The vapour pressure of pure benzene is `22mm` and that of pure toluene is `75mm` at `20^(@)C`. What is the composition of the solution of these two components that has a vapour pressure of 50mm at this temperature ? What is the composition of vapour in equilibrium with this solution ?

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial pressures of each component in the solution. The partial pressure of each component is the product of its mole fraction in the solution and its vapor pressure as a pure substance.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure benzene (P_b) = 22 mm Hg
- Vapor pressure of pure toluene (P_t) = 75 mm Hg
- Vapor pressure of the solution (P_total) = 50 mm Hg

2. **Define Mole Fractions:**
- Let the mole fraction of benzene be \( x_b \).
- Then, the mole fraction of toluene will be \( x_t = 1 - x_b \).

3. **Apply Raoult's Law:**
According to Raoult's Law, the total vapor pressure of the solution can be expressed as:
\[ P_{\text{total}} = P_b \cdot x_b + P_t \cdot x_t \]
Substituting the values we have:
\[ 50 = 22 \cdot x_b + 75 \cdot (1 - x_b) \]

4. **Simplify the Equation:**
Expanding the equation:
\[ 50 = 22 x_b + 75 - 75 x_b \]
Rearranging gives:
\[ 50 = 75 - 53 x_b \]
\[ 53 x_b = 75 - 50 \]
\[ 53 x_b = 25 \]
\[ x_b = \frac{25}{53} \approx 0.47 \]

5. **Calculate Mole Fraction of Toluene:**
Now, we can find the mole fraction of toluene:
\[ x_t = 1 - x_b = 1 - 0.47 = 0.53 \]

6. **Final Composition of the Solution:**
- Mole fraction of benzene \( (x_b) \approx 0.47 \)
- Mole fraction of toluene \( (x_t) \approx 0.53 \)

7. **Calculate Composition of Vapor in Equilibrium:**
To find the composition of the vapor in equilibrium with this solution, we can use the mole fractions and the vapor pressures:
- Partial pressure of benzene in vapor:
\[ P_{b,\text{vapor}} = P_b \cdot x_b = 22 \cdot 0.47 \approx 10.34 \text{ mm Hg} \]
- Partial pressure of toluene in vapor:
\[ P_{t,\text{vapor}} = P_t \cdot x_t = 75 \cdot 0.53 \approx 39.75 \text{ mm Hg} \]
- Total vapor pressure:
\[ P_{\text{vapor}} = P_{b,\text{vapor}} + P_{t,\text{vapor}} \approx 10.34 + 39.75 \approx 50.09 \text{ mm Hg} \]

8. **Calculate Mole Fractions of the Vapor:**
- Mole fraction of benzene in vapor:
\[ y_b = \frac{P_{b,\text{vapor}}}{P_{\text{vapor}}} = \frac{10.34}{50.09} \approx 0.206 \]
- Mole fraction of toluene in vapor:
\[ y_t = \frac{P_{t,\text{vapor}}}{P_{\text{vapor}}} = \frac{39.75}{50.09} \approx 0.794 \]

### Summary of Results:
- Composition of the solution:
- Mole fraction of benzene \( \approx 0.47 \)
- Mole fraction of toluene \( \approx 0.53 \)
- Composition of vapor in equilibrium:
- Mole fraction of benzene \( \approx 0.206 \)
- Mole fraction of toluene \( \approx 0.794 \)