Solutions and Colligative Properties 2 Question 11
11. A solution contain $62 \mathrm{~g}$ of ethylene glycol in $250 \mathrm{~g}$ of water is cooled upto $-10^{\circ} \mathrm{C}$. If $K_{f}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, then amount of water (in $\mathrm{g}$ ) separated as ice is
(2019 Main, 9 Jan II)
(a) 32
(b) 48
(c) 64
(d) 16
Show Answer
Solution:
- Considering the expression of the depression in freezing point of a solution,
$$ \begin{aligned} \Delta T_{f} & =K_{f} \times m \times i \ T_{f}^{\circ}-T_{f} & =K_{f} \times \frac{w_{B} \times 1000}{M_{B} \times w_{A}(\text { ing })} \times i \end{aligned} $$
Here, $T_{f}^{\circ}=0^{\circ} \mathrm{C}, T_{f}=-10^{\circ} \mathrm{C}$
$w_{B}=$ mass of ethylene glycol $=62 \mathrm{~g}$
$M_{B}=$ molar mass of ethylene glycol
$$ =62 \mathrm{~g} \mathrm{~mol}^{-1} $$
$w_{A}=$ mass of water in $\mathrm{g}$ as liquid solvent,
$i=$ van’t-Hoff factor $=1$ (for ethylene glycol in water)
$K_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
On substituting in Eq. (i), we get
$$ \begin{aligned} 0-(-10) & =1.86 \times \frac{62 \times 1000}{62 \times w_{A}} \times 1 \ \Rightarrow \quad w_{A} & =\frac{1.86 \times 62 \times 1000}{10 \times 62}=186 \mathrm{~g} \end{aligned} $$
So, amount of water separated as ice (solid solvent)
$$ =250-w_{A}=(250-186) \mathrm{g}=64 \mathrm{~g} $$
