Question

A solution contains two liquids "A' and 'B' The vapour pressure of the solution is 400 mm of Hg in the vapour phase male fraction A is 0,45. In liquid phase, mole fraction of B 0.35. Calculate the vapour pressure of pure liquids A nd B​

Answer 1

Given that,

Vapour pressure of solution = 400 mm of Hg

Mole fraction A = 0.45

Mole fraction B = 0.35

We need to calculate the partial pressure of A and B

Using formula of partial pressure

For A,

$P'=P\times a$

Where. P'= partial pressure

P = vapour pressure

a = mole fraction

Put the value into the formula

$P_{A}'=400\times0.45$

$P_{A}'=180$

For B,

$P_{B}'=400\times(1-0.45)$

$P_{B}'=220$

We need to calculate the vapour pressure of liquid A and B

Using formula for vapour pressure

For A,

$P_{A}=\dfrac{P_{A}'}{a}$

Put the value into the formula

$P_{A}=\dfrac{180}{0.65}$

$P_{A}=277$

For B,

$P_{B}=\dfrac{220}{1-0.65}$

$P_{B}=629$

Hence, The vapour pressure of liquid A and B are 277 and 629.