ANSWERS TO EXERCISES AND REVISION QUESTIONSvolmolar concentration MnO 4 – (^3) V0.020
amount in moles of FeC 2 O 4 5 0.007
V= 210 cm^3
9G
(i)67 cm^3
volmolar concentration H 2 SO (^4) 1
amount in moles of CuO 1
V1.5
7.95/(63.516)
V0.067 dm^3
(ii)0.29 g (iii)417 cm^3 (iv)3.1 g.
9H
(i)0.1128 g (ii)to ensure they were completely dry
(iii)3.9595 10 –^3 mol dm–^3 (molar mass of Ni (DMG) 2
284.882 and amount of Ni(DMG) 2 in 100 cm^3
0.1128/284.882 mol. Multiply this amount of Ni(DMG) 2
by 10 to get the concentration)
(iv)0.4648 g (mass of Ni^2 +in original solution of volume
2000 cm^3 3.9595 10 –^3 58.692).
9I
(i)1.2 mg (ii)400 ppm.
9J
(i)0.014 mol dm–^3
500 mg in 1 dm–^3 is 500/1000 1/35.5 mol dm–^3
(ii)200 ppm
(the solution has been diluted 250/50 5 times)
(iii)0.5 ppb.
9K
(i)13.4% (ii)42 g (iii)180 cm^3
(iv)425 cm^3.
9L
(i)
XRn nR
PnQnRnS
(ii)
nP nQ
nPnQnRnS nPnQnRnS
nR nS
nPnQnRnSnPnQnRnS
nnPnQnRnS 1
PnQnRnS
(iii)pure P.
9M
(i)3/10 or 0.3
(ii)0.0089, from
37.25/74.5
37.25/74.51000/18
.
9N
(i)weakly acidic (ii)weakly acidic
(iii)strongly basic (iv)strongly basic
(v)weakly basic (vi)weakly acidic
(vii)strongly basic.
Revision questions
9.1 (i)4.20 g (ii)126 g (iii)4.7 g
9.2It absorbs water and carbon dioxide from the air and
therefore would increase in mass when handled in air.
9.3 (i)50 cm^3 (ii)100 cm^3.
9.4 (i)100 cm^3 (ii)23.1 g.
9.5 (i)0.35 g
KOHHNO 3 KNO 3 H 2 O
Amount in mol HNO 3 volmolar conc, i.e.
25/10000.2506.25 10 –^3 mol
The same amount in mol of KOH is needed, i.e.
6.25 10 –^3 (39 16 1) g
(ii)0.39 g, from
CuCO 3 2HNO 3 Cu(NO 3 ) 2 H 2 OCO 2
9.6 (i)0.100 mol dm–^3 (ii)approx. 6 g dm–^3.
9.7100 cm^3.
9.820 g.
9.9KBr: 2000 ppm of K+
0.050 mol dm–^3 of K+is 0.050 39
approx. 2 g dm–^3 or 2000 mg dm–^3
Na 2 SO 4 : 460 ppm of Na+
0.0102 mol dm–^3 Na+is 0.020 23 0.46 g dm–^3
Pb(NO 3 ) 2 : approx. 1200 ppm of Pb^2 +
9.100.020 g
9.110.12 g
20 mg dm–^3 of Na+is 20/1000 1/23 mol dm–^3 Na+.
This corresponds to
or 2the number of mol of Na 2 CO 3 ·10H 2 O
or 20/1000 1/231/2 mol dm–^3 Na 2 CO 3 ·10H 2 O
or 20/1000 1/231/2286 g dm–^3
Na 2 CO 3 ·10H 2 O
9.12 (i)pHlog[H+(aq)]log[4.48 10 –^9 ]8.348
(ii)pHlog[H+(aq)]; 5.93 log[H+(aq)]
[H+(aq)] 10 – 5.93
[H+(aq)]1.2 10 –^6 mol dm–^3
Unit 10
Exercises
10A
Changes in phase are reversible. The key to reversing
such changes is temperature.
10B
(i)Temperature may be thought of as the average kinetic
energy of a large group of molecules. If the same amount
of energy is absorbed by fewer water molecules, the
resulting temperature change is higher.
(ii)Hot days mean a higher temperature, so that more
molecules possess enough energy to break away from
the liquid. Windy days push the air (containing water
vapour) away so that fewer water molecules can
condense back on to the clothes.
(iii)Molecules of a gas are always moving and they
spread out to fill any container (diffusion).
