Cambridge International AS Level Chemistry
Answers to end-of-chapter questions
Answers to EOCQs
Chapter 1
f 0.2 × 24 = 4.8 dm3
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]
+ (81.3 × 11) = 10.8
ii (18.7 × 10)100
[2]
[1 mark for showing masses × % abundance
or 1 error carried forwards from this]
b 2[1]
c i 184.2[1]
ii Fe has several isotopes.
[1]
Total = 6
2
4
a 262.5[1]
72 Hf[1]
b 180
c i (51.5 × 90) + (11.2 × 91) + (17.1 × 92) + (17.4 × 94) + (2.8 × 96)
100
[2]
[1 mark for showing masses × % abundance
or 1 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(aq) + 2HCl(aq)
→ 2NaCl(aq) + CO2(g) + H2O(l)
[1]
b molar mass of sodium carbonate calculated
correctly = 106
[1]
41.5
moles sodium carbonate = 106 = 0.039 mol[1]
moles HCl = 2 × 0.039 = 0.078 mol
[1]
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).[1]
25.0 × 0.0200
d i moles sodium carbonate = 1000
[1]
= 5.0 × 10–4 mol
–4
–3
ii moles HCl = 2 × 5.0 × 10 = 1.0 × 10 mol [1]
1000
concentration of HCl = 1.0 × 10–3 × 12.50
–3
= 0.080 mol dm [1]
e 0.2 mol
[1]
20
a C = 80
12 ; H = 1.0[1]
C = 6.67; H = 20
divide by lowest
20
C = 6.67
[1]
6.67 = 1; H = 6.67 = 3
empirical formula is CH3[1]
b empirical formula mass = 15
15 × n = 30; n = 2, so molecular formula is C2H6
[1]
c Any three explanatory statements from:
volume of gas proportional to number of
moles;[1]
mole ratio is 50 : 300 : 200
so 1 mol hydrocarbon : 6 mol oxygen : 4 mol
carbon dioxide.
[1]
As 4 moles of carbon dioxide from 1 mole
of hydrocarbon, hydrocarbon has 4 carbon
atoms.[1]
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;
[1]
so 4 moles of water formed, meaning 8
hydrogen atoms in hydrocarbon.
[1]
And final deduced equation:
C4H8 + 6O2 → 4CO2 + 4H2O[1]
d moles propane = 24600
000 = 0.025 mol
mass = 0.025 × 44.0 = 1.1 g
5
[1]
Total = 10
[1]
[1]
Total = 10
a 4Na + TiCl4 → 4NaCl + Ti
[1 mark for correct formulae; 1 mark for
balancing]
b 1 mole of TiCl4 gives 1 mole of Ti
189.9 g TiCl4 → 47.9 g Ti
47.9 g Ti
1.0 g TiCl4 → 189.9
47.9 g Ti = 95.9 g Ti
380 g TiCl4 → 380 × 189.9
c 4 moles of Na gives 1 mole of Ti
4 × 23.0 g Na → 47.9 g Ti
1.0 g Na → 4 47.9
× 23.0 g Ti
46.0 g Na → 46 × 4 47.9
× 23.0 g Ti = 24.0 g Ti
[2]
[1]
[1]
[1]
[1]
Total = 6
Cambridge International AS and A Level Chemistry © Cambridge University Press 2014
Cambridge International AS Level Chemistry
Answers to end-of-chapter questions
6 a i 0.0150 dm3[1]
ii 0.0200 dm3[1]
b 0.0200 × 0.0500 = 0.00100 mol
[1] 11
c 0.00100 mol
[1]
0.00100 = 0.0667 mol dm–1[1]
0.0150
Total = 5
7
8
a 80.0 (g mol–1)[1]
b 0.800
80.0 [1]
= 0.0100 mol
[1]
c moles nitrogen(IV) oxide = 0.0100
[1]
3
volume = 0.0100 × 24.0 = 0.024 dm = 240 cm3
[1]
Total = 5
a i moles of HCl = 1.20
24.0 = 0.0500 mol
ii concentration =
c mass of water is 28.05
56.1 × 18.0 = 9.0 g
[1]
Total = 5
a NH3(g) + HCl(g) → NH4Cl(s)[2]
[1 mark for reactants and products; 1 mark
for state symbols]
b NH3 = 17.0 g mol–1[1]
HCl = 36.5 g mol–1[1]
[1]
NH4Cl = 53.5 g mol–1
c 10.7
53.5 g NH4Cl = 10.7 = 0.2 mol
moles of NH3 and of HCl = 0.2 mol
0.2 × 24.0 = 4.8 dm3 of NH3 and HCl
[1]
[1]
[1]
Total = 8
[1]
moles
0.0500 [1]
3 =
0.100
volume in dm
= 0.500 mol dm–3
[1]
25.0
b i 0.500 × 1000 [1]
= 0.0125 mol
[1]
ii moles NaOH = moles of HCl
= 0.0125 mol
[1]
moles
volume =
concentration
3
= 0.0125
0.200 = 0.0625 dm [1]
Total = 7
9
a moles of Cl2 = 4.80
[1]
24.0 = 0.200 mol
b moles of NaOCl = moles of Cl2 = 0.200 mol [1]
mass of NaOCl = 74.5 × 0.200 = 14.9 g
[1]
c moles of NaOH = 2 × moles of chlorine
= 0.400 mol
[1]
3
0.400
volume of NaOH = 2.00 = 0.200 dm [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 g CaO → 111.1 × 28.05
56.1 g CaCl2
= 55.5 g CaCl2[1]
b 1 mole of CaO reacts with 2 moles of HCl
56.1 g CaO reacts with 73.0 g HCl
[1]
28.05 g CaO reacts with 73.0 × 28.05
56.1 = 36.5 g HCl
[1]
Cambridge International AS and A Level Chemistry © Cambridge University Press 2014