Mixture of volatile components A and B has total pressure ( in Torr ) `p=265-130x_(A),` where `X_(A)` is mole fraction of `A` in mixture . Hence `P_(A)^(@)+P_(B)^(@)=` ( in T o r r).

Options

A. 265
B. 135
C. 400
D. 150

Explanation

To solve the problem, we need to find the sum of the vapor pressures of components A and B, denoted as \( P_0^A + P_0^B \).

### Step-by-Step Solution:

1. **Understand the Total Pressure Equation**:
The total pressure \( P \) of the mixture is given by the equation:
\[ P = 265 - 130x_A \]
where \( x_A \) is the mole fraction of component A.

2. **Apply Raoult's Law**:
According to Raoult's law, the total pressure \( P_T \) can also be expressed as:
\[ P_T = P_0^A x_A + P_0^B x_B \]
where \( P_0^A \) and \( P_0^B \) are the vapor pressures of pure components A and B, respectively, and \( x_B = 1 - x_A \).

3. **Substitute \( x_B \)**:
Substitute \( x_B \) in the equation:
\[ P_T = P_0^A x_A + P_0^B (1 - x_A) \]
Simplifying this gives:
\[ P_T = P_0^A x_A + P_0^B - P_0^B x_A \]
Rearranging, we have:
\[ P_T = (P_0^B - P_0^A) x_A + P_0^B \]

4. **Set the Two Expressions for Total Pressure Equal**:
We have two expressions for \( P_T \):
\[ 265 - 130x_A = (P_0^B - P_0^A) x_A + P_0^B \]

5. **Compare Coefficients**:
From the equation above, we can compare coefficients of \( x_A \) and the constant terms:
- Coefficient of \( x_A \):
\[ -130 = P_0^B - P_0^A \quad \text{(1)} \]
- Constant term:
\[ P_0^B = 265 \quad \text{(2)} \]

6. **Substitute \( P_0^B \) into Equation (1)**:
Substitute \( P_0^B = 265 \) into equation (1):
\[ -130 = 265 - P_0^A \]
Rearranging gives:
\[ P_0^A = 265 + 130 = 395 \]

7. **Calculate \( P_0^A + P_0^B \)**:
Now, we can find the sum of the vapor pressures:
\[ P_0^A + P_0^B = 395 + 265 = 660 \]

### Final Answer:
Thus, the sum of the vapor pressures \( P_0^A + P_0^B \) is:
\[ \boxed{660 \text{ Torr}} \]