Complete the following statements by selecting the correct alternative from the choices given :
An aqueous solution of urea freezes at `- 0.186^(@)C, K_(f)` for water = 1.86 K kg. `mol^(-1),K_(b)` for water = 0.512 `"K kg mol"^(-1)`. The boiling point of urea solution will be :

Options

A. 373.065 K
B. 375.186 K
C. 373.512 K
D. 373.0512 K

Explanation

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

### Step 1: Identify Given Values
- Freezing point of the urea solution, \( T_f = -0.186^\circ C \)
- Freezing point depression constant for water, \( K_f = 1.86 \, K \, kg \, mol^{-1} \)
- Boiling point elevation constant for water, \( K_b = 0.512 \, K \, kg \, mol^{-1} \)

### Step 2: Calculate the Depression in Freezing Point
The depression in freezing point (\( \Delta T_f \)) can be calculated as:
\[ \Delta T_f = T_f - T_f^{\text{pure}} = -0.186^\circ C - 0^\circ C = -0.186^\circ C \]

### Step 3: Calculate the Molality (m)
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \cdot m \]
Rearranging gives:
\[ m = \frac{\Delta T_f}{K_f} = \frac{-0.186}{1.86} \]
Calculating this gives:
\[ m = 0.1 \, mol/kg \]

### Step 4: Calculate the Elevation in Boiling Point
Using the molality calculated, we can find the elevation in boiling point (\( \Delta T_b \)):
\[ \Delta T_b = K_b \cdot m = 0.512 \cdot 0.1 = 0.0512 \, K \]

### Step 5: Calculate the Boiling Point of the Urea Solution
The boiling point of the solution can be found by adding the elevation to the boiling point of pure water:
\[ T_b = T_b^{\text{pure}} + \Delta T_b = 373 \, K + 0.0512 \, K = 373.0512 \, K \]

### Final Answer
Thus, the boiling point of the urea solution is:
\[ \boxed{373.0512 \, K} \]

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