Two liquids A and B form ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increase by 10 mm Hg. The vapour pressure of A and B in their pure state (in mm Hg) respectively are

Options

A. 400, 600
B. 500, 500
C. 800, 400
D. 300, 600

Answer

To solve the problem, we need to use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial vapor pressures of each component in the solution. The formula can be expressed as:

\[ P_{solution} = P_{A}^{0} \cdot X_A + P_{B}^{0} \cdot X_B \]

Where:
- \( P_{solution} \) is the vapor pressure of the solution,
- \( P_{A}^{0} \) and \( P_{B}^{0} \) are the vapor pressures of pure components A and B,
- \( X_A \) and \( X_B \) are the mole fractions of components A and B in the solution.

### Step 1: Calculate the mole fractions of A and B in the initial solution
Given:
- Moles of A = 1
- Moles of B = 3

Total moles = 1 + 3 = 4

Mole fraction of A, \( X_A = \frac{1}{4} \)

Mole fraction of B, \( X_B = \frac{3}{4} \)

### Step 2: Write the equation for the initial vapor pressure
Using Raoult's Law for the initial solution:

\[ P_{solution} = P_{A}^{0} \cdot X_A + P_{B}^{0} \cdot X_B \]

Substituting the known values:

\[ 550 = P_{A}^{0} \cdot \frac{1}{4} + P_{B}^{0} \cdot \frac{3}{4} \]

### Step 3: Calculate the new mole fractions after adding one mole of B
When one more mole of B is added, the new moles become:
- Moles of A = 1
- Moles of B = 4

Total moles = 1 + 4 = 5

Mole fraction of A, \( X_A' = \frac{1}{5} \)

Mole fraction of B, \( X_B' = \frac{4}{5} \)

### Step 4: Write the equation for the new vapor pressure
The new vapor pressure increases by 10 mm Hg, so:

\[ P_{new} = 550 + 10 = 560 \]

Using Raoult's Law for the new solution:

\[ 560 = P_{A}^{0} \cdot \frac{1}{5} + P_{B}^{0} \cdot \frac{4}{5} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( 550 = P_{A}^{0} \cdot \frac{1}{4} + P_{B}^{0} \cdot \frac{3}{4} \) (Equation 1)
2. \( 560 = P_{A}^{0} \cdot \frac{1}{5} + P_{B}^{0} \cdot \frac{4}{5} \) (Equation 2)

We can solve these equations simultaneously.

From Equation 1:
\[ 550 = \frac{P_{A}^{0}}{4} + \frac{3P_{B}^{0}}{4} \]
Multiplying through by 4:
\[ 2200 = P_{A}^{0} + 3P_{B}^{0} \] (Equation 3)

From Equation 2:
\[ 560 = \frac{P_{A}^{0}}{5} + \frac{4P_{B}^{0}}{5} \]
Multiplying through by 5:
\[ 2800 = P_{A}^{0} + 4P_{B}^{0} \] (Equation 4)

### Step 6: Solve for \( P_{A}^{0} \) and \( P_{B}^{0} \)
Now we have two equations (3 and 4):
1. \( P_{A}^{0} + 3P_{B}^{0} = 2200 \)
2. \( P_{A}^{0} + 4P_{B}^{0} = 2800 \)

Subtract Equation 3 from Equation 4:
\[ (P_{A}^{0} + 4P_{B}^{0}) - (P_{A}^{0} + 3P_{B}^{0}) = 2800 - 2200 \]
\[ P_{B}^{0} = 600 \]

Substituting \( P_{B}^{0} \) back into Equation 3:
\[ P_{A}^{0} + 3(600) = 2200 \]
\[ P_{A}^{0} + 1800 = 2200 \]
\[ P_{A}^{0} = 400 \]

### Final Answer
The vapor pressures of A and B in their pure states are:
- \( P_{A}^{0} = 400 \) mm Hg
- \( P_{B}^{0} = 600 \) mm Hg