EOC_c01.pdf

Answers to end-of-chapter questionsChapter 1 1 a i The weighted average mass of the atom of an element on a scale where one atom of carbon-12 has a mass of exactly 12 units. [1] (18.7 × 10) + (81.3 × 11) ii = 10.8 [2] 100 c i [1] 100 [2] [1 mark for showing masses × % abundance or one error carried forwards from this] ii The average mass of an atom of a particular isotope on a scale in which an atom of carbon-12 has a mass of exactly 12 units. [1] Total = 5 3 a Na2CO3(s) + 2HCl(aq)  → 2NaCl(aq) + CO2(g) + H2O(l)   b molar mass of sodium carbonate calculated correctly = 106 4.15 moles of sodium carbonate = 0.039 mol 106 moles HCl = 2 × 0.039 = 0.078 mol c The amount of substance that has the same number of specified particles / atoms / molecules, etc. as there are atoms in exactly 12 g of the carbon-12 isotope (or similar wording). 25.0 d i moles sodium carbonate =  × 0.0200 1000 = 5.0 × 10−4 mol ii moles HCl = 2 × 5.0 × 10−4 = 1.0 × 10−3 mol 1000 concentration of HCl = 1.0 × 10−3 ×  12.50 −3 = 0.080 mol dm  AS and A Level Chemistry © Cambridge University Press [1] [1] [1] ( 51.5 × 90 ) + (11.2 × 91) + 17.1 × 92 ) + (17.4 × 94 ) + ( 2.8 × 96 ) = 91.3 20 80 ;H=  1.0 12.0 C = 6.67; H = 20 divide by lowest 4 a C = [1] [1] [1] [1] Total = 6 2 a 262.5 180 b 72 Hf  [1] [1] Total = 10 6.67 20 = 1; H = = 3 6.67 6.67 empirical formula is CH3 b empirical formula mass = 15 15 × n = 30; n = 2, so molecular formula is C2H6 c any 4 of: volume of gas proportional to number of moles; mole ratio is 50 : 300 : 200 1 mol hydrocarbon : 6 mol oxygen : 4 mol carbon dioxide Since 4 moles of carbon dioxide from 1 mole of hydrocarbon, hydrocarbon has 4 carbon atoms 4 carbon atoms will react with 4 moles of oxygen molecules, leaving 2 moles of oxygen molecules (4 moles of oxygen atoms) to react with the hydrogen so 4 moles of water formed, meaning 8 hydrogen atoms in hydrocarbon equation is C4H8 + 6O2 → 4CO2 + 4H2O [1 mark for showing masses × % abundance or one error carried forwards from this] b 2 c i 184.2 ii Fe has several isotopes. e 0.2 mol f 0.2 × 24 = 4.8 dm3 [1] [1] C = 600 = 0.025 mol 24000 0.025 × 44.0 = 1.1 g  [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] d moles propane = [1] [1] Total = 10 5 a 4Na + TiCl4 → 4NaCl + Ti [2] [1 mark for correct formulae; 1 mark for balancing] [1] [1] [1] b 1 mole of TiCl4 gives 1 mole of Ti 189.9 g TiCl4 → 47.9 g Ti 47.9 1.0 g TiCl4 →  g Ti 189.9 47.9 380 g TiCl4 → 380 ×   g Ti = 95.9 g Ti 189.9 [1] [1] [1] Answers to end-of-chapter questions: Chapter 1 1 c 4 moles of Na gives 1 mole of Ti 4 × 23.0 g Na → 47.9 g Ti [1] 47.9 1.0 g Na →  g Ti 4 × 23.0 47.9 46.0 g Na → 46 ×   g Ti = 24.0 g Ti [1] 4 × 23.0 Total = 6 4.80 = 0.200 mol 24.0 b moles of NaOCl = moles of Cl2 = 0.200 mol mass of NaOCl = 74.5 × 0.200 = 14.9 g c moles of NaOH = 2 × moles of chlorine = 0.400 mol 6 a i 0.0150 dm3 ii 0.0200 dm3 b 0.0200 × 0.0500 = 0.001 00 mol c 0.001 00 mol 0.00100 = 0.0667 mol dm−1 0.0150 [1] [1] [1] [1] [1] Total = 5 7 a 80.0 (g mol−1) [1] 0.800 b  [1] 80.0 = 0.0100 mol [1] c moles nitrogen(I) oxide = 0.0100 [1] volume = 0.0100 × 24.0 = 0.024 dm3 / 240 cm3 [1] Total = 5 1.20 8 a i moles of HCl = = 0.0500 mol 24.0 moles 0.0500 ii concentration = =  3 volume in dm 0.100 = 0.500 mol dm−3 25.0 b i 0.500 ×   1000 = 0.0125 mol ii moles NaOH = moles of HCl = 0.0125 mol volume = = [1] [1] volume of NaOH = [1] [1] [1] 0.400 = 0.200 dm3 2.00 [1] [1] [1] [1] [1] d Cl2(g) + 2OH−(aq)  → Cl−(aq) + OCl−(aq) + H2O(l)   [1] Total = 6 10 a 1 mole of CaO gives 1 mole of CaCl2 56.1 g CaO → 111.1 g CaCl2 [1] 28.05 28.05 g CaO → 111.1 ×   g CaCl2 56.1 = 55.5 g CaCl2 b 1 mole of CaO reacts with 2 mole of HCl 56.1 g CaO reacts with 73.0 g HCl 28.05 g CaO reacts with [1] 28.05 73.0 ×  = 36.5 g HCl 56.1 [1] c mass of water is 28.05  × 18.0 = 9.0 g 56.1 [1] [1] Total = 5 11 a NH3(g) + HCl(g) → NH4Cl(s) [2] b NH3 = 17.0 g mol−1 HCl = 36.5 g mol−1 NH4Cl = 53.5 g mol−1 [1] [1] [1]  [1] moles concentration 0.0125 = 0.0625 dm3 0.200 9 a moles of Cl2 = [1 mark for reactants and products; 1 mark for state symbols] 10.7 = 0.2 mol 53.5 moles of NH3 and of HCl = 0.2 mol 0.2 × 24.0 = 4.8 dm3 of NH3 and HCl c 10.7 g NH4Cl = [1] [1] [1] Total = 8 [1] Total = 7 2 Answers to end-of-chapter questions: Chapter 1 AS and A Level Chemistry © Cambridge University Press