Question
Fri July 29, 2011 By: Prashant Jain
 

Vapour pressure of pure liquid A is 60 tore at 300K.It forms an ideal solution with another liquid B.The mole fraction of B is 0.25 and the total pressure of the solution is 82.5 tore at 300K.What is the vapour pressure of pure liquid B at 300K?

Expert Reply
Mon August 01, 2011

According to the Raoult’s law, the total pressure is

PT = P0A XA + P0B XB

where P0A is the vapour pressure of pure liquid A

and, P0is the vapour pressure of pure liquid B.

XA and XB  mole fraction of  liquid A and  liquid B.

Therefore,

82.5 = 60 x 0.75 + P0B x 0.25

82.5 = 45 + P0B x 0.25

 P0B   = 37.5 /0.25 =150 torr

Therefore the vapour pressure of pure liquid B = 150 torr.

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