Answer

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**Hint:**Boiling and freezing point of any substance is altered by the presence of impurities. Freezing point of substance will be lower than that of pure substance when impurities are added in the pure substance.

**Complete step by step answer:**

${K_f}$ is a cryoscopic constant. This constant relates the molality of a substance with depression in freezing point. This depends on the properties of solvent. This constant can be calculated with the following formula:

${K_f} = \dfrac{{RT_f^2M}}{{\Delta {H_{fus}}}}$

Where, $\Delta {H_{fus}}$ is enthalpy of fusion per mole of the solvent, $R$ is gas constant, ${T_f}$ is freezing point of pure solvent and $M$ is molar mass of the solvent

In this question we have given a sample of milk which is diluted and we have to find the amount of water in that sample of milk. We know that about $87\% $ of pure milk is water. Since composition of water is more in milk, we will take water as a solvent and the carbohydrates, proteins, fibers etc. collectively as solute. Since for milk these components or solute are the same, we can say that the number of moles is the same in pure milk and diluted milk (because to dilute milk we will add water in milk not the solute that is components of milk other than water). Let the moles of solute be $x$.

Solvent is water, and the mass of solvent will be different for pure milk and diluted milk since water is added to dilute pure milk. Let the mass (in kilogram) of water that is solvent for milk be ${m_1}$ and for diluted milk be ${m_2}$.

From this we can calculate molality of pure and diluted milk using the formula,

Molality$ = \dfrac{{{\text{number of moles of solute}}}}{{{\text{mass of solvent in kilogram}}}}$

Molality of pure milk$ = \dfrac{x}{{{m_1}}}$

Molality of diluted milk$ = \dfrac{x}{{{m_2}}}$

Freezing point for pure water is ${0^ \circ }C$. As stated above, in addition to impurities, freezing point of substance decreases. Therefore, when components of milk will be added to water the freezing point of water would be decreased. The freezing point of pure milk and diluted milk is given. From this we can calculate the change in freezing point. Freezing point of pure milk is $ - {0.5^ \circ }C$. Therefore $\Delta T$ for pure milk is:

$\Delta T$ for pure milk$ = 0 - 0.5 = {-0.5^ \circ }C$

$\Delta T$ for diluted milk$ = 0 - 0.2 = {-0.2^ \circ }C$

Now, we have calculated $\Delta T$ and molality for pure milk and diluted milk. So, we can apply the formula:

$\Delta T{\text{ = }}{{\text{K}}_f} \times {\text{molality}}$

In this formula ${K_f}$ is a cryoscopic constant for water, which will be the same for diluted milk and pure milk.

For pure milk,

${\text{0}}{\text{.5 = }}{{\text{K}}_f} \times \dfrac{x}{{{m_1}}}$ (equation a)

For diluted milk,

${\text{0}}{\text{.2 = }}{{\text{K}}_f} \times \dfrac{x}{{{m_2}}}$ (equation b)

Taking ratio of equation a and b,

$\dfrac{{0.5}}{{0.2}} = \dfrac{{x \times {m_2}}}{{x \times {m_1}}}$

Solving this we get,

${m_2} = \dfrac{5}{2}{m_1}$

${m_2}$ is mass of water in diluted milk and ${m_1}$ is mass of water in pure milk (stated above).

This means the mass of water in diluted milk is $\dfrac{5}{2}$ times the water in pure milk. Extra water added in pure milk can be calculated by subtracting mass of water in diluted milk from mass of water in pure milk (because mass of water in pure milk is the normal amount of water present in pure milk). Therefore,

Water added in pure milk$ = {m_2} - {m_1}$

${m_2}$ is calculated above in terms of ${m_1}$ and is equal to $\dfrac{5}{2}{m_1}$. So,

Water added in pure milk\[ = \dfrac{5}{2}{m_1} - {m_1} = \dfrac{3}{2}{m_1}\]

From this result we can conclude that $2$ parts of extra water are added in $3$ parts of milk.

**So, the correct answer is option A.**

**Note:**There is depression in freezing point after adding impurities because, on adding impurities vapour pressure becomes lower than that of pure solution. Boiling point of liquid increases on adding impurities because addition of impurities lower the number of molecules of liquid above the surface available to become vapourized during boiling.

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