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Question 8

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.

Answer

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

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