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 3 September 2014 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 18 printed pages and 0 blank page. WORKED SOLUTIONS

2 9647/01/CJC JC2 Preliminary Exam 2014 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. Which one of the following has the same number of the stated partic le as atoms in 127 g of iodine at room conditions? A number of atoms in 79 g of gold B number of anions in 0.5 mol of barium chloride C number of ions in 1 mol of potassium bromide D number of molecules in 10 g of hydrogen fluoride Answer: B 127 g of I2 = 2127 127  mol = 0.5 mol of I2 molecules = 1.0 mol of I atoms = 6.02 × 1023 I atoms A: 79 g of Au = 197 79 mol of Au atoms = 0.401 mol of Au atoms = 2.41 × 1023 Au atoms B: Since there are 2 Cl- anions in 1 BaCl2, there is 1 mol of Cl- in 0.5 mol BaCl2 And thus the no. of anions present = 6.02 × 1023 Cl- C: Since there are 2 ions in 1 KBr, there are 2 mol of ions (K+ and Br-) in 1 mol KBr And thus the no. of ions present = 1.20 × 1024 ions D: 10 g of HF = 191 10  mol of HF molecule = 0.500 mol of HF molecule = 3.01 × 1023 HF molecule

3 9647/01/CJC JC2 Preliminary Exam 2014 [Turn over 2 Use of the Data Booklet is relevant to this question. In polluted air, the white paint pigment in older oi l paintings forms lead(II) sulfide, PbS, that is black in colour. To restore the white colour, an oxidising agent, hydrogen peroxide, H2O2, is used. Given that 0.239 g of PbS requires 25 cm 3 of 0.160 mol dm –3 H2O2 for oxidation, what is the possible identity of the white paint pigment? A S8 B PbS2 C PbSO3 D PbSO4 Answer: D no. of mol of PbS = 1.32207 239.0  = 0.00100 mol no of mol of H2O2 = 1000 25 × 0.160 = 0.00400 mol Thus 4 H2O2 ≡ PbS From data bookl