HCI 2012 ANS

1 C1 HCI Chemistry Block Test ANSWER Paper 1: MCQ 1 B 11 B 2 D 12 A 3 A 13 C 4 B 14 A 5 B 15 C 6 D 16 D 7 B 17 C 8 D 18 A 9 C 19 D 10 D 20 A Paper 2: 1 (a) Mg(NO2)2•xH2O  Mg(NO2)2 + xH2O  mass of water = 2.00 ‒ 1.24 = 0.76 g n(H2O) = 0.76 / 18.0 = 0.04222 mol n(Mg(NO2)2) = 1.24 / 116.3 = 0.01066 mol x = 0.04222 / 0.01066 = 3.96 = 4 (whole no.) (b) The number of water of crystallization is higher than in the given information. The likely reason for the discrepancy is that on heating, the magnesium nitrite decomposed. Therefore the mass loss is higher than the actual value and this results in a higher value of x. (c) (i) 5NO2 – + 2MnO4 – + 6H+  5NO3 – + 2Mn2+ + 3H2O (ii) Using a fixed mass of magnesium nitrite, dissolve in deionised water in a volumetric/standard/graduated flask (iii) The solution in the flask turns from pink to colourless at the endpoint. (iv) Repeat the titration until two titre values within ± 0.10 cm 3 of each other are obtained. 2 (a) (i) 2  and 2  (ii) or or – +

2 (b) (i) CN– + H+  HCN (HCN) = (NaCN) = 14.012.023.0 1.0  = 49.0 1.0 = 0.02041 mol Mass of HCN formed = 0.02041 x 27.0 = 0.551 g (ii) [HCN] = 1000 75 10000.551   mg dm–3 = 7.35 x 10–3 ppm The concentration of HCN has not exceeded the exposure limit. (c) (i) ∆S is positive since the number of moles of gaseous products is more than that of reactants (or there is an increase in the number of gaseous molecules). Hence, there are more ways in which the particles and their energies can be distributed in a larger volume. (ii) (CH4) left = 0.5 – 0.25 = 0.25 mol (HCN) formed = 0.25 mol (H2) formed = 3 x 0.25 = 0.75 mol Total no. of moles of gases = 1.25 mol p = 3-105)(5 0)1(2738.311.25   0 = 387000 Pa = 387 kPa (d) 1. The ions (N3– to Al3+) are isoelectronic. 2. Nuclear charge increases (while shielding effect remains constant). 3. Effective nuclear charge increases Or The electrons are more strongly attracted to the nucleus and hence size of ions decreases. 3 (a) (i) trigonal planar (ii) sp2 (iii) (iv) N O F Ne Na Mg Al Ionic radius

3 (b) (i) Bond energy refers to the average energy required to break 1 mole of a covalent bond in the gaseous atoms. (ii) ΔHr = Σ B.E. (bonds broken) - Σ B.E. (bonds formed) = [(C=C) x 1 + (H-H) x 1] – { [(C-H) x 2] + C-C} = [(610) x 1 + (436) x 1] – { [(410) x 2] +350 } = 1046 – 820 – 350 = – 124 kJ mol–1 (c) (i) Vitamin C can form hydrogen bonding with water molecules in urine. (ii) Dehydroascorbic acid is l ess soluble due to less extensive hydrogen bonding ( or fewer –OH groups available for hydrogen bonding or fewer H atoms that can form hydrogen bonding) 4 (a) (i) (ii) SO2 has two bonding groups of electrons and one lone pair, hence its shape is bent. As lone pair -bond pair repulsion is greater than bond pair -bond pair repulsion, the angle should be 118o. (Accept any angle between 109.5o and 120o.) (b) (i) I2 + SO2 + 2H2O → 2HI + H2SO4 (ii) No. of moles of H2 = 24000 500 = 0.0208 mol No. of moles of HI required = 0.0208 ×2 = 0.0416 mol No. of moles of SO2 required = 0.0208 mol (c) (i) 2HI (g) I2 (s) + H2(g) I2 (g) 2I (g) + 2 H (g) Hrxn = 2(299) – (+151) – (+62.4) – (436) = –51.4 kJ mol–1 (ii) G = H – TS At 400K, G = –51.4 – 400 (–166 ×10–3) G = + 15.0 kJ mol–1 The reaction is not feasible at 400K. +62.4 +151 +436

