CJC Prelim_H2_CHEM_P1_Solutions

1 9647/01/CJC JC2 Preliminary Exam 2016 [Turn over CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2 CHEMISTRY 9647/01 Paper 1 Multiple Choice Wednesday 31 August 2016 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name and HT group on the Answer Sheet in the spaces provided. 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. WORKED SOLUTIONS

2 9647/01/CJC JC2 Preliminary Exam 2016 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 Use of the Data Booklet is relevant to this question. Arsenic(III) sulfide, As2S3, is a bright yellow solid which has been used as a pigment in paintings. When arsenic(III) sulfide is heated in air, it reacts with oxygen to give arsenic(III) oxide and sulfur(IV) oxide. 2As2S3 + 9O2  2As2O3 + 6SO2 What would be the mass of arsenic(III) oxide produced if 550 dm3 oxygen, measured at room temperature and pressure, reacted with arsenic(III) sulfide? A 1010 g B 1070 g C 2010 g D 2160 g Answer: A Number of moles of O2 = 550 24 = 22.9 mol Since 9O2 ≡ 2As2O3 Number of moles of As2O3 = 22.9 9 × 2 = 5.09 mol Mr of As2O3 = 74.9 × 2 + 16.0 × 3 = 197.8 Mass of As2O3 = 5.09 × 197.8 = 1007 g 2 Incomplete combustion of 20 cm 3 of a gaseous hydrocarbon gave carbon dioxide and carbon monoxide in a 6:1 ratio, as well as water vapour. It was found that the carbon dioxide took up a volume of 51.4 cm 3 whereas the water vapour took up a volume of 80 cm3. All gas volumes are measured at the same temperature and pressure. What is the molecular formula of the hydrocarbon? A C2H4 B C2H6 C C3H6 D C3H8 Answer: D 7CxHy + ? O2  6xCO2 + xCO + 7𝑦 2 H2O Reduction of 80 cm3 after passing through dehydrating agent  volume of water vapour was 80 cm3 Reduction of 51.4 cm3 after passing through KOH (aq)  volume of CO2 was 51.4 cm3 Mole ratio of CxHy : CO2 = 7 : 6x = 20 : 51.4

3 9647/01/CJC JC2 Preliminary Exam 2016 [Turn over 6𝑥 7 = 51.4 20 x = 2.99  3 Mole ratio of CxHy : H2O = 7 : 7𝑦 2 = 20 : 80 7𝑦 2 7 = 4 y = 8 Therefore, the molecular formul