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128 5 SIMPLEMIXTURES
�e amount in moles of B is written is nB = bBmA; using this the above expres-
sion for the mole fraction becomes
xB =
nB
mA/MA + nB
= bBmA
mA/MA + bBmA
= bB
1/MA + bB
�e molar mass of A is 74.1 gmol−1, therefore the mole fraction corresponding
to bB = 0.25 mol kg−1 is
xA =
bB
1/MA + bB
= (0.25 mol kg−1)
1/(74.1 × 10−3 kgmol−1) + (0.25 mol kg−1)
= 0.0181...
�e pressure is found by inserting this value into eqn 5.4
pA/(kPa) = 8.41 × 103 × (0.0181...) − 2.784 = 1.50... × 102
�e vapour pressure of B is therefore 1.5 × 102 kPa .
E5B.2(b) Raoult’s law, [5A.22–151], pA = xAp∗A relates the vapour pressure to the mole
fraction of A, therefore from the given data is it possible to compute xA. �e
task is to relate the mole fraction of A to the masses of A (the solvent) and B
(the solute), and to do this the molar massesMA andMB are introduced. With
these nA = mA/MA, wheremA is the mass of A, and similarly for nB. It follows
that
xA =
nA
nA + nB
= mA/MA
mA/MA +mB/MB
= MBmA
MBmA +MAmB
�e �nal form of this expression for xA is rearranged to given an expression for
MB, which is the desired quantity; then xA is replaced by pA/p∗A
MB =
xAMAmB
mA(1 − xA)
= (pA/p∗A)MAmB
mA[1 − (pA/p∗A)]
�e molar mass of the solvent 2-propanol C3H8O, A, is 60.0932 gmol−1, hence
MB =
(pA/p∗A)MAmB
mA[1 − (pA/p∗A)]
= [(49.62 kPa)/(50.00 kPa)] × (60.0932 gmol−1) × (8.69 g)
(250 g) × [1 − (49.62 kPa)/(50.00 kPa)]
= 273 gmol−1
E5B.3(b) �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. From the
data and the known value of Kf it is possible to calculate bB.�e task is then to
relate this to the given masses and the desired molar mass of the solute,MB.
�emolality of B is de�ned as bB = nB/mA, wheremA is themass of the solvent
A in kg. It follows that
bB =
nB
mA
= mB/MB
mA