Question

The vapour pressure of a solvent decreased by 10 mm o f mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm o f mercury?

• A. 0.4
• B. 0.6
• C. 0.8
• D. 0.2

Answer 1

The Correct Option is B

To solve this problem, we need to apply Raoult's Law, which states that the vapour pressure of a solvent decreases when a non-volatile solute is added. The decrease in vapour pressure is proportional to the mole fraction of the solute in the solution.

Let's define the terms:

• P_0: Original vapour pressure of the pure solvent
• \Delta P: Decrease in vapour pressure
• X_{\text{solute}}: Mole fraction of the solute
• X_{\text{solvent}}: Mole fraction of the solvent

According to Raoult's Law:

\Delta P = P_0 \cdot X_{\text{solute}}

Initially, when the vapour pressure decreases by 10 mm of mercury, the mole fraction of the solute is given as 0.2. Using Raoult's Law:

\Delta P = P_0 \cdot 0.2 = 10 \text{ mm of Hg}

We are asked to find the new mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm of mercury:

\Delta P' = 20 \text{ mm of Hg}

Let the new mole fraction of the solute be X'_{\text{solute}}, then:

\Delta P' = P_0 \cdot X'_{\text{solute}}

Given that \Delta P' = 20 \text{ mm of Hg}, we substitute it into the equation:

20 = P_0 \cdot X'_{\text{solute}}

Divide the new decrease by the initial decrease:

\frac{20}{10} = \frac{P_0 \cdot X'_{\text{solute}}}{P_0 \cdot 0.2}

This simplifies to:

\frac{20}{10} = \frac{X'_{\text{solute}}}{0.2}

Solve for X'_{\text{solute}}:

X'_{\text{solute}} = 0.4

The mole fractions of solute and solvent add up to 1:

X'_{\text{solute}} + X'_{\text{solvent}} = 1

Thus, the mole fraction of the solvent is:

X'_{\text{solvent}} = 1 - X'_{\text{solute}} = 1 - 0.4 = 0.6

Therefore, the correct answer is 0.6.