0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If `K_(f)` for water is `1.86K kg mol^(-1)`, the lowering in freezing point of the solution is

Options

A. `-1.12 K`
B. `0.56 K`
C. `1.12 K`
D. `-0.56 K`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- We have a 0.5 molal solution of a weak acid (HX).
- The weak acid is 20% ionized.
- The freezing point depression constant \( K_f \) for water is 1.86 K kg mol\(^{-1}\).

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor \( i \) is calculated using the degree of ionization (α).

1. **Initial concentration of HX**: 1 mol (for calculation purposes).
2. **Degree of ionization (α)**: 20% = 0.2.
3. **At equilibrium**:
- The concentration of \( H^+ \) ions = \( \alpha \) = 0.2 mol.
- The concentration of \( X^- \) ions = \( \alpha \) = 0.2 mol.
- The remaining concentration of un-ionized HX = \( 1 - \alpha = 1 - 0.2 = 0.8 \) mol.

4. **Total number of particles in solution**:
\[ \text{Total particles} = \text{un-ionized} + \text{ionized H}^+ + \text{ionized X}^- = 0.8 + 0.2 + 0.2 = 1.2 \text{ mol} \]

5. **Van 't Hoff factor (i)**:
\[ i = \frac{\text{Total particles}}{\text{Initial particles}} = \frac{1.2}{1} = 1.2 \]

### Step 3: Calculate the depression in freezing point (ΔTf)
Using the formula for freezing point depression:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( i = 1.2 \)
- \( K_f = 1.86 \, \text{K kg mol}^{-1} \)
- \( m = 0.5 \, \text{mol/kg} \)

Substituting the values:
\[ \Delta T_f = 1.2 \cdot 1.86 \cdot 0.5 \]

### Step 4: Perform the calculation
\[ \Delta T_f = 1.2 \cdot 1.86 \cdot 0.5 = 1.116 \, \text{K} \]

### Step 5: Final result
The lowering in freezing point of the solution is approximately:
\[ \Delta T_f \approx 1.12 \, \text{K} \]