TJC Prelim P1 Worked Solution

1 2017 JC2 Prelim H2 CHEMISTRY MCQ Worked Solution 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B D C A D B A D B B D C D A C 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A C A B C B C D B B A A B C C 1 Answer: B Since Ba(NO3)2  N2, Mr of Ba(NO3)2 = 261.3 No. of moles of N2 = No. of moles of Ba(NO3)2 = 1 261.3 = 3.83 x 103 mol Volume of N2 = 3.83 x 103 x 24000 = 91.8 cm3 2 Answer: D Definition – Relative molecular mass is the average mass of one molecule of an element or compound on a scale on which one atom of the 12C isotope of carbon has a mass of 12 units. Option 1 is incorrect. Relative molecular mass is a ratio. Option 2 is incorrect. It should be the ratio of the average mass of a molecule to 1/12 the mass of a 12C atom. Option 3 is correct. 3 Answer: C A is incorrect as SO 2 is an oxidizing agent and oxidises H2S in reaction. B is incorrect as SO 2 is the intermediate and is not regenerated in the reaction. C is correct as H 2S (oxidation state of sulfur is -2) is oxidized to S (oxidation state 0). D is incorrect as reaction II is a comproportionation reaction. 4 Answer: A Angle of deflection 𝛳 ∝ q/m. So 𝛳CamCa/qCa = 𝛳XmX/qX Mass of X = 20(40.1)/2 / 5.08 = 79 (Se) Atomic number of X = 34 5 Answer: D A: Ethene is a planar molecule which has all atoms on the plane B: Tri-iodide has 3 lone pairs and 2 bond pairs, hence the ion is linear and all atoms lie on the same plane C: XeF4 has 4 bond pairs and 2 lone pairs, hence shape is square planar and all atoms lie on the same plane D: BeCl42- has a total of 4 bond pairs (2 covalent bonds and 2 dative bonds) around Be atom. The shape is tetrahedral. 6 Answer: B A: HF has hydrogen bonding between its molecules and hence require a larger energy to overcome compared to pd-pd between HI molecules. B: MgO has a higher boiling point. MgO has a higher lattice energy than NaCl due to larger charge and smaller ionic radii of Mg2+ and O2- ion compared to Na+ and Cl-. C: SiH4 has a higher boiling point as its M r is larger than CH4 and thus the id-id interactions are stronger and more extensive than CH4. D: trans-C2H2Cl2 has a lower boiling point as it has no net dipole moment so the molecule is non-polar and only has id-id interactions between the molecules. cis-C2H2Cl2 has pd-pd interaction between the molecules and more energy is needed to overcome the stronger pd -pd interactions. 7 Answer: A PV = nRT At r.t.p, R = PV/nT = (1 atm x 24000 cm3)/(1 mol x 293K) Mr = mRT / PV = (m x T x 24000) / (p x V x 293) 8 Answer: D A & C: Wrong as the concentration of manganate would decrease slowly at the start of the reaction before decreasing more quickly as more Mn 2+ catalyst is generated. B: Wrong as the volume of CO 2 cannot be increasing rapidly at the start of the reaction due to slow rate of reaction. 9 Answer: B A: HBr will complete dissociate to give free H + ions and Br- ions. Since there is a change in the number of ions as the reaction progresses, so conductivity of the solution increases over time and can be monitored using a conductivity meter. B: Although CO2 gas is produced, the reaction is done at atmospheric press ure and not in a closed system, so pressure of the system will not change over time. C: Br 2(aq) is reddish -brown and all the products are colourless, so the intensity of the solution would decolourise over time. D: As the reaction progresses, more HBr (a strong acid) is produced, so concentration of H+ would increase over time and can be monitoring by quenching and titration against standard NaOH. 10 Answer: B No. of moles of bromine = 32/(79.9 x 2) = 0.200 mol When bromine vapourises, G = 0 since it is an equilibrium reaction. H = TS S = H/T = (30.9 x 103 x 0.2)/(58.8 + 273) = 18.6 J K-1 A: 7 B: 9 C: 8 D: 6

