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.

Given,


PAo = 450 mm Hg


PBo = 700 mm Hg


ptotal = 600 mm of Hg


By using Rault's law,


ptotal = PA + PB


ptotal = PAoxA + PBoxB


ptotal = PAoxA + PBo( 1 - xA )


ptotal = (PAo- PBo)xA + PBo


600 = (450 - 700) xA + 700


-100 = -250 xA


xA = 0.4


∴ xB = 1 - xA


xB = 1 – 0.4


xB = 0.6


Now,


PA = PAoxA


PA = 450 × 0.4


PA = 180 mm of Hg


and


PB = PBox


PB = 700 × 0.6


PB = 420 mm of Hg


Composition in vapour phase is calculated by


Mole fraction of liquid,


=


= 0.30


Mole fraction of liquid,


=


= 0.70


Note: Alternate method to find the Mole fraction of liquid B is


= 1 - Mole fraction of liquid A


= 1 – 0.30


= 0.70


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