Two liquids A and B have vapour pressure of 0.658 bar and 0.264 bar respectively. In an ideal solution of the two , calculate the mole fraction of A at which the two liquids have equal partical pressure .

Explanation

To solve the problem of finding the mole fraction of liquid A in an ideal solution where the partial pressures of liquids A and B are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure A, \( P^0_A = 0.658 \, \text{bar} \)
- Vapor pressure of pure B, \( P^0_B = 0.264 \, \text{bar} \)

2. **Set Up the Equation for Partial Pressures:**
- The partial pressure of A in the solution is given by:
\[ P_A = P^0_A \cdot \chi_A \]
- The partial pressure of B in the solution is given by:
\[ P_B = P^0_B \cdot \chi_B \]

3. **Express Mole Fraction of B:**
- Since the total mole fraction must equal 1, we have:
\[ \chi_B = 1 - \chi_A \]
- Let \( \chi_A = \chi \) (the mole fraction of A). Then, \( \chi_B = 1 - \chi \).

4. **Set the Partial Pressures Equal:**
- We need to find the mole fraction when the partial pressures are equal:
\[ P_A = P_B \]
- Substituting the expressions for \( P_A \) and \( P_B \):
\[ P^0_A \cdot \chi = P^0_B \cdot (1 - \chi) \]

5. **Substitute the Known Values:**
- Plugging in the values of \( P^0_A \) and \( P^0_B \):
\[ 0.658 \chi = 0.264 (1 - \chi) \]

6. **Expand and Rearrange the Equation:**
- Expanding the right side:
\[ 0.658 \chi = 0.264 - 0.264 \chi \]
- Rearranging gives:
\[ 0.658 \chi + 0.264 \chi = 0.264 \]
\[ 0.922 \chi = 0.264 \]

7. **Solve for \( \chi \):**
- Dividing both sides by 0.922:
\[ \chi = \frac{0.264}{0.922} \approx 0.286 \]

8. **Conclusion:**
- The mole fraction of A, \( \chi_A \), is approximately 0.286.
- The mole fraction of B can be calculated as:
\[ \chi_B = 1 - \chi_A = 1 - 0.286 = 0.714 \]

### Final Answer:
- Mole fraction of A, \( \chi_A \approx 0.286 \)
- Mole fraction of B, \( \chi_B \approx 0.714 \)