 × 58.5 will ionise to increase particle concentration in 1000 10 solution. = 0.585 g. 114.45 88. Van’t Hoff factor, i = 65.4 = 1.75 76. For ideal solutions of A and B, the plot of pTotal vs xA or xB is a straight line with slop ≠ 0. α= i − 1 = 1.75 − 1 = 0.75 m −1 1 77. Freezing point will become less than 273 K. 78. M1V1 = M2V2 ⇒ 10 × 10 = 0.1 × V2 or 0.75 × 100 = 75%. V2 = 1000 ml 89. πA = 0.01 × 0.082 × 300 = 0.2463 atm ; Volume of water required πB = 0.001 × 0.082 × 300 = 0.02463 atm = 1000 – 10 = 990 ml. The movement of solvent particles will occur 79. During distillation of immiscible liquids the from dilute solution (B) to A. Hence, the pressure mass of the component in the distillate is directly proportional to the product of its molar mass of 10.2463 – 0.02463 = 0.2216 atm. should be and its vapour pressure. Thus applied on concentrated solution i.e., solution A.

190 CHEMISTRY 90. 3 = 1 ∴ M = 60. Since, both the solutions have same molar 180 M concentration they would have nearly equal boiling points. 91. 0.6% (W/V) solution of urea 100. [NO3–] = [AgNO3] = 0.6 × 1000 = 0.1 M. 0.1 60 × 100 After mixing, [AgNO3] = 2 = 0.05. 101. K4[Fe(CN)6] and Al2(SO4)3 both give 5 mole ions Thus, its concentration is same as 1.8% glucose per mole of the compound. and 0.1 M sucrose. Like urea, glucose and 102. Mass of solute per 1000 g of solvent sucrose are also molecular solutes which do not undergo association or dissociation in solution = 6.8 × 1000 = 68 g and hence produce same number of particles in 100 the solution. Molality = 68 92. I. Normality may be equal to or greater than M molarity. 68 II. Normality of a solution decreases with ΔTf = Kf × m = 1.86 × M increase in temperature. 1.86 × 68 93. Mole fraction of solute = Δp = 0.2 0.93 = M p° M = 136. Mole fraction of solvent = 1 – 0.2 = 0.8 5.85 103. Moles of NaCl = 58.5 = 0.1 ; Moles of solvent = 0.8 × 5 = 4. Moles of water = 90 = 5 94. Mass of NaOH = 0.1 × 250 × 40 = 1 g. 18 1000 Mole fraction of NaCl = 0.1 = 0.0196. 5 + 0.1 95. Strength of solution is 10% or 100 g/litre. 104. Strength per litre = 50 g/litre The volume of solution containing 180 g of the solute Molarity, C = 50 342 1000 50 = 100 × 180 = 1800 ml or 1.8 L. 96. Eq. mass of the acid π = CRT = 342 × 0.082 × 423 = 5.07 atm. = Mol. mass = 200 = 100 Basicity 2 105. Strength per litre = 71 g/litre, mol. mass of Na2SO4 = 142 1000 ml of 0.1 N solution contains Molarity = 71 = 0.5 M. 142 = 100 or 10 g solute 10 106. 2NaOH + H2SO4 ⎯→ Na2SO4 + H2O M1V1 = n1 100 ml of 0.1 N solution contain M2V2 n2 = 10 × 1000 = 1 g. 0.2 × 40 = 2 1000 0.1 × V 1 97. Δp = xB V = 40 ml. pA ° 107. Mass of solvent = V × d = 100 × 1.58 = 158 g 15 = 5 × 18 5 × 18 × 3000 158 3000 M × 100 or M = 15 × 100 = 180. Moles of solvent = 154 ; 98. M1V1 + M2V2 = M3V3 0.5 × 0.5 + 0.1 × 1.5 = 0.2 × 2 0.25 + 0.15 = 0.4. 1 99. Moles of urea = 60 ; Moles of glucose = 3 = 1 180 60

ANSWERS 191 0.5 120. 3.01 × 1023 molecules = 3.01 × 1023 = 0.5 mole Moles of solute = 65 6.02 × 1023 Mole fraction of solute ≈ 0.5 / 65 ∴ ΔTf = 1.86 0.5 × 100 = 3.72 158 / 154 250 ≈ 7.5 × 10–3 ∴ Tf = 273 – 3.72 = 269.28 K. Δp = pA°xB 121. M1V1 + M2V2 + M3V3 = M4V4 pA = pA° – pA°xB = 143 – 143 × 7.5 × 10–3 0.2 × 50 + 0.1 × 50 + 0.05 × 100 = M4 × 200 = 143 – 1.07 = 141.93 mm. M4 = 20 = 0.1 M. 200 10 × d × P 108. Molarity = Mol. mass ∴ [H+] = [HCl] = 1 × 10–1 ∴ pH = – log 1 × 10–1 = 1. = 10 × 1.84 × 98 = 18.4 M. 122. M1V1 + M2V2 = M3V3 98 6 × 250 + 2 × 750 = 2.5 × V3 109. Molar concentration would be inversely 3000 proportional to the molecular mass. V3 = 2.5 = 1200 ml π = CRT Volume of water = 1200 – (250 + 750) Urea having highest conc. would have highest = 200 ml. osmotic pressure whereas sucrose would have least. 123. i = 58.5 ≈ 1.84 31.8 110. pA = pA°xA 1.84 − 1 α = 2 − 1 = 0.84 or 84%.  0.8  60 60 = pA °  0.8 + 0.2  = pA ° = 0.8 = 75 mm. 124. pBenzene = 160 × 0.2 = 32 mm,   pToluene = 68 × 0.8 = 54.4 mm p = 32 + 54.4 = 86.4 mm. 111. Al2(SO4)3 gives maximum number of particles on dissociation and ΔTf would be maximum in 125. In order to produce same lowering in vapour this case. pressure for same solvent, the mole fraction of 112. ΔTf = Kf × m = 1.86 × 0.05 = 0.093°C solute should be same. Freezing point = 0 – 0.093 = – 0.093°C. Thus, WB′MA = WBMA 113. 0.1 M BaCl2 would give maximum no. of particles MB′WA ′ MBWA in solution and hence elevation in boiling point or WB′ = WBMB′WA ′ = 1 × 180 × 1 = 6 g. would be maximum in this case. MBWA 60 × 50 114. π = CRT = 0.05 × 0.082 × 288 = 1.14 atm. 126. On mixing equal volumes of the two solutions 115. Molarity = 10 × d × P = 10 × 1.1 × 20 = 1.36 M. the individual concentrations of the two solutes Mol. mass 162 will become half and hence their individual 116. pTotal = 2.619 + 4.556 = 7.175 kPa osmotic pressures would become half of the initial values. xMeOH = pMeOH = 2.619 = 0.365 The osmotic pressure will be sum of the osmotic pTotal 7.175 pressures of individual components as per xEtOH = 1 – 0.365 = 0.635. Dalton’s law of partial pressure. 5 =1 M = 342 = 68.4. Osmotic pressure = 1.20 + 2.50 = 1.85 bar. 342 M 5 117. ; 2 118. M = 1.86 × 1000 = 60. 127. 2KI + HgI ⎯→ K2 [HgI4] 0.124 × 250 Due to formation of above complex entily, 119. The particle concentration will be highest in K2[HgI4], the number of ions in the solution decreases. Hence, depression in freezing point decimolar Al2(SO4)3. In (d) BaSO4 will be decreases and freezing point gets raised. precipitated as BaCl2 + Na2SO4 ⎯→ BaSO4 + 2Na+ + 2Cl–.

192 CHEMISTRY 128. Concentration C = 0.6 becomes equal to the atmospheric pressure is 0.1 × M the boiling point of the liquid. 137. ΔTf = Kf × m = ΔTb = Kb × m i.e., π = CRT ⇒ 1.23 = 0.6 × 0.082 × 300 ; ∴ ΔTb = Kb 0.1 × M ΔTf K f M = 120 g mol–1. 129. m = 8.1 × 1000 = 1 ΔTb = ΔTf × Kb = 0.186 × 0.512 = 0.0512°. 81 × 100 Kf 1.86 i −1 138. Δp = xB ; 10 = 0.2 ; α = m − 1 or i = α(m – 1) + 1 pA ° pA ° = 0.9(2 – 1) + 1 = 1.9 20 20 xB pA ° 10 0.2 ΔTf = ikf × m = 1.9 × 1.86 × 1 = 3.53 = xB ; = ⇒ xB = 0.4. Freezing point = 0 – ΔTf = – 3.53°C. 139. Let vapour pressure of B be pB 130. ΔTf = Kf × m or m = ΔTf = 0.372 = 0.2 ∴ 84 = 70 × 0.8 + pB × 0.2 ∴ pB = 140 mm. K f 1.86 NaHSO4 ⎯→ Na+ + HSO4– 140. α = i − 1 or 0.3 = i − 1 0.08 0.08 1/m −1 1/4 −1 HSO4– H+ + SO42– ∴ i = 0.775. (0.08 – x) x x 2.5 xB = 2.5 + 55.5 = 0.043. (0.08 – x) + x + x = 0.2 141. x = 0.04 Ka = [H+ ] [SO42− ] = 0.04. 142. pKa = 4 ∴ Ka = 1 × 10–4 [HSO4− ] Also Ka of HX = Cα2 α = (1 − i) α is degree of dissociation of HX 131. 1 1  Ka = 1 × 10−4 m  c 1 × 10−2 − α= = 0.1 or α = 2(1 – i) = 2(1 – 0.52) = 0.96. α = i − 1 Here m = 2 m −1 132. MB = K f WB × 1000 = 2.34 × 28 × 1000 = 123.80 ΔTf × WA 1.68 × 315 ∴ i = 1 + α = 1 + 0.1 = 1.1 143. K2[PtCl6] contains K+, Pt(IV) and Cl– species. It 123.80 Atomicity of phosphorus = 31 ≈ 4. ionises in solution as K2[PtCl6] → 2K+ + [PtCl6]2– Hence formula is P4. 133. π1 = C1RT ; π2 = C2RT ; which pertains to van’t Hoff factor of 3. π1 = C1 or C1 = 405 = 5 or C1 = 5C2. 144. Van’t Hoff factor i and degree of dissociation are π2 C2 C2 81 related as Thus, C1 should be 5 times the C1. Hence, the i − 1 = α or i = 1 + (m – 1)α solution has to diluted 5 times. m −1 K2SO4 : i = 1 + (3 – 1)α = 1 + 2 × 0.4 = 1.8 134. Δp = WBMA or p − 4/5p = WB × 18 KBr : i = 1 + (2 – 1) × 0.5 = 1.5 p MBWA p 75 × 180 K4[Fe(CN)6] : i = 1 + (5 – 1) × 0.2 = 1.6 or WB = 150 g. FeCl3 : i = 1 + (4 – 1) × 0.3 = 1.9. 135. Molality = 0.1 × 1000 = 0.5 1 200 145. ΔTb = Kbm ∴ m = ΔTb/Kb = 0.25/0.52 = 2 ΔTb = Kb × m = 0.513 × 0.5 = 0.2565 Now, mol of urea = 0.5 Tb = (100 + 0.256). mol of water = 1000/10 = 55.5 136. At boiling point vapour pressure of the liquid is equal to the atmospheric pressure. Conversely, the temperature at which vapour pressure

