NYJC 2023 H2 Chemistry 9729 P4 Answer

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 1 H2 Chemistry Prelim Exam Answers Paper 4 Answers 1 (a) (i) Final burette reading / cm3 Initial burette reading / cm3 Volume of FA 2 added / cm3 25.40 25.40 (ii) Volume of FA 2 = 25.40 + 25.40 2 = 25.40 cm3 (b) (i) nNaOH = 25.40 1000 x 0.115 = 0.002921 = 0.00292 mol (ii) Mass of 1 dm3 of FA 1 = 1.75 x 1000 = 1750 g Mass of H3PO4 in 1 dm3 of FA 1 = 8.40 100 x 1750 = 147 g Concentration of H3PO4 in FA 1 = 147 g dm–3 Concentration of H3PO4 in FA 1 = 147 ÷ 98.0 = 1.50 mol dm–3 (iii) Concentration of H3PO4 in FA 3 = 10.0 250 x 1.50 = 0.0600 mol dm–3 Amount of H3PO4 in 25.0 cm3 of FA 3 = 25.0 1000 x 0.0600 = 0.00150 mol (iv) 34 NaOH H PO n n = 0.002921 0.00150 = 1.947  2 (nearest whole number)  Major product is Na2HPO4 H3PO4(aq) + 2NaOH(aq)  Na2HPO4(aq) + 2H2O(l)

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 2 (c) final burette reading 25.60 cm3 initial burette reading 1.35 cm3 volume added 24.25 cm3  The initial burette reading made by student A was 0.05 cm3 greater than the true value but the volume added was exactly 24.25 cm3. The final burette reading was also 0.05 cm3 greater than the true value.  The initial burette reading made by student B was 0.05 cm3 less than the true value and the actual volume added was exactly 24.15 cm3. The final burette reading was 0.05 cm3 greater than the true value.

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 3 2 (a) (i) The temperature, T1, of the hot water at t = 4.0 min is 63.0 °C. t / min T / ºC 0.0 32.0 1.0 32.0 2.0 32.0 3.0 32.0 4.0 –– 5.0 43.0 5.5 42.5 6.0 42.0 6.5 41.5 7.0 41.0 7.5 40.5 8.0 40.0 (ii)

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 4 (iii) Minimum temperature, T2, at t = 4.0 min is 32.0 °C. Maximum temperature, T3, at t = 4.0 min is 44.0 °C. Temperature rise for 50 cm3 of cold water in beaker A, (T3 – T2) is 12.0 °C. Temperature fall for 50 cm3 of hot water from the 250 cm3 beaker, (T1 – T3) is 19.0 °C. (iv) Heat lost by hot water = Heat gained by cold water + Heat gained by beaker A Heat gained by beaker A = Heat lost by hot water – Heat gained by cold water = (50.0 x 4.18 x 19.0) – (50.0 x 4.18 x 12.0) = 1463 = 1460 J (v) Heat absorbed by beaker A = Cbeaker A x (T3 – T2) Heat capacity of beaker A = 1463 12.0 = 121.9 = 122 J oC–1 (b) (i) mass of boiling tube + FA 4 / g mass of empty boiling tube / g mass of boiling tube + residual FA 4 / g mass of FA 4 added / g 9.963 initial temperature of water / oC minimum temperature obtained / oC temperature fall, T / oC 6.0 (ii) Mass of NH4Cl = 9.963 53.5 = 0.1862 = 0.186 mol (iii) Heat absorbed by solution = 100 x 4.3 x 6.0 = 2580 J (iv) Total change in heat energy = 2580 + (121.9 x 6.0) = 3311 J Hsolution = + 3311 0.1862 x 10–3 = +17.78 = +17.8 kJ mol–1 (v) The sign is positive since the reaction is endothermic / heat is absorbed in the reaction as the temperature falls during the reaction.

