A `0.2` molal aqueous solution of a weak acid `HX` is `20%` ionized. The freezing point of the solution is `(k_(f) = 1.86 K kg "mole"^(-1)` for water):

Options

A. `-0.45^(@)C`
B. `-0.9^(@)C`
C. `-0.31^(@)C`
D. `-0.53^(@)C`

Answer

To solve the problem, we need to calculate the freezing point depression of the given solution using the formula for freezing point depression. Here are the steps to arrive at the solution:

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

### Step 2: Calculate the effective molality
Since the acid is weak and only partially ionized, we need to account for the degree of ionization in our calculations. The effective molality (m_eff) can be calculated as:
\[ m_{\text{eff}} = m \times (1 + \alpha) \]
Where:
- \( m \) = molality of the solution
- \( \alpha \) = degree of ionization

Substituting the values:
\[ m_{\text{eff}} = 0.2 \times (1 + 0.2) = 0.2 \times 1.2 = 0.24 \, \text{mol/kg} \]

### Step 3: Use the freezing point depression formula
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = K_f \times m_{\text{eff}} \]

Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 0.24 \, \text{mol/kg} \]
\[ \Delta T_f = 0.4464 \, \text{K} \]

### Step 4: Calculate the new freezing point
The freezing point of pure water is 0°C. Since the freezing point depression is 0.4464 K, the new freezing point (\( T_f \)) of the solution will be:
\[ T_f = 0 - \Delta T_f \]
\[ T_f = 0 - 0.4464 = -0.4464 \, °C \]

### Step 5: Round the answer
Rounding the answer to two decimal places, we get:
\[ T_f \approx -0.45 \, °C \]

### Final Answer
The freezing point of the solution is approximately **-0.45°C**.

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