The freezing point of a diluted milk sample is found to be -0.2^ C,...

Freezing point of diluted milk $=-0.2^{{}^{\circ}} \mathrm{C}\Delta \mathrm{T}_{\mathrm{f}}^{*}=$0.2^{{}^{\circ}} $\mathrm{C}$Free ring point ol pure milk $=-0.5^{{}^{\circ}} \mathrm{C}\Delta \mathrm{T}_{\mathrm{f}}=$0.5^{{}^{\circ}} $\mathrm{C}\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}^{\prime}}=\frac{\mathrm{K}_{\mathrm{f}} \times m}{\mathrm{~K}_{\mathrm{f}} \times m^{\prime}} ; m=\frac{\text{ mole of solute }}{\text{ mass of solvent }(\mathrm{kg})}$ Moles of solute are same in both samples. \therefore $\frac{0.5}{0.2}=\frac{\mathrm{W}^{n}}{\mathrm{~W}}\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{5}{2} ; \mathrm{W}^{\prime}=\frac{5}{2} \mathrm{~W}$ 2 cups of pure milk is mixed with 3 cups of water to make 5 cups of diluted milk.