The depression in freezing point ($\Delta T_f$) for an electrolyte is given by the equation: $\Delta T_f = i \times K_f \times m$.

1. Calculate Molality ($m$): Molar mass of $Na_2SO_4 = 2(23) + 32 + 4(16) = 142 \text{ g mol}^{-1}$. Moles of solute ($n_2$) = $\frac{5.00 \text{ g}}{142 \text{ g mol}^{-1}} \approx 0.0352 \text{ mol}$. Mass of solvent ($w_1$) = $45.0 \text{ g} = 0.045 \text{ kg}$. $m = \frac{0.0352 \text{ mol}}{0.045 \text{ kg}} \approx 0.782 \text{ m}$.

2. Calculate Van't Hoff Factor ($i$): Given $\Delta T_f = 3.82 ^\circ\text{C}$ (magnitude of change) and $K_f = 1.86 ^\circ\text{C m}^{-1}$. Rearranging the formula: $i = \frac{\Delta T_f}{K_f \times m}$.

• $i = \frac{3.82}{1.86 \times 0.782} = \frac{3.82}{1.4545} \approx 2.626$.

Rounding to two decimal places, $i \approx 2.63$.