Solutions and Colligative Properties - Result Question 9-1
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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Answer:
Correct Answer: 9. (c)
Solution:
We know that,
Depression in freezing points $\left(\Delta T_f\right)$
$ \begin{aligned} T^{\circ}{ }f-T_f=K_f & \times m \times i \\ \text { where, } \quad K_f & =\text { molal depression constant } \\ m & =\text { molality }=\dfrac{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 \dfrac{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
$\Delta T_{f_2}=K_f \times m_2 \times i $
$\Rightarrow 0-(-0.5)=0.5=K_f \times \dfrac{w_{\text {milk }} \times 1000}{M_{\text {milk }} \times w_2\left(\mathrm{H}_2 \mathrm{O}\right)} \times 1 $
$\text { So, } \dfrac{0.2}{0.5}=\dfrac{K_f}{K_f} \times \dfrac{w_{\text {milk }} \times 1000}{M_{\text {milk }} \times w_1\left(\mathrm{H}2 \mathrm{O}\right)} \times \dfrac{M{\text {milk }} \times w_2\left(\mathrm{H}2 \mathrm{O}\right)}{w{\text {milk }} \times 1000}=\dfrac{w_2\left(\mathrm{H}_2 \mathrm{O}\right)}{w_1\left(\mathrm{H}_2 \mathrm{O}\right)} $
$\Rightarrow \dfrac{\left.w_2\left(\mathrm{H}_2 \mathrm{O}\right) \text { (in pure milk }\right)}{w_1\left(\mathrm{H}_2 \mathrm{O}\right) \text { (in diluted milk) }}=\dfrac{2}{5}$
i.e. $3$ cups of water has to be added to $2$ cups of pure milk.
BETA
