A 0.1 molal solution of an acid is `4.5%` ionized. Calculate freezing point. (molecular weight of the acid is 300 ). `K_(f)="1.86 K mol"^(-1)"kg".`

Options

A. `-0.199^(@)C`
B. `2.00^(@)C`
C. `0^(@)C`
D. `-0.269^(@)C`

Explanation

To calculate the freezing point of a 0.1 molal solution of an acid that is 4.5% ionized, we can follow these steps:

### Step 1: Calculate the degree of dissociation (α)
Given that the acid is 4.5% ionized, we can express the degree of dissociation (α) as:
\[ \alpha = \frac{4.5}{100} = 0.045 \]

### Step 2: Calculate the molality of the solution
The molality (m) of the solution is given as 0.1 molal.

### Step 3: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:
\[ \Delta T_f = K_f \cdot m \]
where \( K_f = 1.86 \, \text{K kg mol}^{-1} \) and \( m = 0.1 \, \text{molal} \).

Substituting the values:
\[ \Delta T_f = 1.86 \cdot 0.1 = 0.186 \, \text{°C} \]

### Step 4: Calculate the effective molality considering ionization
For a monobasic acid (HA), it dissociates as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
The effective molality (m') after considering the degree of dissociation is given by:
\[ m' = m(1 + \alpha) = 0.1(1 + 0.045) = 0.1 \cdot 1.045 = 0.1045 \, \text{molal} \]

### Step 5: Calculate the new depression in freezing point with effective molality
Now, we can calculate the new depression in freezing point using the effective molality:
\[ \Delta T_f' = K_f \cdot m' = 1.86 \cdot 0.1045 \]
Calculating this gives:
\[ \Delta T_f' = 0.194 \, \text{°C} \]

### Step 6: Determine the freezing point of the solution
The freezing point of pure solvent (water) is 0 °C. The freezing point of the solution will be:
\[ \text{Freezing point} = 0 - \Delta T_f' = 0 - 0.194 = -0.194 \, \text{°C} \]

### Final Answer
The freezing point of the solution is approximately:
\[ -0.194 \, \text{°C} \]