If the vapour pressure of pure A and pure B at 298 K are 60 and 15 torr respectively, what would be the mole fraction of A in vapour phase (at this temperature) in a solution that contains 20 mole per cent of A in the (A + B) binary mixture in liquid phase ?

Options

A. 0.2
B. 0.3
C. 0.5
D. 0.7

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the mole fractions of components A and B in the liquid phase.
Given that the solution contains 20 mole percent of A, we can calculate the mole fraction of A (XA) and B (XB) in the liquid phase.

- Mole fraction of A (XA) = 20/100 = 0.2
- Mole fraction of B (XB) = 1 - XA = 1 - 0.2 = 0.8

### Step 2: Apply Raoult's Law to find the partial pressures of A and B.
Raoult's Law states that the partial pressure of each component in the solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the liquid phase.

- Vapor pressure of pure A (PA°) = 60 torr
- Vapor pressure of pure B (PB°) = 15 torr

Using Raoult's Law:
- Partial pressure of A (PA) = PA° * XA = 60 torr * 0.2 = 12 torr
- Partial pressure of B (PB) = PB° * XB = 15 torr * 0.8 = 12 torr

### Step 3: Calculate the total pressure of the vapor phase.
The total pressure (P) of the vapor phase is the sum of the partial pressures of A and B.

- Total pressure (P) = PA + PB = 12 torr + 12 torr = 24 torr

### Step 4: Determine the mole fraction of A in the vapor phase.
Let the mole fraction of A in the vapor phase be YA. The partial pressure of A in the vapor phase can also be expressed as:

- PA = YA * P

From the previous step, we know PA = 12 torr and P = 24 torr. Therefore, we can set up the equation:

- 12 torr = YA * 24 torr

Now, solve for YA:

- YA = 12 torr / 24 torr = 0.5

### Final Answer:
The mole fraction of A in the vapor phase is 0.5.

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