The composition of vapour when first bubble formed is:

Options

A. `y_(A) = 0.6,y_(b) = 04`
B. `y_(A) = 0.48,y_(b) = 0.52`
C. `y_(A) = 0.52,y_(b) = 0.48`
D. `y_(A) = 0.5,y_(b) = 0.5`

Answer

To solve the problem regarding the composition of vapor when the first bubble is formed from a liquid solution of two pure liquids A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure liquid A (P_A^0) = 500 Torr
- Vapor pressure of pure liquid B (P_B^0) = 800 Torr
- Mole fraction of A in the liquid solution (X_A) = 0.60

2. **Calculate Mole Fraction of B:**
- The mole fraction of B (X_B) can be calculated as:
\[ X_B = 1 - X_A = 1 - 0.60 = 0.40 \]

3. **Calculate Total Pressure (P_total):**
- The total pressure of the system when the first bubble forms can be calculated using Raoult's Law:
\[ P_{total} = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]
- Substituting the values:
\[ P_{total} = (500 \, \text{Torr} \cdot 0.60) + (800 \, \text{Torr} \cdot 0.40) \]
- Calculate each term:
\[ P_{total} = 300 \, \text{Torr} + 320 \, \text{Torr} = 620 \, \text{Torr} \]

4. **Calculate Mole Fraction of A in Vapor Phase (Y_A):**
- The mole fraction of A in the vapor phase can be calculated using:
\[ Y_A = \frac{P_A}{P_{total}} \]
- Where \( P_A \) is the partial pressure of A in the vapor phase:
\[ P_A = P_A^0 \cdot X_A = 500 \, \text{Torr} \cdot 0.60 = 300 \, \text{Torr} \]
- Now substituting into the equation for Y_A:
\[ Y_A = \frac{300 \, \text{Torr}}{620 \, \text{Torr}} \approx 0.48 \]

5. **Calculate Mole Fraction of B in Vapor Phase (Y_B):**
- The mole fraction of B in the vapor phase can be calculated as:
\[ Y_B = 1 - Y_A = 1 - 0.48 = 0.52 \]

6. **Conclusion:**
- Therefore, the composition of the vapor when the first bubble is formed is:
- Mole fraction of A (Y_A) = 0.48
- Mole fraction of B (Y_B) = 0.52

### Final Answer:
The composition of vapor when the first bubble is formed is:
- Mole fraction of A = 0.48
- Mole fraction of B = 0.52