Solution Vapour Pressure and Vapour Composition?

A) At 25°C the vapour pressure of pure pentane is 511 torr and that of hexane is 150. torr.
What is the mole fraction of pentane in a pentane-hexane solution that has a vapour pressure of 387 torr at 25°C?

B) What is the mole fraction of hexane in the vapour that is in equilibrium with this solution?
(Assume ideal gas behaviour.)

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(A) Using Raoult's Law,
The vapour pressure of pentane = mole fraction of pentane(liquid) x pure vapour of pentane
P (p) = X (p) x 511 .................(1)

The vapour pressure of hexane,

P (h) = X (h) x 150 .................. (2)

(1) + (2):

P (p) + P (h) = 511 X(p) + 150 X(h) ..................(3)

Given that the total pressure P(t) = P (p) + P (h) = 387 and
mole fraction, X(h) = 1 - X(p)

Substitute into equation (3),

387 = 511 X(p) + 150 ( 1 - X(p) )
387 = 511 X(p) + 150 - 150 X(p)

X(p) = 0.657

(B)
From equation (1), the vapour pressure of pentane:
P(p) = 0.657 x 511 = 335.7 torr

Mole fraction of hexane in liquid = 1 - 0.657 = 0.343
From equation (2), the vapour pressure of hexane :
P(h) = 0.343 x 150 = 51.5 torr

Using Dalton's Law of Partial Pressure,
Mole fraction of hexane in the vapour = partial pressure of hexane / total pressure = 51.5/387 = 0.133