The vapour pressure of pure liquid A and liquid B at 350 K are
440 mm and 720 mm of Hg. If total vapour pressure of solution is
580 mm of Hg then the mole fraction of liquid A in vapour
phase will be :-

Options

A. 0.31
B. 0.38
C. 0.62
D. 0.76

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
We are given:
- Vapor pressure of pure liquid A, \( P^0_A = 440 \, \text{mm Hg} \)
- Vapor pressure of pure liquid B, \( P^0_B = 720 \, \text{mm Hg} \)
- Total vapor pressure of the solution, \( P_{total} = 580 \, \text{mm Hg} \)

### Step 2: Apply Raoult's Law
According to Raoult's Law, the total vapor pressure of a solution can be expressed as:
\[ P_{total} = P^0_A \cdot x_A + P^0_B \cdot x_B \]
where \( x_A \) and \( x_B \) are the mole fractions of liquids A and B in the liquid phase, respectively.

### Step 3: Express \( x_B \) in Terms of \( x_A \)
Since \( x_A + x_B = 1 \), we can express \( x_B \) as:
\[ x_B = 1 - x_A \]

### Step 4: Substitute into the Raoult's Law Equation
Substituting \( x_B \) into the equation gives:
\[ P_{total} = P^0_A \cdot x_A + P^0_B \cdot (1 - x_A) \]
Substituting the known values:
\[ 580 = 440 \cdot x_A + 720 \cdot (1 - x_A) \]

### Step 5: Simplify the Equation
Expanding the equation:
\[ 580 = 440 \cdot x_A + 720 - 720 \cdot x_A \]
Combining like terms:
\[ 580 = 720 - 280 \cdot x_A \]

### Step 6: Solve for \( x_A \)
Rearranging gives:
\[ 280 \cdot x_A = 720 - 580 \]
\[ 280 \cdot x_A = 140 \]
\[ x_A = \frac{140}{280} = 0.5 \]

### Step 7: Calculate the Mole Fraction of A in the Vapor Phase
Now, we need to find the mole fraction of liquid A in the vapor phase, denoted as \( y_A \). This can be calculated using:
\[ y_A = \frac{P^0_A \cdot x_A}{P_{total}} \]
Substituting the values:
\[ y_A = \frac{440 \cdot 0.5}{580} \]
Calculating:
\[ y_A = \frac{220}{580} = 0.3793 \approx 0.38 \]

### Final Answer
The mole fraction of liquid A in the vapor phase is approximately \( 0.38 \).

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