23MIPrelim Answers (H2 Chem Paper 1)

Class Adm No Candidate Name: This question paper consists of 14 printed pages and 2 blank pages. 2023 Preliminary Examination Pre-University 3 H2 CHEMISTRY 9729/01 Paper 1 Multiple Choice 20th Sep 2023 1 hour Additional materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class and admission number in the spaces provided at the top of this page and on the Multiple Choice Answer Sheet provided. There are thirty questions on this paper. Answer ALL questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the Multiple Choice Answer Sheet provided. Read the instructions on the Multiple Choice Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this question paper. The use of an approved scientific calculator is expected, where appropriate. FOR EXAMINER’S USE TOTAL (30 marks)

2 1 A 6 D 11 C 16 B 21 C 26 B 2 C 7 B 12 D 17 D 22 D 27 C 3 D 8 C 13 A 18 D 23 A 28 C 4 D 9 B 14 B 19 A 24 B 29 B 5 B 10 C 15 D 20 C 25 A 30 A Total: A 6 B 8 C 8 D 8 For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 Which of the following contain the same number of hydrogen atoms? 1 0.10 mol of pentane 2 0.15 mol of but-2-ene 3 0.60 mol of hydrogen gas 4 0.50 mol of steam A 1, 2 and 3 only B 1 and 4 only C 2 and 4 only D 1 and 2 only Pentane is C5H12 amt of H atoms = 0.10 × 12 = 1.2 mol But-2-ene is C4H8 amt of H atoms = 0.15 × 8 = 1.2 mol H2 amt of H atoms = 0.60 × 2 = 1.2 mol H2O amt of H atoms = 0.5 × 2 = 1.0 mol (A)

3 [Turn over 2 Ascorbic acid found in fruits and vegetables is believed to help prevent colds. Analysis of ascorbic acid shows that it contains only carbon, hydrogen and oxygen with 40.91% carbon and 4.55% hydrogen by mass. Ascorbic acid has a molar mass of 176 g mol–1. What is the molecular formula of ascorbic acid? A C2H3O2 B C3H4O3 C C6H8O6 D C6H9O6 Draw up the empirical formula table first C H O % by mass 40.91 4.55 100–40.91– 4.55 = 54.54 amt 40.91/12.0 4.55/1.0 54.54/16.0 =3.409 =4.55 3.409 Simplest ratio 1 = 3 1.33 = 4 3 = 3 Empirical formula = C3H4O3 Let molecular formula be (C3H4O3)x x (3×12.0 + 4 × 1.0 + 3 × 16.0) = 176 x = 2 Hence molecular formula is C6H8O6. Note: 1.33 for H should not be rounded to 1.5 because 1.33 is equivalent to 4/3 hence all values should be multiplied by 3 throughout. (C) 3 Which of the following species is unable to act as a Lewis acid? A AlCl3 B FeCl3 C BeCl2 D CCl4 Lewis acid is an electron pair acceptor. CCl4 does not have empty orbitals to accept any electron pairs. (D)

4 4 In 2020, a massive explosion took place in Beirut at a warehouse which stocked ammonium nitrate. Ammonium nitrate, NH4NO3 decomposes explosively when heated to form nitrous oxide, N2O and water, H2O. What is the change in the oxidation number of the nitrogen atom that is underlined when this reaction happens? A –2 B +1 C +3 D +4 Oxidation number of N in NH4+ = –3 Oxidation number of N in N2O = +1 Change = +4 (D) 5 Use of the Data Booklet is relevant to this question. What do the ions 31P3– and 32S2– have in common? A Both ions have more neutrons than protons. B Both ions have more electrons than neutrons. C Both ions have more protons than electrons. D Both ions have an outer electronic configuration of 2s2 2p6. 31P3– has 15 protons, 16 neutrons and 18 electrons 32S2– has 16 protons, 16 neutrons and 18 electrons Only 31P3– has more neutrons than protons.(16 > 15) A is incorrect Both ions have more electrons than neutrons. (18 > 16) B is correct Both have more electrons than protons (18> 15 or 16) C is incorrect Both P and S are in period 3 so their outer electronic configuration would be 3s2 3p6 D is incorrect (B)

