The pressure of two pure liquid A and B which form an ideal solutions are 400 mm Hg and 800 mm Hg respectively at temperature T.A liquid containing 3:1 molar composition pressure can be varied.The solutions is slowly vapourized at temperature T by decreasing the applied pressure starting with a pressure of 760 mm Hg.A pressure gauge (in mm) Hg is connected which give the reading of pressure applied.
The reading of pressure Gauge at which only vapour phase exists is

Options

A. 501
B. 457.14
C. 425
D. 525

Explanation

To solve the problem, we need to determine the pressure at which only the vapor phase exists when a solution of two liquids A and B is vaporized.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Vapor pressure of pure liquid A, \( P^0_A = 400 \, \text{mm Hg} \)
- Vapor pressure of pure liquid B, \( P^0_B = 800 \, \text{mm Hg} \)
- Molar composition of the solution is 3:1 (A:B).

2. **Calculate Mole Fractions**:
- Let the number of moles of A be \( N_A = 3 \) and the number of moles of B be \( N_B = 1 \).
- Total moles, \( N_{total} = N_A + N_B = 3 + 1 = 4 \).
- Mole fraction of A, \( X_A = \frac{N_A}{N_{total}} = \frac{3}{4} = 0.75 \).
- Mole fraction of B, \( X_B = \frac{N_B}{N_{total}} = \frac{1}{4} = 0.25 \).

3. **Relate Mole Fractions in Vapor Phase**:
- When only vapor phase exists, the mole fractions in the vapor phase are equal to those in the liquid phase:
- \( Y_A = X_A = 0.75 \)
- \( Y_B = X_B = 0.25 \)

4. **Use Raoult's Law**:
- According to Raoult's Law, the total pressure \( P_T \) can be calculated using the formula:
\[ P_T = Y_A \cdot P^0_A + Y_B \cdot P^0_B \]
- Substitute the values:
\[ P_T = (0.75 \cdot 400) + (0.25 \cdot 800) \]

5. **Calculate Total Pressure**:
- Calculate each term:
- \( 0.75 \cdot 400 = 300 \, \text{mm Hg} \)
- \( 0.25 \cdot 800 = 200 \, \text{mm Hg} \)
- Therefore,
\[ P_T = 300 + 200 = 500 \, \text{mm Hg} \]

6. **Determine Pressure at Which Only Vapor Phase Exists**:
- The reading of the pressure gauge at which only the vapor phase exists is the total pressure calculated above. Thus, the pressure gauge reading is:
\[ P_T = 500 \, \text{mm Hg} \]

### Final Answer:
The reading of the pressure gauge at which only vapor phase exists is **500 mm Hg**.