RI 2018 Prelim P1 Ans

1 2018 Y6 H2 Chemistry Preliminary Examinations Paper 1 (Suggested Solutions) Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C A C C C A B B D C A C A D B Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer A C D B D A A B D D C B D D A Q1(C) Since the Ar of Fe is closest to 55.94, the most abundant isotope is 56Fe   abundance of 56Fe = 91.8  abundance of 54Fe & 58Fe = 100 – 91.8 = 8.2% Let the  abundance of 54Fe be x%  abundance of 58Fe = 8.2 – x % = z% Ar of Fe = (53.94)(x) + (55.94)(91.8) + (8.2−x)(57.93) 100 = 55.849  abundance of 54Fe = x = 6.37 %  abundance of 58Fe = 8.2 – x = 1.83 % Q2(A) S2Clx disproportionates completely in water to give S:SO2 = 3:1 = 6:2 Writing and balancing half-equations: [R] 3S2Clx + (3x)e− ⎯→ 6S + (3x)Cl− [O] S2Clx + 4H2O ⎯→ 2SO2 + (8−x)e− + 8H+ + xCl− electrons loss = electrons gain 3x = 8−x x = 2 Q3(C) Angle of deflection ∝ charge mass A Angle of deflection ∝ 1/(15 + 15) = 0.0333 B Angle of deflection ∝ 3/(12 + 16 + 16) = 0.0682 C Angle of deflection ∝ 2/(12 +16) = 0.0714 (greatest value) D Angle of deflection ∝ 2/(16 + 18) = 0.0588 Q4(C) Mass number of E = 241 − 4 = 237 Number of protons in E = 95 − 2 = 93 Number of neutrons in E = 237 − 93 = 144 Q5(C) A Only G has an aldehyde functional group that can be oxidised by potassium dichromate (VI). B Both F and G exhibit intramolecular hydrogen bonding C Both F and G have 8 𝜋 electrons (6 in benzene ring and 2 in C=O). D The carbon on the methyl side chain of F is sp 3 hybridised and tetrahedral in shape, hence F is not planar. G is planar as all the carbon atoms are sp2 hybridised (trigonal planar) and the O atom has a bent shape. Q6(A) Manipulate the ideal gas equation, pV = nRT A p T = nR V = nR (1 V) Since p T ∝ ( 1 V), A is correct. B V T = nR P = constant The graph should be a vertical straight line. C pV = m M RT, where n = m M p = (m V) RT M Since 𝜌 = m V , p = 𝜌 RT M 𝜌 = ( M RT) p Since M RT is constant, gradient is constant and it should be a straight line through origin. D pV = nRT = (nR)T The y-intercept should be 0, i.e. a straight line through origin, as temperature is measured in Kelvins. Q7(B) There are 2 propagation steps that lead to the formation of bromomethane, which explains the presence of 2 peaks (i.e. 2 transition states). The 2 propagation steps are: (a) CH4 + Cl• ⎯→ •CH3 + HCl ∆Ha = [BE(C−H)−BE(H−Cl)] =410−431= −21 kJ mol−1

2 (b) •CH3 + Cl2 ⎯→ CH3Cl + Cl• ∆Hb= [BE(Cl−Cl)−BE(C−Cl)] = 244−340= −96 kJ mol−1 ∆Hprop = ∆Ha + ∆Hb= −117 kJ mol−1 Hence, the propagation step is exothermic. The products formed are at a lower energy level than the reactants. Q8(B) A The maximum attainable oxidation number of each element corresponds to the number of valence electron(s) in each atom of the element, i.e. all the valence electrons can be us