Question

The vapour pressure of pure liquids A and B
are 200 and 350 mm Hg respectively, at
300 K. Mole fraction of liquid A in a mixture
having its vapour pressure 305 mm of Hg is
(1) 0.5
(2) 0.4
(3) 0.3
(4) 0.2​

Answer 1

Pa°=200mm of hg

Pb°=350mm of hg

P=305mm of Hg

X a+X b=1

X b= 1 - X a

we know that,

P=Pa + P b

P=Pa° X a + Pb° X b

P=200× X a + 350 - 350 X a

305=150X a + 350

150 X a= -45

X a=45/150=0.3

I hope this helps