ANSWERS 193 xurea = 0.5 = 8.9 × 10–3. 153. Δp = x2 = 6 / 6 / 60 / 18 = 0.0196. 55.5 + 0.5 p° 60 + 90 146. ΔTb = 100 – 97.4 = 2.6 154. 6% solution of glucose has molality Also ΔTb = iKbm or m = ΔTb = 6 × 1000 ≈ 1.06 m i × Kb 64 60 Thus, it is isotonic with approx. 1 m solution of 2.6 glucose. or m = 2 × 0.52 = 2.5 Δp 1 =1 WNaCl = 2.5 × 58.5 = 146.25 g. 155. p° = Mole fraction of solute = 1+3 4 . 147. Let n molecules of phenol polymerise. 156. X represents elevation in boiling point ΔTb which Now ΔTf = iKf m is equal to Kb m. or i= Δ Tf = 0.93 × 94 × 100 = 0.5 157. Vapour pressure of liquid is only temperature Kfm 1.86 × 9.4 × 103 dependent parameter. Hence, its value is 17.57 Now i − 1 = 1 or i – 1 = 1 −1 or 1 mmHg. 1/n − 1 n i= n 158. Solubility ∝ pressure 1 ∴ S2 = p2 0.5 = or n = 2. S1 p1 n p2 × S1 5.3 × 10−4 × 760 148. YB (mole fraction of B in vapour phase) p1 593 = pB °xB or S2 = = pA °xA + pB°xB = 6.79 ×10–4. 36 × 0.5 159. NaCl is strong electrolyte. Hence, in all the = 108 × 0.5 + 36 × 0.5 = 0.25 solutions, the value of i is same. 149. KCl + AgNO3 ⎯→ KNO3 + AgCl ∴ 25 × M = 40 × 0.5 160. MgCl2 dissociates to give Mg2+ and 2Cl– ions per molecule. Thus, particle concentration becomes or M = 40 × 0.5 = 0.8 mol L–1. three times. 25 Thus ΔTf (MgCl2) = 3 × ΔTf (glucose) As per conditions molality = 0.8 ΔTf = i × Kf × m 161. M = W × 10-3 = d × 10−3 = 2 × 1.86 × 0.8 = 2.97° MB × V(mL) MB Tf = 0 – 2.97 = – 2.97°C. or M ∝ 1 as d is same 150. Degree of dissociation of acetic acid is not given, MB hence answer is unpredictable. Hence, graph between M and MB is given by ‘b’ 162. Dissolution of ionic salt in water occurs if 151. This is as per statement of cryoscopic constant. ΔLH ≤ ΔHydH. 152. H2O and HClO4 show negative deviation. Hence they form maximum boiling azeotrope. ∴ Boiling point will be greater than either of them. 3. ELECTROCHEMISTRY MULTIPLE CHOICE QUESTIONS 1. (a) 2. (c) 3. (b) 4. (c) 5. (d) 6. (c) 7. (b) 8. (a) 12. (b) 13. (c) 14. (b) 15. (d) 16. (d) 9. (b) 10. (c) 11. (c) 20. (d) 21. (c) 22. (a) 23. (b) 24. (d) 28. (d) 29. (a) 30. (d) 31. (a) 32. (b) 17. (d) 18. (c) 19. (a) 36. (c) 37. (a) 38. (d) 39. (c) 40. (b) 25. (a) 26. (d) 27. (a) 33. (b) 34. (d) 35. (d)

194 CHEMISTRY 41. (c) 42. (a) 43. (b) 44. (a) 45. (a) 46. (d) 47. (b) 48. (a) 49. (a) 50. (b) 51. (b) 52. (b) 53. (a) 54. (c) 55. (b) 56. (b) 57. (c) 58. (b) 59. (b) 60. (d) 61. (d) 62. (a) 63. (b) 64. (b) 65. (d). 66. (b) 67. (a) 68. (b) 69. (d) 70. (a) 71. (a) 72. (c) 73. (a) 74. (b) 75. (a) 76. (c) 77. (c) 78. (d) 79. (c) 80. (a) 81. (a) 82. (b) 83. (a) 84. (d) 85. (a) 86. (d) 87. (c) 88. (b) 89. (a) 90. (b) 91. (d) 92. (c) 93. (a) 94. (b) 95. (d) 96. (d) 97. (b) 98. (b) 99. (a) 100. (b) 101. (d) 102. (a) 103. (c) 104. (c) 105. (c) 106. (d) 107. (a) 108. (b) 109. (d) 110. (b) 111. (b) 112. (d) 113. (b) 114. (c) 115. (d) 116. (a) 117. (d) 118. (a) 119. (c) 120. (a) 121. (c) 122. (b) 123. (c) 124. (d) 125. (a) 126. (a) 127. (b) 128. (c) 129. (a) 130. (d) 131. (c) 132. (b) 133. (c) 134. (a) 135. (a) 136. (a) 137. (d) 138. (c) 139. (b) 140. (c) 141. (a) 142. (c) 143. (b) 144. (c) 145. (c) 146. (b) 147. (d) 148. (a) 149. (c) 150. (c) 151. (d) 152. (b) 153. (b) 154. (c) 155. (d) 156. (c) 157. (b) CASE BASED QUESTIONS 4. (b) Case Study 1 1. (c) 2. (d) 3. (d) Case Study 2 1. (a) 2. (d) 3. (c) Case Study 3 1. (d) 2. (a) 3. (a) Case Study 4 1. (b) 2. (d) 3. (d) HINTS/SOLUTIONS 1. The E°red of Ag > E°red of Zn and hence Ag+ ions 9. The equation is, get reduced to Ag by Zn. 2MnO4– + 5SO32– + 6H+ 2. 0 + 2H2O + −2+2 ⎯→ Cr3+ + 3Cl– + 3OH– ⎯→ 5SO42– + 3H2O + 2Mn2+. Cr (OCl)− 10. In each, the electronegative element is oxygen. OCl– gets reduced because O.N. of Cl decreases In BaO2 (barium peroxide) the O.N. of O is – 1 from +1 to –1. while in BaSO4, it is – 2. 11. The given reaction does not involve any change +5 in oxidation number. 3. (CIO3)− + 6H+ + 6e– ⎯→ Cl– + 3H2O. 12. [Pt (C2H4) Cl3]– ; x + 0 + 3(– 1) = – 1 or x = + 2. 13. Cu2+ + 2e– ⎯→ Cu 2 4. Ba(H2PO2)2 ; + 2 + 2(2 + x – 4) = 0 ; x = 2 = +1. 2.0 Faradays can deposit copper = 1 mol 5. Mg2P2O7 ; 4 + 2x + 7(– 2) = 0 or x = + 5. 2.5 Faradays can deposit copper 6. 0 + 6OH− ⎯⎯⎯→ 5Cl− + +5 + 3H2O 3Cl2 (ClO3 ) − Reduction ↑ ↑ = 2.5 = 1.25 mol. Oxidation 2 00 +5 −2 0 14. Ions of the electrolyte act as transportes. 7. I2 + 5O3 + H2O ⎯⎯→ 2HI O3 + 5O2 15. The reactions in two cells are Fe2+ + 2e– ⎯→ Fe O3 is getting reduced to O2– state. and Fe3+ + 3e– ⎯→ Fe. 0 +4 The ratio of iron produced should be 1 : 1 or 3 : 2. 23 8. C12 H22O11 ⎯⎯→ CO2 Thus, sugar molecule is getting oxidized to CO2.

