Two liquids A and B have `P_A^(@)" and P_B^(@)` in the ratio of 1 : 3 and the ratio of number of moles of A and B in liquid phese are 1 : 3 then mole fraction of 'A' in vapour phase in equilibrium with the solution is equal to :

Options

A. 0.1
B. 0.2
C. 0.5
D. 1

Answer

To solve the problem, we need to find the mole fraction of liquid A in the vapor phase at equilibrium with the solution. Here’s a step-by-step solution:

### Step 1: Understand the Given Ratios
We are given:
- The ratio of vapor pressures of A and B: \( P_A^0 : P_B^0 = 1 : 3 \)
- The ratio of moles of A and B in the liquid phase: \( n_A : n_B = 1 : 3 \)

### Step 2: Calculate the Mole Fractions in the Liquid Phase
Let’s denote:
- The number of moles of A, \( n_A = 1 \)
- The number of moles of B, \( n_B = 3 \)

The total number of moles in the liquid phase, \( n_{total} = n_A + n_B = 1 + 3 = 4 \).

Now, we can calculate the mole fractions:
- Mole fraction of A in the liquid phase, \( x_A = \frac{n_A}{n_{total}} = \frac{1}{4} \)
- Mole fraction of B in the liquid phase, \( x_B = \frac{n_B}{n_{total}} = \frac{3}{4} \)

### Step 3: Apply Raoult's Law
According to Raoult's Law, the partial vapor pressure of each component is given by:
- \( P_A = P_A^0 \cdot x_A \)
- \( P_B = P_B^0 \cdot x_B \)

### Step 4: Calculate the Total Vapor Pressure
The total vapor pressure \( P_{total} \) is the sum of the partial pressures:
\[ P_{total} = P_A + P_B = P_A^0 \cdot x_A + P_B^0 \cdot x_B \]

Substituting the values of \( P_A^0 \) and \( P_B^0 \) in terms of \( P_A^0 \):
- Since \( P_B^0 = 3P_A^0 \), we have:
\[ P_{total} = P_A^0 \cdot x_A + 3P_A^0 \cdot x_B \]
\[ = P_A^0 \cdot \left( x_A + 3x_B \right) \]

### Step 5: Substitute the Mole Fractions
Now substituting \( x_A = \frac{1}{4} \) and \( x_B = \frac{3}{4} \):
\[ P_{total} = P_A^0 \cdot \left( \frac{1}{4} + 3 \cdot \frac{3}{4} \right) \]
\[ = P_A^0 \cdot \left( \frac{1}{4} + \frac{9}{4} \right) \]
\[ = P_A^0 \cdot \left( \frac{10}{4} \right) = P_A^0 \cdot \frac{5}{2} \]

### Step 6: Calculate the Mole Fraction of A in the Vapor Phase
The mole fraction of A in the vapor phase \( y_A \) is given by:
\[ y_A = \frac{P_A}{P_{total}} \]
Substituting \( P_A = P_A^0 \cdot x_A \):
\[ y_A = \frac{P_A^0 \cdot x_A}{P_{total}} \]
\[ = \frac{P_A^0 \cdot \frac{1}{4}}{P_A^0 \cdot \frac{5}{2}} \]
\[ = \frac{\frac{1}{4}}{\frac{5}{2}} = \frac{1}{4} \cdot \frac{2}{5} = \frac{2}{20} = \frac{1}{10} \]

### Final Answer
The mole fraction of A in the vapor phase in equilibrium with the solution is:
\[ \boxed{\frac{1}{10}} \]