2022 RI H2Chem Prelims P4 Answers

-1- 2022 RI H2 Chemistry Prelim Paper 4 – Suggested Solutions 1(a) Expt Tinitial / oC Tfinal / oC t / s 1/t / s−1 lg (1/t) Tave / oC TK / K 1/TK / K−1 1 30.4 30.6 136 0.00735 -2.13 30.5 304 0.00329 2 62.4 50.0 22 0.0455 -1.34 56.2 329 0.00304 3 39.2 36.0 72 0.0139 -1.86 37.6 311 0.00322 4 48.2 40.8 45 0.0222 -1.65 44.5 318 0.00315 1(b)(i) (b)(ii) t = 3  60 = 180 s lg (1/t) = lg (1/180) = –2.26 From the graph, 1/TK = 0.00334 K−1 Tave = 26.4 oC (c)(i) Mole ratio of MnO4− : D-glucose : e− = 1 : 2.5 : 5 Hence, mole ratio of D-glucose : e− = 1 : 2 1 mol of D-glucose loses 2 mol of e−. lg (1/t) 1/TK / K-1 -2.80 -2.60 -2.40 -2.20 -2.00 -1.80 -1.60 -1.40 -1.20 -1.00 0.00290 0.00300 0.00310 0.00320 0.00330 0.00340 0.00350 0.00334 –2.26 0.00324 –1.94

-2- (c)(ii) Structure of P: Since 1 mol of D -glucose loses 2 mol of e− and only one carbon atom in each glucose molecule is oxidised, the oxidation state of this carbon must have increased by 2. Hence, the aldehyde group of D-glucose is oxidised to carboxylic acid as the oxidation state of carbon increases from +1 in aldehyde to +3 in carboxylic acid. (d)(i) Maximum total percentage uncertainty = ( 2(0.05)+(0.5)+(0.5) 60 )  100% = 1.83% (d)(ii) Although Tave was calculated, the difference between Tinitial and Tfinal was quite significant, especially for experiments conducted at temperatures above room temperature. This in turn resulted in inaccuracy in the measurement of the temperature of the reaction/measurement of the time required for the reaction to complete at Tave. A modification to the experimental procedure would be to use a water bath maintained at a fixed temperature for each experiment, to equilibrate the reactants at the same fixed temperature before starting the reaction, and to ensure that the temperature of the reaction mixture stayed constant throughout the reaction. OR The human perception of when the reaction mixture turned colourless may be inconsistent/vary across different experiments , hence resulting in inaccuracy in the measurement of the time required for the reaction to complete at Tave. A modification to the experimental procedure would be to use a colorimeter to monitor the absorbance of the reaction mixture, and stopping the stopwatch only when there was no more absorbance (i.e. reaction mixture had turned colourless). 2(a)(i) Mass of capped weighing bottle and FA 4 / g 9.135 Mass of capped weighing bottle after emptying FA 4 / g 5.211 Mass of FA 4 used / g 3.924 Titration number 1 2 Final burette reading / cm3 19.90 39.80 Initial burette reading / cm3 0.00 19.90 Volume of FA 1 used / cm3 19.90 19.90 Values used ✓ ✓ 2(a)(ii) Average volume of FA 1 used = (19.90 + 19.90) / 2 = 19.90 cm3

