2019 NJC Prelim P1 ANS

[Turn over Answer Keys to 2019 SH2 H2 Chemistry Prelim MCQ 1 A 6 C 11 B 16 B 21 B 26 C 2 A 7 A 12 D 17 B 22 C 27 A 3 B 8 C 13 B 18 D 23 C 28 C 4 C 9 D 14 C 19 D 24 A 29 D 5 C 10 C 15 D 20 A 25 C 30 C 1 Amount of Ag deposited = 0.216 107.9 = 0.002002 mol Amount of Ag per cm2 = 0.002002 150 = 0.00001335 mol per cm2 No of Ag atom per cm2 = 0.00001335 × 6.02 × 1023 = 8.034 × 1018 atom per cm2 Ans: A 2 Angle of deflection ∝ charge mass Mass of electron is approximately 1 2000 the mass of a proton, hence electron has the largest angle of deflection. Ans: A 3 Amount of SO32− = 25.0 1000 × 0.10 = 0.0025 mol Amount of electron transferred = 0.0025 × 2 = 0.005 mol Amount of M3+ reacted = 50.0 1000 × 0.10 = 0.005 mol Mol ratio M3+ : e− = 1 : 1 During reduction, 1 mol of M3+ gains 1 mol of e−. Oxidation state of M3+ decreases by 1 unit from +3 to +2. Ans: B 4 The sharp rise between 2 nd and 3 rd I.E. shows that there are 2 valence electron in the outermost shell. Hence it is in Group 2. The next inner quant um shell contains 8 electrons, for d -block element, the next inner quantum shell should contain 3s, 3p and 3d electrons, which would be more than 8. It cannot be a Period 3 Group 2 element as Mg only has 12 electrons. Ans: C

2 NJC/H2 Chem Prelim/01/2019 5 The species are isoelectronic (same number of electrons) with same shielding effect. Nuclear charge increases from Ar < K+ < Ca2+ Hence nuclear attraction towards valence electrons increases from Ar < K+ < Ca2+. Ca2+ would have the smaller radius while Ar has the largest radius. Ans: C 6 Bond angle 1 is 109.5 (4b.p. 0l.p. → tetrahedral) Bond angle 2 is 120 (3b.p. 0l.p. → trigonal planar) Bond angle 3 is 104.5 (2b.p. 2l.p. → bent) Ans: C 7 p1V1 = nRT1 p2V2 = nRT2 𝑝1𝑉1 𝑝2𝑉2 = 𝑇1 𝑇2 (1×105)×𝑉1 𝑝2×4𝑉1 = (20+273) (100+273) p2 = 31830 Pa Ans: A

3 NJC/H2 Chem Prelim/01/2019 [Turn over 8 CO(g) + ½ O2(g) → CO2(g) Combustion is an exothermic process, this means that energy level of CO2 is lower than CO. Hence option 1 is correct. Option 2 is wrong. ΔHf CO2(g) should be more negative than that of CO(g). Since the combustion spontaneous (burns readily), the Kc for the equilibrium would be a very large value. Ans: C 9 ∆Ho1 = 52.2 + 175.8 = +228 kJ mol−1 Ans: D 10 Option A is wrong. Increase in temperature would increase the kinetic energy. Option B is wrong. Catalyst increases the frequency of effective collision. Option D is a correct statement but does not explain “how” fumarase speed up the reaction. Option C is correct. Fumarase is an enzyme that provides alternative reaction pathway with lower activation energy. Ans: C

4 NJC/H2 Chem Prelim/01/2019 11 Rate of reaction = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙 𝑜𝑓 𝑀𝑛𝑂4 − 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 Expt Volume / cm3 Time / s rate C2O42−(aq) MnO4−(aq) Mn2+(aq) water 1 20 30 5 25 30 1 2 20 30 10 20 15 2 3 20 15 10 35 15 1 4 20 20 20 20 10 2 Comparing expt 1 & 2, [C2O42−] and [MnO4−] is kept constant but when [Mn2+] × 2, rate × 2. Hence it is first order w.r.t. Mn2+. Option 1 is correct. Comparing expt 2 & 3, [C2O42−] and [Mn2+] is kept constant but when [MnO4−] ÷ 2, rate ÷ 2. Hence it is first order w.r.t. MnO4−. We cannot determine the order of reaction w.r.t. C2O42−. Rate = k[Mn2+] [MnO4−] [C2O42−]x The slow step would involve one Mn 2+ and one MnO4− ion and maybe C2O42− ion. Option 2 is correct. We cannot deduce the order of reaction w.r.t. C2O42− and hence we cannot conclude the units of the rate constant. Ans: B 12 Lewis base is an electron pair donor. In reaction 1 and 2, NH3 donates electron pair to H+. In reaction 3, NH3 donates electron pair to Ag+. Ans: D

