Two liquids `A`and `B` have vapour pressures in the ratio of `p_(A)@,p_(B)@` = 1 : 2 at a certain temperature. Suppose we have an ideal solution of `A` and `B` in the mole fraction ratio `A:B = 1:2`. What would be the mole fraction of `A` in the vapour in equilibrium with the solution at a given temperature?

Options

A. 0.25
B. 0.2
C. 0.5
D. 0.33

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have two liquids A and B with vapor pressures in the ratio \( P_{A}^{0} : P_{B}^{0} = 1 : 2 \). This means:
\[ \frac{P_{A}^{0}}{P_{B}^{0}} = \frac{1}{2} \]
From this, we can express \( P_{B}^{0} \) in terms of \( P_{A}^{0} \):
\[ P_{B}^{0} = 2 P_{A}^{0} \]

### Step 2: Determine the mole fractions
The mole fraction of A and B in the solution is given as:
\[ X_{A} : X_{B} = 1 : 2 \]
This implies:
\[ X_{A} = \frac{1}{1+2} = \frac{1}{3}, \quad X_{B} = \frac{2}{1+2} = \frac{2}{3} \]

### Step 3: Calculate the partial pressures
Using Raoult's Law, the partial pressures of A and B can be expressed as:
\[ P_{A} = X_{A} \cdot P_{A}^{0} \]
\[ P_{B} = X_{B} \cdot P_{B}^{0} \]
Substituting for \( P_{B}^{0} \):
\[ P_{B} = X_{B} \cdot (2 P_{A}^{0}) = \frac{2}{3} \cdot (2 P_{A}^{0}) = \frac{4}{3} P_{A}^{0} \]

### Step 4: Calculate the total pressure
The total pressure \( P_{total} \) is the sum of the partial pressures:
\[ P_{total} = P_{A} + P_{B} = X_{A} \cdot P_{A}^{0} + \frac{4}{3} P_{A}^{0} \]
Substituting \( X_{A} = \frac{1}{3} \):
\[ P_{total} = \left(\frac{1}{3} P_{A}^{0}\right) + \left(\frac{4}{3} P_{A}^{0}\right) = \frac{5}{3} P_{A}^{0} \]

### Step 5: Calculate the mole fraction of A in the vapor phase
The mole fraction of A in the vapor phase \( Y_{A} \) can be calculated using the formula:
\[ Y_{A} = \frac{P_{A}}{P_{total}} \]
Substituting the expressions for \( P_{A} \) and \( P_{total} \):
\[ Y_{A} = \frac{X_{A} \cdot P_{A}^{0}}{P_{total}} = \frac{\left(\frac{1}{3} P_{A}^{0}\right)}{\left(\frac{5}{3} P_{A}^{0}\right)} \]
The \( P_{A}^{0} \) cancels out:
\[ Y_{A} = \frac{1}{5} \]

### Final Answer
Thus, the mole fraction of A in the vapor in equilibrium with the solution is:
\[ Y_{A} = 0.2 \]

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