2020 YIJC Chem Prelim P2 (Suggested Answers)

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUGGESTED ANWERS CG INDEX NO CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper Additional Materials: Data Booklet 9729/02 1 September 2020 2 hours This document consists of 24 printed pages and 4 blank pages. For Examiner’s Use Paper 1 /30 Paper 2 1 /14 2 /13 3 /15 4 /13 5 /20 Penalty /75 Paper 3 /80 Paper 4 /55 Overall Percentage (%) READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

2 ©YIJC Answer all the questions in this section in the spaces provided. 1 Sodium chlorate(I), NaClO, found in household bleach produces chlorine upon addition of acidified dilute hydrochloric acid according to the equation below. ClO–(aq) + 2H+(aq) + Cl–(aq)  Cl2(aq) + H2O(l) A 15.0 cm 3 sample of a household bleach is diluted to 100 cm 3. 25.0 cm 3 of this solution is then pipetted out into a conical flask containing dilute hydrochloric acid. Excess potassium iodide solution is added which will react with the chlorine present in the sample according to the equation below. Cl2(aq) + 2I–(aq)  I2(aq) + 2Cl–(aq) This resultant solution is then titrated against 0.150 mol dm –3 sodium thiosulfate solution and required 17.20 cm 3 for complete reaction. Sodium thiosulfate reacts with iodine according to the following equation. I2(aq) + 2S2O32–(aq)  2I–(aq) + S4O62–(aq) (a) (i) Calculate the amount of iodine produced in this reaction. Amount of S2O32 = 0.150 0.0172 = 2.58 103 mol Amount of I2 = 0.5 2.58 103 = 1.29 103 mol [1] (ii) Calculate the amount of sodium chlorate(I) present in the 15.0 cm3 sample. Amount of I2 = 1.29 103 mol = Amount of Cl2 Amount of NaClO in 25.0 cm3 of the diluted solution = 1.29 103 mol Amount of NaClO in 15.0 cm3 of the undiluted solution = (1.29 103) (100/25) = 5.16 103 mol [1] (iii) Calculate the concentration of the sodium chlorate(I) in the household bleach in g dm–3. Mass of NaClO in 15.0 cm3 of the undiluted solution = 0.0516 (23.0 + 35.5 +16.0) = 0.38442 g Concentration of NaClO =