ANSWERS 195 16. Cell constant is not given. 1 2 17. Resistance ∝ 1 . 30. AgNO3 + H2O ⎯⎯→ Ag + 2H+ + NO3– + O2 conductivity ∴ pH gradually decreases due to increase in 18. The statement refers to Faraday’s second law H+ conc. of electrolysis. LiCl + H2O ⎯⎯Ele⎯ctri⎯city⎯→ Li+ + OH– 19. The given equation is ECu2+/Cu = − 0.059 log [Cu2+]–1 + E 2+ + 1 Cl 2 + 1 H2 2 2 2 Cu /Cu comparing this with y = mx + c it represents a ∴ pH gradually increases due to increase in OH– ion conc. straight line shown in a graph. 20. Jule/ampere2. second 31. Flow of electrons stops when E° values of two electrodes become same and potential difference = J ×1 become zero. ampere × sec ampere 32. MnO2 is used as depolariser. = J × 1 = volt 33. At anode the reaction is, C ampere ampere Pb + H2SO4 ⎯→ PbSO4 + 2H+ + 2e–. = ohm. 34. Ered of water is more than that of Na+. 35. Reduction of water occurs at cathode. 21. Cathodic reaction involves reduction. 36. CuBr2 ⎯→ Cu2+ + 2Br– 22. X has least reduction potential and hence Anode : 2Br– ⎯→ Br2 + 2e– maximum tendency to oxidize. Cathode : Cu2+ + 2e– ⎯→ Cu. 23. Reaction in hydrogen electrode is 37. E°cell should be +ve or ΔG should be < 0. 2H+ + 2e– ⎯→ H2 38. Electrochemical cell can behave as electrolytic E = E° + 0.059 log [H+ ]2 cell when external emf is greater than Ecell. 2 [H2 ] 39. Ag+ + e– ⎯→ Ag. E can be greater than E° if concentration of H+ ∴ The electricity required for depositing ions is increased. 108 g (1 mol) of Ag = 1 faraday. 24. Large negative reduction potential means large 40. [Fe(H2O)5NO] SO4 positive oxidation potential for X– ⎯→ X + e–. x + 5(0) + 0 – 2 = 0 or x = + 2 Hence X– can be readily oxidized. 41. When weak acid is titrated against strong alkali, 25. E is related to temperature as initially the conductance is low due to poor E = E° + RT ln [Mg 2+ ] 2F [Mg] dissociation of acetic acid. On adding NaOH, highly ionised CH3COONa is formed and conductance increases. 26. Mg2+/Mg couple has reduction potential less than After neutralisation, addition of NaOH causes Hydrogen- electrode. Hence Mg can reduce H+ ions sharp increase in conductance. to H2. 42. Lower the reduction potential, more easily, the 27. The highest cell voltage would be obtained from metal is oxidised and more is its reactivity. the electrode with highest oxidation potential 43. The configuration II, and III can exhibit variable oxidation states. and highest reduction potential because Ecell = Eoxi + Ered. IInd due to partially filled ‘d’-orbitals and IIIrd 28. E°red. of X, Y, W are less than Z. Hence all of due to inert pair effect. them can reduce Z2+ ions. 29. E°cell = E°cathode – E°anode. It is independent of 44. 1.12 L at STP = 1.12 mol = 0.05 mol volume of solutions 22.4 Also, Ecell = E°cell + 0.059 log [cathode] . 1 mol of H2 is obtained by passing electricity n [anode] = 2 × 96500 C

196 CHEMISTRY 0.05 mol of H2 is obtained by electricity E = E° + 0.059 log [Cu2+] = 2 × 96500 × 0.05 2 Now if I is current strength, by decreasing the conc. of Cu2+ by 1/10, E value I × 965 = 2 × 96500 × 0.05 or I = 10 A. will decrease by 0.059 i.e., approx. by 0.03 V. 2 45. O2 is liberated by anodic oxidation of water 2H2O ⎯→ 4H+ + O2 + 4e– 59. (ii) and (iii) combination will give cell voltage = 2.50 V which is maximum out of the given sets. Liberation of 1 mol (22.4 L) require electricity = 4 × 96500 C 60. Zinc will undergo oxidation in preference to iron. ∴ 193 C of electricity will liberate O2 61. Cell constant is constant for a given cell. = 22.4 × 193 L 4 × 96500 62. In XYZ2 the sum of O.N. of all the atoms is zero. = 0.01119 L or ≈ 11.2 cm3. 63. MnO42– ⎯⎯−e−⎯→ MnO4– 46. KBrO4 ; x + (+ 1) + 4(– 2) = 0 or x = +7. It is an oxidation process and occurs at anode. 47. +1 −1 0 +1 −1/3 . Hence, I2 is oxidizing The O.N. of Mn increases from +6 to +7. K I + I2 K I3 64. Conductivity = Cell constant × Conductance agent. +2 +2 +4 +1 65. Ecell = 0.0591 log Keq n 48. SnCl2 + 2Hg Cl2 ⎯→ Sn Cl4 + Hg2Cl2 Thus, HgCl2 is reduced to Hg2Cl2. For a feasible cell reaction Ecell = +ve −1 +4 − −0 ∴ Keq > 1 49. 5H2O2 + 2ClO2 + 2OH ⎯→ 2Cl + 5O2 + 6H2O 66. The balanced equation is, In ClO2, chlorine atom decreases its oxidation number from + 4 to – 1. Hence ClO2 is reduced Cr2O72– + 6Fe2+ + 14H+ and H2O2 is oxidized. ⎯→ Cr3+ + 6Fe3+ + 7H2O. 50. Λ eq and Λm are equal for uni-univalent 67. The balanced equation is electrolytes. Na+ makes same contribution 3MnO4– + 5FeC2O4 + 24H+ irrespective of anion to which it is attached. ⎯→ 5Fe3+ + 3Mn2+ + 10CO2 + 12H2O Now, 5 mol of ferrous oxalate require 51. Anodic reaction involves oxidation of Zn to Zn2+ KMnO4 = 3 mol 1 mol of ferrous oxalate require ions. 52. It is based on Nernst equation. KMnO4 = 3/5 mol = 0.6 mol. 53. E°red X < Y < Z hence Y can displace X and not 3+ 2+ Z. 68. Fe4[Fe (CN)6] . 54. η = − nFE . 69. Two O atoms in H2SO5 are peroxide and have ΔH O.N. = – 1 while the other 3 oxygen atoms have 55. Reactivity depends on the tendency to displace O.N. = – 2. The structure of H2SO5 is the other. Hence the order Y > X > Z. O–2 56. E° value will not depend upon the size of the +1 –1 –1 –2 +1 strip as concentration of solids is taken to be H—O—O—S—O—H unity. O–2 57. The best reducing agent is one whose E°red. is lowest. Thus Mg is the correct choice. Hence O.N. of S is + 6. 70. CaOCl2 contains Ca2+, OCl– and Cl–. Cl has O.N. 58. If solution is diluted to 10 times, molar concentration of Cu2+ is reduced to one tenth of = – 1 in Cl– while its O.N. in OCl– part is + 1. its original value. Nernst equation for Cu2+ + 2e– ⎯→ Cu is 71. Ag+ + e− ⎯→ Ag 1 mol 108 g

ANSWERS 197 54 g of Ag is deposited by electricity = 96500 C 84. The structure of Na2S4O6 is 2 OO Al3+ + 3e− ⎯→ Al + – 1 2 3 4 –+ 27 g NaO—S—S—S—S—ONa Now, 3 × 96500 C deposit Al = 27 g OO 96500 C deposit Al = 27 × 96500 g = 4.5 g. It is evident, that O.N. of two S atoms, i.e., S2 2 2 × 3 × 965000 and S3 is 0. The O.N. of two S atoms, i.e., S1 and S4 is + 5 each. 72. Al3+ + 3e− ⎯→ Al 85. Bromine oxidizes S2O32– to SO42– in which O. N 3 mol 27 g of S changes from +2 to +6. Iodine oxidizes S2O42– where O. N of S changes from +2 to 2.5. In order to deposit 27 g of Al, electricity required = 3F 86. 2 Equivalent of Ag will deposit for every 1 eq. of O2 liberated In order to deposit 9 g of Al, electricity required 1.6 = 3F × 9 = 1 F. ∴ Mass of Ag deposited = 16 × 2 × 208 27 = 21.6 g. +1 − 2 00 87. Copper cannot displace Al+3 ions from the 73. H2O ⎯⎯→ 1 H2 + 2 O2 solution. 1 mol 22.4 L 11.2L 88. Let copper produced = x g. Decomposition of 1 mol of water requires 2 mol According to Faraday’s second law of electron x = 31.75 ∴ 4 mol of electrons produce 2 × 22.4 L 0.504 1 = 48.8 L of H2 and 2 × 11.2 L = 22.4 L of oxygen. Total volume produced at STP or x = 0.504 × 31.75 = 16.002 g. = 44.8 + 22.4 = 67.2 L. 89. The anodic reaction in H2—O2 fuel cell is H2 + 2OH– ⎯→ 2H2O + 2e– 74. Charge in coulombs = 4 × 30 = 120 C. Burning of 1 mole of H2 produces electricity 75. X2+ + 2e– ⎯→ X ; Y+ + e– ⎯→ Y = 2 × 96500 C 2 mol e– produce X = 1 mol = 1 × Mg 2 mol e– produce Y = 2 mol = 2 × M/2 = Mg. Hence, ratio of the masses of Y : X is M : M or 1 : 1. 76. Reduction Means gaining of electrons. 77. H+ conc. should be 1 M. ∴ pH should be zero. Burning of 0.3 mol  6.72  of H2 gives electricity  22.4  78. E°cell = E°red (cathode) + E°oxi. (anode) = – 0.41 + 0.76 = 0.35 V. = 0.3 × 2 × 96500 C 79. E°red of copper = 1.1 – 0.76 = 0.34 V. Current in amperes = 0.3 × 2 × 96500 ∴ E°oxi.. = – 0.34 V. 15 × 60 80. Ecell = 0.34 + 2.37 = 2.71 V. = 64.3 amperes. 90. The liberation of each H+ ion can result in 1 81. From equation (i), EY2 / Y− > EX2 / X− ; NaOH being formed. From (ii), EW2 / W− > EY2 / Y− From eqn. (iii), EX2 / X− > EZ2 /Z− . This accounts ∴ Moles of electrons = 9.65 × 1000 = 0.1 for the sequence (a). 96500 82. ΔG° = – nFE° = – 1 × 96500 × 1.02 = – 98430 J Moles of H+ liberated = 0.1 or – 98.43 kJ. ∴ Moles of NaOH formed = 0.1 Mass of NaOH formed = 0.1 × 40 = 4.0 g. 83. Ionic mobility 91. Q = 0.025 (A) × 60 (s) = 1.5 C 2 × 96500 C deposit Ca atoms = 6.02 × 1023 = λ°Ag+ = 61.92 × 10−4 Sm2 mol−1 = 6.4 × 10–8. F 96500 C mol−1

