At `25^(@)C, 1 mol` of A having a vapour pressure of 100 torr and 1 mole of B having a vapour pressure of 300 torr were mixed. The vapour at equilibrium is removed, condensed and the condensate is heated back to `25^(@)C`. The vapour now formed are again removed, recondensed and analyzed. what is the mole fraction of A in this condensate?

Explanation

To solve the problem, we need to calculate the mole fraction of component A in the final condensate after two rounds of vaporization and condensation. Let's break it down step by step.

### Step 1: Calculate the mole fractions in the liquid phase
We have:
- Moles of A = 1
- Moles of B = 1

The total moles = Moles of A + Moles of B = 1 + 1 = 2.

Now, we can calculate the mole fractions:
- Mole fraction of A (XA) = Moles of A / Total moles = 1 / 2 = 0.5
- Mole fraction of B (XB) = Moles of B / Total moles = 1 / 2 = 0.5

### Step 2: Calculate the total vapor pressure (P_total)
Using Raoult's Law:
\[ P_{total} = P_{A0} \cdot X_A + P_{B0} \cdot X_B \]

Where:
- \( P_{A0} = 100 \) torr (vapor pressure of pure A)
- \( P_{B0} = 300 \) torr (vapor pressure of pure B)

Substituting the values:
\[ P_{total} = (100 \, \text{torr} \cdot 0.5) + (300 \, \text{torr} \cdot 0.5) \]
\[ P_{total} = 50 \, \text{torr} + 150 \, \text{torr} = 200 \, \text{torr} \]

### Step 3: Calculate the mole fraction of A in the vapor phase (Y_A)
Using the formula:
\[ Y_A = \frac{P_A}{P_{total}} \]
Where:
\[ P_A = P_{A0} \cdot X_A = 100 \, \text{torr} \cdot 0.5 = 50 \, \text{torr} \]

Now substituting the values:
\[ Y_A = \frac{50 \, \text{torr}}{200 \, \text{torr}} = \frac{1}{4} = 0.25 \]

### Step 4: Calculate the mole fraction of B in the vapor phase (Y_B)
Since \( Y_B = 1 - Y_A \):
\[ Y_B = 1 - 0.25 = 0.75 \]

### Step 5: First condensation
After the first condensation, the mole fractions in the liquid phase will be:
- Mole fraction of A in the liquid (X_A') = Y_A = 0.25
- Mole fraction of B in the liquid (X_B') = Y_B = 0.75

### Step 6: Calculate the new total vapor pressure (P_total')
Using Raoult's Law again:
\[ P_{total}' = P_{A0} \cdot X_A' + P_{B0} \cdot X_B' \]
Substituting the values:
\[ P_{total}' = (100 \, \text{torr} \cdot 0.25) + (300 \, \text{torr} \cdot 0.75) \]
\[ P_{total}' = 25 \, \text{torr} + 225 \, \text{torr} = 250 \, \text{torr} \]

### Step 7: Calculate the mole fraction of A in the vapor phase after first condensation (Y_A')
Using the formula:
\[ Y_A' = \frac{P_A'}{P_{total}'} \]
Where:
\[ P_A' = P_{A0} \cdot X_A' = 100 \, \text{torr} \cdot 0.25 = 25 \, \text{torr} \]

Now substituting the values:
\[ Y_A' = \frac{25 \, \text{torr}}{250 \, \text{torr}} = \frac{1}{10} = 0.1 \]

### Step 8: Second condensation
After the second condensation, the mole fraction of A in the liquid (X_A'') will be equal to Y_A':
- Mole fraction of A in the liquid (X_A'') = Y_A' = 0.1

### Final Answer
The mole fraction of A in the final condensate is:
\[ \text{Mole fraction of A in the final condensate} = 0.1 \]