The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :

Options

A. 450 mm Hg, 0.5, 0.5
B. 450 mm Hg, 0.4, 0.6
C. 500 mm Hg, 0.4, 0.6
D. 500 mm Hg, 0.5, 0.5

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Given Data
- Vapour pressure of pure liquid A, \( P^0_A = 400 \, \text{mm Hg} \)
- Vapour pressure of pure liquid B, \( P^0_B = 600 \, \text{mm Hg} \)
- Mole fraction of liquid B in the mixture, \( X_B = 0.5 \)
- Therefore, the mole fraction of liquid A, \( X_A = 1 - X_B = 0.5 \)

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