Serangoon Prelim Ans

1 SSS 2024 4E Pure Chemistry Prelim Mark Scheme Paper 1 1 2 3 4 5 6 7 8 9 10 C A C C C B D B A A 11 12 13 14 15 16 17 18 19 20 A B B B C B B A C B 21 22 23 24 25 26 27 28 29 30 D C C A A A A D A A 31 32 33 34 35 36 37 38 39 40 A D C C B B D A A C Paper 2 Section A Marker’s comments 1a A l 1 b N or O 1 c Fe 1 d Ca 1 e C 1 f O 1 2a Yellow and purple Both answers must be correct. 1 b Rf yellow dye = 0.08 (@ 64/200 x 100 = 32% ethanol) 1 c Rf value for yellow dye in pure water is 0, which means that the yellow dye cannot be separated; [1] 2

2 Rf value for red dye in pure ethanol is 0, which means that the red dye cannot be separated;. [1] *both correct but no Rf value mentioned minus 1m *1 correct but no Rf value 0m d At 80% ethanol in the solvent mixture, [1] -All 3 dyes have the same Rf values of 0.4. -All the dyes present travel the same distance/travel at the same speed /rate. -All the 3 dyes have the same solubility. Any one [1] 2 e The order and position of the dyes must be correct 1 3ai Zn + 2HCl → ZnCl2 + H2 1 aii AsH 3 / H3As 1 aiii 2

3 [1] lone (not used for bonding) pair [1] shared electrons btw As and H bi As H percentage / % 97.4 2.6 Ar 75 1 number of moles / mol 97.4/75 = 1.30 2.6/1 = 2.60 simplest ratio 1.30/1.30 = 1 2.60/1.30 = 2 Empirical formula is AsH2 [1] no. of moles expressed to 3 sf [1] ratio expressed to nearest whole no 2 bii n = 154 / 77 = 2 Molecular formula is As2H4 / H4As2 1 biii 1 4a Halogens gain (one) electrons when they react / form ions. R: require an electron 1 b At 2 (aq) Br 2 (aq) Cl2 (aq) At – (aq)   Br – (aq)   Cl – (aq)   one row / one column correct (1) all correct (2) 2 ci colourless to brown 1

4 R: yellow cii Iodide / I – ions  undergo oxidation to become I2, or  lose electrons, or  increase in oxidation state, R: Iodide/potassium iodide is oxidized (already stated in questions) Oxidising agents  gains electrons from potassium iodide  are reduced (any one) 1 1 d  particles move/diffuse from a region of higher concentration to a region of lower concentration  particles dissolve / ionise in water in the damp litmus paper to form H+ ions  The presence of the H+ ions turns the blue litmus paper to red when in contact with litmus (paper). 1 1 1 5ai Factor Ozone levels (Increase or Decrease) Smaller population density Decrease More electrical cars Decrease Sunny skies Increase 1 aii There are more vehicles / more industrial activities emitting [1] oxides of nitrogen and volatile organic compounds during mid-afternoon. Sunlight/UV is strongest [1] during mid-afternoon and will increase the rate of reaction. 2

5 bi Carbon monoxide/unburnt hydrocarbon/ sulfur dioxide/ oxides of nitrogen 1 bii CO and unburnt hydrocarbon- ensure there is sufficient oxygen for complete combustion of petroleum. Sulfur dioxide- flue gas desulfurisation or sulfur dioxide from chimney gases are absorbed by calcium carbonate. 1 6a No effervescence/bubbles were observed when aluminium is placed in sulfuric acid 1 bi Copper is below hydrogen in the metal reactivity series and hence cannot displace hydrogen from sulfuric acid. 1 ii Place a piece of lead metal in a beaker containing aqueous copper(II) nitrate. If lead metal is more reactive than copper, a brown solid will be formed/ blue solution will turn colourless. If lead metal is less reactive than copper, there will be no visible change. (accept any other possible test) 1 1 1 c Example: using zinc or magnesium blocks on ship’s hull / underground pipes / bridge [1] Explain: Zinc or magnesium is more reactive than iron and will corrode in place of iron. [1] 2 7a Experiment B. (no mark w/o reason given) Lumps of lead(II) carbonate has a larger particle size compared to the powder form; thus it has smaller surface area exposed for collisions. [1] 2

