At a given temperature, total vapour pressure in Torr of a mixture of volatile components A and B is given by
` P = 120 - 75X_(B)`
hence vapour pressure ofpure A and B respectively ( in Torr) are :

Options

A. 120,75
B. 120195
C. 120,45
D. 75,45

Answer

To solve the problem, we need to determine the vapor pressures of pure components A and B from the given total vapor pressure equation for their mixture.

### Step-by-Step Solution:

1. **Understand the Given Equation**:
The total vapor pressure \( P \) of the mixture is given by:
\[ P = 120 - 75X_B \]
where \( X_B \) is the mole fraction of component B in the mixture.

2. **Apply Raoult's Law**:
According to Raoult's law, the total vapor pressure \( P_t \) of a mixture is the sum of the partial pressures of its components:
\[ P_t = P^0_A \cdot X_A + P^0_B \cdot X_B \]
where \( P^0_A \) and \( P^0_B \) are the vapor pressures of pure components A and B, respectively, and \( X_A \) and \( X_B \) are their mole fractions.

3. **Express Mole Fractions**:
Since there are only two components, the mole fractions satisfy:
\[ X_A = 1 - X_B \]
Substituting this into the equation gives:
\[ P_t = P^0_A \cdot (1 - X_B) + P^0_B \cdot X_B \]

4. **Rearranging the Equation**:
Rearranging the equation, we have:
\[ P_t = P^0_A - P^0_A \cdot X_B + P^0_B \cdot X_B \]
This can be rearranged to:
\[ P_t = P^0_A - (P^0_A - P^0_B) \cdot X_B \]

5. **Comparing Coefficients**:
Now, we can compare this equation with the given equation \( P = 120 - 75X_B \):
- From the comparison, we can identify:
- \( P^0_A = 120 \)
- \( P^0_A - P^0_B = 75 \)

6. **Calculate \( P^0_B \)**:
From \( P^0_A - P^0_B = 75 \):
\[ 120 - P^0_B = 75 \]
Rearranging gives:
\[ P^0_B = 120 - 75 = 45 \]

7. **Final Results**:
Thus, the vapor pressures of pure components A and B are:
- \( P^0_A = 120 \, \text{Torr} \)
- \( P^0_B = 45 \, \text{Torr} \)

### Summary:
- The vapor pressure of pure A is **120 Torr**.
- The vapor pressure of pure B is **45 Torr**.