The depression in freezing point ($\Delta T_f$) is related to the molar mass of the solute ($M_2$) by the formula:

$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$

Where: $K_f$ = molal freezing point depression constant = $1.2 \text{ K kg mol}^{-1}$ $w_2$ = mass of solute = $5 \text{ g}$ $\Delta T_f$ = depression in freezing point = $0.25 \text{ K}$ $w_1$ = mass of solvent = $100 \text{ g}$

Substituting the values into the equation: $M_2 = \frac{1.2 \times 5 \times 1000}{0.25 \times 100}$ $M_2 = \frac{6000}{25} = 240.0 \text{ g mol}^{-1}$

Therefore, the molar mass of the solute is $240.0 \text{ g mol}^{-1}$.