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. `0.48^@C`
B. `-0.48^@C`
C. `-0.36^@C`
D. `-0.26^@C`

Answer

To solve the problem, we need to find the freezing point of a 0.2 molal aqueous solution of a weak acid \( HX \) with a degree of ionization of \( 0.3 \). We will use the formula for depression in freezing point, which is given by:

\[ \Delta T_f = K_f \cdot m \cdot i \]

where:
- \( \Delta T_f \) = depression in freezing point
- \( K_f \) = freezing point depression constant (for water, \( K_f = 1.85 \, \text{°C kg/mol} \))
- \( m \) = molality of the solution
- \( i \) = van 't Hoff factor

### Step 1: Calculate the van 't Hoff factor \( i \)

The van 't Hoff factor \( i \) for a weak acid that ionizes can be calculated using the formula:

\[ i = 1 + \alpha (n - 1) \]

where:
- \( \alpha \) = degree of ionization
- \( n \) = number of ions produced from one formula unit of the solute.

For the weak acid \( HX \), it ionizes into \( H^+ \) and \( X^- \), producing 2 ions. Thus, \( n = 2 \).

Given \( \alpha = 0.3 \):

\[ i = 1 + 0.3 \times (2 - 1) = 1 + 0.3 = 1.3 \]

### Step 2: Calculate the depression in freezing point \( \Delta T_f \)

Now we can substitute the values into the depression in freezing point formula:

\[ \Delta T_f = K_f \cdot m \cdot i \]

Substituting \( K_f = 1.85 \, \text{°C kg/mol} \), \( m = 0.2 \, \text{mol/kg} \), and \( i = 1.3 \):

\[ \Delta T_f = 1.85 \cdot 0.2 \cdot 1.3 \]

Calculating this:

\[ \Delta T_f = 1.85 \cdot 0.2 = 0.37 \]
\[ \Delta T_f = 0.37 \cdot 1.3 = 0.481 \, \text{°C} \approx 0.48 \, \text{°C} \]

### Step 3: Calculate the freezing point of the solution

The freezing point of the solution can be calculated as follows:

\[ T_f = T_f^{\text{solvent}} - \Delta T_f \]

For water, the freezing point \( T_f^{\text{solvent}} = 0 \, \text{°C} \):

\[ T_f = 0 - 0.48 = -0.48 \, \text{°C} \]

### Final Answer

The freezing point of the solution will be nearest to \( -0.48 \, \text{°C} \).