2024 RI H1Chem Prelims P1 P2 Answers

© Raffles Institution 2024 8873/02/S/24 1 Q1 One mole of any substance contains 6.02 x 10 23 (Avogadro’s constant) particles of the substance. D is correct. Since ethanal exists as a molecule, one mole of ethanal contains 6.02 x 1023 ethanal molecules. Option A is incorrect. Since each molecule of ethanal contains 7 atoms, number of atoms = 6.02 x 1023 x 7 = 4.21 x 1024 Option B is incorrect. Since each molecule of ethanal contains 2 carbon atoms, number of carbon atoms = 6.02 x 1023 x 2 = 1.20 x 1024 Option C is incorrect. Since each molecule of ethanal contains 4 hydrogen atoms, number of hydrogen atoms = 6.02 x 1023 x 4 = 2.41 x 1024 Ans: D Q2 Oxidation: Al ⎯→ Al3+ + 3e One mole of Al will lose 3 mol of electrons. Number of electrons lost from one mole of Al = 3 x 6.02 x 1023 Charge lost from one mole of Al = 3 x 6.02 x 1023 x (−1.60 x 10–19) = −288960 C = −2.89 x 105 C (3 s.f.) Ans: A Q3 As 1 mol of MnO 4– is used, eqns 2 to 4 are each multiplied by a factor to ensure total 5 mol of electrons are involved during each redox reaction. Multiplying and re-writing equations: eqn. 1: MnO4– + 8H+ + 5e– ⇌ Mn2+ + 4H2O eqn.2 x 2.5: 2.5I2 + 5e– ⇌ 5I– eqn.3 x 5: 5Cu2+ + 5e– ⇌ 5Cu+ eqn.4 x 2.5: 2.5SO42– + 5H+ + 5e– ⇌ 2.5SO32– +2.5H2O With reference to the above coefficients, Option 1 is correct as 1 mol of MnO 4– reacts with 5 mol of Cu+ to form 5 mol of Cu2+ ions. Option 2 is incorrect. 1 mol of MnO 4– reacts with 5 mol of I− to form 2.5 mol of I2. If an excess of I− was used (e.g. 7 mol), MnO 4− will be the limiting reagent. Since there is 1 mol of MnO4− present, 2.5 mol of I2 will still be produced. Option 3 is incorrect as MnO4– reacts with sulfite (SO32−) and not sulfate (SO42−). As MnO4– gets reduced, it reacts with sulfite, which get oxidised. Ans: A (1 only) Q4 W has 12 protons and 10 electrons and is hence Mg 2+. Z has 17 protons and 18 electrons and is hence Cl−. Therefore, option D is correct as MgC l2 is an ionic compound. Option A is incorrect as W has 2 more protons than electrons, and hence has a charge of 2+. Option B is incorrect as Z and Y have different numbers of protons and are hence not isotopes. Option C is incorrect as W (Mg2+) and Y (P3−) form W3Y2 (Mg3P2) and not W2Y3. Ans: D Suggested Answers for 2024 Y6 H1 Chemistry Preliminary Examination Paper 1: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 D A A D C D A C B A C C A C B A C C B D 21 22 23 24 25 26 27 28 29 30 D B B B D B A B D A Worked solutions for Paper 1

