A `0.1` molal aqueous solution of a weak acid is `30%` ionized. If `K_(f)` for water is `1.86^(@)C//m`, the freezing point of the solution will be.

Options

A. `-0.24^(@)C`
B. `-0.18^(@)C`
C. `-0.54^(@)C`
D. `-0.36^(@)C`

Explanation

To find the freezing point of a 0.1 molal aqueous solution of a weak acid that is 30% ionized, we can follow these steps:

### Step 1: Determine the degree of ionization
Given that the weak acid is 30% ionized, we can express this as:
\[ \alpha = 0.3 \]

### Step 2: Calculate the van 't Hoff factor (i)
For a weak acid that ionizes into \( H^+ \) and \( A^- \), the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
Initially, we have 1 mol of HA. After ionization:
- Moles of HA remaining = \( 1 - \alpha = 1 - 0.3 = 0.7 \)
- Moles of \( H^+ \) formed = \( \alpha = 0.3 \)
- Moles of \( A^- \) formed = \( \alpha = 0.3 \)

Total moles after dissociation:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 - 0.3 + 0.3 + 0.3 = 1 + 0.3 = 1.3 \]

The van 't Hoff factor \( i \) is given by:
\[ i = \frac{\text{Total moles}}{\text{Initial moles}} = \frac{1.3}{1} = 1.3 \]

### Step 3: Use the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( K_f \) for water = 1.86 °C/m
- \( m \) (molality) = 0.1 mol/kg

Substituting the values:
\[ \Delta T_f = 1.3 \cdot 1.86 \cdot 0.1 \]
Calculating this gives:
\[ \Delta T_f = 1.3 \cdot 1.86 \cdot 0.1 = 0.2418 \text{ °C} \approx 0.241 \text{ °C} \]

### Step 4: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution is:
\[ T_f = T_{f0} - \Delta T_f = 0 - 0.241 = -0.241 \text{ °C} \]

### Final Answer
The freezing point of the solution will be approximately:
\[ \text{Freezing Point} = -0.241 \text{ °C} \]