CIE Nov 2013 v3 Paper 5 Q1 (with explained solutions) |

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CIE Nov 2013 v3 Paper 5 Q1

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Examiners For Use Air, which is 99% nitrogen and oxygen, is slightly soluble in water. At 25 C a saturated solution of air in water has a concentration of 19 cm3 dm3. When water is boiled all the dissolved air is boiled out of solution. (a) (i) The molar enthalpies of solution for nitrogen and oxygen are: Hsoln N2 = 1.04 kJ mol1 and Hsoln O2 = 1.20 kJ mol1 Predict how the solubility of air in water will change as the temperature is increased. Explain this prediction using Le Chateliers principle in terms of the equilibrium between air and the aqueous solution as the temperature is increased. PredictionExplanation(ii) Display your prediction in the form of a sketch graph for the solubility of air between 0 C and 100 C, labelling clearly the axes. Include labelled points to indicate the solubility of air at 25 C and 100 C. [4] (b) If you were to carry out an experiment to investigate how the solubility of air varies as the temperature increases name, (i) the independent variable,(ii) the dependent variable.[1] [Total: 5]UCLES 2013 9701/53/O/N/13<br />

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PMT Page 2 Mark Scheme Syllabus Paper GCE A LEVELOctober/November 2013 9701 53 Question Expected Answer Mark (a) (i) (The solubility of air) decreases (as the temperature is increased). Dissolving is exothermic so an increase in temperature will promote the reverse reaction (or backwards or right to left). (ii) Axes are labelled AND graph is a curve/straight line showing a decrease in solubility with temperature. Graph is scaled and starts at 0 o point (25 o C, 19 cm dm 3 C on the solubility axis AND goes through ), AND ends at point (100 o 3 dm ) AND C, 0cm provided there is no maximum or minimum in the curve. Units required for this mark. (b) (i) temperature (ii) solubility (of air). Total 5 (a) Diagram shows a container with both chemicals named and attached to a syringe connected without leaks. Container shows the catalyst and hydrogen peroxide separated ready to mix. 100 cm of oxygen is 100/24000 = 0.00417 (mol) (0.004166666) Mol of H2O2 is 2mole answer above = 0.00834 (0.00833333) (ecf on alternative volume of oxygen used) Volume of hydrogen peroxide is therefore (answer above1000)/0.30 (27.8 cm Allow 28 cm ) (27.78 cm ) (27.7777773 cm (units are required) (reverse calculation also accepted) The reaction is timed from the moment of mixing to the collection of a stated volume of oxygen. (b) surface area of catalyst. (c) Axes are labelled (vol and time or min or s etc.) AND graph is a curve starting at the origin, shows the steepest slope at the start and slowing down as reaction proceeds x-axis has numerical scale from the origin to at least 3 min and graph shows no change in volume of oxygen after 2 minutes. The time axis must be scaled and have a unit of min or s.Cambridge International Examinations 2013<br />

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