The vapour pressures of pure liquid A and pure liquid B at `20^(@)C` are 22 and 75 mm of Hg respectively. A solution is prepared by mixing equal moles of A and B. Assuming the solution to be ideal, calculate the vapour pressure of the solution.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Vapor pressure of pure liquid A (\(P^0_A\)) = 22 mm of Hg
- Vapor pressure of pure liquid B (\(P^0_B\)) = 75 mm of Hg
- Equal moles of A and B are mixed.

### Step 2: Calculate the mole fractions of A and B
Since equal moles of A and B are mixed, the total moles in the solution is:
- Moles of A = 1
- Moles of B = 1
- Total moles = 1 + 1 = 2

Now, we can calculate the mole fractions:
- Mole fraction of A (\(X_A\)) = \(\frac{\text{Moles of A}}{\text{Total moles}} = \frac{1}{2} = 0.5\)
- Mole fraction of B (\(X_B\)) = \(\frac{\text{Moles of B}}{\text{Total moles}} = \frac{1}{2} = 0.5\)

### Step 3: Calculate the partial vapor pressures of A and B
Using Raoult's Law, the partial vapor pressure of each component in the solution can be calculated as follows:
- Partial vapor pressure of A (\(P_A\)) = \(P^0_A \times X_A = 22 \, \text{mm Hg} \times 0.5 = 11 \, \text{mm Hg}\)
- Partial vapor pressure of B (\(P_B\)) = \(P^0_B \times X_B = 75 \, \text{mm Hg} \times 0.5 = 37.5 \, \text{mm Hg}\)

### Step 4: Calculate the total vapor pressure of the solution
The total vapor pressure of the solution (\(P\)) is the sum of the partial vapor pressures:
- Total vapor pressure (\(P\)) = \(P_A + P_B = 11 \, \text{mm Hg} + 37.5 \, \text{mm Hg} = 48.5 \, \text{mm Hg}\)

### Final Answer
The vapor pressure of the solution is **48.5 mm of Hg**.

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