NYJC Prelim 9729 P2 Answers

[Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME Teachers’ Mark Scheme CLASS TUTOR’S NAME CHEMISTRY 9729/02 Paper 2 Structured 11 September 2018 2 hours 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 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. For Examiner’s Use 1 /24 2 /11 3 /9 4 /13 5 /18 Total /75 This document consists of 17 printed pages and 1 blank page.

2 H2 Chemistry 9729/02 NYJC J2/18 PX [Turn Over For Examiner's Use Answer all questions in the spaces provided. 1(a) An unknown sample was found to contain the anions, Cl −, ClO 3− and NO 3−. A student weighed a sample into a beaker and recorded the following data. mass of beaker and sample / g 68.962 mass of empty beaker / g 67.620 The sample was dissolved and diluted in a 250 cm 3 volumetric flask to obtain solution L. In experiment 1 , a 50 cm 3 portion of solution L was reacted with excess silver nitrate solution. The AgCl precipitated was transferred onto a dry filter paper and was placed under an infra-red lamp. The dry AgCl precipitate was weighed and the following data was obtained. Experiment 1 mass of dry filter paper and AgCl / g 0.737 mass of dry filter paper / g 0.620 In experiment 2 , a gas was bubbled into another 50 cm 3 portion of solution L to convert ClO3− to Cl− before the addition of excess silver nitrate solution. The AgCl precipitated was also dried and weighed. The following data was obtained. Experiment 2 mass of dry filter paper and AgCl / g 0.799 mass of dry filter paper / g 0.651 (i) Write the half equation for the reduction of ClO3− to Cl−. ClO3− + 6H+ + 6e−  Cl− + 3H2O [1] .......................................................................... [1] (ii) Determine the mass of Cl − in 50 cm3 of solution L. mass of AgCl precipitate in expt 1 = 0.737 – 0.620 = 0.117 g mass of Cl− present in 50 cm3 portion = [1] [1] (iii) Determine the mass of Cl − converted from ClO3− in experiment 2. mass of AgCl precipitate in expt 2 = 0.799 – 0.651 = 0.148 g mass of Cl− present in 50 cm3 portion = [1] mass of Cl− from ClO3− in 50 cm3 = 0.03663 − 0.02896 = 0.007674 g ≈ 0.00767 g [1] [2]

3 H2 Chemistry 9729/02 NYJC J2/18 PX [Turn Over For Examiner's Use (iv) Hence, determine the percentage mass of ClO 3− in the unknown sample. Since n(ClO 3−) : n(Cl−) is 1 : 1, mass of ClO 3− present in 50 cm3 = 0.007674 = 0.01805 g [1] mass of ClO3− present in 250 cm3 = 0.01805 = 0.1805 g [1] % ClO3− present in unknown compound = x 100% = 13.45% ≈ 13.5% [1] [3] (v) The E( ClO3−/Cl−) has a value of +1.47 V. From the list of standard electrode potentials in the Data Booklet, identify a gas that would reduce ClO 3− to Cl−. Explain your answer. E( H+/H2) = 0.00 V H2 gas is an appropriate reducing agent. [1] (also accept E( NO3−/NO2) or E( SO42−/SO2) as their E < + 1 . 4 7 V ) E cell = (+1.47) – (0.oo) = +1.47 V > 0 Since, E cell > 0, the reaction is feasible. [1] ......................................................................................................................... ..................................................................................................................... [2] (b) (i) An aqueous solution of HCl has a density of 1.15 g cm −3 and is 30% by mass of HCl. Calculate the concentration in mol dm−3 of this solution of HCl. mass of HCl in 1 cm 3 = x 1.15 = 0.3450 g [1] [HCl] = x 1000 = 9.45 mol dm−3 [1] [2] (ii) Calculate the volume of this solution required to prepare 5 dm 3 of 0.20 mol dm−3 HCl by dilution with water. volume of HCl required = = 0.106 dm−3 = 106 cm3 [1] [1]

4 H2 Chemistry 9729/02 NYJC J2/18 PX [Turn Over For Examiner's Use light (c) The gelatin silver process is the photographic process used with black-and-white films. The following information pertains to the process of taking photographs and developing films. Taking photographs • A 35 mm cartridge of black-and-white print film contains a long strip of plastic that has layered coatings on each side. • On the front side of the film, the layers are made of gelatin which contain grains of silver chloride crystals. • When the shutter of the camera is opened for a fraction of a second to allow the film to be exposed to light, these crystals under go decomposition thereby producing an image on the film. Developing films • After the photographs have been taken, the film is developed in a dark room under a light source that emits low energy light. • Firstly, the film is soaked in water before adding phenidone. Phenidone makes the image more visible by reacting with the ex posed silver chloride crystals to produce silver atoms and two other by-products. • This reaction can only proceed at high pH. • After some time, the reaction will then be quenched. • Finally, the film will be soaked in ammonium thiosulfate, (NH4)2S2O3, which is used as a fixer to make the image permanent and light resistant. This is done through the reaction between the unexposed silver chloride crystals and the fixer. (i) Write the balanced equation for the decomposition of silver chloride crystals when it is exposed to light. 2AgCl 2Ag + Cl2 [1] ..................................................................................................................... [1] (ii) Suggest a suitable colour of the light source that is used in a dark room. Red. (Also accept orange or yellow) [1] ........................................................ [1] (iii) Complete the equation for the reaction between the developing agent, phenidone, and the exposed silver chloride crystals. Hence, state the role of phenidone in this reaction. N NH O phenidone role of phenidone reducing agent [1] ............................................................... [2] + 2AgCl + 2Ag + 2HCl [1] N N OH

5 H2 Chemistry 9729/02 NYJC J2/18 PX [Turn Over For Examiner's Use (iv) Suggest a suitable reagent, other than excess cold water, that can be used to quench the development of the film. Explain. Acetic acid / Citric acid / Any plausible acids. e.g. HCl, HNO3, H2SO4 [1] When acid is added, the pH will be lowered. Hence, the reaction will not proceed at lower pH. [1] ..................................................................................................................... [2] (v)