Two liquid `A` and `B` form an ideal solution at temperature `T`. When the total vapour pressure above the solution is `400` torr , the mole fraction of `A` in the vapour phase is `0.40` and in the liquid phase `0.75`. What are the vapour pressure of pure `A` and pure `B` at temperature `T` ?

Answer

To solve the problem, we need to find the vapor pressures of pure liquids A and B at temperature T, given the total vapor pressure and the mole fractions in both the vapor and liquid phases.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Total vapor pressure, \( P_{total} = 400 \, \text{torr} \)
- Mole fraction of A in the vapor phase, \( y_A = 0.40 \)
- Mole fraction of A in the liquid phase, \( x_A = 0.75 \)

2. **Calculate the Mole Fraction of B in the Vapor Phase:**
\[ y_B = 1 - y_A = 1 - 0.40 = 0.60 \]

3. **Use Raoult's Law to Find the Partial Pressure of A:**
According to Raoult's Law:
\[ P_A = y_A \times P_{total} \]
Substituting the values:
\[ P_A = 0.40 \times 400 = 160 \, \text{torr} \]

4. **Use Raoult's Law to Find the Partial Pressure of B:**
Similarly, for B:
\[ P_B = y_B \times P_{total} \]
Substituting the values:
\[ P_B = 0.60 \times 400 = 240 \, \text{torr} \]

5. **Calculate the Mole Fraction of B in the Liquid Phase:**
\[ x_B = 1 - x_A = 1 - 0.75 = 0.25 \]

6. **Calculate the Vapor Pressure of Pure A:**
Using Raoult's Law:
\[ P_{A}^{\circ} = \frac{P_A}{x_A} \]
Substituting the values:
\[ P_{A}^{\circ} = \frac{160}{0.75} = 213.33 \, \text{torr} \]

7. **Calculate the Vapor Pressure of Pure B:**
Using Raoult's Law:
\[ P_{B}^{\circ} = \frac{P_B}{x_B} \]
Substituting the values:
\[ P_{B}^{\circ} = \frac{240}{0.25} = 960 \, \text{torr} \]

### Final Answers:
- Vapor pressure of pure A, \( P_{A}^{\circ} = 213.33 \, \text{torr} \)
- Vapor pressure of pure B, \( P_{B}^{\circ} = 960 \, \text{torr} \)