An ideal equimolar binary liquid solution of A and B has a vapour pressure of `900 mm` of `Hg`. It is distilled at a constant temperature till `(2)/(3)` (i.e `66.67 %` ) of the original amount. If the mole fraction of A is `0.3` in residue and that of B in condensate `0.4` and also the vapour pressure of residue is `860 mm Hg` which of the following is correct ?

Options

A. Mole fraction of A is more than that of B in vapour phase, above residue as well as above condensate.
B. Vapour pressure of condensate is more than `900 mm Hg` at same temperature.
C. In vapour pressure of residue, vapours of A are more contributing.
D. If the solution would be distilled till `50%` of the original amount then mole fraction of B in liquid condensate would be more than that of A in same solution.

Explanation

To solve the problem step by step, we will analyze the given data and apply Raoult's Law to find the correct statement regarding the binary liquid solution of A and B.

### Step 1: Write down the given data
- Initial vapor pressure of the equimolar solution (P_total) = 900 mm Hg
- Mole fraction of A in the residue (x'_A) = 0.3
- Mole fraction of B in the condensate (x''_B) = 0.4
- Vapor pressure of the residue (P'_residue) = 860 mm Hg
- Amount distilled = 66.67% (or 2/3 of the original amount)

### Step 2: Determine the mole fraction of B in the residue
Since the mole fraction of A in the residue is given as 0.3, we can find the mole fraction of B in the residue:
\[ x'_B = 1 - x'_A = 1 - 0.3 = 0.7 \]

### Step 3: Apply Raoult's Law to the residue
According to Raoult's Law, the vapor pressure of the solution is the sum of the partial pressures of its components:
\[ P'_{residue} = x'_A \cdot P^0_A + x'_B \cdot P^0_B \]
Where \( P^0_A \) and \( P^0_B \) are the vapor pressures of pure components A and B, respectively.

Substituting the known values:
\[ 860 = 0.3 \cdot P^0_A + 0.7 \cdot P^0_B \] (Equation 1)

### Step 4: Determine the mole fraction of A in the condensate
Since the mole fraction of B in the condensate is given as 0.4, we can find the mole fraction of A in the condensate:
\[ x''_A = 1 - x''_B = 1 - 0.4 = 0.6 \]

### Step 5: Apply Raoult's Law to the condensate
For the condensate, we can write:
\[ P_{condensate} = x''_A \cdot P^0_A + x''_B \cdot P^0_B \]
The total vapor pressure of the initial solution is 900 mm Hg, and since the solution is distilled, the vapor pressure of the condensate will be higher than that of the residue.

Let’s denote the vapor pressure of the condensate as \( P_{condensate} \):
\[ P_{condensate} = 0.6 \cdot P^0_A + 0.4 \cdot P^0_B \] (Equation 2)

### Step 6: Solve for \( P^0_A \) and \( P^0_B \)
We can use the total vapor pressure of the original solution to find the pure component pressures. The original solution has a total vapor pressure of:
\[ P_{total} = x_A \cdot P^0_A + x_B \cdot P^0_B = 900 \text{ mm Hg} \]
Since the initial mole fractions of A and B are equal (0.5 each):
\[ 900 = 0.5 \cdot P^0_A + 0.5 \cdot P^0_B \] (Equation 3)

### Step 7: Solve the system of equations
Now we have a system of three equations (1, 2, and 3). We can solve these equations to find \( P^0_A \) and \( P^0_B \).

1. From Equation 3:
\[ P^0_A + P^0_B = 1800 \] (Multiply by 2)

2. Substitute \( P^0_B = 1800 - P^0_A \) into Equation 1:
\[ 860 = 0.3 \cdot P^0_A + 0.7 \cdot (1800 - P^0_A) \]
\[ 860 = 0.3 \cdot P^0_A + 1260 - 0.7 \cdot P^0_A \]
\[ 860 = 1260 - 0.4 \cdot P^0_A \]
\[ 0.4 \cdot P^0_A = 1260 - 860 = 400 \]
\[ P^0_A = 1000 \text{ mm Hg} \]

3. Substitute \( P^0_A \) back into Equation 3 to find \( P^0_B \):
\[ P^0_B = 1800 - 1000 = 800 \text{ mm Hg} \]

### Step 8: Calculate the vapor pressure of the condensate
Now substitute \( P^0_A \) and \( P^0_B \) into Equation 2:
\[ P_{condensate} = 0.6 \cdot 1000 + 0.4 \cdot 800 \]
\[ P_{condensate} = 600 + 320 = 920 \text{ mm Hg} \]

### Step 9: Analyze the statements
1. **Mole fraction of A is more than that of B in vapor phase above residue and condensate.** - This is incorrect as the mole fraction of A in the residue is 0.3 and in the condensate is 0.6.
2. **Vapor pressure of condensate is more than 900 mm Hg at the same temperature.** - This is correct since we found it to be 920 mm Hg.
3. **In vapor pressure of residue, vapors of A are more contributing.** - This is incorrect; the contribution is based on mole fractions.
4. **If the solution would be distilled 50 percent, the mole fraction of B in liquid condensate would be more than that of A in the same solution.** - This is also incorrect.

### Conclusion
The correct statement is that the vapor pressure of the condensate is more than 900 mm Hg at the same temperature.