A solution of urea boils at `100.18^(@)C` at the atmospheric pressure. If `K_(f)` and `K_(b)` for water are `1.86` and `0.512K kg mol^(-1)` respectively, the above solution will freeze at,

Options

A. `-6.54^(@)C`
B. `-0.654^(@)C`
C. `6.54^(@)C`
D. `0.654^(@)C`

Answer

To find the freezing point of the urea solution, we can follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔT_b)

The boiling point elevation can be calculated using the formula:

\[ \Delta T_b = K_b \cdot m \]

Where:
- \( \Delta T_b \) is the elevation in boiling point,
- \( K_b \) is the ebullioscopic constant of the solvent (water in this case),
- \( m \) is the molality of the solution.

Given that the boiling point of the solution is \( 100.18^\circ C \) and the normal boiling point of water is \( 100^\circ C \), we can find \( \Delta T_b \):

\[ \Delta T_b = 100.18^\circ C - 100^\circ C = 0.18^\circ C \]

### Step 2: Calculate the molality (m)

Rearranging the formula for boiling point elevation to find molality:

\[ m = \frac{\Delta T_b}{K_b} \]

Substituting the values we have:

\[ m = \frac{0.18^\circ C}{0.512 \, \text{K kg mol}^{-1}} \approx 0.3516 \, \text{mol/kg} \]

### Step 3: Calculate the depression in freezing point (ΔT_f)

The depression in freezing point can be calculated using the formula:

\[ \Delta T_f = K_f \cdot m \]

Where:
- \( K_f \) is the cryoscopic constant of the solvent (water).

Substituting the values:

\[ \Delta T_f = 1.86 \, \text{K kg mol}^{-1} \cdot 0.3516 \, \text{mol/kg} \approx 0.654 \, \text{K} \]

### Step 4: Calculate the freezing point of the solution

The freezing point of the solution can be calculated as follows:

\[ \text{Freezing Point} = 0^\circ C - \Delta T_f \]

Substituting the value of \( \Delta T_f \):

\[ \text{Freezing Point} = 0^\circ C - 0.654^\circ C \approx -0.654^\circ C \]

### Final Answer

The freezing point of the urea solution is approximately \( -0.654^\circ C \).

---