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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.

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.

Grade:11

1 Answers

Gaurav
askIITians Faculty 164 Points
6 years ago
Given that
Vapour pressure of pure liquid A, PoA= 450 mm of Hg
Vapour pressure of pure liquid A, PoA= 700 mm of Hg
Total vapour pressure, ptotal= 600 mm of Hg
Use the formula of Raoult’s law

600 = (450 – 700) XA+ 700
250 XA= 100
XA= 100/250 = 0.4
Use formula
XB= 1 - XA
Plug the values we get
XB= 1 − 0.4 = 0.6
use formula
PA= PoA× XA= 450 × 0.4= 180 mm of Hg
PB= PoB× XB= 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, =1 – 0 .30 = 0.70

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