The freezing point of a diluted milk sample is found to be –0.2°C, while it should have been–0.5°C for pure milk. How much water has been added to pure milk to make the diluted sample ?

$$\begin{gathered} \text{ Freezing point of milk }=-0.5°\mathrm{C} \\ \to {\mathrm{\Delta T}}_{\mathrm{f}}=0.5°\mathrm{C} \\ \text{ Freezing point of diluted milk }=-0.2°\mathrm{C} \\ \to {\mathrm{\Delta T}}_{\mathrm{f}}=0.2°\mathrm{C} \\ \frac{{\left({\mathrm{\Delta T}}_{\mathrm{t}}\right)}_{\mathrm{i}}}{{\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)}_{\mathrm{ii}}}=\frac{0.5}{0.2}=\frac{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ \mathrm{Both} \mathrm{have} \mathrm{the} \mathrm{same} \mathrm{amount} \mathrm{of} \mathrm{solute}. \\ \mathrm{Let} \mathrm{y} \mathrm{mole} \mathrm{of} \mathrm{solute} \\ \frac{0.5}{0.2}=\frac{\mathrm{y}\times {\mathrm{w}}_2}{{\mathrm{w}}_1\times \mathrm{y}}\Rightarrow \frac{{\mathrm{w}}_2}{{\mathrm{w}}_1}=\frac{5}{2} \\ {\mathrm{w}}_2=\frac{5}{2}{\mathrm{w}}_1 \\ {\mathrm{W}}_{\mathrm{water} }=\frac{5}{2}{\mathrm{w}}_1-{\mathrm{w}}_1=\frac{3}{2}{\mathrm{w}}_1 \end{gathered}$$

Hence,  correct option is: (C) 3 cups of water to 2 cups of pure milk