Question

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 the decrease in vapour pressure is 20 mm of Hg?

• A. 0.6
• B. 0.4
• C. 0.2
• D. 0.8

Hint:

Raoult’s law relates the decrease in vapour pressure to the mole fraction of the solute in the solution. The mole fraction of the solvent can be found by subtracting the solute’s mole fraction from 1.

Answer 1

The Correct Option is A

The relationship between the decrease in vapour pressure and mole fraction of solute is given by Raoult’s Law: \[ \Delta P = P_0 \cdot x_{\text{solute}}, \] where: - \( \Delta P \) is the decrease in vapour pressure, - \( P_0 \) is the initial vapour pressure, - \( x_{\text{solute}} \) is the mole fraction of the solute. Given: - \( \Delta P = 10 \, \text{mm Hg} \) and \( x_{\text{solute}} = 0.2 \), - The new decrease in vapour pressure is 20 mm Hg, so the mole fraction of the solute at this point will be \( x_{\text{solute}} = 0.4 \). Since the sum of mole fractions of solute and solvent must be 1: \[ x_{\text{solvent}} = 1 - x_{\text{solute}} = 1 - 0.4 = 0.6. \] Therefore, the mole fraction of the solvent is 0.6.