Question

Two liquids A and B have vapour pressures in the ratio PA°:PB° =1:3 at a certain temperature. Assuming that A and B form an ideal solution and ratio of mole fractions of A and B in the vapour phase is 4 : 3 , then the mole fraction of A in the solution at the same temperature is :​

Answer 1

Answer:

Here's your answer

Explanation:

Given that =>

PA°:PB° = 1:3 => PB° = PA°

yA:yB = 4:3 =>yA = 4/3yB

where,

PA° = vapour  pressure of liquid A

PB° = vapour pressure of liquid B

yA and yB is a mole fraction of A and B in vapor phase respectively.

Let , x = mole fraction of B in liquid phase

=>(1-x) = mole fraction of A in liquid phase

Now Pt = xApA° + xBPB° = (1-x)PA°+ x3PA°

            = PA° - xPA° + 3xPA° = PA°+ 2xPA°

            = (1+2x)PA°

Now,

yA = yA / yA + yB

    = (4/3yB)/4/3yB + yB

    = 4/3yB/(4yB + 3yB)/3

    = 4yB/4yB + 3yB

    = 4yB/7yB

    = 4/7

yB = 1 - yA

Now ,

PB = xBPB° = yBPt

xB = yBPt/PB°

    = 3/7(1+2x)PA°/3PA°

    = x

=>7x = 1 + 2x

=>5x = 1

=>x = 1/5

So , the mole fraction of B is 1/5

Hope you understand.