2021 TJC H2Chemistry P1 Ans Prelim

1 2021 TJC JC2 H2 Chemistry Prelim MCQ Worked Solutions 1 2 3 4 5 6 7 8 9 10 D C A D A C D B C D 11 12 13 14 15 16 17 18 19 20 A C D C B D D C C B 21 22 23 24 25 26 27 28 29 30 B C B B B B B A D C Qn Worked Solution 1. Answer: D Mr of E2 = (1.18 x 10–22) x (6.02 x 1023) = 71.0 Mr C2H4E2 = 99.0 Amount of C2H4E2 in 49.5 g = (49.5/99.0) = 0.500 mol No. of E atoms in 49.5 g of C2H4E2 = 2 x 0.5 x 6.02 x 1023 = 6.02 x 1023 2. Answer: C + Number of neutrons in Q = 219 – 86 = 133 3. Answer: A The long hydrophobic carbon chains on lecithin can form id-id with non-polar molecules. The bond angle with respect to C for –COO group is 120°. It contains delocalized electrons due to the pi bonds. 4. Answer: D Of the seventeenth carbon atoms in the ring, two are sp2 hybridized while fifteen are sp3 hybridised. Because these carbon atoms are bonded in a ring, the atoms can’t be lying on the same plane (sp3 hybridization requires electron pairs to be arranged in a tetrahedral configuration). The C=C double bond is found in a ring, For the molecule to exhibit cis-trans isomerism at this double bond requires the ring to be twisted and bond will be broken. So cis-trans isomerism for a C=C double bond found in a ring is not possible. Cholesterol has a secondary alcohol which is non-acidic. 5. Answer: A For gas at stp, PV = nRT Ra22 3 88 Q21 9 86 He4 2

2 1x105 x 22700 = 1 x R x 273 R = !"!×$$%""$%& pV = nRT, pV = (mass/M)RT Density = mass/V = pM / RT = '×(×$%&)×!"!×$$%"" 6. Answer: C pV = nRT Since V and n are constant, p µ T and pV µ T (not constant), thus Graph 2 is correct and Graph 1 is incorrect. PV = nRT, concentration = n/V, c = P/R (1/T). Since pressure is constant, c µ 1/T 7. Answer: D In cold/limited amount of water (or when PCl5 : H2O = 1 : 1): PCl5(s) + H2O(l) à POCl3(l) + 2HCl(g) Colourless liquid White fumes 8. Answer: B Since both X and Y form oxides that react with aqueous sodium hydroxide, they cannot be SiO2 (which only react with molten NaOH at high temperature). Oxide of X is likely to be SO2 where S has an oxidation state of +4. 9. Answer: C Iodide ions are better reducing agents since S in SO42-(+6) is reduced to SO2(+4), S(0) and H2S(-2). 10. Answer: D Keeping pressure constant, when temp increases % of N2 increases Þ position of equilibrium shifts left to absorb heat. Backward reaction is endothermic, hence forward reaction is exothermic. Keeping temperature constant, there is an increase in % N2 when pressure changes from P2 to P1. Since equilibrium position shifts left where more gaseous molecules are produced, P1 < P2. 11. Answer: A Statement 1: HClO is a bronsted acid as it donates a proton. Hence the product formed (ClO- ion) is a conjugate base. Statement 2: N2H5+ is the Bronsted acid in Reaction 2 as it donates a proton. Statement 3: Since the POE lies to the right for both reactions, HClO is a stronger acid than N2H5+ from Reaction 1 as it prefers to donate a proton. Likewise for Reaction 2 where N2H5+ is a stronger acid than NH4+. Statement 4: N2H4 is the Lewis base in Reaction 1 as it donates a lone pair of electrons for dative bonding to a proton from HClO. 12. Answer: C Both options A & B have excess sodium hydroxide and HCl respectively after mixing.

