Two liquids `A` and `B` form an ideal solution. At `300 K`, the vapour pressure of a solution containing `1 mol` of `A` and `3 mol` fo `B` is `550 mm Hg`. At the same temperature, if `1 mol`more of `B` is added to this solution, the vapour pressure of the solution increases by `10 mm Hg`. Determine the vapour pressure of `A` and `B` in their pure states.

Explanation

To determine the vapor pressures of liquids A and B in their pure states, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( P^0_A \) = Vapor pressure of pure liquid A
- \( P^0_B \) = Vapor pressure of pure liquid B

### Step 2: Calculate the Total Vapor Pressure of the Initial Solution
Given:
- Moles of A = 1 mol
- Moles of B = 3 mol
- Total vapor pressure \( P_{total} = 550 \, \text{mm Hg} \)

The mole fractions are calculated as follows:
- Total moles = \( 1 + 3 = 4 \)
- Mole fraction of A, \( x_A = \frac{1}{4} \)
- Mole fraction of B, \( x_B = \frac{3}{4} \)

Using Raoult's Law, the total vapor pressure of the solution can be expressed as:
\[ P_{total} = P^0_A \cdot x_A + P^0_B \cdot x_B \]
Substituting the known values:
\[ 550 = P^0_A \cdot \frac{1}{4} + P^0_B \cdot \frac{3}{4} \tag{1} \]

### Step 3: Calculate the Total Vapor Pressure After Adding More B
When 1 mole of B is added:
- New moles of B = \( 3 + 1 = 4 \)
- New 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 total vapor pressure \( P'_{total} = 560 \, \text{mm Hg} \) (increased by 10 mm Hg). Using Raoult's Law again:
\[ P'_{total} = P^0_A \cdot x_A' + P^0_B \cdot x_B' \]
Substituting the new values:
\[ 560 = P^0_A \cdot \frac{1}{5} + P^0_B \cdot \frac{4}{5} \tag{2} \]

### Step 4: Solve the System of Equations
We now have two equations:
1. \( 550 = P^0_A \cdot \frac{1}{4} + P^0_B \cdot \frac{3}{4} \)
2. \( 560 = P^0_A \cdot \frac{1}{5} + P^0_B \cdot \frac{4}{5} \)

Multiply equation (1) by 4:
\[ 2200 = P^0_A + 3P^0_B \tag{3} \]

Multiply equation (2) by 5:
\[ 2800 = P^0_A + 4P^0_B \tag{4} \]

### Step 5: Subtract Equation (3) from Equation (4)
Subtracting (3) from (4):
\[ (2800 - 2200) = (P^0_A + 4P^0_B) - (P^0_A + 3P^0_B) \]
\[ 600 = P^0_B \]

### Step 6: Substitute Back to Find \( P^0_A \)
Substituting \( P^0_B = 600 \) into equation (3):
\[ 2200 = P^0_A + 3(600) \]
\[ 2200 = P^0_A + 1800 \]
\[ P^0_A = 2200 - 1800 = 400 \]

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