NYGH 2013-S3EOY-Chem P1 & P2 Ans

2013 Sec 3 Chemistry EOY Marking Scheme Paper 1 1  A  11  D  21  B  2  A  12  C  22  A  3  A  13  D  23  B  4  B  14  B  24  A  5  D  15  B  25  C  6  A  16  A  26  B  7  B  17  B  27  D  8  C  18  B  28  A  9  C  19  C  29  B  10  C  20  C  30  B  Paper 2 Question Answers Marks A1 (a) Magnesium carbonate / Lead (II) iodide (b) Potassium hydroxide and Ammonium sulfate (c) Ethanoic acid / Ammonium sulfate 1 2 1 A2 (a) Lewis Dot Diagram (b) Number of moles of acid: 20.0/1000 x 0.0625 = 0.00125mol Ratio of methylamine to acid is 20:1 . 1 1 1

A3 (a) 1 : 1 (b) No. of moles of oxygen = 23.9/16 = 1.494 mol Percentage of Y = 100- 23.9 = 76.1% M r of Y = 76.1/1.494 = 50.9 or 51(accept either) Hence Y is Vanadium. (c) Amphoteric Lead (II) oxide / Zinc oxide / Aluminium oxide / Other acceptable amphoteric oxides (d) (i) There are strong electrostatic forces of attraction between the oppositely charged ions in the crystal lattice. Hence a large amount of heat energy is needed to overcome the forces of attraction. (ii) Substance Z is able to conduct electricity in aqueous state / molten state but not in solid state. In the aqueous and molten state, the crystal lattice is broken down, the ions are no longer held in the fixed lattice structure and can move to conduct electricity. [Accepted: any other suitable properties such as soluble in water but not organic solvents / hard and brittle] 1 1 1 1 1 1 1 1 1 1 A4 Relative abundance for X-96 = 2.8% Ar of X = (51.5% × 90 + 11.2% × 91+17.1% × 92 + 17.4% × 94 +2.8% × 96) = 91.4 Identity of X = Zr (zirconium) (ECF possible for identity of metal. However, examples where ECF marks are not given: i. A r = 88.6, rounded off to 88, then identity of element is wrong. ii. If the relative abundance is not calculated but Zr is obtained due to averaging, only 1 mark given) 1 1 A5 (a) When black solid (CuO) stops dissolving, CuO is in excess. (Not accepted: precipitate or white precipitate observed or unreacted CuO) (b) The mass of copper(II)oxide It is added in excess. (c) When it cooled, th e solubility decreases and thus some of the copper(II) nitrate will reappear as crystals. (d) CuO(s) + 2H + (aq) → Cu2+(aq) + H2O(l) 1 1 1 1 1 1

(e) No. of moles of HNO 3 = 250/1000 x 2 = 0.5 mol 2 mol of HNO3 : 1 mol of Cu(NO3)2 No. of mol of Cu(NO3)2 = 0.5x ½ = 0.25 mol. Mass of Cu(NO3)2 = 0.25 x 188 = 47.0 g (f) By lowering the temperature further / allowing the solution more time to cool, more crystals can be formed due to the further decrease in the solubility. (Not accepted: heat longer, heat until saturated or heat until half of it original volume) 1 1 1 1 A6 (a) Number of mol of NO 2 = 0.12 /24 = 0.005 mol Number of molecules = 0.005 X 6X1023 = 3 X 1021 (b) (i) The molecules are closely packed together in a fixed orderly arrangement. The molecules can only vibrate about fixed positions. (ii) Melting 1 1 1 1 1 A7 (a) SO 2(g) + H2O (l) →H2SO3(aq) 2H2SO3(aq) + O2(g)  2H2SO4(aq) (b) Pass the air sample through a solution of acidified potassium manganate (VII)/ or paper. The purple colour will be decolourised / KMnO4 turns from purple to colourless. (c) 1) Add sulfuric acid to excess lead(II) nitrate . 2) Filter and collect the residue. 3) Dry the residue with filter paper. 1 1 1 1 1 1 1 B8a) Bubble the gas into limewater. A wh ite precipitate is formed if carbon dioxide is present. 1 b) Ba 2+, SO4 2− 1 c) The blue ppt dissolves to form a dark blue solution. 1 d) CuSO 4 1 e) Ag +(aq) + Cl−(aq)  AgCl(s) 1 f) There was no effervescence observed when acid was addd. / A precipitate was observed despite the addition of dilute nitric acid./ If carbonate ion was present, the precipitate would have dissolved upon addition of nitric acid 1 g) A is ZnCl 2 or AlCl3. B is CuCO3 1 for each; total 3 (h) Excess aqueous ammonia 1