198 CHEMISTRY 1.5 C deposit Ca atoms = 6.02 × 1023 × 1.5 104. For the reaction, 2H+ + 2e– ⎯→ H2 2 × 96500 E = E° + 0.059 log (H+)2 = 4.68 × 1018. n 92. Total Ag+ ions = 0.2 × 0.1 = 0.02 mol = 0 + 0.059 log (10–10 )2 = – 0.59 V. One fourth of Ag+ = 0.02 = 5 × 10–3 mol n 4 105. Since reduction potential of Ag+/Ag is greater Quantity of electricity required for deposition than Sn. Hence silver electrode must act as of 5 × 10–3 mol of Ag+ = 96500 × 5 × 10–3 C cathode. Now, 96500 × 5 × 103 = 0.1 × t or t = 4825 seconds or 80.4 minutes 106. For concentrations cells E°cell = 0. 107. Electrical work = nFE° 93. Moles of Cu deposited = 3.25 = 0.05 63 For daniel cell n = 2 −3 +6 0 +1 ∴ Work = 2 × 96500 × 1.10 = 212.3 × 103 J = 212.3 kJ. 94. (NH4 )2 Cr2O7 ⎯→ N2 + 4H2O + Cr2O3 108. ΔG° of required reaction = 2 × (– 21.52) Each N atom loses 3 electrons. = – 43.04 kJ. +2 +4 109. According to the given data 95. (C2 O4 )2− ⎯→ 2CO2 + 4e− . C and D are more reactive than A and B. As D does not react with Cn+, therefore C is more +5 −3 reactive than D. Also as B does not react with An+, it is weaker 96. (NO3 )− + 6H2O + 8e− ⎯→ NH3 + 9OH− . oxidising agent than A. Hence 97. No. of coulombs passed = 2 × 5 × 60 × 60 C. C > D > A > B. If the oxidation state is n +, then deposition of 110. Statement I and II are correct but statement 177 g of metal, the electricity required is III is wrong. = n × 96500 C 111. The balanced equation is ∴ The electricity required for 22.2 g of metal 2MnO4– + 5C2O42– + 16H+ ⎯→ 2Mn2+ + 10CO2 + 8H2O n × 96500 × 22.2 Now, 5 moles of oxalic acid (H2C2O4.2H2O) = 177 require KMnO4 = 2 mol n × 96500 × 22.2 126 g i.e., 126 mol of oxalic acid require Now, 177 = 2 × 5 × 60 × 60 126 or n = 2.94 = + 3. 98. Λm = κ × 103 =1 × Cell constant × 103 × 1 M R M = kcell × 103 . MnO4 = 2 = 0.4 mol. RM 5 99. Reaction I is disproportionation where O 112. On calculating E.F. from % age we get the increases its O.N from 0 to + 5 and also decreases formula as XeF6 ∴ Oxidation number of Xe = + 6. from 0 to –1. Reaction IV is comproportionation where S +6 +6e change is O.N from – 2 and + 4 to 0. 113. K2CrO7 ⎯⎯→ 2Cr3+ 100. I2 + SO2 + 2H2O ⎯→SO42– + 2I– + 4H+ Reduction of 1 mol of K2Cr2O7 needs charge = 6 × 96500 C 101. Λ°m (H2O) can be obtained from I, III and also IV. Reduction of 0.4 mol of K2Cr2O7 needs charge = 0.4 × 6 × 96500 C = 2.4 × 96500 C. 102. Lesser is the value of reduction potential, more is the reducing character. 114. According to Faraday’s second law, comparing P, Q cells. 103. Nernst equation should be E = E° + RT ln [Mn+] nF

ANSWERS 199 WAg = EAg or 0.216 = 108 Cu2+ + 2e– ⎯→ Cu ; E°3 = ? WHg EHg WHg 200.6/2 ∴ ΔG°3 = – 2 × E°3 × F = – 2E°3 F Also, ΔG°3 = ΔG°1 + ΔG°2 = – 0.15 F – 0.5 F or WHg = 0.216 × 200.6 = 0.2006 g 108 × 2 = – 0.65 F In P, R cells, 0.216 = 108 Now, – 2E°3F = – 0.65 F WHg 200.6 0.65 or E°3 = 2 = 0.325 V. 0.216 × 200.6 121. pH1 = 3 ∴ [H+]1 = 1 × 10–3 M ; or WHg = 108 = 0.4012 g. pH2 = 5 ∴ [H+]2 = 1 × 10–5 115. 1 mol of electrons Now, Ecell = 0.059 log [1 × 10−3 ] = 96500 C or 96500 C = 1 × 10–16 × t 1 [1 × 10−5 ] or t = 96500 × 1016 s = 9.65 × 1020 s. = 0.059 × 2 = 0.118 V. 116. Au3+ + 3e– ⎯→ Au 122. Molarity of SO42– ions remains the same as they do not participate in reaction (Zn + Cu2+ ⎯→ 197 g of Au is obtained by charge = 3 × 96500 C Cu + Zn2+). 1.234 g of Au is obtained by charge WCu ECu WH2 × 31.75 WH2 EH 1 = 3 × 96500 × 1.234 123. = or WCu = 197 Now, 3×t= 3 × 96500 × 1.234 = 224 × 2 × 31.75 = 6.35 g. 22.4 × 1 197 96500 × 1.234 124. Anodic reaction involves Cu ⎯→ Cu2+ + 2e– 197 or t= ≈ 604 sec. Cathodic reaction involves Cu2+ + 2e– ⎯→ Cu 117. The reaction I2 + 2Br– ⎯→ 2I– + Br2 is non- Thus, amount of copper dissolved and that spontaneous because reduction potential of Br2/ deposited would be same i.e., Br– is larger than that of I2/I–. 3.2 118. Apply Nernst equation, to the reaction 3.2 g = 63.5 mol = 0.05 mol. Pb + Sn2+ ⎯→ Pb2+ + Sn 125. In conversion of C6H5NO2 ⎯→ C6H5NH2, no. of electrons per mole of substance involved = 6 mol. 0.059 [Sn2+ ] or E° + 2 log [Pb2+ ] ∴ 0.2 mol will require = 0.2 × 6 = 1.2 mol or log [Sn2+ ] = 0.01 × 2 = 0.3 126. In conversion of NH3 to NO, number of electrons [Pb2+ ] 0.059 involved = 5 mol per mol of NH3. Thus, for 0.5 mole of NH3, the molecules of oxidant required [Sn2+ ] = 1.25 NA. or [Pb2+ ] = antilog (0.3). 127. Cd required for 3 g of Hg = 10 × 3 = 0.33 g 90 119. Cu2+ + 2e– ⎯→ Cu ; E° = 0.34 V Cd2+ + 2e– ⎯→ Cd E = 0.34 + 0.059 log [Cu2+] Electricity required for depositing 0.33 g of 2 cadmium or E = 0.34 – (0.0296) (9) = 0.34 – 0.266 = 2 × 96500 × 0.33 = 568.6 C. 112 = 0.07 V. 128. Oxidation state = 119 × 5 × 60 × 60 × 2 = +2.0. 120. Cu2+ + e ⎯→ Cu+ ; E1° = 0.15 V 96500 × 22.2 ∴ ΔG°1 = – 1 × 0.15 × F = – 0.15 F Cu+ + e– ⎯→ Cu ; E°2 = 0.5 V 129. O.N. of N in N2H4 is –2. Since 1 mol of N2H4 ∴ ΔG°2 = – 1 × 0.50 × F = – 0.5 F contains 2 mol of N atom, therefore each mole of N atom lose 5 mol of electrons. Hence O.N. of N in the new compound will be +3.

200 CHEMISTRY 130. Λm = κ × 1000 140. In NH4NO3, N atom exhibit oxidation number M of – 3(in NH+4) part and + 5 (in NO3–) part. or M= κ × 1000 = 3.06 × 10−6 × 103 CASE BASED QUESTIONS Λm 3.06 Case Study 1 = 1 × 10–3 mol L–1 1. This relationship is based upon Kohlrausch’s law Ksp = [X2+] [Y2–] = (1 × 10–3) (1 × 10–3) 2. The statements II and III are correct as they = 1 × 10–6. are based on facts. 131. Cu +2 + Zn 2e− + Zn2+ 3. Λ°m (H2O) = λ°H+ + λ°OH− = 34.95 + 19.9 (0.1 mol) 100/65 = 54.85 S m2 mol–1 ⎯→ Cu = (1.54) Λm (H2O) = κ = 3.5 × 10−3 Thus, Cu2+ ions are limiting reagent 103 × Cm 1000 × 55.5 Now, 0.1 mole of Cu+2 transfers electrons = 0.2 mol or Amount of electricity = 0.2 × 96500 C. = 6.306 × 10–8 S m2 mol–1. Now, 0.2 × 96500 = 2 × t or t = 9650 C = 2.68 hr. 4. H3PO4 is weakest acid among the given. Hence, 0.059 its Λ°m is lowest. n 132. Apply, E°cell = log K . Case Study 2 133. E°cell = E°cathode – E°anode 1. ΔGÈ = – nFEÈcell = – 4 × 96500 × 1.23 = – 474780 J = 1.50V – (– 0.25 V) = + 1.75 V. or – 474.78 kJ. 134. Larger the value of E°red larger is tendency for 2. Combination of I and IV electrode will give cell reduction and consequently stronger will be the oxidant. Similarly, smaller the value of E°red with largest emf. larger the tendency for oxidation and consequently stronger will be the reductant. 3. For spontaneity of cell reaction ΔG < 0 and Ecell > 0 135. Add the two values of electrode potential to get the desired value. Case Study 3 1. EÈcell = EÈRed (cathode) – EÈRed (anode) 136. Ecal. – EH2 = 0.67 or EH2 = – 0.67 + 0.28 = EÈRed (cathode) + EÈOX (anode) = – 0.39 V = 0.80 + (– 0.50) = 0.75 V Now, – 0.39 = 0+ 0.059 log H+ ln K = nFEscell = 2 × 0.75 V × F 1 RT RT = 2 × 0.75 (V) × 38.9 ( V ) = 58.35 or – log H+ = 0.39 = 6.6. 2. (a) C6H12O6 + H2O —→ C6 H12 O7 + 2H+ + 2e 0.059 E = EÈ + RT ln (H+)–2 2F 137. According to given data, = EÈ + 2.303  RT  (– 2 log [H+]) C is most reactive. A is least reactive because it 2  F  cannot reduce Bn+ and Dn+ ions. D is less reactive than B. = EÈ + 2.303 × 2 × pH 2 38.9 Hence, C > B > D > A. = EÈ + 2.303 × 11 = EÈ + 0.65 138. In the balanced reaction, 38.9 − 4e− The EÈOX increases by 0.65. 2Zn(s) ⎯⎯→ 2Zn2+(aq) 3. Since EsAg+ /Ag > Es ]+ . ∴ 4e– are transferred. 139. From the given values of reduction potentials [Ag(NH3 )2 / Ag the statements (i) and (iii) are both correct. Hence (Ag (NH3)2)+ is a weaker oxidising agent then Ag+.

