At `300 K`, the vapour pressure of an ideal solution containing one mole of `A` and `3` mole of `B` is `550 mm` of `Hg`. At the same temperature, if one mole of `B` is added to this solution, the vapour pressure of solution increases by `10 mm` of `Hg`. Calculate the `V.P.` of `A` and `B` in their pure state.

Options

A. 400 & 600 mm respectively
B. 600 & 400 mm respectively
C. 200 & 400 mm respectively
D. 400 & 200 mm respectively

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution is equal to the vapor pressure of the pure components multiplied by their respective mole fractions in the solution.

### Step-by-Step Solution:

1. **Identify the given data:**
- Moles of A = 1
- Moles of B = 3
- Total moles in the initial solution = 1 + 3 = 4
- Vapor pressure of the solution (P_solution) = 550 mm Hg
- When 1 mole of B is added, the new vapor pressure (P_solution_new) = 560 mm Hg (an increase of 10 mm Hg).

2. **Set up the equation using Raoult's Law for the initial solution:**
\[ P_{\text{solution}} = P_{A}^0 \cdot X_A + P_{B}^0 \cdot X_B \]
Where:
- \( X_A = \frac{1}{4} \) (mole fraction of A)
- \( X_B = \frac{3}{4} \) (mole fraction of B)

Substituting the values:
\[ 550 = P_{A}^0 \cdot \frac{1}{4} + P_{B}^0 \cdot \frac{3}{4} \quad \text{(Equation 1)} \]

3. **Set up the equation for the new solution after adding 1 mole of B:**
- New moles of B = 3 + 1 = 4
- Total moles = 1 + 4 = 5
- New mole fraction of A, \( X_A' = \frac{1}{5} \)
- New mole fraction of B, \( X_B' = \frac{4}{5} \)

The new equation becomes:
\[ 560 = P_{A}^0 \cdot \frac{1}{5} + P_{B}^0 \cdot \frac{4}{5} \quad \text{(Equation 2)} \]

4. **Solve the equations simultaneously:**
- From Equation 1:
\[ 550 = \frac{1}{4} P_{A}^0 + \frac{3}{4} P_{B}^0 \]
Multiply through by 4:
\[ 2200 = P_{A}^0 + 3 P_{B}^0 \quad \text{(Equation 1')} \]

- From Equation 2:
\[ 560 = \frac{1}{5} P_{A}^0 + \frac{4}{5} P_{B}^0 \]
Multiply through by 5:
\[ 2800 = P_{A}^0 + 4 P_{B}^0 \quad \text{(Equation 2')} \]

5. **Subtract Equation 1' from Equation 2':**
\[ (2800 - 2200) = (P_{A}^0 + 4 P_{B}^0) - (P_{A}^0 + 3 P_{B}^0) \]
Simplifying gives:
\[ 600 = P_{B}^0 \]

6. **Substitute \( P_{B}^0 \) back into Equation 1' to find \( P_{A}^0 \):**
\[ 2200 = P_{A}^0 + 3(600) \]
\[ 2200 = P_{A}^0 + 1800 \]
\[ P_{A}^0 = 2200 - 1800 = 400 \text{ mm Hg} \]

### Final Results:
- Vapor pressure of pure A, \( P_{A}^0 = 400 \text{ mm Hg} \)
- Vapor pressure of pure B, \( P_{B}^0 = 600 \text{ mm Hg} \)