Two liquids `A` and `B` form an ideal soluton. The vapour pressure of pure `A` and pure `B` are `66 mm Hg` and `88 mm Hg`, respectively. Calculate the composition of vapour `A` in the solution which is equilbrium and whose molar volume is `36%`.

Options

A. 0.43
B. 0.7
C. 0.3
D. 0.5

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Vapor pressure of pure A, \( P_A^0 = 66 \, \text{mm Hg} \)
- Vapor pressure of pure B, \( P_B^0 = 88 \, \text{mm Hg} \)
- Mole fraction of vapor A in the solution, \( X_{A,V} = 0.36 \)

### Step 2: Understand the relationship between vapor pressure and mole fraction
For an ideal solution, the partial vapor pressure of each component can be calculated using Raoult's Law:
\[ P_A = P_A^0 \cdot X_A \]
\[ P_B = P_B^0 \cdot X_B \]
where \( X_A \) and \( X_B \) are the mole fractions of A and B in the liquid phase, respectively.

### Step 3: Use the total vapor pressure equation
The total pressure \( P_{total} \) in the system can be expressed as:
\[ P_{total} = P_A + P_B = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]
Since \( X_B = 1 - X_A \) (because the sum of mole fractions in a binary solution is 1), we can rewrite the total pressure as:
\[ P_{total} = P_A^0 \cdot X_A + P_B^0 \cdot (1 - X_A) \]

### Step 4: Relate the mole fraction of vapor A to the total pressure
The mole fraction of vapor A in the total vapor can also be expressed as:
\[ X_{A,V} = \frac{P_A}{P_{total}} \]
Substituting \( P_A \) from Raoult's Law:
\[ X_{A,V} = \frac{P_A^0 \cdot X_A}{P_A^0 \cdot X_A + P_B^0 \cdot (1 - X_A)} \]

### Step 5: Substitute the known values and simplify
Given that \( X_{A,V} = 0.36 \):
\[ 0.36 = \frac{66 \cdot X_A}{66 \cdot X_A + 88 \cdot (1 - X_A)} \]

### Step 6: Solve for \( X_A \)
Cross-multiplying gives:
\[ 0.36 \cdot (66 \cdot X_A + 88 \cdot (1 - X_A)) = 66 \cdot X_A \]
Expanding and rearranging:
\[ 0.36 \cdot 88 - 0.36 \cdot 88 \cdot X_A = 66 \cdot X_A - 0.36 \cdot 66 \cdot X_A \]
Combining like terms:
\[ 0.36 \cdot 88 = (66 - 0.36 \cdot 66 + 0.36 \cdot 88) \cdot X_A \]

### Step 7: Calculate \( X_A \)
Now we can substitute the values:
\[ 0.36 \cdot 88 = (66 - 23.76) \cdot X_A \]
\[ 31.68 = 42.24 \cdot X_A \]
\[ X_A = \frac{31.68}{42.24} \approx 0.75 \]

### Step 8: Conclusion
Thus, the mole fraction of A in the liquid phase is approximately \( X_A = 0.75 \).