6g of urea is dissolved in 90g of water. The relative lowering of vapour pressure is equal to

Options

A. 0.02
B. 0.05
C. 0.1
D. 0.04

Explanation

To solve the problem of finding the relative lowering of vapor pressure when 6 g of urea is dissolved in 90 g of water, we can follow these steps:

### Step 1: Identify the given data
- Mass of solute (urea, W2) = 6 g
- Mass of solvent (water, W1) = 90 g
- Molar mass of urea (NH2CO) = 60 g/mol
- Molar mass of water (H2O) = 18 g/mol

### Step 2: Calculate the number of moles of solute (urea)
To find the number of moles of urea, we use the formula:
\[ \text{Number of moles of solute (n2)} = \frac{\text{mass of solute (W2)}}{\text{molar mass of solute (M2)}} \]
Substituting the values:
\[ n2 = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]

### Step 3: Calculate the number of moles of solvent (water)
To find the number of moles of water, we use the formula:
\[ \text{Number of moles of solvent (n1)} = \frac{\text{mass of solvent (W1)}}{\text{molar mass of solvent (M1)}} \]
Substituting the values:
\[ n1 = \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ mol} \]

### Step 4: Calculate the relative lowering of vapor pressure
The relative lowering of vapor pressure (ΔP/P0) can be expressed as:
\[ \frac{\Delta P}{P_0} = \frac{n2}{n1} \]
Substituting the values we calculated:
\[ \frac{\Delta P}{P_0} = \frac{0.1 \text{ mol}}{5 \text{ mol}} = 0.02 \]

### Final Answer
The relative lowering of vapor pressure is 0.02.

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