NJC Physical Chemistry Revision Part 1 Answers

National Junior College 2024 H2 Chemistry Revision Package 0 1) Atoms, Moles and Stoichiometry Answers: 1) Atoms, Molecules and Stoichiometry 1 A 2 C 3 D 4 D 5 A 6 D 7 C 8 C 9 D 10 B 11 B 12 D 13 A 14 D 15 B 16 A 17 B 1) Atoms, Moles and Stoichiometry 1 s. Amount of KOH used for complete neutralisation = 25 1000 × 1.0 × 10-2 = 2.5 × 10-4 mol Amount of H+ in the resin = 2.5 × 10-4 mol Ca2+ Ξ 2H+ Amount of Ca2+ in 50 cm3 sample = 2.5 × 10-4 ÷ 2 = 1.25 × 10-4 mol Concentration of Ca2+ in original sample = 1000 50 × 1.25 × 10-4 = 2.5 × 10-3 mol dm−3 Ans: A 2 Amount of NaOH used = 25.80 1000 × 0.100 = 2.58 × 10-3 mol Total amount of H+ used = 50 1000 × 0.050 = 2.5 × 10-3 × 2 = 5 × 10-3 mol Amount of H+ that reacted with NH3 = 5 × 10-3 − 2.58 × 10-3 = 2.42 × 10-3 mol NH3 + H+ 🡪 NH4+ Amount of NH3 passed into H2SO4 = 2.42 × 10-3 mol (NH4)2SO4 Ξ 2 NH3 Amount of (NH4)2SO4 in 25.0 cm3 = 2.42 × 10-3 ÷ 2 = 1.21 × 10-3 mol Amount of (NH4)2SO4 in 250 cm3 = 1.21 × 10-3 × 10 = 1.21 × 10-2 mol

National Junior College 2024 H2 Chemistry Revision Package 1 Mass of (NH4)2SO4 in 250 cm3 = 1.21 × 10-2 × 132.1 = 1.598g Percentage by mass of (NH4)2SO4 = 1.598 5 × 100% = 31.96% ≈ 32.0% Ans: C 3 A) If water evaporated from the solution, the concentration of the barium hydroxide will increase and a larger volume of HCl will be used to titrate instead. B) A possible answer, but we will look for more scientific answers. C) Having less water of crystalisation than stated does not affect the amount of Ba(OH)2 D) Barium hydroxide is basic and it reacts with carbon dioxide to form barium carbonate which will be seen as a solids in the water. Ans: D 4 At rtp, 1 mole of oxygen gas contains 24 000 cm3 Amount of O2 present in 24 cm3 = 24 24000 = 1 × 10-3 mol Each O2 contains 2 atoms of O, and 1 mole contains 6.02 x 1023 atoms Number of oxygen atoms present in 24 cm3 of oxygen gas = 6.02 x 1023 × 2 × 1 × 10-3 = 1.204 x 1021 Ans: D 5 Given the density of mixture is 1.82 g dm −3 , and that Argon’s density is 1.78 g dm−3, the gas mixed should have a higher molar mass of Argon at 39.9 g mol−1. Hence, the answer is CO2 which has a molar mass of 44 g mol−1. Ans: A

National Junior College 2024 H2 Chemistry Revision Package 2 6 [Hint: C in CO2 and H in H2O come from the iron organic compound only] Mass of C in the iron organic compound = 12 44 × 2.23 = 0.6082 g (4.s.f) Mass of H in the organic compound = 2 18 × 0.457 = 0.05078 g (4.s.f) Mass of Fe in the organic compound = 0.944 − 0.6082− 0.05078= 0.2850 g C H Fe Mass 0.6082 0.05078 0.2850 Ar 12 1.0 55.8 mole 0.05068 0.05078 0.005108 ÷ by smallest no. 10 10 1 Hence empirical formula of X = FeC10H10 Ans: D 7 PV=nRT 100000 × 20 × 10−6 = 0.0337 𝑀𝑟 × 8.31 × 300 Mr = 42.0 CxHy(g) + (x + 𝑦 4) O2(g)  x CO2(g) + 𝑦 2