NYGH 2012-S3EOY-Chem Ans

2012 Secondary Three End-of-Year Paper 1 and 2 (Chemistry) Paper 1 1 D 11 C 21 D 2 A 12 C 22 A 3 A 13 B 23 B 4 B 14 A 24 A 5 D 15 B 25 D 6 B 16 A 26 B 7 C 17 A 27 C 8 A 18 C 28 A 9 B 19 B 29 D 10 B 20 B 30 B Paper 2 Section A No Answers Marks A1(a) Element V has a lower melting point. Element V has simple molecular structure with weak intermolecular forces of attraction while element W is a metallic element with strong metallic bonds. During melting, less energy is required to overcome the weak intermolecular forces of attraction between molecules in V than strong metallic bonds in W. [3] b(i) It does not conduct electricity in molten state. [1] (ii) Start with 2 marks, deduct 1 mark for the any of the errors listed (max 2 marks). 1. Wrong chemical symbols. (e.g. Uses C l instead of R) 2. Shares wrong number of electrons./ Wrong number of non bonding electrons 3. Non bonding electrons are not paired. 4. Uses symbols other than dot and cross. 5. Brackets Max 1 mark if ionic compound is drawn. [e.c.f. from bi] [2] c(i) Gp II [1] (ii) U is able to conduct electricity in solid state as it is a metal with ‘sea of delocalized electrons’. [2] A2(a) Any excess aqueous nitric acid in the reaction is difficult to separate from the soluble salt product while excess insoluble magnesium hydroxide may be removed easily by filtration. [2] R  RO 1

No Answers Marks b(i) 3 ions [1] (ii) Covalent and ionic bonds [1] A3(a) (i) K2CO3(aq) + 2HNO3 (aq) →2KNO3 (aq) + H2O(l) + CO2(g) [2] (ii) Amount of K2CO3 added = 4.80÷ Molar mass of K2CO3 = 4.80÷138 = 0.03478 mol Amount of HNO3 used = 0.024x 1.67 = 0.04008 mol Mole ratio of K 2CO3 : HNO3 is 1:2. Thus, 0.03478 x 2 = 0.06956 mole HNO3 is required to completely react with 4.80 g of K2CO3. HNO3 is the limiting reagent. [2] (iii) Since HNO3 is the limiting reagent, and mole ratio of HNO3: KNO3 is 1:1, theoretical yield of KNO3 is 0.04008 mole. Percentage yield = mass of actual yield ÷ mass of theoretical yield x 100 1.70 = mass of actual yield÷(0.04008 x 101) Mass of actual yield = 6.88g (to 3 s.f.) [1] (iv) Excess potassium carbonate added dissolved in the solution and thus could not be removed by filtration. [1] b(i) OH-(aq) + H+(aq) → H2O(l) [2] (ii) M1V1 ÷ M2V2 = Mole ratio of KOH is to HNO3 (0.400 x 0.0300) ÷ (concentration of HNO3 x 0.0200) = 1 Concentration of HNO3 = 0.600 mol/dm3 [2] (c) Student B’s method is more suitable. [1] A4(a) Dibasic/ 2 [1] b(i) Every mole of HA will produce 2 mole of ions. Moles of HA that dissociated = 2 moles HA [1] (ii) Every mole of H2B will produce 3 mole of ions. Moles of H2B that dissociated = 2 moles H2B [1] (iii) H A is the stronger acid. [1] (c) 1. The solid may be zinc as reaction did not occur due to the absence of water. 2. Not all metals will react with acid. Unreactive metals such as copper and silver will not. [2] A5a(i) D, C, B, A [1] (ii) Particles at point Q are closely packed while those at point P are far apart from each other. Particles at point P moves around at a faster speed than those at point Q. / Particles at P move randomly by bombarding with one another at higher speed and particles at Q move by sliding over one another. [2] b(i) 2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(g) (ii) From the equation, gas volume ratio of CH4 :CO is 1:1. Volume of methane burnt = 30cm3 Mass of methane = Moles of methane x molar mass of methane = 0.03 ÷ 24 x (12+4) = 0.02 g [2] (iii) Mole ratio = volume ratio Moles of water vapor = 0.05 x ÷ 24 = 0.001667mole No. of water vapor molecules = 0.001667 x 6.02 x1023 molecules = 1.00 x 1021 molecules (3s.f.) [2] 2

