Question

At \(298\ K\), vapour pressures of two pure liquids A and B are \(200\) and \(400\ \text{mm Hg}\) respectively. If mole fractions of A and B in solution are \(0.7\) and \(0.3\) respectively, what is the mole fraction of B in vapour phase?

• A. \(0.279\)
• B. \(0.721\)
• C. \(0.538\)
• D. \(0.462\)

Hint:

For an ideal binary liquid solution, use Raoult's law: \[ p_i=x_ip_i^\circ \] and mole fraction in vapour phase: \[ y_i=\frac{p_i}{p_{\text{total}}} \]

Answer 1

The Correct Option is D

Step 1: List the given data.
Vapour pressure of pure A: \(P_A^* = 200\) mmHg
Vapour pressure of pure B: \(P_B^* = 400\) mmHg
Mole fraction of A in the liquid phase: \(x_A = 0.7\)
Mole fraction of B in the liquid phase: \(x_B = 0.3\) (since \(x_A + x_B = 1\))

Step 2: Apply Raoult's Law to find partial vapour pressures.
Raoult's Law states that the partial vapour pressure of each component is proportional to its mole fraction in the liquid phase: \[ P_A = x_A \times P_A^* = 0.7 \times 200 = 140\ \text{mmHg} \] \[ P_B = x_B \times P_B^* = 0.3 \times 400 = 120\ \text{mmHg} \]

Step 3: Calculate the total vapour pressure of the solution.
\[ P_{\text{total}} = P_A + P_B = 140 + 120 = 260\ \text{mmHg} \]

Step 4: Apply Dalton's Law to find the mole fraction of B in vapour phase.
In the vapour phase, the mole fraction of any component equals the ratio of its partial pressure to the total vapour pressure: \[ y_B = \frac{P_B}{P_{\text{total}}} = \frac{120}{260} \]

Step 5: Calculate the numerical value.
\[ y_B = \frac{120}{260} = \frac{6}{13} \approx 0.4615 \approx 0.462 \]

Step 6: Interpret the result and state the answer.
The mole fraction of B in the vapour (0.462) is greater than its mole fraction in the liquid (0.3). This makes physical sense because B has a higher vapour pressure (400 mmHg vs 200 mmHg), so it is more volatile and becomes enriched in the vapour phase.
\[ \boxed{y_B = 0.462} \]