Ratio of Observed to Theoretical Freezing Point Depression

To determine the ratio of the observed depression of freezing point to the depression of freezing point in the absence of ionic dissociation, we need to calculate the van 't Hoff factor ($i$).

Step 1: Write the dissociation reaction

The salt $\text{MX}_2$ dissociates in an aqueous solution as follows: $\text{MX}_2(\text{aq}) \rightleftharpoons \text{M}^{2+}(\text{aq}) + 2\text{X}^-(\text{aq})$

The number of ions produced per formula unit of $\text{MX}_2$ upon complete dissociation is: $n = 1 + 2 = 3$

Step 2: Calculate the van 't Hoff factor ($i$)

The relationship between the degree of dissociation ($\alpha$) and the van 't Hoff factor ($i$) is given by: $i = 1 + (n - 1)\alpha$

Given that the degree of dissociation $\alpha = 0.5$ and $n = 3$: $i = 1 + (3 - 1) \times 0.5$ $i = 1 + 2 \times 0.5 = 1 + 1 = 2$

Step 3: Determine the ratio of freezing point depressions

The observed freezing point depression ($\Delta T_{f, \text{observed}}$) is given by: $\Delta T_{f, \text{observed}} = i \cdot K_f \cdot m$

The freezing point depression in the absence of ionic dissociation ($\Delta T_{f, \text{theoretical}}$ where $i = 1$) is: $\Delta T_{f, \text{theoretical}} = K_f \cdot m$

Taking the ratio: $\text{Ratio} = \frac{\Delta T_{f, \text{observed}}}{\Delta T_{f, \text{theoretical}}} = \frac{i \cdot K_f \cdot m}{K_f \cdot m} = i = 2$

Thus, the ratio of the observed freezing point depression to the theoretical freezing point depression is 2.