ANSWERS TO EXERCISES AND REVISION QUESTIONSpressure. Therefore, 0.122 mol of O 2 occupies 0.122
24 2.9 dm^3. Summarizing, 10.0 g of potassium
chlorate decomposes to make about 2.9 dm^3 of oxygen
gas.
10.4PT 25 50 75 kPa 75 000 Pa.
Rearrangement of the equation PV = nRTgivesnRTPVTo work out the number of moles of chlorine gas and
oxygen gas, substitute into this equation using the
partial pressure of each gas for P:
number of moles of Cl 2
(25 000 0.0100)/8.3145 293 0.103
number of moles of O 2
(50 000 0.0100)/8.3145 293 0.205
Total number of moles of gas 0.1030.205
0.308, so
mole fraction of O 2 0.2050.3080.666
(Quicker way! The mole fraction equals the ratio
pO 2 /PT50.0 kPa/75.0 kPa 0.666.)
10.5Number of moles of N 2 2.0/280.071
number of moles of H 2 6.0/2.03.0
total number of moles of gas 3.071PTnRTV3.0710.00208.3145^473
PT6.0 106 Pa60 atm
10.6Start by calculating the number of moles of He in
the balloon:nPV1.00^10(^5) 1.00
RT 8.3145 293 41.05
Then apply the equation
TPVnR
At the new altitude
T5.00^10
(^4) 1.66
41.058.3145 243 K
We conclude that the temperature at the new altitude
was 243 K.
10.7First, calculate the molar volume of an ideal gas at
273 K and 1 atm pressure:
VnRTP
V1.000101 3258.3145^273 0.0224 m^3 or 22.4 dm^3
The closer the experimentally determined molar volume
of a real gas is to 22.4 dm^3 , the more ideally behaved is
the gas. The experimental molar volumes suggest that
H 2 behaves most ideally, followed by CO 2 and (least
ideally) NH 3. This reflects the greater degree of
intermolecular force between NH 3 molecules.
10.8At 40 °C, the vapour pressure of water 7370 Pa.
The vapour occupies half the flask so that
V1.0 dm^3 0.0010 m^3.
nPVRT^7370 8.31450.0010 313 2.8 10 –^3 mol
To convert the number of moles of water to molecules of
water, we multiply by NA:
number of molecules
2.8 10 –^3 6.022 1023
1.7 1021
10.9If a solvent has a high vapour pressure at room
temperature, it is likely it has a sufficiently high vapour
pressure at lower temperatures to be ignited by a naked
flame.
Unit 11
Exercises
11A
Oil floats on water – this was concealed by the opaque
earthenware jug. The captain poured the oil layer into his
opponent’s tankard.
11B
Cyclohexane is less dense than water (see Table 11.1)
and floats to the top. Table 11.1 shows the density of
cyclohexane as 0.77 g cm–^3 , so that 130 cm^3 of
cyclohexane has a mass of 100 g. Table 11.3 shows
that 100 g of cyclohexane saturated with water contains
0.01 g of water. Therefore, 130 cm^3 of cyclohexane
contains 0.01 g of water.
11C
All the molecules are very similar and differ only in their
chain length. Butan-1-ol, possessing the shortest chain,
is predicted to be the most soluble in water.
11D
1 dm^3 of saturated solution contains 1.6 10 –^2 mol, so
that 100 cm^3 of solution contains 1.6 10 –^3 mol.
M(PbCl 2 )278 g mol–^1. The mass of 1.6 10 –^3 mol is
(1.6 10 –^3 278)0.45 g. The equation for the
dissolving of solid lead chloride is
PbCl 2 (s) Pb^2 +(aq)2Cl–(aq)
Any lead chloride that dissolves subsequently
dissociates into ions. The concentration of dissolved
lead chloride is 1.6 10 –^2 mol dm–^3 , so that the
concentration of lead ions is also
1.6 10 –^2 mol dm–^3. The concentration of chloride ions
is double this, i.e. 3.2 10 –^2 mol dm–^3.
11E
For copper(I) iodide,
CuI(s) Cu+(aq)I–(aq)
Ks[Cu+(aq)][I–(aq)] mol^2 dm–^6
For aluminium hydroxide,
Al(OH) 3 (s) Al^3 +(aq)3OH–(aq)
Ks[Al^3 +(aq)][OH–(aq)]^3 mol^4 dm–^12
11F
BaSO 4 (s) Ba^2 +(aq)SO 42 – (aq)
Ks[Ba^2 +(aq)] [SO 42 – (aq)]s ss^2
or,
s Ks(1.1 10 –^10 )1.0 10 –^5 mol dm–^3
MgF 2 (s) Mg^2 +(aq)2F–(aq)
The concentration of fluoride ions is twice the molar
solubility of MgF 2 , i.e.
