`MX_(2)` dissociates into `M^(2+)` and `X^(ө)` ion in an aqueous solution, with a degree of dissociation `(alpha)` of `0.5`. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in absence of ionic dissociation is

Explanation

To solve the problem, we need to find the ratio of the observed depression of freezing point of the solution to the value of the depression of freezing point in the absence of ionic dissociation.

### Step-by-step Solution:

1. **Understanding the Dissociation**:
The compound \( MX_2 \) dissociates in an aqueous solution as follows:
\[ MX_2 \rightarrow M^{2+} + 2X^{-} \]
This means that for every mole of \( MX_2 \) that dissociates, it produces 1 mole of \( M^{2+} \) and 2 moles of \( X^{-} \).

2. **Degree of Dissociation**:
Given that the degree of dissociation \( \alpha \) is 0.5, we can calculate the concentrations of the ions produced from the dissociation.

3. **Initial Concentration**:
Let the initial concentration of \( MX_2 \) be \( C \) mol/L. After dissociation:
- The concentration of \( M^{2+} \) will be \( C \alpha = C \times 0.5 = 0.5C \).
- The concentration of \( X^{-} \) will be \( 2C \alpha = 2C \times 0.5 = C \).

4. **Total Ion Concentration**:
The total concentration of ions in the solution (van 't Hoff factor \( i \)) can be calculated as:
\[ i = 1 + \alpha + 2\alpha = 1 + 0.5 + 2(0.5) = 1 + 0.5 + 1 = 2.5 \]

5. **Depression of Freezing Point**:
The depression of freezing point \( \Delta T_F \) is given by the formula:
\[ \Delta T_F = i \cdot K_F \cdot m \]
where \( K_F \) is the cryoscopic constant and \( m \) is the molality of the solution.

6. **Without Ionic Dissociation**:
In the absence of ionic dissociation, the van 't Hoff factor \( i \) is simply 1. Therefore, the depression of freezing point without dissociation is:
\[ \Delta T_F^{\text{no dissociation}} = 1 \cdot K_F \cdot m = K_F \cdot m \]

7. **Calculating the Ratio**:
Now we can find the ratio of the observed depression of freezing point to the depression in the absence of ionic dissociation:
\[ \text{Ratio} = \frac{\Delta T_F^{\text{dissociation}}}{\Delta T_F^{\text{no dissociation}}} = \frac{2.5 \cdot K_F \cdot m}{K_F \cdot m} = 2.5 \]

### Final Answer:
The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is **2.5**.