Question

The vapour pressure of pure solvent and solution are 120 mm Hg and 108 mm Hg respectively. The mole fraction of the solvent in the solution is ____________.

Options

• 0.1

• 0.9

• 1.2

• 1.08

Answer 1

The vapour pressure of pure solvent and solution are 120 mm Hg and 108 mm Hg respectively. The mole fraction of the solvent in the solution is 0.9.

Explanation:

The vapour pressure of pure solvent = `"P"_1^0` = 120 mm Hg

The vapour pressure of solution = P = 108 mm Hg

Mole fraction of solute = X₂

According to Raoult's law, `("P"_1^0 - "P")/("P"_1^0)` = X₂

`(120  "mm Hg" - 108  "mm Hg")/(120  "mm Hg")` = X₂ = 0.1

Mole fraction of solvent = 1 − X₂ = 1 − 0.1 = 0.9