1. Use freezing point depression relation

For dilute solutions, $\Delta T_f \propto m$ where $m$ is the molality of solute particles.

For milk, assume the amount of solute originally present remains the same, and only water is added. Then the freezing point depression becomes inversely proportional to the amount of solvent.

2. Compare pure milk and diluted milk

For pure milk: $\Delta T_f^{\text{pure}} = 0.5^\circ C$

For diluted milk: $\Delta T_f^{\text{diluted}} = 0.2^\circ C$

Thus, $\frac{\Delta T_f^{\text{pure}}}{\Delta T_f^{\text{diluted}}} = \frac{0.5}{0.2} = 2.5$

Since $\Delta T_f \propto \frac{1}{\text{mass of solvent}}$ we get $\frac{\text{solvent in diluted sample}}{\text{solvent in pure milk}} = 2.5$

3. Find added water

Let the water originally present in pure milk be $W$. After dilution, total water becomes $2.5W$

So added water is $2.5W - W = 1.5W$

Thus, $\frac{\text{added water}}{\text{original water}} = 1.5 = \frac{3}{2}$

So the ratio is: $\text{water added} : \text{pure milk} = 3:2$

4. Match with options

This corresponds to:

C: 3 cups of water to 2 cups of pure milk

5. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They match.