In a `0.2` molal aqueous solution of a weak acid `HX` the degree of ionization is `0.3`. Taking `k_(f)` for water as `1.85` the freezing point of the solution will be nearest to-

Options

A. `-260^(@)C`
B. `+0.480^(@)C`
C. `-0.480^(@)C`
D. `-0.360^(@)C`

Explanation

To solve the problem step by step, we need to find the freezing point of the given solution using the provided data.

### Step 1: Understand the given data
We have:
- Molality (m) = 0.2 mol/kg
- Degree of ionization (α) = 0.3
- Freezing point depression constant (k_f) for water = 1.85 °C kg/mol

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) can be calculated using the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
where \( n \) is the number of particles the solute dissociates into. For the weak acid \( HX \), it dissociates into \( H^+ \) and \( X^- \), so \( n = 2 \).

Substituting the values:
\[ i = 1 + (2 - 1) \cdot 0.3 \]
\[ i = 1 + 1 \cdot 0.3 \]
\[ i = 1 + 0.3 = 1.3 \]

### Step 3: Calculate the freezing point depression (ΔTf)
The freezing point depression can be calculated using the formula:
\[ \Delta T_f = k_f \cdot m \cdot i \]
Substituting the values:
\[ \Delta T_f = 1.85 \cdot 0.2 \cdot 1.3 \]

Calculating:
\[ \Delta T_f = 1.85 \cdot 0.2 = 0.37 \]
\[ \Delta T_f = 0.37 \cdot 1.3 = 0.481 \]

### Step 4: Calculate the new freezing point (Tf)
The freezing point of the solution can be calculated using the formula:
\[ T_f = T_{f, \text{water}} - \Delta T_f \]
Given that the freezing point of pure water \( T_{f, \text{water}} = 0 \) °C:
\[ T_f = 0 - 0.481 \]
\[ T_f = -0.481 \text{ °C} \]

### Step 5: Conclusion
The freezing point of the solution is approximately -0.481 °C. Therefore, the nearest option is:
**Answer: -0.480 °C**

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