Vapor Pressure
329
SOLUTION:
To use Raoult's law, we must compute the mole fraction of A in the solution (see p
191).
Moles of A =
3 °'°
g
= 0.250 mole
120
8
mole
Moles of C 6 H 4 C1 2 = '
0 '°
g
= 0.0680 mole
147—
8 —
mole
0.250 mole
Mole fraction of A = ......
——
— = 0.786
(0.250 + 0.0680) mole
P = (0.786)(70.0 torr) = 55.0 torr
PROBLEM:
When 6.00 g of substance Z are dissolved in 20.0 g of C 2 H 4 Br 2 , the solution
has a vapor pressure of 9.00 torr at 22.0°C. Pure C 2 H 4 Br 2 has a vapor pressure
of 12.70 torr at 22.0°C. What is the molecular weight of Z?
SOLUTION:
Let n = the number of moles of Z dissolved in
'
8
=0.106 moles of C 2 H 4 Br 2
187.8
g
mole
Substituting the known quantities into Raoult's law,
P = \Po
we obtain
9.00 torr = x (12.70 torr)
X ^
= 0.709
12.70 torr
By definition,
moles of C 2 H 4 Br 2
X = (moles of Z) + (moles of C 2 H 4 Br 2 )
0 . 709 =
°106
n + 0.106
(0.106X1 - 0.709)
—
= 0.0435 mole Z