CJC Prelim P3_SOL

[Turn over 9729/03/CJC JC2 Preliminary Examination 2017 CHEMISTRY 9729/03 Paper 3 Free Response 25 August 2017 Candidates answer on separate paper. 2 hours Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. 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. Instruction Catholic Junior College JC2 Preliminary Examinations Higher 2 Suggested Solutions

2 9729/03/CJC JC2 Preliminary Examination 2017 Section A Answer all the questions in this section. 1 Use of the Data Booklet is relevant to this question. (a) Under acidic conditions, ethanol reacts with dichromate( VI) ions quantitatively to give ethanoic acid and chromium(III) ions. 3CH3CH2OH + 2Cr2O72– + 16H+  3CH3CO2H + 4Cr3+ + 11H2O A student carried out the following experiment to determine the concentration of ethanol in a brand of wine. A 10.0 cm3 sample of the wine was diluted to 250 cm3. He then added 25.0 cm3 of 0.156 mol dm –3 K2Cr2O7(aq) and excess dilute H 2SO4 to a 25.0 cm 3 portion of the diluted solution. After allowing the mixture to stand for about an hour at room temperature, the excess K2Cr2O7 in the mixture was then titrated with 0.118 mol dm–3 (NH4)2Fe(SO4)2(aq) using an appropriate indicator. At the end-point of the titration, he used 12.25 cm3 of the Fe2+(aq) solution. (i) Write an ionic equation for the reaction that occurs during titration. [1] Cr2O72– + 6Fe2+ + 14H+  2Cr3+ + 6Fe3+ + 7H2O (ii) Calculate the concentration, in mol dm–3, of ethanol in this brand of wine. [4] mol of excess Cr2O72– = 6 1  mol of Fe2+ used in titration = 6 1  0.118  1000 25.12 = 2.41  10–4 mol mol of Cr2O72– added initially = cV = 0.156  1000 0.25 = 3.90  10–3 mol mol of Cr2O72– that react with CH3CH2OH = (3.90  10–3) – (2.41  10–4) = 3.66  10–3 mol 3CH3CH2OH + 2Cr2O72– + 16H+  3CH3CO2H + 4Cr3+ + 11H2O mol of CH3CH2OH in 25.0 cm3 of diluted soln = 2 3  mol of Cr2O72– reacted = 2 3  3.66  10–3 = 5.49  10–3 mol mol of CH3CH2OH in 250 cm3 of diluted soln = 0.25 250  5.49  10–3 = 0.0549 mol Since 10.0 cm3 of the sample of spirit was diluted to 250 cm3, mol of CH3CH2OH in 10.0 cm3 of spirit = 0.0549 mol  [CH3CH2OH] in the brand of spirit = 0.10 1000  0.0549 = 5.49 mol dm–3

3 [Turn over 9729/03/CJC JC2 Preliminary Examination 2017 (b) The relatively high alcohol content in wine makes it difficult to pair win