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 of finding the lowering in freezing point of a 0.5 molal aqueous solution of a weak acid (HX) that is 20% ionized, we will follow these steps:

### Step 1: Understand the given data
- Molality (m) of the solution = 0.5 mol/kg
- Degree of ionization (α) = 20% = 0.2
- Freezing point depression constant (Kf) for water = 1.86 K kg/mol

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) can be calculated using the degree of ionization (α). For a weak acid (HX), it dissociates as follows:
\[ \text{HX} \rightleftharpoons \text{H}^+ + \text{X}^- \]

The number of particles produced from one formula unit of HX is 2 (one H⁺ and one X⁻).

Using the formula:
\[ \alpha = \frac{i - 1}{n - 1} \]
where \( n \) is the number of particles produced (which is 2 for HX).

Substituting the values:
\[ 0.2 = \frac{i - 1}{2 - 1} \]
\[ 0.2 = i - 1 \]
\[ i = 1 + 0.2 = 1.2 \]

### Step 3: Calculate the freezing point depression (ΔTf)
The formula for the freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]

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

### Step 4: Perform the calculation
Calculating the above expression:
\[ \Delta T_f = 1.2 \cdot 1.86 \cdot 0.5 \]
\[ \Delta T_f = 1.2 \cdot 0.93 \]
\[ \Delta T_f = 1.116 \, \text{K} \]

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

### Conclusion
Thus, the lowering in freezing point of the solution is **1.12 K**.

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