If an ideal solution is made by mixing 2 moles of benzene `(p^(@)=266mm)` and 3 moles of another liquid `(p^(@)=236 mm)`. The total vapour pressure of the solution at the same temperature would be

Options

A. 502 mm
B. 248 mm
C. 600 mm
D. 250.6 mm

Answer

To find the total vapor pressure of the solution made by mixing benzene and another liquid, we can use Raoult's Law. Here's a step-by-step solution:

### Step 1: Identify the given data
- Moles of benzene (A) = 2 moles
- Vapor pressure of pure benzene (P₀A) = 266 mm
- Moles of the other liquid (B) = 3 moles
- Vapor pressure of pure liquid B (P₀B) = 236 mm

### Step 2: Calculate the total moles of the solution
Total moles = Moles of benzene + Moles of liquid B
Total moles = 2 moles + 3 moles = 5 moles

### Step 3: Calculate the mole fractions of benzene and liquid B
- Mole fraction of benzene (XA) = Moles of benzene / Total moles
XA = 2 moles / 5 moles = 0.4

- Mole fraction of liquid B (XB) = Moles of liquid B / Total moles
XB = 3 moles / 5 moles = 0.6

### Step 4: Apply Raoult's Law to find the total vapor pressure
According to Raoult's Law, the total vapor pressure (P) of the solution is given by:
\[ P = P₀A \cdot XA + P₀B \cdot XB \]

Substituting the values:
\[ P = (266 \, \text{mm} \cdot 0.4) + (236 \, \text{mm} \cdot 0.6) \]

### Step 5: Perform the calculations
- For benzene:
\[ 266 \, \text{mm} \cdot 0.4 = 106.4 \, \text{mm} \]

- For liquid B:
\[ 236 \, \text{mm} \cdot 0.6 = 141.6 \, \text{mm} \]

### Step 6: Add the contributions to find the total vapor pressure
\[ P = 106.4 \, \text{mm} + 141.6 \, \text{mm} = 248 \, \text{mm} \]

### Final Answer
The total vapor pressure of the solution is **248 mm**.

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