`10` mole of liquid `'A'` and `10` mole of liquid `'B'` is mixed in a cylindrical vessel containing a piston arrangement. Initially a pressure of `2` atm is maintained on the solution. Now, the piston is raised slowly and isothermally. Assuming A and B to be completely miscible and forming an ideal solution.
`P_(A)^(@) = 0.6 ` atm and `P_(B)^(@) = 0.9` atm
The pressure below which the evaporation of liquid solution will start is :

Options

A. `0.6` atm
B. `0.8` atm
C. `0.75` atm
D. `0.9` atm

Answer

To solve the problem, we need to determine the pressure below which the evaporation of the liquid solution will start. Given that we have 10 moles of liquid A and 10 moles of liquid B, we can follow these steps:

### Step 1: Calculate the mole fractions of A and B
The total number of moles in the solution is:
\[ \text{Total moles} = n_A + n_B = 10 + 10 = 20 \text{ moles} \]

The mole fraction of A (\(X_A\)) is given by:
\[ X_A = \frac{n_A}{\text{Total moles}} = \frac{10}{20} = 0.5 \]

The mole fraction of B (\(X_B\)) is given by:
\[ X_B = \frac{n_B}{\text{Total moles}} = \frac{10}{20} = 0.5 \]

### Step 2: Use Raoult's Law to calculate the total vapor pressure
According to Raoult's Law, the partial vapor pressure of each component in an ideal solution is given by:
\[ P_A = X_A \cdot P_A^0 \]
\[ P_B = X_B \cdot P_B^0 \]

Where:
- \(P_A^0 = 0.6 \text{ atm}\) (vapor pressure of pure A)
- \(P_B^0 = 0.9 \text{ atm}\) (vapor pressure of pure B)

Calculating the partial pressures:
\[ P_A = 0.5 \cdot 0.6 = 0.3 \text{ atm} \]
\[ P_B = 0.5 \cdot 0.9 = 0.45 \text{ atm} \]

### Step 3: Calculate the total vapor pressure of the solution
The total vapor pressure (\(P_{total}\)) of the solution is the sum of the partial pressures:
\[ P_{total} = P_A + P_B = 0.3 + 0.45 = 0.75 \text{ atm} \]

### Conclusion
The pressure below which the evaporation of the liquid solution will start is:
\[ \boxed{0.75 \text{ atm}} \]