Given: Mass of solute ($w_2$) = $1.00 \text{ g}$ Molar mass of solute ($M_2$) = $250 \text{ g/mol}$ Mass of solvent (benzene, $w_1$) = $51.2 \text{ g} = 0.0512 \text{ kg}$ Freezing point depression constant ($K_f$) = $5.12 \text{ K kg mol}^{-1}$

The depression in freezing point ($\Delta T_f$) is given by the formula: $\Delta T_f = K_f \times m$ Where $m$ is the molality of the solution. $m = \frac{w_2}{M_2 \times w_1 \text{ (in kg)}} = \frac{1.00}{250 \times 0.0512} = \frac{1}{12.8} \text{ m}$

Now, substitute the values into the equation: $\Delta T_f = 5.12 \times \frac{1}{12.8} = 0.4 \text{ K}$

Therefore, the freezing point of benzene will be lowered by $0.4 \text{ K}$.