Question

two liquids A and B have vapour pressure in the ratio 2:3 at a certain temperature .Assuming that A and B forms an ideal solution and ratio of mole fraction of A and B in the vapour phase is 1:3,then the mole fraction of A in the solution at the same temperature is

Answer 1

Answer:

Mole fraction of A is=$\frac{1}{3}$

Explanation:

given condition of total vapour pressure is=2+3=5

mole fraction of A=$\frac{1}{4}$

mole fraction of B=$\frac{3}{4}$(moles are in the ratio 1:3)

$p_{A}$=$X_{A}$×$P_{Tot}$

                       =$\frac{1}{4}$×5

                       =1.25

$p_{B}$=$X_{B}$×$p_{Tot}$

                       =$\frac{3}{4}$×5

                       =3.75

Then the mole fraction of A is

$\frac{p_{A} }{p_{B} }$α$X_{A}$

$X_{A}$=$\frac{1.25}{3.75} =\frac{1}{3}$

Answer 2

The mole fraction of A is $\frac{1}{3}.$

To find:

The “mole fraction” of the substance A.

Given:

Let us consider that,  

The given solution is ideal.

The “mole fraction” of A is $\mathrm{Y}_{\mathrm{A}}$

The “mole fraction” of B is$\mathrm{Y}_{\mathrm{B}}$

In vapour phase the “mole fraction” of the ratio  of A and B is 1:3.

Solution:

We know that,  

$\frac{Y_{A}}{Y_{B}}=\frac{1}{3}$

In vapour phase, $\frac{p_{A}}{p_{B}}=\frac{1}{3}$

Hence, $Y_{A}=\frac{P_{A}}{p} and Y_{B}=\frac{P_{B}}{p}$

$\frac{P_{A}^{o} X_{A}}{P_{B}^{o} X_{B}}=\frac{1}{3}$

$\frac{X_{A}}{X_{B}}=\frac{1}{3} \times \frac{3}{2}=\frac{1}{2}$

$\frac{X_{A}}{1-X_{A}}=\frac{1}{2}$

$X_{A}= \frac{1}{3}$

Therefore, the “Mole fraction” of substance A is $\frac{1}{3}.$