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 5 (c) major source of error suggested improvement explanation heat loss / gain lid prevents convection or evaporation insulation prevents conduction polystyrene cup provides insulation

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 6 3 (a) 1. Fill a burette with sodium thiosulfate. 2. Using separate 100 cm 3 measuring cylinders measure 75 cm 3 of potassium peroxodisulfate and potassium iodide. Transfer both solutions into a 250 cm 3 beaker and start the stopwatch. 3. Using a 25 cm 3 pipette transfer 25 cm 3 of the solution into a clean 250 cm 3 conical flask and add about 150 cm 3 of deionised water when timing is 2 minutes. 4. Titrate the iodine in the conical flask against sodium thiosulfate from t he burette 5. Add 3 –5 drops of starch when the colour of the colour of the solution turns yellow and continue titrating till the blue–black colour decolorised. 6. Repeat steps 4 to 7 for another 4 times at 5, 8, 11 and 15 minutes (approximate) (b) n(I–) present in 150 cm3 reaction mixture = 75/1000 x 0.800 = 0.0600 mol n(I–) present in 25.0 cm3 pipetted volume = 0.0600 x (25.0 / 150) = 0.0100 mol max n(I2) produced in 25 cm3 pipetted volume = ½ x 0.0100 = 0.0050 mol n(S2O3 2–) need to react with max I2 produced = 2 x 0.0050 = 0.0100 mol V(S2O3 2–) required for compete reaction = 0.0100 / 0.200 = 0.050 dm3 (c) Since 1st t/2 = 2nd t/2, order with respect to iodide ion concentration is first order. (d) There is high activation energy between S 2O8 2– and I– ions as they are both negatively charged hence. (e) Adding water to the extracted volumes will decrease the concentration of the reactants, causing the reactant particles to be further apart hence frequency of effective collision and rate of reaction decreases significantly. 1st t/2 2nd t/2 t/2 x V(S2O3 2–) / cm3 t / min 25.0 37.5 43.75 50.0 0

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 7 4 (a) (i) Table 4.1 test observations FA 5 (i) To 1 cm depth of the FA 5 solution in a test–tube, add aqueous sodium hydroxide, with shaking, until no further change is seen. White ppt formed is soluble in excess aq NaOH to give a colourless solution. (ii) To 1 cm depth of the FA 5 solution in a test–tube, add aqueous ammonia, with shaking, until no further change is seen. White ppt formed is insoluble in excess aq NH3. Identity of cation in FA 5: Al3+

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 8 (ii) test observations (i) Test for CO3 2 To 1 cm depth of FA 5 solution in a test –tube, add 1 cm depth of HCl(aq). (H2SO4 and HNO3 accepted) No effervescence. No observable change. (ii) Test for halide ions To 1 cm depth of FA 5 solution in a test –tube, add 1 cm depth of HNO3(aq), followed by 1 cm depth of AgNO3(aq). Add excess NH3(aq) to the resulting solution. White ppt formed is soluble in NH3(aq). (iii) Test for SO4 2 To 1 cm depth of FA 5 solution in a test –tube, add 1 cm depth of BaCl2(aq) followed excess HCl(aq). White ppt formed insoluble in excess HCl(aq). Anions present: SO4 2 and Cl (b) (i) Table 4.2

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 9 test observations (i) To 1 cm depth of FA 6, add 1 cm depth of dilute sulfuric acid followed by 3 drops of aq potassium manganate(VII). Warm the test–tube in a hot water bath. Purple aq KMnO4 decolourised. (ii) To 1 cm depth of FA 6 , add 5 drops of 2,4 – dinitrophenylhydrazine and warm. No orange ppt formed / Yellow solution remains. (iii) To 6 drops of FA 6 , add 1 cm depth of aqueous iodine followed by 1 drop of aqueous sodium hydroxide. Warm the mixture in a water bath for 1 minute. Cool the mixture. Yellow ppt of CHI3 formed. (ii) C CH 3 OH H