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 prefer