The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K.

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Solution

Given that

Vapour pressure of pure liquid A, P^o_A= 450 mm of Hg

Vapour pressure of pure liquid A, P^o_A= 700 mm of Hg

Total vapour pressure, p_total = 600 mm of Hg

Use the formula of Raoult’s law

 

600 = (450 – 700) X_A  + 700

250 X_A = 100

X_A = 100/250 = 0.4

Use formula

X_B = 1 - X_A

Plug the values we get

X_B= 1 − 0.4 = 0.6

use formula

P_A = P^o_A × X_A = 450 × 0.4  = 180 mm of Hg

P_B = P^o_B × X_B = 700 × 0.6 = 420 mm of Hg

Now, in the vapour phase:

Mole fraction of liquid A

= 180 /(180+ 420))= 0.30

Mole fraction of liquid B, Y_B  = 1 −  Y_A =1 – 0 .30 = 0.70