© Raffles Institution 2024 8873/02/S/24 2 Q5 Option C is correct. There is a big jump from the 12 th to 13th ionisation energies. Since a lot of energy is required to remove the 13th electron, the 13th electron must be in an inner electronic shell. Therefore, G has an electronic configuration of 1s22s22p63s23p2 since a lot of energy is required to remove the 13 th electron from the innermost first electronic shell (n = 1). G is hence Si, which forms the chloride SiCl4. For the same reason, option B is incorrect. Option A is incorrect as Si is a Period 3 element. G cannot be carbon as carbon only has 6 electrons. Option D is incorrect as SiCl4 has a simple molecular structure. This is not to be confused with SiO 2, which has a giant molecular structure. Ans: C Q6 shape structure polar molecule? SF4 see-saw (4 bond pairs, 1 lone pair) yes CH2Br2 tetrahedral (4 bond pairs) yes NF3 trigonal pyramidal (3 bond pairs, 1 lone pair) yes SiCl4 tetrahedral (4 bond pairs) no Ans: D Q7 Option A is correct as H2O forms a dative covalent bond with H + to form H 3O+, which has a trigonal pyramidal shape about central atom O. Option B is incorrect. A dative covalent bond is also formed in option B, but this results in a tetrahedral shape about the central atoms B and N. Option C is incorrect as bonds are being broken and not formed. Option D is incorrect. Although the shape about N in NH3 is trigonal pyramidal, the N -H bonds formed are covalent bonds, not co-ordinate bonds. Ans: A Q8 1 is correct as Br2 has a larger electron cloud size than NH3 and hence there are stronger instantaneous dipole- induced dipole (id -id) interactions between Br 2 molecules than NH 3 molecules. The id -id interactions between Br2 molecules are stronger than the hydrogen bonding and id-id between NH3 molecules. 2 is incorrect. The electron cloud size of H 2O (10 electrons) is similar to that of NH 3 (10 electrons), which does not explain the significant difference in boiling points of H2O and NH3. 3 is correct. Since hydrogen bonds exist between H 2O molecules and between NH3 molecules, more extensive hydrogen bonds in H 2O than in NH 3 would explain why the boiling point of H2O is much higher than that of NH3. H2O molecules form an average of 2 H −bonds per molecule of H2O while NH3 molecules form an average of 1 H−bond per molecule of NH3. Ans: C (1 and 3 only)

© Raffles Institution 2024 8873/02/S/24 3 Q9 Re-drawing the structure of A IBN to include lone pairs of electrons: Option B is correct since there are 4 lone pairs of electrons in AIBN. Option A is incorrect since the N atoms in the N=N bond have 2 bond pairs and one lone pair each, so they have a bent shape. Option C is incorrect as it is linear (2 bond pairs) about C in CN. Option D is incorrect as A IBN contains 5  bonds and 23  bonds. Ans: B Q10 Option A is correct. It is an endothermic reaction as hydrogen bonds between water molecules are broken when water is vaporised. Option B is an exothermic reaction as more hydrogen bonds are formed between H2O molecules to freeze H2O. Option C is an exothermic reaction as O –H bonds are being formed. Option D is exothermic as combustion of hydrocarbons is always exothermic. Ans: A Q11 H2SO4 + 2NaOH ⎯→ Na2SO4 + 2H2O Amount of H2SO4 = 50 x 10−3 x 2.0 = 0.1 mol Amount of NaOH used = 100 x 10−3 x 1.5 = 0.15 mol Since H2SO4 and NaOH react in a 1:2 ratio, NaOH is the limiting reagent. Therefore, amount of H 2O formed = amount of NaOH = 0.15 mol. Δ𝐻𝑛𝑒𝑢𝑡 = − 𝑞 𝑛𝐻2𝑂 = − 𝑚𝑐ΔT 0.15 = − (150)(4.2)(10−3)(12.6) 0.15 = –52.9 kJ mol–1 Ans: C Q12 When given enthalpy change of formation data, ∆Hr can be calculated using the following formula:  rH n = fH (products) m − fH (reactants) Applying the formula, Enthalpy change of combustion of 2 mol of H2S(g) = 2(–286) + 2(–297) – 2(–20) = –1126 kJ mol–1 Enthalpy change of combustion of H2S(g) = –1126 kJ mol–1 2 = –563 kJ mol–1 Ans: C Q13 Let n be the number of half-lives passed. Initial concentration x (1/2)n = final concentration 0.3 x (1/2)n = 0.0375 (1/2)n = 0.125 n = 3 t1/2 = 900/3 = 300 s From 0.0375 to 0.01875 mol dm –3, another half-life has passed. i.e. n = 4 Total time taken = n x t1/2 = 4 x 300 s = 1200 s Ans: A Q14 For both experiments, since rate (gradient) remains constant when [X] decreases, the reaction is zero order with respect to X. When [Y] = 1.0 mol dm−3, gradient = 0.10−0.06 0−4 = −0.01. Since rate is always positive, rate = 0.01 When [Y] = 2.0 mol dm−3, gradient = 0.10−0.02 0−2 = −0.04. Since rate is always positive, rate = 0.04 When [Y] was doubled, rate quadrupled from 0.01 to 0.04 mol dm −3 min−1, showing that the reaction is 2 nd order with respect to Y. Therefore, rate = k [Y]2. Ans: C

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