10C
No. Cooking is a chemical reaction and the rate of
reaction depends upon the temperature. The
temperature of boiling water is the same (normally
100°C) whether the water boils slowly or rapidly.
10D
(i)(20/760) 101 325 2666 Pa 2700 Pa
(ii) 10 101 325 1 013 250 Pa (iii)107 000 Pa,
(iv)1000 Pa.
10E
The temperature has increased by a factor of
(2500/500)5.00. This means that the volume
increases by fivefold, from 73.0 cm^3 to 365 cm^3
(370 cm^3 to two sig figs).
10F
Pressure has increased by a factor of 90.0/30.0 3.00,
causing the volume to reduce by a factor of one-third:
1
3.00200.066.7 cm
3
10G
The number of moles of Zn (mass/molar mass)
0.50/650.0077 mol. The equation is:
Zn(s)H 2 SO 4 (aq)ZnSO 4 (aq)H 2 (g)
Therefore, 1 mol of Zn produces 1 mol of H 2 , so that
0.0077 mol of Zn produces 0.0077 mol of H 2. 1 mol of
H 2 occupies 24 dm^3 at room temperature and pressure,
so that 0.0077 mol of H 2 occupies 0.0077 24
0.185 dm^3 ( 190 cm^3 ).
10H
Volume of argon in 1.0 dm^3 (0.94/100) 1.00
0.0094 dm^3 (9.4 cm^3 ). Mass of argon in 1 kg of air is
(1.28/100) 1.00.0128 kg (≈13 g)
10I
T20.0273.15293.2 K
V0.0020 m^3
PnRT/V(1.0 8.3145 293.2)/0.0020
1 218 698 Pa 1.2 106 Pa (to two sig figs).
10J
The greater the intermolecular forces, the greater the
deviations from ideal behaviour. CO 2 is a larger molecule
than H 2 or N 2 , and so the London dispersion forces
between its molecules will be greater. (The vapour of
polar molecules, such as ethanol or water, show
considerable deviations from ideal behaviour because
even stronger intermolecular forces are present.)
10K
(i)A liquid boils when its vapour pressure equals the
external (atmospheric) pressure. Atmospheric pressure
decreases with increasing altitude, so that the vapour
pressure required in order for boiling to occur is lower
than is needed at sea level. The (reduced) vapour
pressure is therefore achieved at a lower liquid
temperature, i.e. the boiling point is lower.
(ii)Salty water contains ions from the salt at the surface
of the solution. The ions occupy sites that would be
taken by water molecules in pure water, and so the
number of water molecules in the surface that can
escape is reduced in comparison with pure water at the
same temperature. A higher temperature than 100 °C is
needed in order for the water molecules in the surface of
the solution to provide a vapour pressure of 1 atm
(boiling point).
10L
Total gas pressure air pressure vapour pressure of
ethanol, so
101 ?7.9
and air pressure 101 7.9 93 kPa.
10M
(i)Treat the gases separately. 119 °C is at the critical
temperature of oxygen so that liquid oxygen will
condense if subjected to its critical pressure. However,
119 °C is above the critical temperature of hydrogen;
hydrogen will not condense out, whatever the applied
pressure.
(ii)The problem with liquefying hydrogen gas is getting
its temperature at (or below) the critical temperature of
240 °C. This is well below the normal boiling point of
liquid oxygen or liquid nitrogen. Liquid helium (bp
269 °C) could be used as coolant.
Revision questions
10.1 (i)(0.124/760)1.63 10 –^4 atm
(ii)1.5 101 325 151 988 Pa
(iii) 89 1000 89 000 N m–^2
(iv)100.0 kPa is 1 bar so that 101.325 kPa is 1.013 25
bar
(v) 100 273.15173.15 K.
10.2Avogadro’s law states that equal volumes of gases
(at the same pressure and temperature) contain the
same number of molecules. Therefore, if (within
experimental error) equal volumes of H 2 and Cl 2 react
together, producing twice that volume of HCl, the ratio in
which the molecules react must be 1:1 and the ratio of
HCl product molecules to H 2 or Cl 2 molecules must be
2:1. This confirms the ratio contained in the equation.
10.3M(KClO 3 )122.5 g mol–^1 , so that the number of
moles of KClO 3 10.0/122.50.0816.
According to the chemical equation, 2 mol of KClO 3
gives 3 mol of O 2. Therefore, 0.0816 mol of KClO 3 gives
(3/2) 0.08160.122 mol of O 2.
1 mol of O 2 occupies 24 dm^3 at room temperature and
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