The liquid 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 increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg).

Options

A. 400, 600
B. 500 , 500
C. 600 , 400
D. none of these

Answer

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

### Step 1: Write down the given information
- Vapor pressure of the solution (P_total) = 550 mm Hg
- Moles of A (n_A) = 1
- Moles of B (n_B) = 3
- Increase in vapor pressure after adding 1 mole of B = 10 mm Hg
- New vapor pressure of the solution = 550 mm Hg + 10 mm Hg = 560 mm Hg

### Step 2: Calculate the mole fractions
- Total moles in the initial solution = n_A + n_B = 1 + 3 = 4
- Mole fraction of A (X_A) = n_A / (n_A + n_B) = 1 / 4 = 0.25
- Mole fraction of B (X_B) = n_B / (n_A + n_B) = 3 / 4 = 0.75

### Step 3: Apply Raoult's Law for the first solution
According to Raoult's Law:
\[ P_{total} = P^0_A \cdot X_A + P^0_B \cdot X_B \]
Substituting the known values:
\[ 550 = P^0_A \cdot 0.25 + P^0_B \cdot 0.75 \]
Multiplying through by 4 to eliminate the fraction:
\[ 2200 = P^0_A + 3P^0_B \]
This is our **Equation 1**.

### Step 4: Calculate the mole fractions 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') = 1 / 5 = 0.20
- New mole fraction of B (X_B') = 4 / 5 = 0.80

### Step 5: Apply Raoult's Law for the new solution
The new total vapor pressure is 560 mm Hg:
\[ P_{total}' = P^0_A \cdot X_A' + P^0_B \cdot X_B' \]
Substituting the known values:
\[ 560 = P^0_A \cdot 0.20 + P^0_B \cdot 0.80 \]
Multiplying through by 5:
\[ 2800 = P^0_A + 4P^0_B \]
This is our **Equation 2**.

### Step 6: Solve the system of equations
Now we have two equations:
1. \( 2200 = P^0_A + 3P^0_B \) (Equation 1)
2. \( 2800 = P^0_A + 4P^0_B \) (Equation 2)

Subtract Equation 1 from Equation 2:
\[ (2800 - 2200) = (P^0_A + 4P^0_B) - (P^0_A + 3P^0_B) \]
\[ 600 = P^0_B \]

### Step 7: Substitute back to find \( P^0_A \)
Substituting \( P^0_B = 600 \) mm Hg into Equation 1:
\[ 2200 = P^0_A + 3(600) \]
\[ 2200 = P^0_A + 1800 \]
\[ P^0_A = 2200 - 1800 = 400 \text{ mm Hg} \]

### Final Answer:
- Vapor pressure of A in pure state \( P^0_A = 400 \) mm Hg
- Vapor pressure of B in pure state \( P^0_B = 600 \) mm Hg