Solutions and Colligative Properties 2 Question 9
9. The freezing point of a diluted milk sample is found to be $-0.2^{\circ} \mathrm{C}$, while it should have been $-0.5^{\circ} \mathrm{C}$ for pure milk. How much water has been added to pure milk to make the diluted sample?
(2019 Main, 11 Jan I)
(a) 2 cups of water to 3 cups of pure milk
(b) 1 cup of water to 3 cups of pure milk
(c) 3 cups of water to 2 cups of pure milk
(d) 1 cup of water to 2 cups of pure milk
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Solution:
- We know that,
Depression in freezing points $\left(\Delta T_{f}\right)$
$T^{\circ}{ }{f}-T{f}=K_{f} \times m \times i$
where, $\quad K_{f}=$ molal depression constant
$$ \begin{aligned} m & =\text { molality }=\frac{w_{\text {solute }} \times 1000}{M_{\text {solute }} \times w_{\text {solvent (in g) }}} \ i & =\text { van’t Hoff factor } \end{aligned} $$
For diluted milk
$$ \begin{gathered} \Delta T_{f_{1}}=K_{f} \times m_{1} \times i \ \Rightarrow 0-(0.2) \Rightarrow 0.2=K_{f} \times \frac{w_{\text {milk }} \times 1000}{M_{\text {milk }} \times w_{1}\left(\mathrm{H}_{2} \mathrm{O}\right)} \times 1 \end{gathered} $$
For pure milk
$$ \begin{aligned} \Delta T_{f_{2}} & =K_{f} \times m_{2} \times i \ \Rightarrow 0-(-0.5) & =0.5=K_{f} \times \frac{w_{\text {milk }} \times 1000}{M_{\text {milk }} \times w_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)} \times 1 \end{aligned} $$
So, $\frac{0.2}{0.5}=\frac{K_{f}}{K_{f}} \times \frac{w_{\text {milk }} \times 1000}{M_{\text {milk }} \times w_{1}\left(\mathrm{H}{2} \mathrm{O}\right)} \times \frac{M{\text {milk }} \times w_{2}\left(\mathrm{H}{2} \mathrm{O}\right)}{w{\text {milk }} \times 1000}=\frac{w_{2}\left(\mathrm{H}{2} \mathrm{O}\right)}{w{1}\left(\mathrm{H}_{2} \mathrm{O}\right)}$
$\Rightarrow \frac{w_{2}\left(\mathrm{H}{2} \mathrm{O}\right) \text { (in pure milk) }}{w{1}\left(\mathrm{H}_{2} \mathrm{O}\right) \text { (in diluted milk) }}=\frac{2}{5}$
i.e. 3 cups of water has to be added to 2 cups of pure milk.
