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 solve the problem, we will follow these steps:

### Step 1: Understand the given data
- We have a 0.1 molal aqueous solution of a weak acid.
- The weak acid is 30% ionized.
- The cryoscopic constant \( K_f \) for water is \( 1.86 \, \text{°C/m} \).

### Step 2: Calculate the degree of ionization
The degree of ionization \( \alpha \) is given as 30%, which can be expressed as:
\[ \alpha = \frac{30}{100} = 0.3 \]

### Step 3: Determine the van 't Hoff factor \( i \)
The van 't Hoff factor \( i \) for a weak acid that dissociates can be calculated using the formula:
\[ i = 1 + \alpha(n - 1) \]
where \( n \) is the number of particles the solute dissociates into. For a weak acid \( HA \) that dissociates into \( H^+ \) and \( A^- \), \( n = 2 \). Thus:
\[ i = 1 + 0.3(2 - 1) = 1 + 0.3 = 1.3 \]

### Step 4: Calculate the depression in freezing point \( \Delta T_f \)
Using the formula for freezing point depression:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( i = 1.3 \)
- \( K_f = 1.86 \, \text{°C/m} \)
- \( m = 0.1 \, \text{molal} \)

Substituting the values:
\[ \Delta T_f = 1.3 \cdot 1.86 \cdot 0.1 = 0.2418 \, \text{°C} \]

### Step 5: Calculate the freezing point of the solution
The freezing point of the solution \( T_f \) can be calculated using:
\[ T_f = T_{f, \text{solvent}} - \Delta T_f \]
Since the freezing point of pure water (the solvent) is \( 0 \, \text{°C} \):
\[ T_f = 0 - 0.2418 = -0.2418 \, \text{°C} \]
Rounding this, we can say:
\[ T_f \approx -0.24 \, \text{°C} \]

### Final Answer:
The freezing point of the solution will be approximately \( -0.24 \, \text{°C} \).

---