DHS H2 Chem 2013 Prelim P2 Soln

_______________________________________________________________________________________________ This question paper consists of 16 printed pages and 1 blank page. © DHS 2012 [Turn over Name: Suggested Solutions Index Number: Class: DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY 9647/02 Paper 2 Structured 19 September 2013 2 hours Additional Materials: Data Booklet INSTRUCTIONS TO CANDIDATES 1 Write your name, index number and class on this cover page. 2 Answer all questions. 3 Write your answers in the spaces provided on the question paper. 4 A Data Booklet is provided. 5 The number of marks is given in brackets [ ] at the end of each question or part question. 6 You may use a calculator. FOR EXAMINER’S USE Question No. 1 2 3 4 5 6 Total % Marks 12 15 9 10 15 11 [72] 510

2 © DHS 2013 9647/02 [Turn over Answer all questions in the space provided. 1 Planning Solubility is defined as the mass of solid that will dissolve in and just saturate 100 g of solvent at a particular temperature. When potassium nitrate dissolves in water, the temperature of the solution decreases as the enthalpy change of solution is endothermic. You are to plan an experiment to investigate how the solubility of potassium nitrate varies with temperature. The units of solubility are grams per one hundred grams of water (g/100 g of water). (a) (i) Predict how the solubility of potassium nitrate will change if the solution temperature is increased. By Le Chatelier’s principle, when temperature is increased, the equilibrium will shift to the right to absorb the excess heat and favour the endothermic reaction. Hence the solubility will increase with increasing temperature. (ii) Display your prediction in the form of a sketch, labeling clearly the y-axis. solubility 25 temperature/ °C [3] 511

3 © DHS 2013 9647/02 [Turn over (b) You are to design an experiment to test your prediction in (a). (i) Given that the solubility of potassium nitrate at 70 °C = 11.3 mol dm–3 Calculate the maximum mass of potassium nitrate that can be dissolved in 100 g of water at 70 °C. Maximum mass = (100/1000) x 11.3 x (39.1 x 14.0 x 3(16.0)) = 114 g (ii) Describe how you would carry out the experiment. You should • include the apparatus used; • ensure a wide range of results suitable for analysis by graph; • decide on the amount of water and potassium nitrate to be used; • describe how the temperature of the solution is to be maintained. You may also assume that standard laboratory apparatus are available. 1. Measure approximately 120 g of potassium nitrate using a weighing balance. [1] Measure 100 ml of water using a measuring cylinder/burette and transfer the water to a 250 cm3 beaker. (total mass of solid + water should not exceed 250 g, mass of solid added should be more than 114g in every 100 g of water) 2. Add potassium nitrate to the beaker containing water and stir the mixture. 3. Heat the mixture to 70 °C in a thermostat controlled water bath. 4. Stir the mixture gently throughout the heating. 5. Allow to stand/ leave for a long time to establish equilibrium 6. Filter the solution (not cooled or decanted) . 7. Heat the residue to dryness and weigh the residue . 8. Repeat step 1 – 7 for another four experiments (at least) but changing the temperature in step 3 to 30 °C, 40 °C, 50 °C and 60 °C respectively. (accept any temperature range of at least 40 °C)   512

4 © DHS 2013 9647/02 [Turn over (iii) Draw a table with appropriate headings to show the data you would record when carrying out your experiments and the values you would calculate in order to construct a graph to support or reject your prediction in (a) . The headings must include the appropriate units. Mass of potassium nitrate /g Mass/Volume of water /g or cm3 Mass of dry residue /g Solubility = Initial mass of potassium nitrate – mass of dry residue Temperature (°C) Solubility (g /100g water) 30 40 50 60 70 [7] (c) Potassium nitrate is used as an excellent additive in fertilisers. It dissolves in water with an enthalpy change of solution of +34.9 kJ mol–1. It is given that the lattice energy of potassium nitrate is –687 kJ mol –1 while the enthalpy change of hydration of potassium ion is –322 kJ mol –1. With the aid of an energy level diagram, calculate the enthalpy change of hydration of nitrate ion. endothermic dissolution( ΔH soln θ > 0) By Hess’ Law, L.E = ∑(∆Hhyd of ions) – ∆Hsol –687 = (–322 + Enthalpy change of hydration of nitrate ion) – (+34.9) Enthalpy change of hydration of nitrate = –330 kJ mol –1 [2] energy K+ (g) + NO3 – (g) K+ (aq) + NO3 – (aq) KNO3 (s) ΔHlatt θ ΔHhyd θ ΔHsoln θ 513

5 © DHS 2013 9647/02 [Turn over 2 The use of Data Booket is relevant to this question. The nitrates, carbonates and hydroxides of Group II elements can undergo thermal decomposition. The nitrates of lead and zinc can undergo thermal decomposition similar to calcium nitrate. The decomposition temperature of the three nitrates are given in the table below. Compound Decomposition temperature / 0C Lead(II) nitrate, Pb(NO3)2 290 Zinc nitrate, Zn(NO3)2 105 Calcium nitrate, Ca(NO3)2 132 (a) Explain the data above by quoting relevant values from the Data Booklet and using your knowledge and understanding of the trend in the decomposition temperature of the Group II nitrates. Cation Ionic radius /nm Pb2+ 0.120 Zn2+ 0.074 Ca2+ 0.099 The decomposition temperature increases in the order from zinc nitrate to calcium nitrate to lead(II) nitrate. This is because the ionic radius of Zn 2+ is smaller than Ca 2+ which in turn is smaller than Pb2+. Zn2+ has highest charge density and hence greatest polarising power. Zn2+ is able to distort the electron cloud of the nitrate anion to the most and weaken the N–O bond to the greatest extent. Hence zinc nitrate has a lowest thermal stability and can decompose at the lowest temperature. (b) The mineral hydromagnesite is a hydrated ca rbonate of magnesium, with the formula Mgx(CO3)y(OH)z.nH2O and a molar mass of 466 g mol -1. Group II carbonates and hydroxides decompose to give the same solid product. When 1.00 g of a pure sample of hydromagnesite was heated to constant mass, 0.378 g of carbon dioxide was given off. Steam was also produced. The remaining white solid from the above decomposition was completely dissolved in 50 cm3 of a 1.0 mol dm-3 solution of hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 250 cm3 with deionised water. A 25.0 cm3 aliquot of this diluted solution required 28.50 cm3 of a 0.10 mol dm-3 solution of sodium hydroxide for complete neutralisation. (i) Calculate the value of y. moles of hydromagnesite = 1 / 466 = 2.15 x 10–3 mol moles of CO2 given off = 0.378 / 44 = 8.59 x 10–3 mol y = 8.59 x 10 –3 ÷ 2.15 x 10–3 = 4 514

6 © DHS 2013 9647/02 [Turn over (ii) Calculate the value of x. moles of HCl(aq) used = 50/1000 x 1 = 0.050 mol moles of NaOH used to neutralize remaining HCl(aq) in 25 cm3 = 28.5/1000 x 0.10 = 0.00285 mol moles of HCl reacted with MgO = 0.050 – (0.00285 x 10) = 0.0215 mol MgO + 2HCl → MgCl 2 + H2O moles of MgO formed = 0.0215 ÷ 2 = 0.0107 mol = moles of Mg present x = 0.0107 / 2.15 x 10 –3 = 5 (iii) Hence or otherwise, deduce the values of z and n. For a neutral solid Mg5(CO3)4(OH)z.nH2O, +2 x 5 – 2 x 4 + (–1 x z) = 0 therefore, z = 2 to balance the overall charge. Mass of water per mole of hydromagnesite = 466 – (5 x 24.3 + 4 x