NJC phy chem rev 2025 ans

National Junior College 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 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 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 H2O(l) Reacting vol ratio / cm3 20 excess 40 20 Mole ratio 20 20 = 1 40 20 = 2 20 20 = 1 x=2 , 𝑦 2=1 🡪 y = 2 Hence, empirical formula = C2H2 However, as Mr is 42, there is short of 16, which constitutes an O. Hence, ans is C2H2O Ans: C 8 Mass of CO2 in CaCO3 = 44 100.1 × 0.05 = 0.02198 g Amount of CO2 = 0.02198 44 = 5 × 10−4 mol

National Junior College H2 Chemistry Revision Package 3 Volume of CO2 at r.t.p = 5 × 10−4 × 24 = 0.012 dm3 Percentage by volume of CO2 = 0.012 10 × 100 = 0.12% Ans: C 9 Cu2+ has been reduced to Cu+, which reacts with the unreacted I− to for the white precipitate CuI. Cu2+ + e 🡪 Cu+ 2I− 🡪 I2 + 2e Overall: 2Cu2+ + 2I− 🡪 2Cu+ + I2 Assuming only half of the I- reacts =[ 10 1000 × 0.2] ÷ 2 = 1 × 10−3 mol Amount of I2 that reacts with S2O32− = 1 × 10−3 ÷ 2 = 5 × 10−4 2S2O32– + I2  S4O62– + 2I – Amount of S2O32− required for reaction = 2 × 5 × 10−4 = 1 × 10−3 mol Volume of required S2O32−= 1 × 10−3 0.04 ×1000 = 25.0 cm3 Ans: D 10 No. of I2 particles = 600 253.8 × 6.02 × 1023 = 1.423 × 1024 No. of H2O particles = 50 18 × 6.02 × 1023 = 1.672 × 1025 No. of HCl particles = 50 22.7 × 6.02 × 1023 = 1.326 × 1024 No. of CH3CO2H particles = 550 1000 × 2.5 × 6.02 × 1023 =3.311 × 1023 Ans: B

National Junior College H2 Chemistry Revision Package 4 11 Option A: 91.1 100 × 28 + 7.9 100 × 29 + 1 100 × 30 = 28.099 Option B: 92.2 100 × 28 + 4.7 100 × 29 + 3.1 100 × 30 = 28.109 Option C: 95 100 × 28 + 3.2 100 × 29 + 1.8 100 × 30 = 28.039 Option D: 96.3 100 × 28 + 0.3 100 × 29 + 3.4 100 × 30 = 28.071 Ans: B 12 No. of molecules in 28.0 g sample of 14N2 = 6.02 × 1023 No. of atoms in 28.0 g sample of 14N2 = 6.02 × 1023 × 2 =1.20 × 1024 No. of atoms in 24.0 g of 12C = 6.02 × 1023 × 2 =1.20 × 1024 No. of atoms in 4.0 g of 4He = 6.02 × 1023 No. of atoms in 32.0 g of 16O2 = 6.02 × 1023 Ans: D 13 One mole of metal atoms = 6.02 × 1023 Option A: No. of atoms of mol of hydrogen gas = 6.02 × 1023 Option B: No. of atoms of mol of 12C = 1 12 × 6.02 × 1023 = 5.02 × 1022 Option C: False. Mass of 1 mol of hydrogen atoms 2 g Option D: False. There are different metals with different oxidation states Ans: A 14 Option A: Electronic Configuration of Fe2+: 1s22s22p63s23p63d8 Option B: angle of deflection α 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 False as is heavier than hence the angle of deflection is lesser for . Option C: Both have same no. of protons. Option D: Total no. of protons and electrons in Fe2+ = 26 + 24 = 50 Nucleon number = 56

National Junior College H2 Chemistry Revision Package 5 Ans: D 15 Volume of H2O = 24 cm3 Final volume of residual gases = 64 – 24 =40 cm3 Residual gas consist of CO2 + H2O Volume of CO2 = 60 100 × 40 = 24 cm3 By comparing volume of H2O to CO2, Empirical formula = CH2 Ans: B 16 Amount of NaOH that reacted with H2SO4 = 40 1000 × 0.875 × 2 = 0.07 mol Amount of NaOH that reacted with both CO2 and NO2 = 0.25 – 0.07 = 0.18 mol Amount of NO2 is 0.02 mol Amount of NaOH that react with CO2 = 0.18 – 0.02 = 0.16 mol Amount of CO2 = 0.16 ÷ 2 = 0.08 x= 0.08 ÷ 0.02 = 4 Phosphorus pentoxide reacts with the H2O The increase in mass of 1.15g = H2O Amount of H2O = 1.15 18 = 0.06389 mol y/2 = 0.06389 ÷ 0.02 ≈ 3.1945 🡪 y=6.389 4:6.4 ≈ 2:3 Ans:A 17 Amount of O2 = 26 24000 = 1.083 × 10−3 mol H2O2 🡪 O2 + 2H+ + 2e Amount of MnO4−= 1.083 × 10−3 mol Concentration of MnO4− = 1000 25 × 1.083 × 10−3 = 4.33 x 10–2 mol dm–3 Ans: B

National Junior College H2 Chemistry Revision Package 6 Section B 1(a) Amount of Fe2+ = 0.50 55.8+32.1+16.0 ×4+7×18 = 1.80 × 10−3 mol Fe2+ → Fe3+ + e MnO4− + 8H+ + 5e 🡪 Mn2+ + 4H2O MnO4− + 8H+ + 5 Fe2+ 🡪 Mn2+ + 5 Fe3+ + 4H2O Amount of MnO4− = 1.80 × 10−3 ÷ 5 = 3.6 × 10−4 mol Volume of KMnO4 required = 3.6 ×10−4 0.02 ×1000 = 18 cm3 1 (b) Mass of Cu2S present = 0.77 100 × 5000 × 1000 = 38500g Max mass of copper produced = 63.5 ×2 63.5 ×2+32.1 × 38500 = 30732 g ≈ 30 700g 2 (i) Flow rate of HCl= 3.65 35.5+1.0 = 0.1 mol min−1 Flow rate of ethene = 0.1 ÷ 2 = 0.05 mol min−1 (ii) Amount of HNO3 present = 2.5 1000 × 0.100 = 2.5 × 10−4 mol Amount of NaOH in 10 cm3 sample = 2.5 × 10−4 mol min−1 No. of moles of unreacted NaOH = 1000 10 × 2.5 × 10−4 =0.025 mol (iii) No. of moles of NaOH that react with the HCl = 0.100 – 0.025 = 0.075 mol No. of moles of HCl that emerged from reaction chamber = 0.075 mol

National Junior College H2 Chemistry Revision Package 7 (iv) Based on the amount of HCl that reacted. Conversion percentage = 0.025 0.100 × 100% = 25% (v) Any one of the following: - All the HCl emerged from chamber was absorbed by NaOH - C2H4 does not undergo combustion inside the chamber 3 (a) Mn2+ / observe colour change from colourless to pale pink. (b) [R] MnO4− + 8H+ + 5e→Mn2+ + 4H2O [O] C