The molal elevation constant of water =`0.52 K m^(-1)`. The boiling point of `1.0 molal` aqueous `KCl` solution (assuming complete dissociation of `KCl`) should be

Options

A. `100.52^(@)C`
B. `101.04^(@)C`
C. `99.48^(@)C`
D. `98.96^(@)C`

Explanation

To solve the problem, we need to calculate the boiling point elevation of a 1.0 molal aqueous KCl solution, given the molal elevation constant of water (K_b = 0.52 K kg/mol) and the complete dissociation of KCl into its ions.

### Step-by-Step Solution:

1. **Identify the given data:**
- Molal elevation constant (K_b) = 0.52 K kg/mol
- Molality (m) of KCl solution = 1.0 molal
- Van't Hoff factor (i) for KCl = 2 (since KCl dissociates into K⁺ and Cl⁻ ions)

2. **Use the formula for boiling point elevation:**
The formula for boiling point elevation (ΔT_b) is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]

3. **Substitute the values into the formula:**
\[ \Delta T_b = 2 \cdot 0.52 \, \text{K kg/mol} \cdot 1.0 \, \text{molal} \]
\[ \Delta T_b = 2 \cdot 0.52 = 1.04 \, \text{K} \]

4. **Calculate the boiling point of the solution:**
The boiling point of pure water (T_b⁰) is 100 °C. Therefore, the boiling point of the KCl solution (T_b) can be calculated as:
\[ T_b = T_b⁰ + \Delta T_b \]
\[ T_b = 100 °C + 1.04 °C = 101.04 °C \]

5. **Final answer:**
The boiling point of the 1.0 molal aqueous KCl solution is **101.04 °C**.