B9a) 124 g/mol 1 b) 24 / 24.0 dm 3 1 c) No. of moles of CuCO 3 = 31.0 ÷ 124 = 0.250 mol 1 d) CuCO 3(s) + 2HCl(aq)  CuCl2(aq) + H2O(l) + CO2(g) 0.250 moles of HCl would react with 0.125 moles of CuCO3. Hence HCl is the limiting reactant. 1 1 1 e) Mole ratio of HCl to CO 2 is 2:1 Maximum volume of CO2 produced = ½ × 0.250 × 24000 = 3000 cm3 1 for ratio, 1 for volume of gas f) Percentage yield of CO 2 = 2800 ÷ 3000 × 100% = 93.3% 1 g) Carbon dioxide is slightly soluble in water. 1 B10 EITHER (a) Titration number 1 2 3 4 Final burette reading/ cm3 26.50 26.90 27.60 26.00 Initial burette reading / cm3 0.00 0.60 1.30 0.20 Volume of ferrous solution used /cm3 26.50 26.30 26.30 25.80 Best titration ()   Average volume of ferrous ammonium sulfate: (26.30 + 26.30 )/2 = 26.30 cm 3 1 (subtraction correct, 2dp) 1 1 (b) Purple 1

(c) Molar mass of (NH 4)2SO4.FeSO4.6H2O = 392.0 g/mol. Concentration of FeSO4 solution = = 0.0800 mol dm -3 Therefore no of moles of FeSO4 which reacted = = 0.00210 mol (3sf) 7.84 x 1000 392 x 250 26.30 x 0.0800 = 0.002104 mol 1000 1 1 (d) No of moles of KMnO 4 reacted =     = 0.000420 mol (3sf) 25.0 x 0.0168 = 0.0004200 mol 1000 1 (e) x =5 , y = 1 z = 8 No of moles of FeSO4 No of moles of KMnO4 = 0.002104 = 5.00 0.0004200 1 1 1

B10 OR a) 2NaOH+ H 2SO4  Na2SO4 + 2H2O No of moles of sodium hydroxide which reacted: 12.6/1000 x 0.250 = 0.00315 mol From equation, NaOH:H 2SO4 is 2:1 No of moles of excess sulfuric acid: 0.000315 x ½ = 0.0001575 = 0.00158 mol (3sf) (Note: any method which calculates the “number of moles” of the impure substance will be penalized) 1 1 for ratio, 1 for no of moles b) No of moles of acid originally: = 0.3125 mol No of moles of acid which reacted with carbonate: 0.03125-0.001575= 0.02968 mol = 0.0297mol (3sf) 25.0 x 1.25 1000 1 1 c) K 2CO3 + H2SO4  K2SO4 + CO2 + H2O No of moles of potassium carbonate reacted = 0.02968 x 1 = 0.02968 mol Therefore mass of potassium carbonate reacted = 0.02968 x 138 = 4.096g Percentage purity = 4.096 x 100% = 85.3% 4.80 1 1 1 d) Calcium carbonate will react with sulfuric acid to form calcium sulfate which is insoluble in water 1 1