3g urea is dissolved in 45g of water. The relative lowering of vapour pressure is :

Options

A. 0.05
B. 0.04
C. 0.02
D. 0.01

Explanation

To solve the problem of finding the relative lowering of vapor pressure when 3g of urea is dissolved in 45g of water, we can follow these steps:

### Step 1: Calculate the number of moles of urea (B)
- **Formula**: Number of moles = Mass (g) / Molecular mass (g/mol)
- **Given**: Mass of urea = 3g; Molecular mass of urea = 60 g/mol
- **Calculation**:
\[ \text{Number of moles of urea} = \frac{3 \text{ g}}{60 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 2: Calculate the number of moles of water (A)
- **Given**: Mass of water = 45g; Molecular mass of water = 18 g/mol
- **Calculation**:
\[ \text{Number of moles of water} = \frac{45 \text{ g}}{18 \text{ g/mol}} = 2.5 \text{ moles} \]

### Step 3: Calculate the mole fraction of urea (XB)
- **Formula**: Mole fraction (XB) = Number of moles of B / (Number of moles of A + Number of moles of B)
- **Calculation**:
\[ \text{Mole fraction of urea} (X_B) = \frac{0.05}{0.05 + 2.5} = \frac{0.05}{2.55} \approx 0.0196 \]

### Step 4: Calculate the relative lowering of vapor pressure (ΔP/P0)
- **Relation**: ΔP/P0 = XB (mole fraction of solute)
- **Calculation**:
\[ \text{Relative lowering of vapor pressure} = 0.0196 \approx 0.02 \]

### Conclusion
The relative lowering of vapor pressure when 3g of urea is dissolved in 45g of water is approximately **0.02**.

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