NJC 2016 A Level Chem Solution

H2 Chemistry TYS 2016 suggested answers Questions that are no longer in syllabus Paper 1 Paper 2 5(b)(ii) Paper 3 Paper 1 Answer Key 1 B 11 C 21 B 31 B 2 D 12 A 22 B 32 A 3 C 13 C 23 B 33 C 4 D 14 B 24 A 34 D 5 A 15 D 25 D 35 C 6 C 16 A 26 C 36 B 7 C 17 A 27 A 37 C 8 D 18 A 28 B 38 B 9 B 19 D 29 B 39 A 10 D 20 D 30 B 40 C 2016 A level Paper 1 suggested answers 1) Amount of molecule = 1.0 𝑥 mol Amount of atom = (1.0 𝑥 × 2)mol No of atoms = 2.0 𝑥 × L Ans: B 2) 2H2S + CS2 + 6O2  CO2 + 4SO2 + 2H2O Ans: D 3) From the data, the isotopes of copper have mass number of 63 and 65. Note: mass number 197 is Au Ar of Cu = 63×65+65×29 94 = 63.6 Ans: C 4) No. of protons No. of neutrons No. of electrons 36S2 16 20 18 37Cl 17 20 18 Nucleons = protons + neutrons (different nucleon no. for the 2 ions) The ions have the same outer electronic configuration of 3s23p6 Ans: D

5) Ans: A 6) Possible structures of C2N Ans: C 7) pV = nRT = 𝑚 𝑀𝑟 RT (Note: Vol of gas collected = 782 = 76 cm3) Mr = 𝑚𝑅𝑇 𝑝𝑉 = 0.293×8.31×(97+273) (101000)×(76×10−6) = 117 Ans: C 8) Lattice Energy is the amount of energy released when 1 mole of ionic solid is formed from its gaseous ions. Ans: D 9) By Hess' Law, Hf KCl(s) = Hat K(s) + Hat Cl2(g) + 1st I.E.of K(g) + 1st E.A. of Cl(g) + L.E. of KCl(s) Hf KCl(s) = +90 + ½ ×(+242) + (418) + (355) + (710) = 436 kJmol1 Ans: B Energy / kJ mol1 0 K (s) + ½ Cl2 (g) KCl (s) (s)KCl ΔH θ f K (g) + ½ Cl2 (g) (s)K ΔHθ at (g)K of I.E. 1st K (g) + Cl (g) K+ (g) + e + Cl (g) K+ (g) + Cl − (g) (g) Cl E.A. 1st (s)KC of L.E. l (g) Cl H 2 θ at

10) S > 0 as there is an increase in number of gaseous molecules after the reaction G = H  TS For G to be > 0 at 378 K, H must be a positive value. Ans: D 11) Eocell = Eo (Ag+/Ag)  Eo (Fe3+/ Fe2+) = + 0.03V. When [Fe3+] increases, position of eqm Fe3+ + e Fe2+ shifts to the right, E(Fe3+/ Fe2+) increases and thus Eocell could become 0.00V. Ans: C (Note: Surface area of electrode would not affect Eo values) 12) [O] 2O2  O2 + 4e Q = It = neF ne = 8×60×100 96500 = 0.4974 mol nO2 = 0.4974 4 = 0.12435 mol VO2 = 0.12435 × 22.7 = 2.82 dm3 Ans: A 13) Let solubility of ZnF2(s) be s mol dm3 ZnF2(s) + aq Zn2+(aq) + 2 F (aq) Initial / mol dm3  0 0 Change / mol dm3 s +s +2s Eqm / mol dm3  s 2s Ksp = [Zn2+][F]2 = 4s3 s = 0.2 mol dm3 [F] = 2s = 0.4 mol dm3 For precipitation of BaF2, [Ba2+][F]2 = Ksp and [F] = 0.4 mol dm3 [Ba2+] = 1.6×10−7 (0.4)2 = 1×106 mol dm3 Ans: C 14) For weak acid, [H+] = √𝐾𝑎×𝑐 = 1×103 mol dm3 pH = lg [H+] = 3 Ans: B 15) Rate = k[IO3][I]2[H+]2 Expt 1: y = k (a)(a)2(a)2  eqn (1) Expt 2: rate 2 = k (1 2 a)(2a)2(3a)2  eqn (2) 𝑒𝑞𝑛 2 𝑒𝑞𝑛 1 : 𝑟𝑎𝑡𝑒 2 𝑦 = 𝑘×18𝑎5 𝑘×𝑎5 Rate 2 = 18y Ans: D

16) Both Al2O3 and MgO