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 t