An aqueous solution of urea is found to boil at `100.52^(@)C`. Given `K_(b)` for water is 0.52 K kg `mol^(-1)`, the mole fraction of urea in the solution is

Options

A. 1
B. 0.5
C. 0.018
D. 0.25

Explanation

To solve the problem of finding the mole fraction of urea in an aqueous solution that boils at 100.52°C, we can follow these steps:

### Step 1: Calculate the boiling point elevation (ΔTb)
The boiling point elevation is given by the formula:
\[ \Delta T_b = T_b(\text{solution}) - T_b(\text{solvent}) \]
Where:
- \( T_b(\text{solution}) = 100.52°C \)
- \( T_b(\text{solvent}) = 100°C \) (boiling point of pure water)

Calculating ΔTb:
\[ \Delta T_b = 100.52°C - 100°C = 0.52°C \]

### Step 2: Use the boiling point elevation formula to find molality (m)
The relationship between boiling point elevation and molality is given by:
\[ \Delta T_b = K_b \cdot m \]
Where:
- \( K_b = 0.52 \, \text{K kg mol}^{-1} \)

Rearranging the formula to find molality:
\[ m = \frac{\Delta T_b}{K_b} \]

Substituting the values:
\[ m = \frac{0.52}{0.52} = 1 \, \text{mol/kg} \]

### Step 3: Determine the number of moles of urea
Since the molality is defined as the number of moles of solute per kilogram of solvent, we have:
- 1 mole of urea in 1 kg of water.

### Step 4: Calculate the number of moles of water
To find the mole fraction, we need the number of moles of water. The molecular weight of water (H2O) is approximately 18 g/mol.

Given that we have 1 kg (or 1000 g) of water:
\[ \text{Number of moles of water} = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.55 \, \text{mol} \]

### Step 5: Calculate the mole fraction of urea
The mole fraction (X) of urea is given by:
\[ X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} \]

Substituting the values:
\[ X_{\text{urea}} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \approx 0.0177 \]

### Final Answer
Thus, the mole fraction of urea in the solution is approximately:
\[ X_{\text{urea}} \approx 0.018 \]