Let $Xg$ of the compound is to be dissolved. Number of moles of compound, $n=\frac{\text{molar mass}}{\text{given mass}}=\frac{256}{W}$ Mass of Benzene $=75\,\mathrm{g}=0.075\,\mathrm{Kg}$ Molality $=\frac{\text{moles of solute }}{\text{mass of solvent in }\mathrm{kg}}$ $=\frac{W}{256\times 0.075}=\frac{W}{19.2}$ Freezing point depression $\Delta T_f=0.48\,\mathrm{K}$ Molar depression of freezing point constant $=5.12\,\mathrm{K}\,\mathrm{kg}/\mathrm{mol}$ It is defined as the total number of molecules of reactants. $\Rightarrow T_f=K_f 0.48$ $=5.12\,\mathrm{W}=1.8\,\mathrm{g}$ Hence, $1.8\,\mathrm{g}$ solute has been dissolved.