An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:

Options

A. `0.0585 K`
B. `0.0512 K`
C. `1.864 K`
D. `0.0265K`

Explanation

To solve the problem, we need to calculate the elevation in boiling point (\( \Delta T_b \)) of the aqueous solution of NaCl, given that the freezing point depression (\( \Delta T_f \)) is \( -0.186^\circ C \).

### Step-by-Step Solution:

1. **Calculate the Depression in Freezing Point (\( \Delta T_f \))**:
\[ \Delta T_f = T_f^0 - T_f \]
Where:
- \( T_f^0 \) is the freezing point of pure water, which is \( 0^\circ C \).
- \( T_f \) is the freezing point of the NaCl solution, which is \( -0.186^\circ C \).

Substituting the values:
\[ \Delta T_f = 0 - (-0.186) = 0.186^\circ C \]

2. **Use the Freezing Point Depression Formula**:
The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \( K_f \) is the freezing point depression constant for water, given as \( 1.86 \, K \cdot kg \cdot mol^{-1} \).
- \( m \) is the molality of the solution.

Rearranging for molality (\( m \)):
\[ m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} \]

3. **Calculate the Molality (\( m \))**:
\[ m = \frac{0.186}{1.86} \approx 0.1 \, mol/kg \]

4. **Use the Boiling Point Elevation Formula**:
The formula for boiling point elevation is:
\[ \Delta T_b = K_b \cdot m \]
Where:
- \( K_b \) is the boiling point elevation constant for water, given as \( 0.512 \, K \cdot kg \cdot mol^{-1} \).

Substituting the values:
\[ \Delta T_b = 0.512 \cdot m = 0.512 \cdot 0.1 \]

5. **Calculate the Elevation in Boiling Point (\( \Delta T_b \))**:
\[ \Delta T_b = 0.0512 \, K \]

### Final Answer:
The elevation in boiling point of the NaCl solution is \( 0.0512 \, K \).

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