Vapour pressure of pure `A(p_(A)^(@))=100` mm Hg
Vapour pressure of pure `B(p_(B)^(@))= 150` mm Hg
2 mol of liquid A and 3 mol of liquid B are mixed to form an ideal solution. The vapour pressure of solution will be:

Options

A. `a.) 185 mm`
B. `b.) 130 mm`
C. `c.) 148 mm`
D. `d.) 145 mm`

Explanation

To find the vapor pressure of the solution formed by mixing 2 moles of liquid A and 3 moles of liquid B, we can use Raoult's Law. Here are the steps to solve the problem:

### Step 1: Identify the given data
- Vapor pressure of pure A, \( P^0_A = 100 \, \text{mm Hg} \)
- Vapor pressure of pure B, \( P^0_B = 150 \, \text{mm Hg} \)
- Moles of A, \( n_A = 2 \)
- Moles of B, \( n_B = 3 \)

### Step 2: Calculate the total moles in the solution
Total moles \( n_{total} = n_A + n_B = 2 + 3 = 5 \)

### Step 3: Calculate the mole fractions of A and B
- Mole fraction of A, \( X_A = \frac{n_A}{n_{total}} = \frac{2}{5} \)
- Mole fraction of B, \( X_B = \frac{n_B}{n_{total}} = \frac{3}{5} \)

### Step 4: Apply Raoult's Law
According to Raoult's Law, the total vapor pressure of the solution \( P_{total} \) is given by:
\[ P_{total} = P^0_A \cdot X_A + P^0_B \cdot X_B \]

### Step 5: Substitute the values into the equation
Substituting the known values:
\[ P_{total} = (100 \, \text{mm Hg} \cdot \frac{2}{5}) + (150 \, \text{mm Hg} \cdot \frac{3}{5}) \]

### Step 6: Calculate the individual contributions
Calculating the contributions:
- Contribution from A:
\[ 100 \cdot \frac{2}{5} = 40 \, \text{mm Hg} \]
- Contribution from B:
\[ 150 \cdot \frac{3}{5} = 90 \, \text{mm Hg} \]

### Step 7: Find the total vapor pressure
Now, add the contributions from A and B:
\[ P_{total} = 40 \, \text{mm Hg} + 90 \, \text{mm Hg} = 130 \, \text{mm Hg} \]

### Final Answer
The vapor pressure of the solution will be \( P_{total} = 130 \, \text{mm Hg} \).

---