ANSWERS 201 Case Study 4 2. 2 mol of NaCl produces 2 mol of sodium and 1. 500 mL of 4 M aqueous NaCl contains hence 2 mol of Na-Hg amalgam. = 2 (23 + 200) g = 446 g NaCl = M × V(L) = 4 × 500 = 2 mol 1000 3. 2 Na+ + 2e– ⎯⎯→ 2Na or 2 Cl– ⎯⎯→ Cl2 + 2e– 2 NaCl ⎯⎯→ 2 Na + Cl2 Thus two moles of NaCl requires 2F or 2 × 96500 C of charge for complete electrolysis. Thus, 2 mol of NaCl on electrolysis will give one mol of Cl2. 4. CHEMICAL KINETICS MULTIPLE CHOICE QUESTIONS 1. (c) 2. (a) 3. (d) 4. (d) 5. (c) 6. (d) 7. (c) 8. (d) 12. (d) 13. (c) 14. (d) 15. (c) 16. (a) 9. (c) 10. (c) 11. (d) 20. (c) 21. (d) 22. (a) 23. (a) 24. (c) 28. (b) 29. (a) 30. (d) 31. (c) 32. (b) 17. (c) 18. (d) 19. (c) 36. (b) 37. (c) 38. (b) 39. (d) 40. (b) 44. (d) 45. (c) 46. (d) 47. (c) 48. (d) 25. (c) 26. (c) 27. (d) 52. (b) 53. (d) 54. (c) 55. (c) 56. (a) 60. (d) 61. (b) 62. (b) 63. (b) 64. (b) 33. (d) 34. (b) 35. (a) 68. (d) 69. (c) 70. (c) 71. (a) 72. (b) 76. (c) 77. (c) 78. (c) 79. (b) 80. (d) 41. (d) 42. (b) 43. (c) 84. (b) 85. (c) 86. (d) 87. (b) 88. (b) 92. (a) 93. (d) 94. (c) 95. (b) 96. (d) 49. (b) 50. (c) 51. (b) 100. (a) 101. (a) 102. (c) 103. (c) 104. (c) 108. (d) 109. (b) 110. (d) 111. (d) 112. (c) 57. (a) 58. (d) 59. (c) 116. (b) 117. (d) 118. (c) 119. (c) 120. (a) 124. (c) 125. (d) 126. (d) 65. (c) 66. (c) 67. (a) 73. (b) 74. (d) 75. (b) 81. (a) 82. (d) 83. (d) 89. (b) 90. (b) 91. (c) 97. (c) 98. (c) 99. (b) 105. (c) 106. (c) 107. (d) 113. (d) 114. (b) 115. (d) 121. (a) 122. (c) 123. (c) CASE BASED QUESTIONS Case Study 1 1. (c) 2. (d) 3. (c) 4. (d) 4. (d) Case Study 2 1. (b) 2. (a) 3. (c) HINTS/SOLUTIONS 1. In terms of product, rate expression is 5. Autocatalysis is by virtue of Mn2+ ion. + 1 Δ[NO2 ] . 6. Rate constant of all reactions is independent of 4 Δt conc. of reactants. 2. Order = 3/2 + (– 1/2) = 1. 7. A catalyst provides a path of low actuation energy and hence enhances the rate. It cannot 3. rate = k [CH3COOH] [C2H5OH] change free energy. When each conc. is reduced to half, the rate of 8. Increase in temperature by 10° doubles the 1 effective numbers of collisions for most of the reaction would become 4 times. reactions. 4. rate of reaction depends on many factors. Changing temperature changes the rate.

202 CHEMISTRY 9. Increase in temperature enhances in the rate of 23. r = k[A]x [B]y ...(i) both exothermic and endothermic reaction. 2r = k[A]x [2B]y ...(ii) 10. Greater the order of the reaction, more is the 8r = k[2A]x [2B]y ...(iii) percent increase in the rate of the reaction for a given increase in conc. Dividing (ii) by (i) we get y = 1 11. Order of reaction is determined experimentally and dividing (iii) by (ii) we get x = 2 and slowest step in mechanism gives the order of reaction. Hence r = k[A]2 [B]. 12. Energy of colliding molecules is equal to or 24. For elementary or single step reactions rate is greater than threshold frequency. given by the balanced equation 13. Rate law tells us about the order of the reaction. r = k[A2] [B2]. 25. By observing the rate equation we come to know 14. Rise in temp. enhances the rate of a reaction i.e., value of k increases. that the rate determining step involves 1 molecule of A and 1 molecule of B. 15. k = A e−Ea /RT 26. 1 d[H2 ] = 1 d[NH3 ] −3 dt 2 dt Ea Ea ln k = ln A – RT or ln A – ln k = RT . d[NH3 ] 2 d[H2 ] 2 dt 3 dt 3 1 = − = (0.01) = 0.007 mol L–1 s–1. [ A ]n−1 16. t1/2 ∝ , where n is order of the reaction. 27. r = k(pN2O5)  conc. of N2O5 ∝ pN2O5 For zero order reaction, − d[ pN2O5 ] dt ∝ 1 ∝ [A]. = k pN2O5 [ A ]−1 t1/2 d[ pN2O5 ] pN2O5 2 × 0.693 1.386 = – kdt ...(1) k k 17. t3/4 = 2 (t1/2) = = . Integrating (1), we get 18. Increase in rate of reaction is maximum for the log (pN2O5) = (– k)t + C This is an equation of straight line with slope reaction having the maximum activation energy. 19. Ea > ΔH equal to – k 28. E Ea > DH Ea The Eab > Eaf Ea H2 + I2 Eab DH HI 20. In KMnO4 titration with oxalic acid, Mn+2 ions 29. The percent increase in the rate for a given rise act as autocatalyst. in temperature is more at lower temperature because at low temperature a large fraction of 21. Greater the order of the reaction, greater is the molecules does not have energy sufficient to increase in the rate of the reaction for a given cross over the barrier. increase in the concentration. 22. In the reaction, r ∝ [A] ∝ 1 and independent 30. log k2 = Ea  T2 − T1  [B] k1 2.303R  T1T2    of C hence increasing conc of A in the rate. For Ea to be same k2/k1 should be equal. 31. Arr. Equation is k = A. e−Ea /RT .

ANSWERS 203 32. For the given reaction, From (ii) and (iii) log k = Ea  T2 − T1  r3 = 4 = (4)n (4)m 2.303R  T1 × T2  r1   = 25 × 103  10  = 0.14 But n = 1 m = 0.  300  4 = (4) (4)m or 2.303 × 8.314 × 310  Hence k = antilog 0.14 = 1.38. 0.693 43. t0.5 = k and hence it is independent of conc. 33. k = A e−Ea /RT of reactants. ln k = ln A –  Ea  1 . 44. Arrhenus equation is k = A e−Ea /RT  R   T  or ln k = ln A – Ea RT 34. By reducing the volume to one-fourth, the concentrations of both A and B would become differentiating it w.r.t. temp. we get 4 times. d ln k = Ea 35. Knowing value of k at two different dt RT2 temperatures Ea can be calculated by using . Arrhenus equation. 45. Decrease in temperature decrease value of k. k2 Ea  T2 − T1  46. Conc. of both the reactants after dilution would k1 2.303R  T1 × T2  be reduced to one-half of the initial conc.  log = . 47. A zero order reaction takes place at constant rate and hence conc. of A decreases with time at  constant rate. 36. Because its apparent molarity is more than 48. All the three given statements are in correct. three. 37. Higher the order of the reaction, sharper is the 49. By observing the dimensions of k we come to decrease in the rate of the reaction with progress know it is second order reaction. of the reaction (due to fall in conc.). 50. ΔE = Ea( f ) − Ea(b) 1 d[NH3 ] 1 d[H2O] 38. − 4 dt = 6 dt – 20 = 30 − Ea(b) Ea(b) = 50 kJ. d[H2O] = − 6 d[NH3 ] = 6 × 0.02 dt 4 dt 4 51. ΔE = Ea( f ) − Ea(b) = 0.03 mol L–1 min–1. y = x − Ea(b) 39. This is an ionic reaction. 40. r1 = k(a)3/2 (b)–1/2 Ea(b) = x – y. r2 = k(4a)3/2 (4b)–1/2 52. t0.75 = 1386s = 2 × t0.5 r2 = (4)3/2 (4)–1/2 = 4. t0.5 = 1386 = 693 s r1 2 41. k (rate constant) increases with increase in 0.693 k = 693 s = 1 × 10–3 s–1. temperature whereas K (equilibrium constant) increases for endothermic reactions and 53. t7/8 = 3t1/2 54. For a zero order reaction rate does not change decreases for exothermic reactions. with time. 42. r1 = k(a)n (b)m ...(i) r2 = k(2a)n (b)m ...(ii) 55. Rate = k[X] r3 = k(4a)n (4b)m ...(iii) rate 7.5 × 10−4 = 1.5 × 10–3 s–1. From (i) and (ii) k= [X] = 0.5 r2 = 2 = (2a)n = (2)n or n=1 56. t0.75 = 2t0.5 r1 (a)n Hence t0.5 = 1.386 = 0.693 h 2

