Liquids A`(p_(A)^(@)=360" mm Hg")` and `B(p_(B)^(@)=320" mm Hg")` are mixed. If solution has vapour pressure 340 mm Hg, then mole fraction of B will be

Options

A. `33/34`
B. `1/34`
C. `1/2`
D. `1/4`

Answer

To find the mole fraction of component B in a solution of liquids A and B, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial vapor pressures of its components.

### Step-by-step solution:

1. **Identify Given Values:**
- Vapor pressure of pure A, \( P_{A}^0 = 360 \, \text{mm Hg} \)
- Vapor pressure of pure B, \( P_{B}^0 = 320 \, \text{mm Hg} \)
- Vapor pressure of the solution, \( P_{solution} = 340 \, \text{mm Hg} \)

2. **Use Raoult's Law:**
According to Raoult's Law, the vapor pressure of the solution can be expressed as:
\[ P_{solution} = P_{A}^0 \cdot X_A + P_{B}^0 \cdot X_B \]
where \( X_A \) and \( X_B \) are the mole fractions of A and B, respectively.

3. **Express Mole Fractions:**
Since \( X_A + X_B = 1 \), we can express \( X_A \) in terms of \( X_B \):
\[ X_A = 1 - X_B \]

4. **Substitute into Raoult's Law:**
Substitute \( X_A \) into the equation:
\[ 340 = 360(1 - X_B) + 320X_B \]

5. **Expand and Rearrange:**
Expanding the equation gives:
\[ 340 = 360 - 360X_B + 320X_B \]
Combine like terms:
\[ 340 = 360 - 40X_B \]

6. **Isolate \( X_B \):**
Rearranging the equation to isolate \( X_B \):
\[ 340 - 360 = -40X_B \]
\[ -20 = -40X_B \]
\[ X_B = \frac{1}{2} \]

7. **Conclusion:**
The mole fraction of component B, \( X_B \), is \( \frac{1}{2} \).

### Final Answer:
The mole fraction of B is \( \frac{1}{2} \).