The freezing point depression constant for water is $1.86^\circ\text{C m}^{-1}$. If $5.00 \text{ g Na}_2\text{SO}_4$ is dissolved in $45.0 \text{ g H}_2\text{O}$, the freezing point is changed by $-3.82^\circ\text{C}$. Calculate the van't Hoff factor for $\text{Na}_2\text{SO}_4$.

Given: Freezing point depression constant ($K_f$) = $1.86^\circ\text{C m}^{-1}$ Mass of solute ($\text{Na}_2\text{SO}_4$, $w_2$) = $5.00 \text{ g}$ Mass of solvent ($\text{H}_2\text{O}$, $w_1$) = $45.0 \text{ g} = 0.045 \text{ kg}$ Change in freezing point ($\Delta T_f$) = $3.82^\circ\text{C}$ (The magnitude of depression is $3.82^\circ\text{C}$) Molar mass of $\text{Na}_2\text{SO}_4$ ($M_2$) = $2(23) + 32 + 4(16) = 142 \text{ g/mol}$

The depression in freezing point is given by the formula: $\Delta T_f = i \times K_f \times m$ Where $m$ is the molality of the solution. $m = \frac{w_2}{M_2 \times w_1 \text{ (in kg)}} = \frac{5.00}{142 \times 0.045} = 0.7825 \text{ m}$

Now, substituting the values into the depression of freezing point equation: $3.82 = i \times 1.86 \times 0.7825$ $3.82 = i \times 1.45545$ $i = \frac{3.82}{1.45545} \approx 2.624 \approx 2.63$

Thus, the van't Hoff factor ($i$) for $\text{Na}_2\text{SO}_4$ is approximately $2.63$.