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

Options

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

Explanation

To solve the problem, we will use Raoult's Law, which states that the total pressure of a mixture of volatile components is equal to the sum of the partial pressures of each component. The partial pressure of each component can be calculated using the formula:

\[ P_T = P_{A}^0 \cdot X_A + P_{B}^0 \cdot X_B \]

Where:
- \( P_T \) = total pressure of the mixture
- \( P_{A}^0 \) = vapor pressure of component A
- \( P_{B}^0 \) = vapor pressure of component B
- \( X_A \) = mole fraction of component A
- \( X_B \) = mole fraction of component B

Given:
- \( P_T = 760 \, \text{torr} \)
- \( P_{A}^0 = 300 \, \text{torr} \)
- \( P_{B}^0 = 800 \, \text{torr} \)

### Step-by-step Solution:

1. **Set up the equation using Raoult's Law:**
\[ P_T = P_{A}^0 \cdot X_A + P_{B}^0 \cdot X_B \]
Since \( X_A + X_B = 1 \), we can express \( X_B \) as \( 1 - X_A \).

2. **Substitute \( X_B \) in the equation:**
\[ 760 = 300 \cdot X_A + 800 \cdot (1 - X_A) \]

3. **Expand the equation:**
\[ 760 = 300 \cdot X_A + 800 - 800 \cdot X_A \]
\[ 760 = 800 - 500 \cdot X_A \]

4. **Rearrange the equation to isolate \( X_A \):**
\[ 500 \cdot X_A = 800 - 760 \]
\[ 500 \cdot X_A = 40 \]

5. **Solve for \( X_A \):**
\[ X_A = \frac{40}{500} = 0.08 \]

6. **Find \( X_B \):**
\[ X_B = 1 - X_A = 1 - 0.08 = 0.92 \]

### Final Answer:
The composition of the mixture at the normal boiling point of 350 K is:
- \( X_A = 0.08 \)
- \( X_B = 0.92 \)