204 CHEMISTRY R= 0.693 0.693 = 1h–1 = 1 × 60 × 60 [A]2 t0.5 = 0.693 From first step we can write, K = [A2 ] = 3.6 × 10–3 s–1. [A]2 = K[A2] 57. By observing the dimensions of k, we come to [A] = K1/2 [A2]1/2 ...(ii) know that the given reaction is a zero order Substituting the value of [A] from equation (ii) reaction. For a zero order reaction, in equation (i) Rate = Rate constant = 1.0 × 10–2 mol L–1 s–1 . 58. ΔE = Ea(f ) − Ea(b) = 60 – 20 = 40 kJ r = k . K1/2 [A]1/2 [B2] = k′ [A2]1/2 [B2] ⇒ ΔE = EP – ER 59. 40 = EP – 20 ; EP = 60 kJ. 1 ...(i) ∴ The order of the reaction is 1 2 . ΔE = + 20 kJ 66. r = k[A]x ΔE = Eact( f ) − Eact(b) 2r = k[4A]x ...(ii) 20 = 30 − Eact(b) Dividing (ii) by (i) we get 2 = (4)x or (2) = (2 × 2)x Eact(b) = 30 – 20 = 10 kJ. 60. k = A e−Ea /RT Hence x = 1/2. ln k = ln A –  Ea  1 67. Here, partial pressure is to proportional to  R  T concentration. ln k =  − Ea  1 + ln A 68. In first two hours the conc. falls by 2 mol L–1. In  K  T the next four hours the conc. falls by 4 mol L–1. Thus, the reaction is proceeding at constant y = mx + c. speed that is independent of the initial conc. of the reactant. Thus, the order of the reaction is 61. t0.75 = 2t0.5 zero. ∴ t0.5 = t0.75 32 = 16 min. 69. For first order reaction t1/2 is independent of 2 =2 initial concentration of the reactants. Hence, slope of a graph log t1/2 vs. log [A]0 would be zero. 62. Suppose 2p is the pressure of C after 10 min. 70. t3/4 = 2t1/2. Fall in pressure of A = 2p 71. 4 hrs. = 4 half lives Fall in pressure of B = p Total fall in pressure = (2p + p) – 2p = p = 20 mm 1 ⎯⎯t1/2⎯→ 1 ⎯⎯t1 / 2⎯→ 1 ⎯⎯t1/2⎯→ 1 ⎯⎯t1 / 2⎯→ 1 2 4 8 16 Pressure of C = 2p = 40 mm Rate of appearance of C = 40 = 4 mm/min. Fraction left after 4 half lives = 1 or  1 4 10 16  2  63. rapp of SO3 = 2 rdiss of O2 Fraction reacted in 4 half lives = 1 − 1 = 15 . ∴ 2 × 2 × 10–4 = 4 × 10–4 mol L–1 s–1. 16 16 64. Fall in conc. = 4 – 1 = 3 mol L–1 1 3 Time interval = 30 min 72. In the first hour fraction reacted = For a zero order reaction, 2 After the first hour fraction left = 3 Rate constant = Rate of reaction = 3 mol L−1 = 0.1 mol L–1 min–1. In the second hour one-third of 2 reacts. 30 min 3 65. From the slow step, the rate law may be written 73. For zero order reaction, t1/2 ∝ [A]. as : 1 r = k[A] [B2] ...(i) 74. t1/2 ∝ [A]n−1 .

ANSWERS 205 75. − d[A] = k [A]0, – d[A] = kdt When 50% of the reactants are converted into dt products Integrating from t = 0 to t = t rate = 1 × 10–2 (0.5)2 = 2.5 × 10–3 mol L–1 s–1. [A]0 – [A] = kt. 88. r1 = k(a) (b)1/2 76. ΔE = Ea(f ) − Ea(b) = 30 – 50 = – 20 kJ. When conc. of both the reactants is increased four times 77. − d[ A ] = k[A]0 r2 = k(4a) (4b)1/2 = 8k(a(b))1/2 = 8r1. dt 89. At equilibrium, the concentration of A falls from – d[A] = kdt 0.6 mol L–1 to 0.3 mol L–1 and the concentration of B rises from zero to 0.6 mol L–1. Thus, 0.3 mol Integrating from [A]0 to [A] and t = 0 to t = t L–1 of A on decomposition give 0.6 mol L–1 of B. [A]0 – [A] = kt. Therefore, n = 2. 78. By observing the dimensions of k, we come to know that the given reaction is of second order. 90. k = 1 × 10–2 mol L–1 s–1 ∴ r = k[A]2 = 1 × 10–2 × (2)2 When 90% reaction is complete, [A] = 0.1 mol L–1 and [B] = 0.1 mol L–1. = 4 × 10–2 mol L–1 s–1. Rate = k[A] [B] = 1 × 10–2 (0.1) (0.1) 79. Starting with 4 mol L–1, the conc. was reduced = 1 × 10–4 mol L–1 s–1. to half in 1 hour whereas starting with 2 mol L–1 the conc. was reduced to half in 0.5 hour. Thus, 91. k= 0.693 0.693 min–1 t1/2 here is directly proportional to the initial t1/2 = 10 conc. of the reactants. Hence, order of the reaction is zero. t= 2.303 log 100 = 2.303 log 10 = 33 min. k 10 0.0693 80. t1/2 = 100 minutes. The reaction cannot goto 92. In first case conc. is reduced to half in 1 hour. In 100% completion and hence time required is second case the conc. is reduced to one-fourth in infinity. two hours (i.e., 2 half lives). This indicates that [O2 ][O] [O3 ] the half-life in the present reaction is [O3 ] [O2 ] 81. K= , [O] = ...(i) independent of the initial conc. of the reactants. This is true for first order reactions. rate = k[O3] [O] ...(ii) 93. Suppose order of the reaction is n [O3 ] [O3 ]2 From (i) and (ii) rate = k[O3] . K [O2 ] = K′ [O2 ] t1/2 ∝ [CH3CHO]1–n 0.693 0.693 410 =  364 1−n k = 1 × 10−2 880  170  82. t1/2 = = 69.3 s 1 mole would be reduced to 0.25 mole in two half log 410 = (1 − n) log  364  lives, i.e., 69.3 × 2 = 138.6 s. 880  170  83. rate = k [X]3 n = 2. k= rate = 5 × 10−2 = 50 L2 mol–2 s–1. 94. After long time whole of A would change into B [X]3 (0.1)3 and C. On decomposition one mole of A changes into three moles of B and C. Therefore, 84. As the rate of the reaction is increasing four Initial pressure of A = 270 = 90 mm. times when conc. of HI is doubled, the order of 3 the reaction is 2. Suppose after 10 minutes pressure of A falls by 85. Since EP is less than ER, the reaction must be x mm. In that case pressure of B and C would exothermic. be 2x and x mm respectively. 86. As the reaction proceeds at constant speed with Total pressure = (90 – x) + 2x + x = (90 + 2x) mm the progress, it must be a zero order reaction. 90 + 2x = 176 mm 2x = (176 – 90) mm 87. k = 1 × 10–2 L mol–1 s–1 x = 43 mm rate = k [A]2 [B]0

206 CHEMISTRY Pressure of A after 10 minutes CASE BASED QUESTIONS = (90 – x) mm Case Study 1 = (90 – 43) = 47 mm. 1. Let the rate of reaction be expressed as, r = k[A]x[B]y 95. t3/4 = 2t1/2 = 0.693 × 2 . k Accordingly, 96. k = 3.0 × 10–5 s–1. The units of suggests that the 3.0 × 10–4 = k[0.1]x[0.1]y exp1; 7.2 × 10–3 = k[0.2]x[0.3]y reaction is first order exp2; 8.1 × 10–3 = k[0.3]x[0.1]y exp3; ∴ r = k[N2O5] ∴ [N2O5] = 2.4 × 10−5 = 0.8 mol L–1. Now, exp3 = 8.1 × 10−3 = k[0.3]x [0.1]y 3 × 10−5 exp1 3.0 × 10−4 k[0.1]x [0.1]y 97. From the dimensions of rate constant it is clear or 2.7 × 10–1 = (3)x that the reaction is a zero order reaction. It or (3)3 = (3)x. Hence x = 3 would proceed at constant rate of 10–2 mol L–1 s–1. In 10 minutes (600 seconds) the amount of Similarly, y=1 A reacted would be Order w.r.t A = 3, B = 1 2. Rate =k[A]3[B]1 = 10–2 × 600 mol L–1 or 3.0 × 10–4 = k[0.1]3[0.1] = 6 mol L–1 No. of moles of A left unreacted = 10 – 6 = 4 or k= 3.0 × 10−4 = 3.0 moles. 10−4 98. As the rate becomes 3 times more when conc. 3. rinitial = 3.0 × (0.2)3(0.6)1 changes or increases by 9 times or 3.0 × 0.2 × 0.2 × 0.2 × 0.6 ∴ 3 = (3 × 3)x or x = 1/2. = 1.44 × 10–2 mol L–1s–1 99. Rate = k[A] [B] 4. Pre-exponential factor is 1018 When concentrations of both A and B become Case Study 2 half, the rate of reaction would become one- fourth of the initial rate. 1. For the given reaction, t1/2 = 0.693 k 100. A ⎯→ 2B + C. or k = 0.693 = 0.693 = 0.0315 y–1 —Insted conc. p0 0 0 r = k[A] k 22 —in pressure after 10 mm (p0 – x) 2x x Since, k = k1 + k2 or k2 = 0.0315 – k1 —After long time 0 2B & C Now we know that for the parallel reactions, (p0 – x) + 2x + x = 160 mm ...(i) After long time there as 3 moles (2B & k) k1 = x = 2 k2 y 98 ∴ pressure due to 3 moles = 300 mm or p due to 1 mole = 100 mm k1 = 2 0.0315 − k1 98 Hence p0 = 100 mm ∴ or 98k = 0.063 – 2k Substituting this valve in eq. (i) 11 100 – x + 2x + x = 160 or k1 = 63 × 10–5 y–1 or 100 + 2x = 160 2. We know that, k = k1 + k2 ∴ k2 = k – k1 = 0.0315 – 63 × 10–5 160 − 100 60 x= 2 = 2 = 30 mm = 0.03087 = 0.031 y–1 Hence pressure of A at 10 mm 3. t1/2 = 22 yrs (given) = 100 – 30 = 70 mm. ∴ Cn = C0 = 2 2n 22 101. rate = k[A] The rate vs. [A] graph would have slope equal (n = number of self lines in 44 yrs = 2) to k. or Cn = 2 = 0.5 kg 4 (Amt. of reactant left)

