Question

The vapour pressure of pure liquids a and b are 450 and 700 mm of hg respectively at 350 k. Calculate mole fraction of a in vapour phase if total vapour pressure of solution is 600 mm of hg.

Answer 1

It is given that:
PAo = 450 mm of Hg
PBo = 700 mm of Hg
ptotal = 600 mm of Hg
From Raoult's law, we have:
ptotal = PA + PB

 
Therefore, xB = 1 - xA
= 1 - 0.4
= 0.6
 
Now,  PA = PAo xA
= 450 × 0.4
= 180 mm of Hg
and PB = PBo xB
= 700 × 0.6
= 420 mm of Hg
Now, in the vapour phase: Mole fraction of liquid A =  PA/ (PA + PB )
=180 / (180+420)
= 180/600
= 0.30
And, mole fraction of liquid B = 1 - 0.30
= 0.70