At 300 K, the vapour pressure of an ideal solution containing 1 mole of A and 3 moles of B is 500 mm Hg. At the same temperature, 2 moles of B are added to this solution. The vapour pressure of solution increases by 10% of the original vapour pressure. Correct statements about the vapour pressure are

Options

A. Vapour pressure of A in the pure state is 50 mm Hg
B. Vapour pressure of B in the pure state is 650 mm Hg
C. Ratio of final pressure to the initial vapour pressure is 1:0.5
D. Ratio of vapour pressure of pure B to the vapour pressure of pure A is `13:1`

Answer

To solve the problem step-by-step, we will use Raoult's Law and the information provided in the question.

### Step 1: Determine the initial vapor pressure of the solution
Given:
- Moles of A = 1
- Moles of B = 3
- Total moles = 1 + 3 = 4

Using Raoult's Law:
\[ P = P^0_A \cdot X_A + P^0_B \cdot X_B \]
Where:
- \(P\) = vapor pressure of the solution = 500 mm Hg
- \(X_A\) = mole fraction of A = \(\frac{1}{4} = 0.25\)
- \(X_B\) = mole fraction of B = \(\frac{3}{4} = 0.75\)

The equation becomes:
\[ 500 = P^0_A \cdot 0.25 + P^0_B \cdot 0.75 \quad (1) \]

### Step 2: Determine the new vapor pressure after adding B
When 2 moles of B are added:
- New moles of B = 3 + 2 = 5
- Total moles = 1 + 5 = 6

The new mole fractions are:
- \(X_A = \frac{1}{6}\)
- \(X_B = \frac{5}{6}\)

The new vapor pressure \(P_1\) increases by 10%:
\[ P_1 = 500 + 0.1 \times 500 = 550 \text{ mm Hg} \]

Using Raoult's Law again:
\[ 550 = P^0_A \cdot \frac{1}{6} + P^0_B \cdot \frac{5}{6} \quad (2) \]

### Step 3: Solve the equations
We now have two equations:
1. \(500 = P^0_A \cdot 0.25 + P^0_B \cdot 0.75\)
2. \(550 = P^0_A \cdot \frac{1}{6} + P^0_B \cdot \frac{5}{6}\)

#### From Equation (1):
\[ 500 = 0.25P^0_A + 0.75P^0_B \]
Multiply through by 4:
\[ 2000 = P^0_A + 3P^0_B \quad (3) \]

#### From Equation (2):
\[ 550 = \frac{1}{6}P^0_A + \frac{5}{6}P^0_B \]
Multiply through by 6:
\[ 3300 = P^0_A + 5P^0_B \quad (4) \]

### Step 4: Solve equations (3) and (4)
Subtract equation (3) from equation (4):
\[ 3300 - 2000 = (P^0_A + 5P^0_B) - (P^0_A + 3P^0_B) \]
\[ 1300 = 2P^0_B \]
\[ P^0_B = 650 \text{ mm Hg} \]

Substituting \(P^0_B\) back into equation (3):
\[ 2000 = P^0_A + 3(650) \]
\[ 2000 = P^0_A + 1950 \]
\[ P^0_A = 50 \text{ mm Hg} \]

### Step 5: Conclusion
- The vapor pressure of pure A, \(P^0_A\), is 50 mm Hg.
- The vapor pressure of pure B, \(P^0_B\), is 650 mm Hg.
- The ratio of the final pressure to the initial pressure is \(\frac{550}{500} = 1.1\).
- The ratio of vapor pressures \(P^0_B : P^0_A = 650 : 50 = 13 : 1\).

### Correct Statements:
1. Vapor pressure of A in pure state is 50 mm Hg (Correct).
2. Vapor pressure of B in pure state is 650 mm Hg (Correct).
3. The ratio of vapor pressure of pure B to pure A is 13:1 (Correct).
4. The ratio between final pressure to initial pressure is not 1:0.5 (Incorrect).