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

To solve this problem, we begin by understanding Raoult's law, which states that the vapour pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent. If a solute is non-volatile, it does not contribute to the vapour pressure.

Initially, when the vapour pressure decreases by 10 mm of Hg, the mole fraction of the solute \(x_{\text{solute}}\) is 0.2. Let's denote the initial vapour pressure of the pure solvent as \(P^0\).

According to Raoult's law: 

\(P_{\text{solution}} = x_{\text{solvent}} \times P^0\)

Given that the decrease in vapour pressure is 10 mm of Hg, which implies:

\(P^0 - P_{\text{solution}} = 10 \text{ mm Hg}\)

From the formula of Raoult's law, the mole fraction of the solvent \(x_{\text{solvent}}\) can be expressed as:

\(x_{\text{solvent}} = 1 - x_{\text{solute}} = 1 - 0.2 = 0.8\)

Now consider the change in vapour pressure to 20 mm of Hg. Again applying Raoult's law:

\(P^0 - P_{\text{solution}} = 20 \text{ mm Hg}\)

The mole fraction of the solvent in the new situation is denoted by \(x'_{\text{solvent}}\).

Using the relation of proportional decrease:

\(x'_{\text{solvent}} \times P^0 = P^0 - 20\)

And since the total pressure decrease is doubled, the new mole fraction \(x_{\text{solute}}'\) will be inverse to solvent coverage:

Using the equation:

\(P_{\text{solution}} = x'_{\text{solvent}} \times P^0 = (1 - x_{\text{solute}}') \times P^0\)

Solving for the new condition:

Since mole fraction of solute \(x'_{\text{solute}} = \frac{0.2 \times 2}{1} = 0.4\). Thus:

\(x'_{\text{solvent}} = 1 - 0.4 = 0.6\)

Thus, the mole fraction of the solvent when the vapour pressure decrease is 20 mm of Hg is 0.6. This confirms that the correct answer is 0.6.

Answer 2

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.