RI 2024 Prelims P4 Solutions

-1- 2024 RI H2 Chemistry Paper 4 – Suggested Solutions 1(a)(i) Table 1.1 temperature of room temperature water, T1 / °C 30.4 temperature of hot water, T2 / °C 70.1 maximum temperature of combined water, T3 / °C 48.2 1(a)(ii) Heat energy lost by the hot water Heat energy gained by the room temperature = mc∆T water = mc∆T = 50 × 4.18 × (T2 – T3) = 50 × 4.18 × (T3 – T1) = 50 × 4.18 × (70.1 – 48.2) = 50 × 4.18 × (48.2 – 30.4) = 4577.1 J = 3720.2 J = 4580 J (3 s.f.) = 3720 J (3 s.f.) 1(a)(iii) Heat energy absorbed by the calorimeter Heat capacity of calorimeter, Ccal = heat energy lost by hot water = 856.9 T3 – T1 = 856.9 48.2 – 30.4 – heat energy gained by room temp water = 48.14 J oC−1 = 4577.1 – 3720.2 = 48.1 J oC−1 (3 s.f.) = 856.9 J = 857 J (3 s.f.) 1(b)(i) mass of capped weighing bottle and FA 1 / g 8.782 mass of capped weighing bottle after emptying FA 1 / g 4.745 mass of FA 1 used / g 4.037 time / min temperature / °C 0.0 30.4 0.5 30.4 1.0 30.4 1.5 30.4 2.0 30.4 2.5  3.0 36.2 3.5 36.4 4.0 36.5 4.5 36.3 5.0 36.2 5.5 36.1 6.0 36.0 6.5 36.0 7.0 35.9 7.5 35.8 8.0 35.6

-2- 1(b)(ii) 1(b)(iii) Tmin = 30.4 oC Tmax = 36.8 oC ∆T = +6.4 oC 1(b)(iv) Total heat change, q = heat absorbed by solution + heat absorbed by calorimeter = (mc∆T) + (Ccal ∆T) = (100 × 4.18 × 6.4) + (48.14 × 6.4) = +2983 J = +2980 J (3 s.f.) 1(b)(v) Amount of MgSO4 used = 4.037 24.3 + 32.1 + 4(16.0) = 0.03353 mol ∆Hsol = – q n(MgSO4) = – +2983 0.03353 = –89000 J mol−1 (or –89.0 kJ mol−1) (3 s.f.)

-3- 1(c) The heat energy absorbed by and heat capacity of calorimeter obtained in (a)(iii) is higher than expected. Hence, the ∆Hsol of FA 1 obtained in (b)(v) is more exothermic than expected (or the magnitude of ∆ Hsol of FA 1 obtained in (b)(v) is higher than expected). 2(a)(i) tests observations 1 Add 1 cm depth of FA 2 into a clean test-tube. To this test -tube add 10 drops of sodium hydroxide followed by iodine solution, dropwise, until a permanent orange colour is present. Warm the test -tube in the beaker of hot water for 2 minutes. (pale) Yellow ppt formed. 2 Add 1 cm depth of FA 2 into a clean test-tube. Add 8 drops of Fehling’s solution. Warm the test -tube in the beaker of hot water for 3 minutes. Brick-red ppt formed. 3 Test solution FA 2 with Universal Indicator paper. Universal Indicator paper turned dark yellow-green. pH is 7. 2(a)(ii) C C H O C O H H H H H 2(b)(i) tests observations 1 Using a 10 cm 3 measuring cylinder, add 2 cm3 of FA 3 into a clean boiling tube. Using another 10 cm3 measuring cylinder, measure out 7 cm 3 of aqueous sodium hydroxide. Slowly with shaking, add this completely to FA 3. Stir the contents of the boiling tube with a glass rod. Filter the mixture into a clean test -tube. The filtrate will be used for test 4. W