2022 HCI Prelim P2 Mark Scheme

2022 HCI C2 H2 Chemistry Prelims / Paper 2 HWA CHONG INSTITUTION 2022 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS (PAPER 2) 1 (a) (i) Hydrocarbons / Alkanes / Alkenes / Arenes [1] 1 (a) (ii) N2 + O2 → 2NO [0.5] 2NO + O2 → 2NO2 [0.5] 1 (a) (iii) Forms photochemical smog / leads to acid rain [1] 1 (b) (i) pV = nRT n(gas) = pV ÷ RT = (10 × 105)(63.2 ÷ 1000) ÷ (8.31)(900+273) = 6.48 mol [1] correct substitution and conversions [1] answer (ecf) 1 (b) (ii) Amount of SO2 = 0.0183 ÷ 64.1 = 2.85 × 10–4 mol Number of SO2 particles = 2.85 ×10–4 × L (where L is Avogadro constant) Number of particles in fuel exhaust = 6.48 × L Amount of SO2 in ppm = [(2.85 × 10–4 × L) ÷ (6.48 × L)] × 1 × 106 = 44.1 ppm The sample meets the sulfur emission standards [1] Finding the number of SO2 particles [1] Calculating the concentration of SO2 in ppm and correct conclusion (ecf given from part (b)(i)). ([–0.5] if comparison is correct but answer in ppm is absent)

1 (c) • SO3 dissolves in water in a violent / very exothermic reaction • forming a strongly acidic solution around pH 1 • SO3(l) + H2O(l) → H2SO4(aq) • Na2O dissolves completely in water in a vigorous / exothermic reaction • forming a strongly alkaline colourless solution around pH 13 • Na2O(s) + H2O(l) → 2Na+(aq) + 2OH−(aq) • Al2O3 is insoluble / has no reaction in water • The resulting mixture is neutral / pH 7 8 points – [4] 5 to 7 points – [3] 4 points – [2] 2 to 3 points – [1] 1 point – [0.5] [4] total 2 (a) (i) Intermediate A [1], step 1: C l2(aq) [1], step 2: excess conc H 2SO4, heat OR concentrated H3PO4, heat OR Al2O3, heat [1] WRONG INTERMEDIATE A LOSES MARKS FOR STEP 1 AND STEP 2. Intermediate B [1]

2022 HCI C2 H2 Chemistry Prelims / Paper 2 2 (a) (ii) The –CyO2H is nearer the electron withdrawing amine group, that better disperses the negative charge on the conjugate carboxylate (–CyO2–), stabilising th is conjugate carboxylate more, this makes the –CyO2H group more acidic and hence is deprotonated first. All 4 points below = [2]. 2 out of 4 points = [1] - amine group (NH2) is nearer –CyO2H than to –CxO2H (compares distance) - amine group is electron withdrawing - negative charge on the conjugate base (–CyO2–) is better dispersed - conjugate base of –CyO2H is more stable (than that of –CxO2H) 2 (a) (iii) Naturally occurring MSG consists of only 1 enantiomer. MSG synthesised in the laboratory will be a racemic mixture, the enantiomer produced in lab synthesis may interact with the taste bud differently. [1] for realising the lab synthesis results in a racemic mixture, or at least acknowledge different ratios of enantiomers may be produced in lab synthesis vs the natural process. 2 (b) Relative mass of MSG = 183 Amount of Na+ in 1 g of MSG = 1 ÷ 183 = 5.46 × 10–3 mol [1] Relative mass of NaCl = 58.5 Amount of Na+ in 1