5 [Turn over 6 The table below shows the ionisation energy (in kJ mol–1) of four elements labelled A, B, C and D. Element 1st I.E. 2nd I.E. 3rd I.E. 4th I.E. 5th I.E. 6th I.E. A 1012 1907 2914 4964 6274 21 267 B 787 1577 3232 4356 16 091 19 805 C 1521 2666 3931 5771 7238 8781 D 738 1451 7733 10 543 13 630 18 020 Which element above has no electrons in the valence p subshell? A element A B element B C element C D element D Look for the large jump that indicates removal of electron from the inner principal quantum shell. Element A large jump between 5th and 6th ➔From group 15 Element B large jump between 4th and 5th ➔From group 14 Element C no visible large jump from first 6 electrons ➔From group 17 or 18 Element D large jump between 2nd and 3rd ➔From group 2 (does not have p electrons) (D) 7 Which molecules contain at least one bond angle of 120°? 1 C2H4 2 PCl5 3 NF3 A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only C2H4 is trigonal planar about both C atoms ➔ bond angle is 120° PCl5 is trigonal bipyramidal about P ➔ bond angle is 120° at the equatorial plane NF3 is trigonal pyramidal about N ➔ bond angle is 107° (B)

6 8 Elements E, F and G are all in the first two periods of the Periodic Table. Their Pauling electronegativity values, EN, are given in the table. element EN E 1.0 F 2.5 G 3.0 Compounds exist with formulae EG, FG and G2. What is the correct order of increasing boiling point of the three compounds? A EG → FG → G2 B EG → G2 → FG C G2 → FG → EG D G2 → EG → FG EG with big difference in electronegativity ➔ ionic compound with ionic bonds FG with small difference in electronegativity ➔ covalent compound. Polar ➔ pd–pd G2 with no difference in electronegativity ➔ covalent compound. Non polar ➔ id–id G2 lowest followed by FG followed by EG as the highest. (C) 9 When an evacuated glass bulb of volume 300 cm3 is filled with a gas at 450 K and 101 kPa, the mass of the bulb increases by 0.679 g. The gas obeys the ideal gas equation. What is the identity of the gas? A argon B krypton C neon D nitrogen Using pV = nRT pV = (M/Mr)RT Mr = MRT/pV Mr = 0.679 × 8.31 × 450/ (101 × 103 × 300 × 10–6) = 83.8 (B)

7 [Turn over 10 Which graph shows the correct plot of pV against p for a fixed mass of ideal gas at two temperatures, T1 and T2, in which T1 > T2? A B C D For a fixed mass of gas, n is constant

8 pV = nRT for T1 and T2, pV = constant x (fixed value of T) pV will be a constant for all values of p ➔ horizontal straight line since T1>T2 ➔ p1V1 > p2V2 T1 line will be higher than T2 line (C) 11 A reaction pathway diagram for the reaction of aqueous barium hydroxide and dilute hydrochloric acid is shown. What is the value of the enthalpy change of neutralisation, Hneut? A c B c – d C c 2 D (c – d) 2 c is the H for the reaction and d is the activation energy so we should focus on c. We are interested in Hneut which is for one mole of water formed. Since the reaction forms two moles of water, the value of Hneut will be half of c or c 2. (C)

9 [Turn over 12 The enthalpy change of reaction, ∆Hro, for the following reaction is e kJ mol–1. 2C(s) + 3H2(g) + 3½O2(g) → 2CO2(g) + 3H2O(l) ∆Hro = e kJ mol–1 What is the correct expression to calculate the value of e? A 2 × ∆Hco [CO2(g)] – 3 × ∆Hfo [H2(g)] B 2 × ∆Hco [CO2(g)] + 3 × ∆Hfo [H2O(g)] C 2 × ∆Hfo [CO2(g)] – 3 × ∆Hfo [H2(g)] D 2 × ∆Hfo [CO2(g)]