CJC 2019 Paper 2 Promo Answers

9729 / CJC JC1 Promotional Examination 2019 CANDIDATE NAME CLASS 1T CHEMISTRY 9729/02 Paper 2 Structured Questions Wednesday 2 October 2019 1 hour Candidates answer on the Question Paper. Additional Materials: 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. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculat or is expected, where appropriate. A Data Booklet is provided. At the end of examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 10 printed pages. For Examiner’s Use Paper 1 15 Paper 2 Q1 / 8 Q2 / 4 Q3 / 3 Q4 / 5 Q5 / 6 Q6 / 9 35 Paper 3 Q1 /20 Q2 /20 40 Paper 4 Q1 /15 Q2 /15 30 TOTAL 120 OVERALL/% Grade Catholic Junior College JC1 Promotional Examinations Higher 2 Answers

2 [Turn over 9729 / CJC JC1 Promotional Examination 2019 Paper 2 Answer all the questions. Write your answers in the spaces provided. You are advised to spend not more than one hour on this section. 1 Ethanedioic acid, is a dibasic, organic acid with the formula HO 2CCO2H. It is commonly found in many leafy vegetables, fruits, nuts and seeds. It is able to react with a base such as sodium hydroxide, NaOH. (a) FA 1 is a solution containing 5.00 g dm–3 of a similar dibasic, organic acid, HO2C(CH2)nCO2H. When 25.0 cm 3 of FA 1 is titrated against NaOH of co ncentration 0.125 mol dm–3, 17.00 cm3 of NaOH is required. The equation for this reaction is given as follows. HO2C(CH2)nCO2H(aq) + 2NaOH(aq) → Na2(O2C(CH2)nCO2) (aq) + 2H2O(l) (i) Calculate the amount in moles of NaOH required to react with 25.0 cm 3 of the acid solution in FA 1. Amount of NaOH = ( ૚ૠ.૙૙ ૚૙૙૙ x 0.125) = 0.00213 mol [1] (ii) Calculate the amount in moles of the dibasic acid, HO2C(CH2)nCO2H in 25.0 cm3 of FA 1 that has reacted. Amount of acid = (½ x 0.00213) = 0.00106 mol [1] (iii) Calculate the concentration of the acid, HO2C(CH2)nCO2H in mol dm-3 of solution in FA 1. Concentration of acid = ( ૚૙૙૙ ૛૞.૙ x