In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be (`K_(f)` = 1.86 K kg `mol^(-1)`):

Options

A. 274.674 K
B. 271.60 K
C. 273 K
D. none of these

Answer

To solve the problem, we need to calculate the freezing point of a 0.5 molal KCl solution that is 50% dissociated. We'll use the formula for freezing point depression and the Van't Hoff factor.

### Step-by-Step Solution:

1. **Identify the given values:**
- Molality (m) = 0.5 mol/kg
- Degree of dissociation (α) = 50% = 0.5
- Freezing point depression constant (Kf) = 1.86 K kg/mol

2. **Calculate the Van't Hoff factor (i):**
- KCl dissociates into K⁺ and Cl⁻ ions.
- The dissociation can be represented as:
\[ KCl \rightarrow K^+ + Cl^- \]
- For a 50% dissociation, the total number of moles at equilibrium will be:
\[ \text{Total moles} = 1 - \alpha + \alpha + \alpha = 1 + \alpha \]
- Since α = 0.5:
\[ \text{Total moles} = 1 + 0.5 = 1.5 \]
- The Van't Hoff factor (i) is given by:
\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{1 + \alpha}{1} = 1 + 0.5 = 1.5 \]

3. **Calculate the freezing point depression (ΔTf):**
- The formula for freezing point depression is:
\[ \Delta T_f = K_f \times m \times i \]
- Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 0.5 \, \text{mol/kg} \times 1.5 \]
- Calculating:
\[ \Delta T_f = 1.86 \times 0.5 \times 1.5 = 1.395 \, \text{K} \]

4. **Calculate the freezing point of the solution (Tf):**
- The freezing point of pure water (T0f) is 0°C or 273 K.
- The freezing point of the solution can be calculated as:
\[ T_f = T_{0f} - \Delta T_f \]
- Substituting the values:
\[ T_f = 273 \, \text{K} - 1.395 \, \text{K} = 271.605 \, \text{K} \]
- Rounding to three significant figures, we get:
\[ T_f \approx 271.6 \, \text{K} \]

### Final Answer:
The freezing point of the solution is approximately **271.6 K**.