Depression in `f.p.` of an aqueous urea solutio is `1.86^(@) .(K_(f)=1.86^(@) mol^(-1) kg)` Hence, mole`-` fraction of urea in this soluiton is `:

Options

A. `0.982`
B. `0.567`
C. `0.943`
D. `0.018`

Explanation

To find the mole fraction of urea in the solution, we can follow these steps:

### Step 1: Understand the relationship between freezing point depression and molality
The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \cdot m \]
where:
- \(\Delta T_f\) is the depression in freezing point,
- \(K_f\) is the cryoscopic constant,
- \(m\) is the molality of the solution.

### Step 2: Substitute the known values
From the problem, we know:
- \(\Delta T_f = 1.86^\circ C\)
- \(K_f = 1.86 \, \text{mol}^{-1} \, \text{kg}\)

Substituting these values into the equation:
\[ 1.86 = 1.86 \cdot m \]

### Step 3: Solve for molality (m)
To find \(m\), we can rearrange the equation:
\[ m = \frac{1.86}{1.86} = 1 \, \text{mol/kg} \]

### Step 4: Calculate the number of moles of urea
Since the molality is 1 mol/kg, this means there is 1 mole of urea in 1 kg of water.

### Step 5: Calculate the number of moles of water
The molar mass of water (H₂O) is calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- O: 16 g/mol × 1 = 16 g/mol
- Total: 2 + 16 = 18 g/mol

In 1 kg (1000 g) of water, the number of moles of water is:
\[ \text{Moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 6: Calculate the mole fraction of urea
The mole fraction of urea (\(X_{\text{urea}}\)) is given by the formula:
\[ X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} \]
Substituting the values we have:
\[ X_{\text{urea}} = \frac{1}{1 + 55.56} = \frac{1}{56.56} \approx 0.0177 \]

### Step 7: Final answer
Thus, the mole fraction of urea in the solution is approximately:
\[ X_{\text{urea}} \approx 0.018 \]