1.

Vapour pressure of pure liquids P and Q are 700 and 450 mm Hg respectively at 330K. What is the composition of the liquid mixture at 330 K, if the total vapour pressure is 600 mm Hg?

Answer»

Solution :Vapour PRESSURE of the mixture = 600 mm HG
= 700 X mole fraction of P + 450 x mole fraction of Q = (700 x X) + 450(1-X) Mole fraction of component liquid, P = X = 0.6 Mole fraction of component liquid,
Q=1-X=0.4


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