Mixture of two liquids A and B is placed in cylinder containing piston. Piston is pulled out isotehrmally so that volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining the mole fraction of A in vapour is `0.4`. Given `P_(A)^(@) = 0.4` atm and `P_(B)^(@) = 1.2` atm at the experimental temperature. Calculate the total pressure at which the liquid has almost evaporated. (Assume ideal behaviour)

Explanation

To solve the problem, we need to calculate the total pressure at which the liquid has almost evaporated, given the mole fraction of component A in the vapor phase and the vapor pressures of both components A and B.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mole fraction of A in vapor phase, \( Y_A = 0.4 \)
- Vapor pressure of A, \( P_A^0 = 0.4 \) atm
- Vapor pressure of B, \( P_B^0 = 1.2 \) atm

2. **Express the Total Pressure**:
The total pressure \( P_T \) can be expressed using the mole fractions of A and B in the liquid phase:
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]
where \( X_A \) and \( X_B \) are the mole fractions of A and B in the liquid phase, respectively.

3. **Relationship Between Mole Fractions**:
Since \( X_A + X_B = 1 \), we can express \( X_B \) in terms of \( X_A \):
\[ X_B = 1 - X_A \]

4. **Substituting \( X_B \) into the Total Pressure Equation**:
Substitute \( X_B \) into the total pressure equation:
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot (1 - X_A) \]
This simplifies to:
\[ P_T = P_A^0 \cdot X_A + P_B^0 - P_B^0 \cdot X_A \]
\[ P_T = (P_A^0 - P_B^0) \cdot X_A + P_B^0 \]

5. **Using the Relationship Between Mole Fractions in Vapor and Liquid**:
We know that:
\[ Y_A = \frac{P_A^0 \cdot X_A}{P_T} \]
Rearranging gives:
\[ P_T = \frac{P_A^0 \cdot X_A}{Y_A} \]

6. **Substituting Known Values**:
Now we can substitute \( Y_A = 0.4 \), \( P_A^0 = 0.4 \) atm into the equation:
\[ P_T = \frac{0.4 \cdot X_A}{0.4} \]
This simplifies to:
\[ P_T = X_A \]

7. **Finding \( X_A \)**:
From the earlier expression for \( P_T \):
\[ P_T = (0.4 - 1.2) \cdot X_A + 1.2 \]
\[ P_T = -0.8 \cdot X_A + 1.2 \]
Setting the two expressions for \( P_T \) equal gives:
\[ X_A = -0.8 \cdot X_A + 1.2 \]
Rearranging:
\[ 1.8 \cdot X_A = 1.2 \]
\[ X_A = \frac{1.2}{1.8} = \frac{2}{3} \]

8. **Finding \( X_B \)**:
Now, we can find \( X_B \):
\[ X_B = 1 - X_A = 1 - \frac{2}{3} = \frac{1}{3} \]

9. **Calculating Total Pressure**:
Finally, substituting \( X_A \) and \( X_B \) back into the total pressure equation:
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]
\[ P_T = 0.4 \cdot \frac{2}{3} + 1.2 \cdot \frac{1}{3} \]
\[ P_T = \frac{0.8}{3} + \frac{1.2}{3} = \frac{2}{3} \text{ atm} \approx 0.66 \text{ atm} \]

### Final Answer:
The total pressure at which the liquid has almost evaporated is approximately **0.66 atm**.