Given at `350K, p_(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 need to find the composition of the mixture at the normal boiling point (350 K) given the vapor pressures of components A and B.

### Step-by-step Solution:

1. **Understand the Normal Boiling Point**:
The normal boiling point of a mixture occurs when the total vapor pressure equals 1 atm (760 torr).

2. **Identify Given Values**:
- Vapor pressure of A, \( p_A^0 = 300 \, \text{torr} \)
- Vapor pressure of B, \( p_B^0 = 800 \, \text{torr} \)

3. **Use Raoult's Law**:
According to Raoult's Law, the total vapor pressure \( P \) of a mixture is given by:
\[ P = p_A^0 \cdot X_A + p_B^0 \cdot X_B \]
where \( X_A \) and \( X_B \) are the mole fractions of A and B, respectively.

4. **Set Up the Equation**:
Since \( X_A + X_B = 1 \), we can express \( X_B \) as \( 1 - X_A \). Substituting this into the equation gives:
\[ 760 = 300 \cdot X_A + 800 \cdot (1 - X_A) \]

5. **Simplify the Equation**:
Expanding the equation:
\[ 760 = 300X_A + 800 - 800X_A \]
Rearranging gives:
\[ 760 = 800 - 500X_A \]
\[ 500X_A = 800 - 760 \]
\[ 500X_A = 40 \]
\[ X_A = \frac{40}{500} = 0.08 \]

6. **Calculate \( X_B \)**:
Now, using \( X_A + X_B = 1 \):
\[ X_B = 1 - X_A = 1 - 0.08 = 0.92 \]

7. **Conclusion**:
The composition of the mixture at the normal boiling point is:
- Mole fraction of A, \( X_A = 0.08 \)
- Mole fraction of B, \( X_B = 0.92 \)

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