6 Hence lumps will have a lower frequency of effective collisions between reactant particles and thus have a lower rate of reaction. [1] b Experiment using CaCO3 (x/100 mol) will have a larger amount of carbonate present compared to PbCO3 ( x/267 mol), as CaCO3 has a smaller molar mass than PbCO3. [1] Thus, greater volume CO2 will be produced in the reaction that uses CaCO3 [1] 2 ci steeper gradient than A; lower volume of gas 1 cii The rate of reaction is fast initially as sulfuric acid is a dibasic acid and hence has a twice the number of H+ per unit volume / compared to HNO3 in experiment A [1] The small volume of gas produced quickly remains constant as PbCO3 reacts with sulfuric acid to form insoluble PbSO4 which forms a protective layer around the carbonate and prevents further reaction from taking place. [1] 2

7 8a Cu 2+ + 2e → Cu 1 b Filter the mixture; [1] Rinse the copper with water and allow to dry Weigh the pure, dry copper using an electronic weighing balance [1] Add the mass of this pure dry copper to the mass of the increase of copper at the negative electrode to obtain the total mass of the copper produced. Or find the gain in mass of copper at the negative electrode and add it to the mass of pure, dry copper obtained after filtration. [1] 3 c (i) Yes, I agree. Keeping the current constant e.g at 0.3A, when time is doubled (from 10 mins to 20 min), total mass of copper produced is doubled (0.06g to 0.12g) [1] (ii) Yes, I agree. Keeping time constant e.g. 20 minutes, when current is doubled (from 0.3 A to 0.6 A), the total mass of copper produced is double (0.12 g to 0.24 g) [1] 2 d 3 answers: The solution becomes acidic; [1] Copper(II) ions are selectively discharged at the cathode while hydroxide ions are discharged at the anode. [1] As a result, the concentration of hydrogen ions is greater than hydroxide ions. [1] Or The blue electrolyte fades/colourless. [1] This is because copper(II) ions which gives the blue colour of the electrolyte are selectively discharged at the cathode. [1] As a result, the concentration of copper(II) ions in the electrolyte decreases overtime. [1] 3

8 e Mass of copper = 0.24 (from the graph) No. of moles of copper = 0.2464 = 0.00375 mol [1] No. of copper atoms = 0.00375 x 6 x 1023 = 2.25x 1021 atoms [1] 2 9a Helium is a noble gas with a stable electronic configuration of 2. [1] OR Helium has a fully filled valence electron shell. Hence, helium does not need to gain, lose or share electrons and will not react with other substances. [1] 2 accept “stable electronic configuration of noble gas” b Any two of the following  Heating is involved in both separation methods.  Both separation methods separate miscible liquids.  The substances with the lower boiling point will be separated out / collected first for both methods. OR Both separation methods are based on difference in boiling points of the substances. 2 reject “Both are separation techniques” reject “Both can obtain pure substances” reject “both methods involves use of tubes” ci Signal A – toluene Signal B – mesitylene 1 both must correct; reject if spelling incorrect cii amount of toluene : amount of mesitylene = Signal A height : Signal B height = 3 : 5 2 Allow error carry forward if student has identify the signals wrongly in 9c(i)

9 Percentage amount of toluene = 3/ (3+5) ×100 = 37.5 % [1] Percentage amount of mesitylene = 5/ (3+5) ×100 = 62.5 % [1] OR 100% - 37.5% = 62.5% di A higher flow rate will result in shorter retention time. [1] For example, with hexane when the flow rate in