At `27^(@)C`.the vapour pressure of an ideal solution containing 1 mole of `A` and 1 mole and `B` is `500 mm` of `Hg`. At the same temperature, if `2` mol of `B` is added to this solution the vapour pressure of solution increases by `50 mm` of `Hg`. The vapour pressure of `A` and `B` in their pure states is respectively.

Options

A. `600 mm`,`400 mm`
B. `400 mm`,`600 mm`
C. `300 mm`,`700 mm`
D. `200 mm`,`800 mm`

Explanation

To solve the problem, we need to find the vapor pressures of pure substances A and B, given the conditions of an ideal solution. Let’s break down the solution step by step.

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

### Step 2: Initial Conditions
Initially, we have:
- 1 mole of A
- 1 mole of B

Total moles = 1 + 1 = 2 moles.

The mole fractions are:
- Mole fraction of A, \( x_A = \frac{1}{2} \)
- Mole fraction of B, \( x_B = \frac{1}{2} \)

### Step 3: Apply Raoult's Law
According to Raoult's law, the vapor pressure of the solution \( P_t \) is given by:
\[ P_t = P^0_A \cdot x_A + P^0_B \cdot x_B \]
Substituting the values we have:
\[ 500 = P^0_A \cdot \frac{1}{2} + P^0_B \cdot \frac{1}{2} \]
Multiplying through by 2:
\[ 1000 = P^0_A + P^0_B \quad \text{(Equation 1)} \]

### Step 4: New Conditions After Adding 2 Moles of B
When 2 moles of B are added, the new composition is:
- 1 mole of A
- 3 moles of B (1 original + 2 added)

Total moles = 1 + 3 = 4 moles.

The new mole fractions are:
- Mole fraction of A, \( x'_A = \frac{1}{4} \)
- Mole fraction of B, \( x'_B = \frac{3}{4} \)

### Step 5: New Vapor Pressure
The new vapor pressure \( P'_t \) is given to be 550 mm Hg (an increase of 50 mm Hg from the initial 500 mm Hg):
\[ P'_t = P^0_A \cdot x'_A + P^0_B \cdot x'_B \]
Substituting the new values:
\[ 550 = P^0_A \cdot \frac{1}{4} + P^0_B \cdot \frac{3}{4} \]
Multiplying through by 4:
\[ 2200 = P^0_A + 3P^0_B \quad \text{(Equation 2)} \]

### Step 6: Solve the Equations
Now we have two equations:
1. \( P^0_A + P^0_B = 1000 \)
2. \( P^0_A + 3P^0_B = 2200 \)

We can solve these equations simultaneously. From Equation 1, we can express \( P^0_A \):
\[ P^0_A = 1000 - P^0_B \]
Substituting this into Equation 2:
\[ 1000 - P^0_B + 3P^0_B = 2200 \]
\[ 1000 + 2P^0_B = 2200 \]
\[ 2P^0_B = 1200 \]
\[ P^0_B = 600 \text{ mm Hg} \]

Now substitute \( P^0_B \) back into Equation 1 to find \( P^0_A \):
\[ P^0_A + 600 = 1000 \]
\[ P^0_A = 400 \text{ mm Hg} \]

### Final Answer
The vapor pressures of pure A and B are:
- \( P^0_A = 400 \text{ mm Hg} \)
- \( P^0_B = 600 \text{ mm Hg} \)