3 Option C forms phenylamine and unreacted phenylammonium chloride after mixing which is an alkaline buffer. Option D forms CH3CO2H and unreacted HCl. 13. Answer: D Enthalpy change of solution = -457 – 390 + 918 kJ mol-1 = +71.0 kJ mol-1 Amount of NaF = 8.4/42 mol = 0.200 mol Heat absorbed by 0.200 mol NaF = 14.2 kJ 14.2 x 1000 = mcDT 14.2 x 1000 = 250 x 4.2 x (Tinitial – 20) Tinitial = 33.52 oC 14. Answer: C DHreaction = 2(-243) – 2(-20.5) kJ mol-1 = -445 kJ mol-1 As T increases, position of equilibrium shifts left, decreasing yield of sulfur(s). 15. Answer: B Consider a pseudo 1st order reaction: rate = k[A]1[B]b, where B is in excess. So rate = k’[A]1, where k’ = k[B]b Hence, k’ depends on temperature (which affect k), conc of B (excess reactant) and presence of catalyst (which affects activation energy). 16. Answer: D Total volume is constant, so vol is proportional to conc of reactants. Relative initial rate of reaction is proportional to (vol of T / time). Hence, Experiment Volume of S / cm3 Volume of T / cm3 Volume of U / cm3 Initial rate 1 10 5 5 0.250 2 5 5 5 0.125 3 10 5 2.5 0.125 4 10 2.5 5 0.250 Comparing experiment 1 and 2, when [S] is halved, initial rate is halved too à order wrt [S] = 1 Comparing experiment 1 and 4, when [T] is halved, initial rate is remains constant à order wrt [T] = 0 Comparing experiment 1 and 3, when [U] is halved, initial rate is halved too à order wrt [U] = 1 17. Answer: D Option 1: wrong as there are 4 bond pairs around B atom and should be tetrahedral shape (sp3 hybridised).

4 Option 2: Each boron uses two electrons in bonding to the terminal hydrogen atoms and has one valence electron remaining for additional bonding. The bridging hydrogen atoms provide one electron each. Option 3: Due to 3-center 2-electron bonds, the B-H bond in the ring is a “half-bond” and thus is weaker than the terminal B-H bonds. 18. Answer: C A & D: achiral, so does not exist as enantiomers. B: C: 19. Answer: C 20. Answer: B There are 5 different types of hydrogen for substitution. 21. Answer: B A: Acidified potassium manganate would oxidise the alkene. The ester would be hydrolysed too. Ethanedioic acid is produced which can be further oxidised to form carbon dioxide. B: Carboxylic acid needs to be converted to a more reactive acyl chloride. C: X can undergo neutralisation with the carboxylic acid group under room conditions. When heated, the ester group would be hydrolysed. D: Na metal reacts with both the carboxylic acid and alcohol groups to produce 2 moles of hydrogen gas. 22. Answer: C P contains either alcohol or carboxylic acid since it gives white fumes with PCl5. * * * * * * * a b b c c d e e

5 A is incorrect as tertiary alcohols cannot be oxidised. B is incorrect as ketones cannot be oxidised. D is incorrect as side chain oxidation of benzene ring requires acidified potassium manganate(VII) instead. 23. Answer: B Statement 2 is incorrect as LiAlH4 would also reduce the ester group. Hydrogen with platinum catalyst can be used instead to only reduce the nitrile. Statement 3 is incorrect as W reacts with excess bromoethane to form RN+(CH2CH3)3 which is not basic. Statement 4 is correct. Y is ethanol and gives a positive iodoform test. 24. Answer: B According to Markovnikov’s rule, H would be added to the carbon atom in the C=C with more H. The structures of A and D are incorrect. Step 2: From molecular formula of E, bromine is substituted by –CN. 25. Answer: B A is incorrect as the reaction does not involve the ester. C is incorrect as it has one missing -CH2. D has the double bond at the wrong position. 26. Answer: B Propagation: CH2Cl2 + Cl • à • CHCl2 + HCl • CHCl2 + Cl2 à CHCl3 + Cl • CHCl3 + Cl • à • CCl3 + HCl • CCl3 + Cl2 à CCl4 + Cl • Termination: • CCl3 + • CHCl2 à CCl3CHCl2 27. Answer: B (CLT for 2021 exam) 28. Answer: A A: 2-nitrophenol or 4-nitrophenol is formed which is a pale yellow precipitate. B & C: -COCH3 is ring deactivating while –CH2CH3 and –NH2 are ring activating. Hence, the electron density in phenylethanone is lower. It would undergo nitration at a slower rate and higher temperature. D: As –OH is ring activating, it does not require concentrated sulfuric acid to generate the electrophile. 29. Answer: D During discharging, the cell behaves as a galvanic cell, hence reaction is spontaneous, ∆G is negative. At electrode X, Pb is oxidised to PbSO4, hence X is anode, and Eo2 is more positive than Eo1.

6 For 2 mol of electrons transferred during discharge, the overall equation is Pb + PbO2 + 2H+ + 2HSO4– à 2PbSO4 + 2H2O. As the cell discharges, [H+] decreases, hence pH incr