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