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 step-by-step, we will follow the reasoning laid out in the video transcript.

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

### Step 2: Calculate Mole Fractions in the Liquid Phase
Let’s denote:
- \( N_A = 1 \) (mole of A)
- \( N_B = 3 \) (moles of B)

Now, we can calculate the total number of moles in the liquid phase:
\[ N_{total} = N_A + N_B = 1 + 3 = 4 \]

Next, we calculate the mole fractions:
- Mole fraction of A in the liquid phase (\( x_A \)):
\[ x_A = \frac{N_A}{N_{total}} = \frac{1}{4} = 0.25 \]
- Mole fraction of B in the liquid phase (\( x_B \)):
\[ x_B = \frac{N_B}{N_{total}} = \frac{3}{4} = 0.75 \]

### Step 3: Relate Mole Fractions in Vapor Phase to Liquid Phase
The mole fraction of A in the vapor phase (\( Y_A \)) can be expressed using Raoult's Law:
\[ Y_A = \frac{P_A}{P_{total}} \]
Where \( P_A \) is the partial pressure of A, and \( P_{total} \) is the total pressure.

Using Raoult's Law, the partial pressure of A is given by:
\[ P_A = x_A \cdot P_A^0 \]
And the total pressure \( P_{total} \) is:
\[ P_{total} = P_A + P_B = P_A + P_B^0 \cdot x_B \]

### Step 4: Calculate Partial Pressures
From the given ratio of vapor pressures:
\[ P_A^0 : P_B^0 = 1 : 3 \implies P_A^0 = k \quad \text{and} \quad P_B^0 = 3k \]

Thus, we can express:
\[ P_A = x_A \cdot P_A^0 = 0.25 \cdot k \]
\[ P_B = x_B \cdot P_B^0 = 0.75 \cdot 3k = 2.25k \]

### Step 5: Calculate Total Pressure
Now, we calculate the total pressure:
\[ P_{total} = P_A + P_B = 0.25k + 2.25k = 2.5k \]

### Step 6: Calculate Mole Fraction of A in Vapor Phase
Now substituting back into the equation for \( Y_A \):
\[ Y_A = \frac{P_A}{P_{total}} = \frac{0.25k}{2.5k} = \frac{0.25}{2.5} = \frac{1}{10} = 0.1 \]

### Conclusion
The mole fraction of A in the vapor phase in equilibrium with the solution is:
\[ \boxed{0.1} \]