ANSWERS 207 Amt. of produced formed after 44 yrs = 1.5 kg d[A] or – dt = (k1 + k2) [A] 98 Hence amt. of C formed = 1.5 × 100 = 1.47 kg. d[A] 4. For the given reaction, or – [A] = (k1 + k2) dt On integrating the above equation we get k1 B ; conc. at time t = x A k2 (k1 + k2)t = 2.303 log [A0 ] [ A] C ; conc. at time t = y The rate of formation of B, (i) For conversion of A→B – d[A] = d[B] = k1[A] r = d[B] = k [A] = x dt dt B dt 1 dt (ii) For conversion of A→C– d[A] = d[C] = k2[A] Also d[C] x dt dt rC = dt = k2 [A] = dt Hence Nitrate of disappearance of rB k1 = x rC k2 y Hence, = . d[A] A = – dt = k1[A] + k2[A] = (k1 + k2) [A] 5. SURFACE CHEMISTRY MULTIPLE CHOICE QUESTIONS 1. (a) 2. (c) 3. (c) 4. (a) 5. (c) 6. (b) 7. (b) 8. (d) 12. (a) 13. (c) 14. (b) 15. (b) 16. (b) 9. (b) 10. (b) 11. (b) 20. (b) 21. (b) 22. (c) 23. (d) 24. (a) 28. (b) 29. (b) 30. (d) 31. (a) 32. (a) 17. (d) 18. (a) 19. (d) 36. (a) 37. (a) 38. (c) 39. (d) 40. (a) 44. (c) 45. (a) 46. (d) 47. (c) 48. (c) 25. (b) 26. (d) 27. (b) 52. (c) 53. (c) 54. (c) 55. (b) 56. (b) 60. (d) 61. (b) 62. (a) 63. (c) 64. (d) 33. (b) 34. (b) 35. (c) 68. (d) 69. (b) 70. (c) 71. (a) 72. (b) 76. (d) 77. (b) 78. (d) 79. (d) 80. (b) 41. (a) 42. (d) 43. (d) 84. (c) 85. (c). 86. (a) 87. (a) 88. (c) 92. (d) 93. (b) 94. (a) 95. (c) 96. (d) 49. (a) 50. (a) 51. (a) 100. (c) 101. (c) 102. (a) 103. (d) 104. (d) 108. (d) 109. (d) 110. (b) 111. (d) 112. (c) 57. (c) 58. (b) 59. (a) 116. (b) 117. (b) 118. (a) 119. (a) 120. (c) 124. (a) 125. (d) 126. (a) 127. (a) 128. (d) 65. (c) 66. (a) 67. (d) 132. (d) 133. (d) 134. (d) 135. (c) 136. (a) 140. (c) 141. (c) 142. (c) 143. (d) 144. (c) 73. (d) 74. (c) 75. (b) 148. (a) 149. (a) 150. (c) 151. (b) 152. (c) 156. (a) 157. (b) 158. (a) 159. (d) 160. (b) 81. (d) 82. (c) 83. (d) 89. (b) 90. (c) 91. (d) 97. (c) 98. (c) 99. (d) 105. (b) 106. (d) 107. (d) 113. (c). 114. (c) 115. (d) 121. (a) 122. (d) 123. (c) 129. (c) 130. (b) 131. (d) 137. (a) 138. (c) 139. (b) 145. (a) 146. (d) 147. (b) 153. (a) 154. (a) 155. (d)

208 CHEMISTRY CASE BASED QUESTIONS Case Study 1 1. (b) 2. (c) 3. (b) 4. (c) 5. (a) 4. (c) 5. (b) Case Study 2 1. (d) 2. (b) 3. (b) HINTS/SOLUTIONS 1. Tyndall effect is due to scattering of light by 25. The average molecular mass of bigger particles colloidal particles. is determined by study of osmotic pressure as it is the only colligative property which is 2. It is water in fat type. measurable for solutions containing such macro 3. Physisorption does not need any activation particles. energy. 26. H2 due to small size of molecules readily adsorbs. 4. Soap sol contains negatively charged ionic 27. It is less than that of dispersion medium. 28. Chemisorption needs activation energy and micelles. 5. Colloidal solutions pass through the filter paper. hence does not occur at very low temperature. 6. The colloidal formed has a fluid like appearance. 29. Both are differentiated on the basis of particle 7. Protective power is given in terms of gold size. number. 30. It is due to similar crystal structure as a result 8. Statement is the reason itself. 9. Adsorption is exothermic process due to of which seeding occurs. 31. The light scattered can be seen with high attractive forces between adsorbate and adsorbent. resolution ultramicroscope. 10. Lower is gold number more is stability of colloid. 11. In fog water (liquid) is dispersed in air (gas). 32. Metal hydroxide sols have +ve charge on 12. Lyophillic means affinity for dispersion medium. colloidal particles generally. 13. Colloidal sols have particle size in the range 1–100 nm. 33. Starch sol is macro molecular. 14. During electrophoresis colloidal particles may go to cathode or anode depending on the charge 34. These are purified by ultrafilteration, dialysis on them. or electrodialysis. 15. Sodium reacts with water violently. 16. It is called adsorption isotherm. 35. Starch has affinity for dispersion medium. 17. Charcoal is good adsorbent for easily liquifiable gases. 36. It increase with rise in pressure. 18. Physisorption is due to weak forces usually van der waal’s forces and hence has low heat of 37. Adding hyophilic sol will not neutralise the adsorption. charge on colloidals and hence no precipitation. 19. Lyophilic sols are reversible. 20. The correct sequence ‘b’ is in accordance with 38. Adsorption is exothermic (ΔH < 0) and modern theory of heterogeneous catalysis. spontaneous (ΔG < 0). 21. Molecules move haphazardly. 22. Adsorption is mono molecular due formation of 39. Ferric hydroxide sol is hydrophobic. chemical bonds. 23. There are many methods for coagulation. 40. Peptisation is the process of changing freshly 24. Adsorption decreases with increase in prepared ppt into colloidal solution. temperature. 41. These are heterogeneous. 42. As2S3 sol is hydrophobic. 43. Urea dissolves to form a true solution. 44. It is solid in gas or air type. 45. It is known as foam. 46. The substance that gets adsorbed. 47. 1 nm (= 10–7 cm) – 100 nm = (10–5 cm) 48. Tyndall effect. 49. A substance is classified on the basis of molecular size as macromolecular, multimolecular and associated colloids.

ANSWERS 209 50. PO43– has maximum coagulating power for 77. Easily liquifiable gases adsorb more. Fe(OH)3, a positively charged sol. 78. Hydrophobic sols are irreversible. 79. It is due to double layer formation. 51. Crystallisation cannot seperate colloidal 80. Gold sol is negatively charged. Mg2+ has substances from impurities. maximum charge among the given cations. 52. Solid in gas is Aerosol. 81. Colloidal silver is used for curing eye disease. 82. It is water in oil emulsion. 53. Gel has a rigid appearance and is liquid in solid 83. According to Hardy Schulze rules coagulating type of collidal system. power of an ion depends on type of charge and 54. Statement itself is an explanation. magnitude of charge. 84. CMC is critical micelle concentration, the 55. Electrolytes coagulate colloidal solutions. particles in solution associate and form micelles. 85. Heterogeneous catalysis can be explained on the 56. Soap emulsifies greasy impurities. basis of adsorption. 86. Arsenious sulphide sol is negatively charged sol. 57. Sulphur (S8) is insoluble in water. Smaller the charge on the positive ion, smaller 58. It is determined by the movement of boundary is the coagulating power. during electrophoresis experiment. 87. Fe(OH)3 is a positively charged sol. Comparing the charge on negative ions, Cl– has least charge 59. Hyophobic means hatred for. and hence least coagulating power and maximum coagulation value. 60. Apply Hardy Schulze rule the one with more charge will be the best. 88. In Bredig’s arc method a spark is passed thro’ metal strips dipped in ice cold bath. Hence it 61. AgI particles acquire positive charge due to involves dispersion as well as condensation. adsorption of Ag+ ions from excess AgNO3. 89. Salts in the sea-water coagulate the colloidal 62. Apply Hardy Schulze rules. Phosphate ion has particles present in the river-water. 3 unit negative charge. 90. It is used as a stabilizer. 63. Gas in liquid is example of foam. 91. Soap is an example of associated colloid. 64. Colloidal particles in lyophobic sols have little interaction with the dispersion medium and 92. Gel is a system with liquid as disperse phase hence their particles are not solvated. and solid as dispersion medium. 65. It is called electrophoresis. 93. It is called electro phoreses. 66. These cannot be changed to colloidal solutions 94. It is blue in colour. by simple methods. 95. Emulsifying agent for milk is factose. 67. Emulsion, sol and suspension are all heterogeneous. 96. Minimum amount of protective colloid that should be added to prevent coagulation. 68. Physical adsorption is also called van der Waal adsorption. 97. It arises due to collision between dispersed phase and dispersion medium particles. 69. A liquid in solid gives or rigid appearance colloid called gel. 98. Hydroxide sols carry a positive charge due to adsorption of +ve ions on the precipitate. 70. Glucose NaCl + Ba(NO3)2 in water produces a true solution. 99. According to Hardy Schulze rules coagulating power of an ion depends on type of charge and 71. Freundlich proposed theory for unimolecular magnitude of charge. adsorption. 100. Fog is an example of aerosol (liquid water in air). 72. It has a rigid appearance. 101. Gum sol can easily mix with dispersion medium 73. Due to adsorption of Fe3+ ions, colloidal particles like water etc. would acquire positive charge. 74. In Bredig’s arc method, the metal first gets vaporised (dispersed) and then the vapours are condensed in cold water to form colloidal solution. 75. It is given as millimole/litre. 76. Adsorption is exothermic. ∴ ΔH < 0 and also ΔG < 0 and also ΔS < 0.

