Given at 350 K, `P_(A)^(@)` = 300 torr and `P_(B)^(@)`=800 torr the composition of the mixture having a normal boiling point of 350 K is :

Options

A. `X_(A)=0.08`
B. `X_(A)=0.06`
C. `X_(A)=0.04`
D. `X_(A)=0.02`

Answer

To solve the problem, we need to find the composition of the mixture at its normal boiling point of 350 K, given the vapor pressures of the pure components A and B.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure A, \( P_A^0 = 300 \) torr
- Vapor pressure of pure B, \( P_B^0 = 800 \) torr
- Total pressure at boiling point, \( P_{total} = 760 \) torr (standard atmospheric pressure)

2. **Apply Raoult's Law:**
According to Raoult's Law, the total vapor pressure of a solution is given by:
\[ P_{total} = x_A \cdot P_A^0 + x_B \cdot P_B^0 \]
where \( x_A \) and \( x_B \) are the mole fractions of components A and B, respectively.

3. **Express \( x_B \) in terms of \( x_A \):**
Since the sum of mole fractions is equal to 1:
\[ x_B = 1 - x_A \]

4. **Substitute \( x_B \) into Raoult's Law:**
Substitute \( x_B \) into the equation:
\[ P_{total} = x_A \cdot P_A^0 + (1 - x_A) \cdot P_B^0 \]
This simplifies to:
\[ 760 = x_A \cdot 300 + (1 - x_A) \cdot 800 \]

5. **Expand and Rearrange the Equation:**
Expanding the equation gives:
\[ 760 = 300x_A + 800 - 800x_A \]
Rearranging this results in:
\[ 760 = 800 - 500x_A \]
\[ 500x_A = 800 - 760 \]
\[ 500x_A = 40 \]

6. **Solve for \( x_A \):**
Dividing both sides by 500:
\[ x_A = \frac{40}{500} = 0.08 \]

7. **Find \( x_B \):**
Now, using the relationship \( x_B = 1 - x_A \):
\[ x_B = 1 - 0.08 = 0.92 \]

8. **Conclusion:**
The composition of the mixture is:
- Mole fraction of A, \( x_A = 0.08 \)
- Mole fraction of B, \( x_B = 0.92 \)

### Final Answer:
The composition of the mixture is:
- \( x_A = 0.08 \)
- \( x_B = 0.92 \)