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Question: The freezing point of a diluted milk sample is found to be \( -{{0.3}^{\circ }}C \) , while it shoul...

The freezing point of a diluted milk sample is found to be −0.3∘C-{{0.3}^{\circ }}C , while it should have been −0.7∘C-{{0.7}^{\circ }}C for pure milk. How much water has been added to pure milk to make the diluted sample?
(A) 1 cup of water to 2 cups of pure milk
(B) 3 cups of water to 2 cups of pure milk
(C) 4 cups of water to 3 cups of pure milk
(D) 7 cups of water to 4 cups of pure milk

Explanation

Solution

Freezing point of the substance is defined as the temperature at which the vapour pressure of its liquid is equal to the vapour pressure of corresponding solid.

Formula used: ΔTf=Kf×m\Delta {{T}_{f}}={{K}_{f}}\times m
Where ΔTf=\Delta {{T}_{f}}= Depression in freezing point
m=m= Molality of solution
Kf={{K}_{f}}= Molal depression constant .

Complete step by step solution
Given that
Freezing point of pure milk =−0⋅7∘C=-0\cdot 7{}^\circ C
Tf1=−0⋅7∘C{{T}_{f1}}=-0\cdot 7{}^\circ C
ΔTf1=0⋅7∘C\Delta {{T}_{f1}}=0\cdot 7{}^\circ C
Freezing point of diluted milk
Tf2=−0⋅3∘C ΔTf2=0⋅3∘C \begin{aligned} & T{{f}_{2}}=-0\cdot 3{}^\circ C \\\ & \Delta {{T}_{f2}}=0\cdot 3{}^\circ C \\\ \end{aligned}
Depression in freezing point == Molal Depression constant molality of solution
ΔTf=Kf×m\Delta {{T}_{f}}={{K}_{f}}\times m
(ΔTf1)\left( \Delta {{T}_{f1}} \right) for pure milk is given as
=Kf×x+1000w1..........(1)={{K}_{f}}\times \dfrac{x+1000}{{{w}_{1}}}..........(1)
ΔTf2 for diluted milk=Kf×x×1000w2........(2)\Delta {{T}_{f2}}\text{ for diluted milk}={{K}_{f}}\times \dfrac{x\times 1000}{{{w}_{2}}}........(2)
Divide (1) by (2)
(ΔTf)1(ΔTf)2=Kf×x×1000w1Kf×x×1000w2\dfrac{{{\left( \Delta {{T}_{f}} \right)}_{1}}}{{{\left( \Delta {{T}_{f}} \right)}_{2}}}=\dfrac{{{K}_{f}}\times \dfrac{x\times 1000}{{{w}_{1}}}}{{{K}_{f}}\dfrac{\times x\times 1000}{{{w}_{2}}}}
0⋅70⋅3=Kf×x×1000×w2Kf×x×1000×w1\dfrac{0\cdot 7}{0\cdot 3}=\dfrac{{{K}_{f}}\times x\times 1000\times {{w}_{2}}}{{{K}_{f}}\times x\times 1000\times {{w}_{1}}}
7×1010×3=w2w1 73w1=w2 \begin{aligned} & \dfrac{7\times 10}{10\times 3}=\dfrac{{{w}_{2}}}{{{w}_{1}}} \\\ & \dfrac{7}{3}{{w}_{1}}={{w}_{2}} \\\ \end{aligned}
. W(water)=73w1−w1 =43w1 \begin{aligned} & W(water)=\dfrac{7}{3}{{w}_{1}}-{{w}_{1}} \\\ & =\dfrac{4}{3}{{w}_{1}} \\\ \end{aligned}
Thus we require 3 cup of water and 4 cup of milk.
Therefore the correct option is (B).

Additional information
Obelia scopic and cryoscopic methods effective when
Solutions are dilute. Solutions obeys Raoult's law.
The solute is non volatile.
There is no association and dissociation of solute molecules in the solution.
Solute doesn’t form a solid solution with solvent in frozen state, i.e., only solvent separates in solid state on freezing the solution.

Note
The knowledge of freezing point is must. Depression in freezing point is a colligative property. Overall we have four colligative properties. The student must be familiar with colligative properties and their effects. Depression in freezing point is also called cryoscopic constant. Depression of freezing point is directly proportional to the molal concentration of solution i.e it depends on the concentration. Molal depression constant may be defined as depression in the freezing point of a solvent when 1 mol of unknown ionisable solute is dissolved in 1 kg of the solvent.The depression in freezing point temperature is determined experimentally by Beckmann's method in the laboratory.