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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 133
�ese two expressions for ptot are set equal and the resulting equation rear-
ranged to �nd xA
p∗AxA
yA
= p
∗
B(1 − xA)
1 − yA
hence xA =
p∗B yA
p∗A(1 − yA) + p∗B yA
With the data given
xA =
p∗B yA
p∗A(1 − yA) + p∗B yA
= (82.1 kPa) × (0.612)
(68.8 kPa)∗(1 − 0.612) + (82.1 kPa) × (0.612)
= 0.653... and xB = 1 − 0.653... = 0.346...
�e composition of the liquid is therefore xA = 0.653 and xB = 0.347 .
�e total pressure is computed from pA = ptot yA and pA = p∗AxA to give ptot =
xAp∗A/yA
ptot =
xAp∗A
yA
= (0.653...) × (68.8 kPa)
0.612
= 73.4 kPa
E5B.12(b) If the solution is ideal, the vapour pressure of component J in the solution
obeys Raoult’s law, [5A.22–151], pJ = p∗J xJ, where xJ is the mole fraction in
the solution. In the gas the partial pressure is pJ = yJptot, where yJ is the mole
fraction in the vapour.
Assuming ideality, the total pressure is computed as
ptot = pA + pB = p∗AxA + p∗B(1 − xA)
= (110.1 kPa) × (0.4217) + (76.5 kPa) × (1 − 0.4217) = 90.7 kPa
�e normal boiling point is when the total pressure is 1 atm, but the pressure
predicted by Raoult’s law is signi�cantly di�erent from this. Raoult’s law there-
fore does not apply and the solution is not ideal .
It is not possible to �nd yA, the composition of the vapour, from pA = ptot yA
because pA cannot be computed from the data given.
Solutions to problems
P5B.2 �e freezing point depression ∆Tf is related to the molality of the solute B,
bB, by [5B.12–161], ∆Tf = KfbB, where Kf is the freezing-point constant. �e
molality corresponding to the given freezing-point depression is therefore
bB,app =
∆Tf
Kf
= 0.0703 K
1.86 K kg mol−1
= 0.0377... mol kg−1
�e stated molality of the solution is 0.0096 mol kg−1, so it is evident that there
are more species in the solution than the number of molecules of�(NO3)4
added. �e ratio of bB,app to bB gives an indication of the number of species
present per molecule dissolved
bB,app
bB
= 0.0377... mol kg
−1
0.0096 mol kg−1
= 3.93...

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