2021 MI Prelim H2 Chemistry P1 (9729 01) Answers

Class Adm No Candidate Name: This question paper consists of 12 printed pages and 2 blank pages. 2021 Preliminary Examination Pre-University 3 H2 CHEMISTRY 9729/01 Paper 1 Multiple Choice xxnd Sep 2021 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 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 contains the same number of stated particles as the number of atoms in 19.0 g of fluorine gas? 1 Number of molecules in 2.0 g of hydrogen gas 2 Number of ions in 58.5 g of sodium chloride solid 3 Number of atoms in 14.0 g of nitrogen gas 4 Number of atoms in 12.0 g of graphite A 3 only B 1, 3 and 4 only C 2, 3 and 4 only D 1, 2, 3 and 4 Number of atoms in 19.0 g of fluorine gas = 1 mole of F atoms 🗸 1) Number of molecules in 2.0 g of hydrogen gas = 1 mole of H2 molecules ✗ 2) Number of ions in 58.5 g of sodium chloride = 2 moles of ions (both Na+ and Cl–) 🗸 3) Number of atoms in 14.0 g of nitrogen gas = 1 mole of N atoms 🗸 4) Number of atoms in 12.0 g of graphite solid = 1 mole of C atoms. 2 20 cm3 of an unknown hydrocarbon was completely combusted in excess oxygen gas. Upon cooling back to room temperature and pressure, the total volume of gases was 50 cm3 less than the initial total volume. After passing the remaining gases through KOH(aq), the volume of gases further decreased by 80 cm3. What is the molecular formula of the unknown hydrocarbon? A C2H4 B C4H4 C C4H6 D C4H8 CxHy + (x + y/4) O2 à xCO2 + (y/2) H2O Total vol / cm3 Initial vol / cm3 20 A 0 - A + 20 Change in vol / cm3 -20 -20x – 20y/4 +20x - Final vol / cm3 0 A – 20x – 5y 20x - A – 5y From qn, 80 cm3 of CO2 gas reacted with KOH(aq) 20x = 80 x = 4 From initial to final volume, volume dropped by 50 cm3 A + 20 – 50 = A – 5y 5y = 30 y = 6 Answer: C4H6

3 [Turn over 3 Which of the following statements is incorrect about the following reaction? 3VO2– + 2H+ → VO3– + 2VO + H2O A It is a displacement reaction. B The oxidation number of V in VO3– is +5. C VO2– is acting as a reducing agent. D VO2– is acting as an oxidising agent. Oxidation number of V in the reactant VO2– = +3 B) TRUE. Let x be the oxidation number of V in VO3–. x + 3(-2) = -1. x = +5 Options C & D – TRUE. Oxidation number of V is oxidised from +3 in VO2– to +5 in VO3– and simultaneously reduced to +2 in VO. It is a disproportionation reaction where VO2– has acted as a reducing agent (itself oxidised) and also an oxidising agent (itself reduced) Option A is incorrect as a displacement reaction is when a metal is oxidised when placed in a solution containing the cation of another metal which is reduced. 4 A copper rod weighing 15 g was dipped into 50 cm3 of 2.00 mol dm–3 AgNO3 solution. Silver crystals were observed to grow on the copper rod after a while. Which of the following statements are correct? 1 The EcellO of the reaction is +1.14V. 2 The solution turns blue. 3 6.35 g of copper would have reacted upon completion of reaction. A 1 only B 2 only C 2 and 3 only D 1, 2 and 3 ✗ A) EcellO = (+0.80) – (+0.34) = +0.46 V 🗸 B) Cu metal is oxidised to Cu2+(aq), which gives the solution a blue colour ✗ C) Amount of Ag+ = 0.050 × 2.00 = 0.100 mol Amount of Cu = 15 / 63.5 = 0.236 mol 2 mol of Ag+ reacts with 1 mol of Cu. Ag+ is the limiting reagent. 0.100 mol of Ag+ reacts with ½ × 0.100 = 0.0500 mol of Cu Mass of Cu reacted = 0.0500 × 63.5 = 3.175 g

4 5 Which of the following statements is not true about lithium? A It has a naturally occurring isotope which contains 3 neutrons. B It has a naturally occurring isotope which contains 4 neutrons. C It has the full electronic configuration 2s1. D It forms an ion with a +1 oxidation number. ✗ A & B) Both true as lithium as a relative atomic mass of 6.9 hence it is likely that there exists the 6Li isotope with 3 neutrons and 7Li isotope with 4 neutrons. 🗸 C) FALSE. The full electronic configuration is 1s2 2s1 and not just 2s1. ✗ D) Group 1 metals tends to lose an electron to form Li+, which has +1 oxidation number. 6 Which of the following properties can be explained by hydrogen bonding? 1 The higher density of ice compared to liquid water. 2 The ability of water glider insects to float on water. 3 Ethanoic acid having an apparent molecular mass of 120.0 in non-polar solvents. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 ✗ 1) FALSE as ice has lower density due to the open tetrahedral arrangement of H2O molecules in ice (due to hydrogen bonding, but not higher density) 🗸 2) TRUE as hydrogen bonding is considered strong intermolecular force which results in high surface tension of water 🗸 3) TRUE as the -COOH groups have H bonded to O and lone pair on electronegative O which forms hydrogen bonding. As energy released from the formation of id-id of -COOH groups with non-polar solvents is insufficient to overcome the hydrogen bonding between -COOH groups, the ethanoic acid molecules dimerise through the formation of hydrogen bonding to give an apparent Mr which is double 2 × 60.0 = 120.0

5 [Turn over 7 Two elements D and E have the following properties. D E Melting point / °C 3600 –38 Appearance Dull Shiny Electrical conductivity when solid Yes Yes What are the likely identities of D and E? D E A SiO2 H2 B MgO Si C Si Na D C Hg D has very high melting point yet dull appearance – means it is likely to be giant covalent instead of metals which have a shiny appearance. The only giant covalent structure we learnt that conducts electricity when solid is Carbon in the form of Graphite. E has a low melting point indicating it could likely be a simple molecular structure. However, its shiny appearance and electrical conductivity contradicts that of a simple molecular structure. Instead we should be looking at metals which have a melting point lower than room temperature. Prior knowledge tells us one such metal is mercury which is a liquid metal. 8 Which of the following statements best explains the following observation? Melting point of NaBr / °C 747 Melting point of MgBr2 / °C 711 A The size of the anion electron cloud is large and easily distorted. B NaBr has a larger magnitude of lattice energy than MgBr2. C There are more bromide ions in MgBr2 than in NaBr. D Magnesium has a higher charge and smaller size than sodium. Both have giant ionic lattice structures. As the MgBr2 theoretically has a larger magnitude lattice energy than NaBr, the melting point of MgBr2 is predicted to be higher than NaBr yet in reality it is lower. This is due to partial covalent character, which can be attributed to distortion of anion charge cloud by the cation. Since the anion is the same in both cases, the factor that changed is the charge density of the cation, which is α !"#$%&'()& ((,-(! $#.(/')

6 9 Which of the following statements about the standard enthalpy change of neutralisati