NJC_Prelim_P1 P2 ANS

NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS Higher 1 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions 8872/02 17 September 2009 2 hours For Examiner’s Use Section A 1 2 3 4 5 Section B B6 B7 B8 READ THE INSTRUCTION FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use paper clips, highlighters, glue or correction fluid. Section A Answer all questions . Section B Answer any two questions. 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. Total This question paper consists of 13 printed pages (including this page).

Section A Answer all the question s in this section in the spaces provided. 1 5.00 g of impure iron solid was dissolved in an excess of sulfuric acid and the resulting solution containing Fe2+ was made up to 250 cm 3. 10.0 cm3 of this solution was titrated against a solution of KMnO 4 of concentration 0.01 mol dm 3. It was found that 24.00 cm3 of KMnO4 was required. (a) Suggest a reason why an indicator was not required for this titration. There is a distinct colour change from purple to pale pink /pale orange at end point. [1] (b) Write a balanced redox equation for the reaction between Fe 2+ and MnO4 . 5 Fe2+(aq) + MnO4 (aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) [2] (c) A student claims that the percentage purity of iron in the sample is 33.5%. Justify if the statement made by the student is true. No. of moles of MnO 4  used = 24.00 x 103 x 0.01 = 2.4 x 104 mol No. of moles of Fe2+ in 10.0 cm3 = 5 x 2.4 x 104 = 1.2 x 103 mol No. of moles in 250 cm3 =1.2 x 103 x 10 250 = 0.03 mol Mass of Fe in sample = 0.03 x 55.8 = 1.674 g % of Fe in sample = 00 . 5 674 . 1x 100 = 33.5% Student’s claim is true. [3] [Total: 6] 2

2 The graphs below shows the rate of hydrolysis of ethyl ethanoate, CH3COOC2H5 in an acidic medium. The reaction was followed twice with different concentrations of HCl and the following results were obtained. 77.5, 0.004 155, 0.002 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0 50 100 150 200 250 (a) Using the graphs, determine the order of reaction with respect to both ethylethanoate and HCl. From the graph of [HCl] = 0.15 mol dm 3, t ½ (1) = t ½ (2) = 77.5 min, thus the reaction is 1st order with respect to ethylethanoate. To find order with respect to HCl, compare initial rates by drawing tangents at t = 0 when the graphs are the most linear. For [HCl] = 0.15 mol dm 3, rate = 0 50 006 . 0 01 . 0   = 8.00 x 105 mol dm3 min1 For [HCl] = 0.10 mol dm3, rate = 0 75 006 . 0 01 .