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?

We begin with the well-known relation for freezing-point depression,

$$\Delta T_f = iK_f m,$$

where $$\Delta T_f$$ is the lowering in freezing point, $$i$$ is the van’t Hoff factor, $$K_f$$ is the cryoscopic constant, and $$m$$ is the molality of the solute present. For a given solvent (water in this case) and a fixed solute (the dissolved milk solids), the factors $$i$$ and $$K_f$$ remain the same. Hence, for two samples prepared from the same original milk, the ratio of their freezing-point lowerings equals the ratio of the concentrations of milk solids in the two samples.

Pure milk shows a freezing point of

$$T_{\text{pure}} = 0 \,^ \circ\! \text{C} - 0.5 \,^ \circ\! \text{C} = -0.5 \,^ \circ\! \text{C},$$

so its depression is

$$\Delta T_{f,\text{pure}} = 0.5 \,^\circ\! \text{C}.$$

The diluted sample freezes at

$$T_{\text{diluted}} = 0 \,^\circ\! \text{C} - 0.2 \,^\circ\! \text{C} = -0.2 \,^\circ\! \text{C},$$

so its depression is

$$\Delta T_{f,\text{diluted}} = 0.2 \,^\circ\! \text{C}.$$

Using the proportionality of depression to concentration, we write

$$\frac{\Delta T_{f,\text{diluted}}}{\Delta T_{f,\text{pure}}} \;=\; \frac{m_{\text{diluted}}}{m_{\text{pure}}} \;=\; \frac{C_{\text{diluted}}}{C_{\text{pure}}},$$

where $$C$$ denotes the concentration of milk solids (proportional to the volume of pure milk present per unit total volume).

Substituting the numerical values,

$$\frac{0.2}{0.5} = \frac{C_{\text{diluted}}}{C_{\text{pure}}} \quad\Longrightarrow\quad \frac{C_{\text{diluted}}}{C_{\text{pure}}} = 0.4.$$

Thus the diluted milk contains only 40 % of the concentration of milk solids that pure milk does.

Let us suppose we started with $$x$$ cups of pure milk and then added $$y$$ cups of water. The amount of milk solids remains proportional to $$x$$, while the total volume becomes $$x + y$$. Therefore, the concentration of milk solids in the mixture is

$$C_{\text{diluted}} = \frac{x}{x + y}.$$

Because the concentration of pure milk itself is, by definition, 1 (or 100 %), we have

$$\frac{x}{x + y} = 0.4.$$

Now we solve this equation step by step:

$$x = 0.4(x + y)$$

$$x = 0.4x + 0.4y$$

$$x - 0.4x = 0.4y$$

$$0.6x = 0.4y$$

$$y = \frac{0.6}{0.4}x$$

$$y = 1.5x.$$

This tells us that the volume of water added is 1.5 times the volume of the original milk. Converting 1.5 : 1 into whole-number cups, we multiply by 2 to avoid fractions:

$$\text{Water} : \text{Milk} \;=\; 1.5 : 1 \;=\; \frac{3}{2} : 1 \;=\; 3 : 2.$$

So, 3 cups of water have been mixed with 2 cups of pure milk.

Hence, the correct answer is Option B.