# 2.8   The vapour pressure of pure liquids A and B are    and   respectively, at . Find out the composition of the liquid mixture if total vapour pressure is . Also find the composition of the vapour phase.

Let the composition of liquid A (mole fraction) be xA.

So mole fraction of B will be xB = 1 - xA.

Given that,

Using Raoult’s law ,

Putting values of ptotal and vapour pressure of pure liquids in the above equation, we get :

600 =   450.xA  +    700.(1 -  xA)

or                           600 - 700  =  450xA - 700xA

or                                  xA   =   0.4

and                               xB    =   0.6

Now pressure in vapour phase :

=   450(0.4)  = 180 mm of Hg

=    700(0.6)   = 420 mm of Hg

And mole fraction of liquid B = 0.70

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