The depression of freezing point is an important colligative property, which is influenced by the number of solute particles in a solution. When a solute dissociates or ionizes in a solution, it increases the number of particles in the solution, thereby affecting colligative properties such as the depression of freezing point. The degree of dissociation ($$\alpha$$) gives us a measure of the extent to which a compound dissociates into its ions. For our case, MX₂ dissociates into one M²⁺ ion and two X^- ions.

Let's initially consider 1 mole of MX₂ is present. Since the degree of dissociation ($$\alpha$$) is 0.5, this means half of the MX₂ dissociates into its ions, and half remains undissociated.

For the dissociation reaction:

$$\text{MX}_{2} \rightarrow \text{M}^{2+} + 2\text{X}^{-}$$

If the initial amount of MX₂ is 1 mole, then:

• The amount of MX₂ that dissociates = $$\alpha = 0.5$$ moles
• The amount of M²⁺ formed = $$\alpha = 0.5$$ moles (since for every mole of MX₂ that dissociates, 1 mole of M²⁺ is formed)
• The amount of X^- formed = $$2\alpha = 2 \times 0.5 = 1$$ mole (since for every mole of MX₂ that dissociates, 2 moles of X^- are formed)

The Van't Hoff factor (i) quantifies the effect of solute particles on the colligative properties of a solution. It is defined as the ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved in the solvent.

$$i = \frac{\text{Total number of particles after dissociation}}{\text{Number of moles of solute originally dissolved}}$$

Considering the dissociation and the amounts calculated:

• Total number of particles after dissociation = (undissociated MX₂) + (M²⁺) + (X^-) = $$(1 - \alpha) + \alpha + 2\alpha$$
• Putting the value of $$\alpha = 0.5$$, we get:

  $$i = \frac{(1 - 0.5) + 0.5 + 2(0.5)}{1} = \frac{1 + 1}{1} = 2$$

Now, the depression in freezing point ($$\Delta T_f$$) is directly proportional to the molal concentration of the solute particles and Van't Hoff factor (i):

$$\Delta T_f = i \cdot K_f \cdot m$$

Where $$K_f$$ is the cryoscopic constant and $$m$$ is the molality of the solution.

Without ionic dissociation, the value of $$i$$ would have been 1 (since the solute would not have dissociated into multiple particles). However, due to the dissociation, the value of $$i$$ has increased to 2. 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, therefore, would be directly equal to the ratio of the Van't Hoff factors:

$$\frac{\text{Observed } \Delta T_f}{\text{In the absence of ionic dissociation}} = \frac{2 \cdot K_f \cdot m}{1 \cdot K_f \cdot m} = \frac{2}{1} = 2$$

This ratio shows that due to the ionic dissociation of MX₂ into M²⁺ and X^- ions with a degree of dissociation ($$\alpha$$) of 0.5, the observed depression in freezing point is twice the value it would have been in the absence of ionic dissociation.