CJC_H2_CHEM_P1_SOL Prelim

1 [Turn over CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2 CHEMISTRY 9647/01 Paper 1 Multiple Choice Wednesday 2 September 2015 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and HT group on the Answer Sheet in the spaces provided. Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. There are forty 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 separate Answer Sheet. Read the instructions on the 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 booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 30 printed pages and 0 blank page. WORKED SOLUTIONS

2 9647/01/CJC JC2 Preliminary Exam 2015 Section A For each question there are four possible answers, A, B, C and D. Choose the one you consider to be correct and record your choice in soft pencil on the separate Answer Sheet provided. 1 When 20 cm 3 of a gaseous hydrocarbon was completely burnt in 130 cm 3 of oxygen, the volume of ga s remaining after the reaction wa s 100 cm 3. This volume was decreased to 40 cm 3 when the resulting mixture wa s passed through aqueous sodium hydroxide. All measurements were made at room temperature and pressure. What is the formula of this hydrocarbon? A C2H2 B C3H6 C C3H8 D C4H10 Answer: B Since temperature and pressure are constant, mol ratio of gas ∝ vol ratio CxHy (g) + )( 4 yx O2 (g) → xCO2 (g) + 2 y H2O (l) Initial vol/ cm3: 20 130 0 0 Final vol/ cm3: 0 40 (100-40) Vol reacted or formed/ cm3: 20 (130-40) 60 = 90 1 vol 4.5 vol 3 vol  By inspection, x = 3. 2 9) 4 y(3   y = 6  molecular formula of the hydrocarbon is C3H6 2 Consider the following half-equations: C2O4 2–  2CO2 + 2e– Fe2+  Fe3+ + e– MnO4 – + 8H+ + 5e–  Mn2+ + 4H2O What volume of 0.01 mol dm –3 potassium manganate( VII) is required to oxidise completely 25.0 cm3 of an acidified solution of 0.01 mol dm–3 FeC2O4? A 10 cm3 B 15 cm3 C 25 cm3 D 42 cm3 Answer: B Both Fe2+ and C2O4 2– ions can be oxidised by MnO4 – ions which is reduced. Hence, there are two oxidation half-equations to consider: Fe2+  Fe3+ + e– --- 1 and C2O4 2–  2CO2 + 2e– --- 2 So combine 1+2 first by summing up both half-equations to get: Fe2+ + C2O4 2–  Fe3+ + 2CO2 + 3e– --- 3

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