-3- 2(b)(i) n(MnO4−) = 19.90 1000  0.0100 =0.0001990 mol n(Fe2+) in 25.0 cm3 of FA 5 = 5(MnO4−) = 5  0.0001990 = 0.0009950 mol [Fe2+] = 0.0009950 0.025 = 0.0398 mol dm−3 2b(ii) n(Fe2+) in 250 cm3 of FA 5 = 0.25  0.0398 = 0.009950 mol n(FA 4) in 250 cm3 of FA 5 = n(Fe2+) = 0.009950 mol Mr of FA 4 = 3.924 0.00995 = 394 2(b)(iii) Total number of positive charge = total number of negative charge (+1)a + 2 = 2b a = 2b − 2 2(b)(iv) Using the formula (NH4)aFe(SO4)b.6H2O, Mr of FA 4 = a(14.0 + 4  1.0) + 55.8 + b(32.1 + 16.0  4) + 6(2  1.0 + 16.0) = 18.0a + 96.1b + 163.8 From (b)(ii), Mr = 394.4  18.0a + 96.1b + 163.8 = 394.4 -----------(1) Substitute a = 2b−2 in (1), 18.0(2b − 2) + 96.1b + 163.8 = 394.4 36.0b – 36.0 + 96.1b + 163.8 = 394.4 132.1b = 266.6  b = 2.018 = 2 (whole number) a = 2(2) – 2 = 2 2(c) The student’s titre value would be higher. Due to insufficient amount of H 2SO4, MnO4− is reduced to MnO 2 instead of Mn 2+. As MnO4− accepts 3 e− when it is reduced to MnO 2 compared to accepting 5 e− when it is reduced to Mn 2+, more MnO 4− is needed to oxidise the same amount of Fe 2+, so titre value to reach end-point is higher. 3(a) test observations Using a 10 cm3 measuring cylinder, add 5.0 cm3 of FA 3 to a boiling tube. To this boiling tube, add 5.0 cm3 of FA 8 using the same 10 cm3 measuring cylinder. Leave this boiling tube in the hot water bath for about 3 minutes. Filter the reaction mixture into another boiling tube. Wash the residue with some deionised water. The residue is FA 9. Retain the filter funnel containing FA 9 for 3(c)(iii). No observable change. Red/brown ppt formed. Red/brown residue. Brown filtrate.

-4- 3(b) test observations with FA 6 observations with FA 7 (i) Add about 2 cm depth of FA 6 to a test-tube. To this test-tube, add 2 cm depth of dilute sulfuric acid, followed by 1 drop of potassium manganate(VII) solution and shake well. Leave the test-tube to stand in a beaker of hot water for about 5 minutes. Decolourisation of purple KMnO4 no observable change (ii) Add 1 cm depth of aqueous silver nitrate to a test-tube. Then slowly add 1 cm depth of aqueous sodium hydroxide. Add aqueous ammonia slowly, with shaking, until the precipitate just dissolves. You may use a clean glass rod to stir the mixture and help dissolve the precipitate. Add 1 cm depth of FA 6 to this mixture, shake the tube and place it in the test-tube rack to stand. Silver mirror formed. no observable change (iii) Add about 1 cm depth of FA 7 to a test-tube. To this test-tube, add 4 drops of sodium hydroxide solution followed by iodine solution, dropwise, until a permanent orange/red colour is present. Warm the mixture in a beaker of hot water for about 2 minutes. solution FA 6 turns orange Yellow ppt formed in orange / red solution. Add sodium hydroxide solution using a teat pipette until no further change is seen. solution FA 6 turns colourless Solution turns colourless. Yellow ppt remained.

-5- 3(b)(iv) functional group evidence FA 6 aldehyde In test (b)(ii), FA 6 gives a silver mirror with Tollens’ reagent. FA 7 ketone In test (b)(i), FA 7 did not decolourise purple acidified KMnO4. In test (b)(iii), FA 7 gave yellow ppt, CH I3, with alkaline aq iodine. 3(c) test observations (i) Test solution FA 8 with Universal Indicator paper. Universal indicator turns dark blue. pH is 12. (ii) Add about 1 cm depth of FA 8 to a test-tube. Then add 1 cm depth of dilute nitric acid. To this test-tube, add about 1 cm depth of aqueous barium nitrate slowly, with shaking, until no further change is seen. Solution FA 8 turns from dark blue to light blue. No effervescence. White ppt formed in light blue solution. (iii) Place the filter funnel containing FA 9 into a clean test-tube. Using a 10 cm3 measuring cylinder, carefully add 2.0 cm3 of dilute sulfuric acid to the filter funnel. The filtrate will collect in the test-tube. The residue is FA 10. The filtrate is FA 11. The residue turned from red/brown to red- brown/dark brown/black . Red-brown/dark brown/black residue. Blue filtrate. (iv) Add 1 cm depth of FA 11 to a test-tube. Add aqueous ammonia slowly, with shaking, until no further change is seen. Blue ppt formed, soluble in excess NH3(aq) to form a dark blue solution. 3(c)(v) Cu2+ 3(c)(vi) Oxidation state of element X: +1 Type of reaction: Disproportionation 3(c)(vii) SO42– is present. In test (c)(ii), FA 8 reacted with Ba(NO3)2 in the presence of HNO3(aq) to form white ppt of BaSO4.

-6- 4(a)(i) Let Vneut cm3 be the volume of barium hydroxide and (100 - Vneut) cm3 be the volume of hydrochloric aci