[F–(aq)] 2 s
and
[F–(aq)]^2 [2s]^2 4 s^2
Apart from this complication we follow the same pattern
as for BaSO 4 :
Ks[Mg^2 +(aq)] [F–(aq)]^2 s 4 s^2 4 s^3
or,
s^3 √(Ks)^3 √(6.4^10
- 9
44 )
s 3
√(1.6^10
– (^9) )1.2 10 – (^3) mol dm– 3
11G
(i)The product of ion concentrations is
[Mg^2 +(aq)][F–(aq)]^2 [0.0010][0.0030]^2 9.0 10 –^9
Table 11.4 shows that Ks6.4 10 –^9 mol^3 dm–^9.
Therefore, the solubility product is exceeded and
magnesium fluoride precipitates.
Addition of 90 cm^3 of water to 10 cm^3 of mixture
dilutes the ion concentrations by a factor of ten. Then,
the product of ion concentrations is
[Mg^2 +(aq)][F–(aq)]^2 [0.00010][0.00030]^2
9.0 10 –^12
This is well below the solubility product and so the
magnesium fluoride will not precipitate out.
(ii)[Ba^2 +(aq)] from BaCl 2 is 2.0 10 –^4 mol dm–^3 before
mixing and 1.0 10 –^4 mol dm–^3 after mixing.
[SO 42 – (aq)] from Na 2 SO 4 is 4.0 10 –^4 mol dm–^3 before
mixing and 2.0 10 –^4 mol dm–^3 after dilution.
The product of ion concentrations is
[Ba^2 +(aq)] [SO 42 – (aq)][1.0 10 –^4 ][2.0 10 –^4 ]
2.0 10 –^8
This is greater than Ks(1.1 10 –^10 mol^2 dm–^6 ), and so
the barium sulfate will precipitate out.
11H
The concentration of SO 2 in the water layer is
0.900.100.090 mol dm–^3.
11I
cRT, so that
c/RT91/8.3145 300
c0.036 mol m–^3 3.6 10 –^5 mol dm–^3
Therefore, 1 dm^3 of solution contains 3.6 10 –^5 moles.
Using
moles mass
molar mass
the molar mass is
mass 0.25
moles3.6 10 –^5 6900 g mol
- 1
The average molecular mass is therefore 6900 u.
Revision questions
11.1The true answers are (i)and(iii).
11.2 (i)Miscible, presumably because strong hydrogen
bonding occurs between the water and propan-2-ol
molecules.
(ii)Miscible, presumably because the intermolecular
forces between propanone and methyl benzene (a big
molecule) are at least as strong as those between pairs
of molecules of methyl benzene and between pairs of
propanone molecules.
11.3C is the least polar solvent. Ionic solids are at best
only slightly soluble in non-polar solvents.
11.4 (i)
(a)Ks[Ca^2 +(aq)][CO 32 – (aq)]s^2 wheresis the molar
solubility.
s Ks9.3 10 –^5 mol dm–^3
(b) The solubility in g dm–^3 is obtained by multiplying the
molar solubility by the molar mass of calcium carbonate
(100 g mol–^1 ), giving the solubility as 9.3 10 –^3 g dm–^3
at that temperature.
(ii)The mass of CaCl 2 in 20.0 cm^3 of solution
2.0 10 –^5 g. The molar mass of CaCl 2 is
111 g mol–^1 , so the number of moles of CaCl 2 is
2.0 10 –^5 /1111.8 10 –^7 mol
[CaCl 2 (aq)][Ca^2 +(aq)]
1.8 10 –^7 1000/209.0 10 –^6 mol dm–^3.
The sodium carbonate solution has been diluted to
1.0 10 –^5 mol dm–^3.
[Na 2 CO 3 (aq)][CO 32 – (aq)]1.0 10 –^5 mol dm–^3.
The ionic product is
[Ca^2 +(aq)] [CO 32 – (aq)]
(9.00 10 –^6 ) ( 1.0 10 –^5 )9.0 10 –^11
This is less than Ks, so that calcium carbonate will not
precipitate out.4390230_000118_2 9 _Ans.qxd 3/2/06 2:22 pm Page 439