No Answers Marks Accept answer if student use 6 x 1023 as Avogadro’s number. Alternative answer: Limiting reagent is oxygen. Moles of water = ( (20/3) x 4)/ 24000 =0.001111 Moles of water molecules = 0.001111 x 6.02 x 1023 = 6.69 x 1020 (3 s.f.) (c) C H Mass ratio 85.7 14.3 Mole ratio 85.7 ÷ 12 = 7.14 14.3 ÷ 1 = 14.3 Simplest ratio 1 2 Empirical formula is CH2 [2] Section B B6 (a) Ammonia 1 (b) Use a piece of moist red litmus paper to test for the gas. The moist red litmus paper will turn blue. 2 (c) A – Copper (II) oxide B- copper(II) sulfate C- copper (II) carbonate D- Carbon dioxide 4 (d) CuO(s) + 2H+(aq) → Cu2+(aq) + H2O(l) 1 (e) Carbonic acid 1 (f) Her conclusion is incorrect as there is a neutralization reaction taking place between the sodium hydroxide and carbonic acid. OR Her conclusion is incorrect as sodium carbonate formed is soluble in water and appears as a colourless solution and hence it seems there is no reaction when there is a reaction occurring. 1 B7 (a) CuO(s) + 2HNO3 (aq) → Cu(NO3)2(aq) + H2O (l) 2 (b) Number of moles of CuO= 8.00/ (64+16) = 0.100 mol CuO is the limiting reagent as 0.500 mol of nitric acid is in excess. 2 (c) According to the equation, the number of moles of CuO= the number of moles of Cu(NO3)2 = 0.100 mol Theoretical yield of copper(II) nitrate = 0.100 x (64+2x14+16x6) =18.8 g % yield = 9.45/18.8 x 100% = 50.27% =50.3 % (3 s.f.) 3 (d) As the the impurity of the copper(II) oxide is the only factor affecting the yield, the % purity will be equal to the % yield = 50.3% OR Number of mole of copper(II) nitrate =9.45/ (64+16x6+14x2) =9.45 / 188 =0.05027 mol Number of mole of pure copper(II) oxide = number of mole of copper(II) nitrate = 0.05027 mol Mass of pure copper(II) oxide = 0.05027 x (64+16) = 4.0216g % purity = 4.0216/ 8.00 x100 = 50.27 % = 50.3% (3 s.f.) 1 (e) Theoretical mass of copper(II) nitrate = 12.0 / (80/100) = 15.0g Number of moles of theoretical copper(II) nitrate = 15.0/ (64+2x14+16x6) = 0.07979 mol According to the equation, the number of moles of CuO= the number of moles of Cu(NO3)2 = 0.07979 mol Mass of CuO= 0.07979 x (64+16) =6.38g (3s.f.) 2 3

4    No Answers Marks B8 Either (ai) No. of moles of sodium hydroxide e = 31.7 / 1000 x 0.1 = 0.00317 mol. us The mole ratio of NaOH: acid=1:1 Hence, no. of moles of acid used = 0.00317 mol. Thus the concentration of the acid in mol/dm3 = 0.00317 x1000/25 = 0.1268 mol/dm3 = 0.127 mol/dm3 (3.s.f) 3 (aii) Relative Molecular mass of the acid = 8.00 / 0.1268 = 63.091= 63.1 (3.s.f) 2 (aiii) From part (aii), Mr(HXO3) = 63.1 Ar(H) + Ar(X) + 3[Ar(O)] = 63.1 Ar(X) = 14.1. From the periodic table, X is nitrogen. Hence the acid is nitric acid, HNO 3. 1 (b) Number of moles of NaOH= 0120 x 0.0208 = 0.002496 Number of moles of acid = 0.002496 / 3 = 0.000832 The concentration of the acid in mol/dm3 = 0.000832/ 0.025= 0.03328 mol/dm3. Concentration of the acid in g/dm3 = conc. in mol/dm3 x Mr = 0.03328 x 98=3.26144 g/dm3 Thus, percentage purity = % 10000 . 5 26144 . 3× = 65.2% 4 B8 Or (ai) No. of moles of KMnO4 that reacted = 02 . 01000 27 × = 5.4 x 10-4 or 0.00054 mol. 1 (aii) From the equation, 2 moles of KMnO4 reacted with 10 moles of FeSO4. 0.00054 moles of KMnO4 reacted with moles0027 . 0 00054 . 02 10 = × of FeSO4. Thus, in 25.0 cm3, there are 0.0027 moles of FeSO4. 1 (aiii) 25.0 cm3 of solution contains 0.0027 moles of FeSO4. 1000 cm3 of solution contains 0.108 moles of FeSO4. 1 (aiv) Mr(FeSO4) = 56+32+4x16 = 152 Mass of anhydrous iron(II) sulfate in 1 dm3 of solution = 0.108 x 152=16.416 g= 16.4 g 1 (av) Mass of water in 30.0 g of hydrated iron(II) sulfate = 30.0 - 16.4 = 13.6 g 4 . 16 6 . 13 4 =FeSO of M moleculeswater ALL of M r r 4 . 16 6 . 13 152 ) ( 18=X X = 7 2 (bi) H2X + 2NaOH → Na2X + 2H2O 1 mole of the acid reacts with 2 moles of NaOH No. of moles of NaOH = 0.004 moles. No. of moles of H 2X= 0.5 x 0.004= 0.002 moles. Hence concentration of the acid= 0.002 x