When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg. The mole fraction of the solute in the solution is 0.2. What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm

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Solution

Q.
When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg. The mole fraction of the solute in the solution is 0.2. What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg [2025]

1 0.2

2 0.4

3 0.8

4 0.6

Ans.
(4)

$\frac{P_{A}^{°} - P_{A}}{P_{A}^{°}} = x_{B}$

Where $P_{A}^{°}$ is vapour pressure of pure solvent, $P_{A}$ is vapour pressure of solution and $x_{B}$ is mole fraction of solute.

As per two cases given in question:

$\frac{10}{P_{A}^{°}} = 0.2 and \frac{20}{P_{A}^{°}} = x_{B}$

Solving these two gives $x_{B} = 0.4$ .

Mole fraction of solvent $(x_{A})$ when $x_{B} = 0.4$ is

$x_{A} = 1 - x_{B} = 1 - 0.4 = 0.6$

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