210 CHEMISTRY 102. It is less than water. 130. It is used to determine size of colloidal particles. 103. Due to different size light of different colours is 131. Addition of electrolytes will neutralise charge scattered. and hence can cause precipitation. 104. ΔH adsorption of chemisorption > physisorption. 105. Soap sol contains negatively charged ionic 132. Lyophilic sol particles cannot be easily detected by ultramicroscope. micelles. 106. The charge on smoke particle is neutralised and 133. Among gelatin, gum arabic and starch, gelatin has maximum protecting power and hence least these particles are precipitated. gold number. 107. Physical adsorption is non specific. 108. Cheese is not a sol. 134. x = kp1/n or log x = log k + 1 log p 109. For spontaneous process ΔH = –ve but as the m mn gas molecules adsorb therefore ΔS is also –ve. Plot of log x versus log p gives a slope = 1 110. It is called synerisis. m n 111. Hardy Schulze rules. 112. This type of phenomenon is called Thixotropy. and intercept = log k. 113. Sugar solution is a true solution. 114. Sulphide sol is –vely charged. 135. Apply Hardy Schulze rules. 115. High pressure and low temperature. 136. Glucose has aldehydic group therefore shall produce colloidal Ag. 116. Hydrophilic sols are reversible in nature. 137. Statement ‘a’ is a definition itself. 117. Charge on colloidal particles would be negative due to adsorption of I– ions on AgI. 138. Liquid in liquid colloidal is an emulsion and not a gel. 118. Apply Hardy Schulze rules. 139. A gas with high critical temperature has greater 119. Flocculation value is as per option ‘a’ by extent of adsorption. definition. 140. Graph ‘c’ correctly depicts the Freundlich 120. 0.25 g = 250 mg. adsorption isotherm. 121. Soap is sod salt of fatty acids. In aqueous CASE BASED QUESTIONS solution the hydrophobic part of soap interlocks Case Study 1 to form micelles. 1. Higher the critical temperature of the gas larger 122. [Fe(CN)6]4– is most effective according to Hardy will be the extent of its adsorption. Schulze rule. 2. The attractive forces in physisorption are van 123. Apply Hardy Schulze rules. As2C3 is a negative der Waal forces. sol. 3. The statement (b) is against the facts. 124. Applying Fe3+ seals the blood vessel. aP 125. Al(OH)3 sol is a positively charged sol. 4. The equation, x/m = 1 + bP is called Langmuir [Fe(CN)6]4– has maximum charge and hence has maximum coagulating power and minimum absorption isotherm. coagulation value. 5. The plot a gives the correct variation of 126. Precipitation is also called flocculation. physisorption with temperature. 127. Blood is a negatively charged sol. Case Study 2 1. Particle size greater than 103 refer to 128. The expression is x = kp1/n suspension. m 2. The correct name is micelle. 3. Both As2S3 and gold sol are lyophobic sols. or log x = log k + 1 log p. 4. Lower the gold number higher is the protecting mn power. 5. Amount of protective colloids in mg required to 129. Chemisorption needs activation energy. prevent precipitation of 10 ml of standard gold Therefore, at lower temperatures, it shows an increase in extent of adsorption with increase sol = – 0.2 × 1000 × 10 = 25 mg. in temperature. 80

ANSWERS 211 6. GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS MULTIPLE CHOICE QUESTIONS 1. (d) 2. (a) 3. (b) 4. (b) 5. (d) 6. (b) 7. (a) 8. (c) 12. (d) 13. (a) 14. (c) 15. (b) 16. (c) 9. (c) 10. (a) 11. (d) 20. (d) 21. (b) 22. (a) 23. (b) 24. (c) 28. (a) 29. (b) 30. (b) 31. (b) 32. (d) 17. (b) 18. (c) 19. (c) 36. (c) 37. (d) 38. (d) 39. (c) 40. (d) 44. (c) 45. (b) 46. (b) 47. (c) 48. (d) 25. (d) 26. (a) 27. (b) 52. (d) 53. (a) 54. (a) 55. (c) 56. (d) 60. (b) 61. (c) 62. (b) 63. (c) 64. (a) 33. (d) 34. (a) 35. (c) 68. (c) 69. (b) 70. (c) 71. (a) 72. (d) 76. (a) 77. (a) 78. (b) 79. (c) 80. (c) 41. (c) 42. (b) 43. (d) 84. (b) 85. (c) 86. (d) 87. (b) 88. (b) 92. (d) 93. (a) 94. (b) 95. (b) 96. (c) 49. (b) 50. (c) 51. (a) 4. (d) 5. (a) 6. (a) 57. (c) 58. (c) 59. (b) 65. (d) 66. (a) 67. (d) 73. (c) 74. (c) 75. (c) 81. (b) 82. (c) 83. (b) 89. (b) 90. (c) 91. (d) 97. (d) CASE BASED QUESTIONS Case Study 1 1. (b) 2. (c) 3. (b) HINTS/SOLUTIONS 1. The impurities associated with ore after mining 18. Highly electropositive metals such as are called gangue or matrix. aluminium, magnesium (sodium, potassium, calcium, etc.) cannot be reduced with carbon so 2. Igneous rocks. they are obtained by electrolytic reduction and 3. Aluminothermy. not by smelting. 4. It is an example of hydrometallurgy because the 19. Flux. ore is first converted into aqueous solution form 20. Blast furnace. and the metal is precipitated. 21. Roasting. 5. Sodium being highly reactive metal never occurs 22. Alkali metals are very reactive. in free state. 23. KCl . MgCl2 . 6H2O. 6. Froth-floatation process is applied for sulphide 24. Zone refining method is used for obtaining ores. 7. Ores are the minerals from which metals can metals in high state of purity. be extracted conveniently and economically. 25. Fluorspar is CaF2. 8. Impurities + flux → slag. 26. Al reacts with air to form Al2O3 which protects 9. Dolomite is CaCO3 . MgCO3. 10. CaSO4 . 2H2O is gypsum. the metal from further corrosion. 11. Flux combines with impurities to form slag. 27. Leaching. 12. Flux. 28. Microcosmic salt is Na(NH4)HPO4.4H2O. 13. Bauxite. 29. Roasting. 14. Hoope’s process. 15. Pyrolusite is MnO2. 30. Al2O3 + 2NaOH ⎯→ 2NaAlO2 + H2O. 16. Haematite, an oxide ore of iron is concentrated by gravity separation method. (Soluble) 17. During smelting, ore in molten state is reduced with coke. 31. CaCO3 on heating gives CaO, which is basic in nature. It can react with acidic impurities to form slag. 32. Bauxite ore is concentrated by leaching. 33. Malachite, Cu(OH)2 . CuCO3, is an ore of copper.

212 CHEMISTRY 34. Nitriding is the process employed for surface 62. During electrorefining of copper, the less hardening of articles of alloy steels. During the reactive metals such as silver and gold fall below process articles are heated in an atmosphere of anode in the form of anode mud. ammonia, which results in the formation of nitrides of metals such as Cr, Mo, etc. on the 63. Anode mud contains less reactive metals such surface. as gold and silver whereas more reactive metals such as Fe and Mg go into solution as their ions. 35. Quenched steel has high carbon content and hence is very hard and brittle. 64. Pig iron is converted into steel using a Bessemer converter. 36. Iron is protected from rusting by coating it with a thin layer of zinc. The process is known as 65. 0.1 – 1.5 %. galvanization. 66. CaCO3 ⎯→ CaO + CO2 37. Conc. HNO3. CaO + SiO2 ⎯→ CSalSaigO3 38. Cu(OH)2 . CuCO3. 39. Silver occurs in combined state (as argentite, 67. Gold has highest density among the given metals. Ag2S and horn silver, AgCl) as well as in native state. 68. Alumina dissolves in molten cryolite to form a melt at lower temperature. 40. Brass contains about 60% copper and 40% zinc. 69. CaO + SiO2 ⎯→ CaSiO3 41. Zincite is ZnO. FeO + SiO2 ⎯→ FeSiO3. 42. Wrought iron is the most pure form of iron. 70. Highly electropositive metals such as Al cannot be purified by this method. 43. Collectors actually increase the non wettability of the mineral particles. 71. Silver. 44. Nickel is purified by Mond’s process and not by 72. They are hydrophobic. Van Arkel process. 73. Alkali metals are extracted by electrolysis of 45. Cu2S + 2Cu2O ⎯→ 6 Cu + SO2 their fused halides. 46. 2[Au(CN)2]– (aq) + Zn (s) ⎯→ [Zn(CN)4]2– (aq) + 74. Cr2O3 + Al ⎯→ Al2O3 + Cr 2 Au(s). Here, Al is oxidized to Al2O3 and acts as reducing agent. 47. Na[Ag(CN)2]. 48. Aluminium bronze contains about 90% copper 75. 2Cu2O + Cu2S ⎯→ 6Cu + SO2. 76. Ignition mixture contains about 15 parts of and 10% aluminium. barium peroxide and 2 parts of magnesium 49. Aluminium and iron are the two most abundant powder. metals in the earth’s crust. 77. The statement (a) is correct 50. Titanium and Zirconium are purified by Van Arkel method. 78. CaF2 makes the mixture more conducting. 79. The statement (c) is correct. 51. Al3+ is reduced to Al at cathode. 80. Lime stone on heating yields CaO, which is basic 52. 2Cu2O + Cu2S ⎯→ 6Cu + SO2 in character. CaO combines with acidic Copper (I) oxide Copper (I) sulphide impurities (such as SiO2) to form slag. 53. Copper matte is mainly Cu2S and FeS. 81. Highly electropositive metals are obtained by 54. Copper and tin. electrolytic reduction because they, themselves being very strong reducing agents, cannot be 55. While vitriol is ZnSO4 . 7H2O. obtained by chemical reduction. 56. Cassiterite is SnO2. 57. Cinnabar on roasting is directly reduced to 82. This is Van-Arkel process for purification of metals. mercury. 83. FeO + SiO2 ⎯⎯→ FeSSlaigO3 58. During extraction of iron limestone is used as 84. Argentite is Ag2S. flux. It removes SiO2 (impurity) as slag. 85. The statement (c) is correct. 59. Silica (flux) removes FeO as FeSiO3 (slag). FeO + SiO2 ⎯→ FeSiO3. 60. Ag2S + 4NaCN ⎯→ Na[Ag(CN)2] + Na2S Na2S + 2O2 + H2O ⎯→ Na2S2O3 + 2NaOH The complex Na[Ag(CN)2] contains [Ag(CN)2]– ions. 61. Tempering reduces the hardness of quenched steel.