5 NJC/H2 Chem Prelim/01/2019 [Turn over 13 CuS (s) ⇌ Cu2+ (aq) + S2− (aq) s s Ksp = s2 s = √(8.5 × 1045) = 9.22 × 1023 mol dm−3 Ag2S (s) ⇌ 2Ag+ (aq) + S2− (aq) 2s s Ksp = (2s)2(s) 4s3 = 1.6 × 1049 s = 3.42 × 1017 mol dm−3 Bi2S3 (s) ⇌ 2Bi3+ (aq) + 3S2− (aq) 2s 3s Ksp = (2s)2(3s)3 108s5 = 1.1 × 1073 s = 1.00 × 1015 mol dm−3 Ans: B 14 At cathode (−ve electrode): Species present: Na+ and H2O [R] 2H2O + 2e− → H2 + 2OH− Blue litmus remains blue At anode (+ve electrode): Species present: Cl− (concentrated) and H2O Note: Concentrated Cl− will be oxidized in preference over H2O. [O] 2Cl−→ Cl2 + 2e− Chlorine bleaches the litmus paper. Ans: C 15 Statement A is correct. SiCl4 + 2H2O → SiO2 + 4HCl Statement B is correct. Al2O3 is amphoteric oxide that can react with both acid and base to give soluble products. Statement C is correct. Na2O + H2O → 2NaOH SO3 + H2O → H2SO4 2NaOH + H2SO4 → Na2SO4 + 2H2O Statement D is wrong. Acidity of Period 3 chlorides in water increase from NaCl to PCl5. pH of the resultant solution decreases from pH 7 to pH 1. Ans: D

6 NJC/H2 Chem Prelim/01/2019 16 M(NO3)2(s) → MO(s) + 2NO2(g) + ½ O2(g) Mass loss is due to NO2 and O2. We can assume 3.29 g of N2O5(g) is loss. M(NO3)2(s) → MO(s) + N2O5(g) Amount of N2O5 = 3.29 14.0×2+16.0×5 = 0.03046 mol Amount of M(NO3)2 = 0.03046 mol Mr of M(NO3)2 = 5.00 0.03046 = 164.1 Ar of M = 164.1 – 2×(14.0+16.0×3) = 40.1 M is calcium. Ans: B 17 Statement A is wrong. Ag+ is reduced. Statement B is correct. Cu+ is used up in reaction 2 and regenerated in reaction 3. Statement C is wrong. Cu+ acts as a reducing agent in reaction 2. Statement D is wrong. The depth of colour should be related to concentration of Ag(s). Ans: B 18 Oxides of nitrogen can contribute to  formation of smog through reaction with other air pollut ants. PAN is a major component of smog which is harmful to plants and humans. Hydrocarbons + O2 + NO2 + light  peroxyacetyl nitrate (PAN)  formation of ozone at lower atmosphere. High concentration of ozone at lower atmosphere causes respiratory problems. NO2  NO + O O + O2  O3  NO2 acts as a catalyst for the oxidation of SO2 to SO3, SO2 + ½ O2 SO3 NO2 + SO2  SO3 + NO NO + ½ O2  NO2 (regenerated) SO3 dissolves in rainwater to cause acid rain (H2SO4). Ans: D

7 NJC/H2 Chem Prelim/01/2019 [Turn over 19 Option 1 is correct. The molecule contains a secondary alcohol group. Option 2 is wrong. The C=C bond consists of one σ and one  bond. Option 3 is wrong. Most of the C atoms are sp 3 hybridized with tetrahedral geometry. This shows that the C atoms are not in the same plane. Ans: D 20 The C=C bond between C11 and C12 is in a cis arrangement. Considering the following structure with aldehyde and cyclohexene ring, O H R The long aliphatic side chain has a formula of C13H19. When drawing out, there will need to have 4 C=C in the long aliphatic side chain. Total number of C=C in the molecule = 5 Ans: A 21 CH3 CH3CH3 O OH O O CH3 O OH O 2 4 + 8CO2 + 4H2O+ [O] Ans: B 22 The example shows that CN− can undergo nucleophilic substitution with C−Br at a faster rate than I−. Option C is the best explanation. Ans: C

8 NJC/H2 Chem Prelim/01/2019 23 Br CH2CH2 Br Nu Sub with CN C CH2CH2 C NN acidic hydrolysis C CH2CH2 C OHHO O O Ans: C 24 Reaction A can occur at room temperature. Reaction B requires heating with ethanolic KCN for nucleophilic substitution. Reaction C requires heating with concentrated H2SO4 for elimination. Reaction D requires heating for hydrolysis. Ans: A 25 Ethanol, C 2H5OH (M r = 46.0), reacts with hot acidified KMnO 4 to gives ethanoic acid, CH3COOH (Mr = 60.0). Amount of ethanol = 2.76 46.0 = 0.06 mol Theoretical yield of ethanoic acid = 0.06 × 60.0 = 3.6 g Mass of 75% yield of ethanoic acid = 0.75 × 3.6 = 2.7 g Ans: C 26 Compound N  does not contain carboxylic acid/p