Question

Wo liquid a and b form an ideal solution at temperature t. When the total vapour pressure above the solution is 400 torr the mole fraction of a in the vapour phase is 0.40 and liquid phase 0.75, the vapour pressure of pure a at temperature t will be:

Answer 1

Solution :

Given in vapour phase mole fraction of liquid A=50A=50

Therefore , mole fraction of liquid B=1−0.5=0.5B=1−0.5=0.5

now

0.5=PAPA+PB0.5=PAPA+PB-------------(1)

and

0.5=PBPA+PB0.5=PBPA+PB-------------(2)

Dividing (1) and (2)

1=PAPB1=PAPB

or PA=PBPA=PB

Now,

PA=P∘A×XAPA=PA∘×XA

and

PB=P∘B×XBPB=PB∘×XB

Since ,PA=PBPA=PB

P∘A×XA=P∘B×XBPA∘×XA=PB∘×XB

200×XA=75×XB200×XA=75×XB

XA=0.375XBXA=0.375XB

We know

XA+XB=1XA+XB=1

0.375XB+XB=10.375XB+XB=1

XB=0.73XB=0.73

=0.27=0.27

hence mole percent of A in the liquid =27%=27%

 HOPE IT HELPS...