RI 2023 Y5 Promo Solutions

© Raffles Institution 2023 9729/S/23 Page 1 2023 Y5 H2 Chemistry Promotion Examination – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B C C D A B B D B C A C D C A MCQ worked solutions A1 (B) After cooling back to room temperature and pressure, H2O condenses back to liquid state. 4CH3NH2(g) + 9O2(g) → 4CO2(g) + 2N2(g) + 10H2O(l) Initial V/ cm3 20 60 0 0 0 Change in V/cm3 −20 −45 +20 +10 − Final V/ cm3 0 15 20 10 − Final total volume = 15 + 20 + 10 = 45 cm3 A2 (C) Mol ratio of MnO42− to MnO4− = 3 : 2 2 mol of MnO42− is oxidised to 2 mol of MnO4−. Since 3 mol of MnO 42− underwent disproportionation, 3−2 = 1 mol of MnO 42− is reduced to the manganese- containing product. Oxidation no. increased from +6 to +7 from MnO42− to MnO4−. Hence, amt of e − lost by 2 mol of MnO 42− = 2 mol Total amt of e− lost = total amt of e− gained, Amt of e− gained by 1 mol of MnO42− = 2 mol Hence, the oxidation number of Mn decreases by 2 to form the manganese- containing product. Oxidation number of the product = +6 − 2 = +4 The only manganese- containing product in the options with a oxidation number of +4 is MnO2. A3 (C) The data for gold should be ignored as the question is about the Ar value for copper. A r = (63×0.62 + 65×0.32) ÷ 0.94 = 63.7 A4 (D) Option 1 is incorrect as 36S has a nucleon number of 36 while 37Cl has a nucleon number of 37 . The charge of the ion indicates the number of electrons lost or accepted and does not affect the nucleon number. Option 2 is correct as the S atom accepts 2 electrons to form an ion with an outer electronic configuration of 3s 2 3p6, while the Cl atom accepts 1 electron to form an ion with an outer electronic configuration of 3s2 3p6 Option 3 is in correct. 36S2− has 16+2 = 18 electrons and 36−16= 20 neutrons . 37Cl− has 17+1 = 18 electrons and 37−17= 20 neutrons. A5 (A) As Ba is in group 2, it loses 2 valence electrons to form Ba2+. Hence, the anion has a charge of 2−. In the O22− anion, each of the 2 electrons from Ba is accepted by each of the 2 oxygen atoms. A6 (B) Recall that a gas deviates from ideal gas behavior at high pressure and low temperature. The higher the pressure and the lower the temperature, the greater the deviation from ideality.

© Raffles Institution 2023 9729/S/23 Page 2 A7 (B) Assuming ideal gas behaviour, pV = nRT n = pV RT Since pressure is kept constant at 1 atm, no. of moles of air in the heated balloon initial no. of moles of air in the balloon = 101325 × 5500 R × (65 + 273) × R × (25 + 273) 101325 × 5000 = 0.97 Note: Since p and R are constants, students can also make use of n ∝ V T and thus n2 n1 = V2 T2 × T1 V1 . A8 (D) COCl2(g) ⇌ CO(g) + Cl2(g) Kc = a = [CO][Cl2] [COCl2] Let Kc’ be the equilibrium constant of the reaction below.