2 11 Answer : D Heat released when E is burnt = 200 x 4.18 x 26.4 = 22070 J No. of moles of E = 22076 ÷ (3290 x 103) = 0.00671 mol Mr of E = 0.47/0.00671 = 70 (C5H10) Option 1 is incorrect as it is short of a CH2. Options 2 & 3 are isomers of C5H10. 12 Answer : C From reaction I, Kc = [X2Y] [X2][Y2] 1 2 = 2 For reaction II, Kc’ = [X2]2[Y2] [X2Y]2 = 1 Kc 2 = 1 4 13 Answer : D A: P + Q R + 3S Initial amt/mol - 0.5 0 - Change in amt/mol - -0.1 +0.1 - Eqm amt/mol - 0.4 0.1 - Kc = [𝑅] [𝑄] = 0.1 0.4 = 0.25. B: When temperature increases, the forward reaction is favoured by Le Chatelier’s Principle to absorb heat. At equilibrium, there will be greater no. of moles of R and lesser no. of moles of S. Thus, Kp increases. C: There is equal number of moles of gas on both sides of the equation. Increasing pressure will not affect equilibrium. D: P is a solid and is not included in the equilibrium constant. Changing its concentration will not affect the equilibrium. 14 Answer: A Buffers are formed when roughly equal amount of the conjugate acid-base pair is present. A is not a buffer as the resultant mixture only contains CH3CH2CO2H from the reaction between HCl and CH3CH2CO2. B is buffer as it contains HCO3 and CO32. C is a buffer as CH 3CH2NH2 is in excess and CH3CH2NH3+ is formed from the reaction between CH3CH2NH2. D is a buffer as NH4+ is in excess and NH3 is formed from the reaction between NH4+ and NaOH. 15 Answer: C A & B : During rusting, oxygen is being reduced to hydroxide ions and iron being oxidised to iron(II) ions. C: O2 + 4H+ + 4e- 2H2O +1.23 V EOcell = +1.23 – (-0.44) = +1.67 V Since EOcell > 0, the reaction is spontaneous. Hence, it is not inhibited a low pH. D: EoFe2+/Fe = - 0.44 V EoMg2+/Mg = -2.38 V Since E oMg2+/Mg is more negative than E oFe2+/Fe, magnesium will be oxidised instead of iron. Hence, magnesium acts as a sacrificial metal and prevents corrosion. 16 Answer: A 1: E oCu2+/Cu = +0.34 V EoMg2+/Mg = -2.38 V EoAg+/Ag = +0.80 V At the anode, copper is preferentially oxidised over water to form copper(II) ions. As Eᶿ(Mg2+/Mg) is more negative than Eᶿ(Cu2+/Cu), magnesium will also be preferentially oxidised. Ag will not be oxidised as E ᶿ(Ag+/Ag) is more positive than Eᶿ(Cu2+/Cu). Ag will be collected below the anode as ‘anode sludge’. The decreased in mass at the anode is due to oxidation of copper and magnesium. Silver is not being oxidised but is collected as anode sludge. 2: At the cathode, only copper(II) ions were redu ced to copper. As E ᶿ(Mg2+/Mg) is more negative than Eᶿ(Cu2+/Cu), magnesium will not be preferentially reduced together with copper. 3: The amount of copper deposited is dependent on the magnitude of the current and time in which the current was supplied. It is not affected by the size of the electrode. 17 Answer: C A The covalent character increase across the period. B The melting point increase then decrease across the period. C The pH change from basic to neutral to acidic across the period. D Only Al2O3, P 4O10 and SO 3 are soluble in NaOH, so there is actually no trend. 18 Answer: A The charge density decrease s with increasing atomic no., hence the tendency to form complexes, acidity of the aqueous chloride solution (because ease of hydrolysis decreases) and magnitude of hydration energy decreases The reduction potential becomes more negative with increasing atomic no., hence the metal’s reducing power increases. 19 Answer: B Since X- can react with both Y2 and Z2 in the process forming X2, it is the strongest reducing agent. Since Y- cannot react with both X2 and Z2 to form Y2, it is the weakest reducing agent.

3 20 Ans