4 Paper 3 Question 1 (a) CaF2 has a giant ionic structure, with strong ionic bonding (strong electrostatic forces of att raction b/w cations and anions). When force applied, ions of like charges are brought next to each other and the repulsive forces crack the compound. (b)(i) E.A. of F = (-1220 – 177 – 590 – 1150 -158 + 2613) / 2 = -341 kJ mol-1 (ii) From Data booklet, r+(Ca2+) = 0.099nm, r+(Mg2+) = 0.065nm L.E. r+(Ca2+) > r+(Mg2+) Hence, magnitude of L.E. of MgF2 is greater than magnitude of L.E. of CaF2. (c) X has a net dipole moment (or is polar) and permanent dipole- permanent dipole interactions exist. Y has net zero dipole moment (or is non-polar) and dispersion forces exist between molecules of Y More energy is required to overcome the stronger permanent dipole- permanent dipole interactions, hence Y has higher boiling point. (d) H-F bond is highly polarized (large difference in electronegativity between H and F) while P-H bond is non-polar. Partial ionic character in H-F bond leads to increase in strength. (e)(i) HF forms hydrogen bonding with ethanol molecules. NaF dissolves in water to form ion-dipole interaction with water molecules. 158

5 Question 2 (a)(i) O+(g)  O2+(g) + e (a)(ii) ∆H = +3390 – 1310 = +2080 kJ mol‒1 (a)(iii) There is a big jump in ionization energy from the 7 th IE to the 8th IE. Therefore the 8th electron to be removed is in an inner quantum shell, so A has 7 valence electrons. (a)(iv) The first IE of A is smaller than oxygen. Therefore it cannot be fluorine because fluorine is in the same period and has a higher effective nuclear charge, so its first IE would be greater. or A should be chlorine because Cl has a larger atomic radius. The electron to be removed is further from the nucleus and not attracted as strongly, therefore less energy is required to remove it and the first IE is lower compared to oxygen. (b)(i) pV=nRT pV=(m/Mr) RT Mr = (RT/P) (m/V) which is in the form Molar mass = y x density i.e. y = RT/P At stp, y = 8.31 x 273 / 101300 = 2.24 x 10−2 (shown) The second mark is awarded if student recognizes y as the molar volume at stp and converts 22.4 dm3 to 2.24 x 10−2 m3. (b)(ii) Apparent molar mass of mixture = 2.24 × 10−2 x 1.83 × 103 = 41.0 g mol‒1 The shielding gas is E. The apparent molar mass of mixture is higher than that of argon (Mr = 39.9), therefore the second gas must have Mr higher than argon, therefore it is carbon dioxide (Mr = 44.0). (b)(iii) Let the number of moles of argon be x. 39.9x + 44.0(1-x) = 41.0 x = 0.732 (b)(iv) Shielding gas E shows the most deviation from ideal gas behaviour. Out of the 4 gases H2, He, O2 and CO2, the CO2 has the largest electron cloud, so the strength of the dispersion forces between its molecules (and also between its molecules and Ar atoms) should be the strongest. (accept volume is more significant as electron cloud size is larger) (c)(i) Number of moles of Sn2+ in 25.0 cm3 of solution G = 0.0135 x 0.0200 x 2.5 = 6.75 x 10-4

6 (c)(ii) Number of moles of Sn2+ in second titration = 0.0203 x 0.0200 x 2.5 = 1.015 x 10-3 Number of moles of Sn4+ in 25.0 cm3 of solution G = 1.015 x 10-3 – 6.75 x 10-4 = 3.40 x 10-4 Sn2+/Sn4+ ratio = 2 : 1 (c)(iii) SnO SnO2 (c)(iv) Ratio of SnO : SnO2 = 2 : 1  2SnO + SnO2 = Sn3O4 Formula of F = Sn3O4 or Formula of F = 2SnO•SnO2 or (c)(v) Fe(II) will be further oxidized to Fe(III) by KMnO4 in addition to the oxidation of Sn(II) to Sn(IV). Th