If two substances A and B have `p_(A)^(@): p_(B)^(@) = 1:2` and have mole fraction in solution as 1:2 then mole fraction of A in vapour phase is

Options

A. `0.33`
B. `0.25`
C. `0.52`
D. `0.2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given:
- The ratio of the vapor pressures of substances A and B in their pure states:
\[ \frac{P^0_A}{P^0_B} = \frac{1}{2} \]
- The mole fraction of A and B in the solution:
\[ \frac{X_A}{X_B} = \frac{1}{2} \]

### Step 2: Express the Mole Fractions
From the mole fraction ratio, we can express the mole fractions as:
- Let \( X_A = 1 \) and \( X_B = 2 \).
- The total mole fraction \( X_{total} = X_A + X_B = 1 + 2 = 3 \).
- Therefore, the mole fractions can be calculated as:
\[ X_A = \frac{1}{3}, \quad X_B = \frac{2}{3} \]

### Step 3: Calculate the Partial Pressures
According to Raoult's Law, the partial pressures of A and B can be calculated as:
\[ P_A = X_A \cdot P^0_A \]
\[ P_B = X_B \cdot P^0_B \]

Substituting the values of \( P^0_A \) and \( P^0_B \):
\[ P_A = \frac{1}{3} \cdot P^0_A \]
\[ P_B = \frac{2}{3} \cdot P^0_B \]

### Step 4: Substitute the Vapor Pressure Ratio
From the ratio of vapor pressures:
\[ P^0_B = 2 \cdot P^0_A \]
Substituting \( P^0_B \) in the equation for \( P_B \):
\[ P_B = \frac{2}{3} \cdot (2 \cdot P^0_A) = \frac{4}{3} \cdot P^0_A \]

### Step 5: Calculate Total Pressure
The total pressure \( P_{total} \) is the sum of the partial pressures:
\[ P_{total} = P_A + P_B = \frac{1}{3} P^0_A + \frac{4}{3} P^0_A = \frac{5}{3} P^0_A \]

### Step 6: Calculate the Mole Fraction of A in the Vapor Phase
Using Dalton's Law, the mole fraction of A in the vapor phase \( Y_A \) can be expressed as:
\[ Y_A = \frac{P_A}{P_{total}} \]

Substituting the values:
\[ Y_A = \frac{\frac{1}{3} P^0_A}{\frac{5}{3} P^0_A} = \frac{1}{5} \]

### Step 7: Conclusion
Thus, the mole fraction of A in the vapor phase is:
\[ Y_A = 0.2 \]

### Final Answer
The mole fraction of A in the vapor phase is **0.2**.

---