Solutions and Colligative Properties - Result Question 11-1
11. A solution contain $62$ g of ethylene glycol in $250$ 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 g) separated as ice is
(2019 Main, 9 Jan II)
(a) $32$
(b) $48$
(c) $64$
(d) $16$
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Answer:
Correct Answer: 11. (c)
Solution:
Considering the expression of the depression in freezing point of a solution,
$\Delta T_f =K_f \times m \times i $
$T_f^{\circ}-T_f =K_f \times \dfrac{w_B \times 1000}{M_B \times w_A(in \hspace{1mm} g)} \times i\quad$ ……(i)
Here, $T_f^{\circ}=0^{\circ} \mathrm{C}, T_f=-10^{\circ} \mathrm{C}$
$w_B =\text { mass of ethylene } \mathrm{glycol}=62 \mathrm{~g} $
$M_B =\text { molar mass of ethylene glycol } $
$ =62 \mathrm{~g} \mathrm{~mol}^{-1} $
$w_A =\text { mass of water in } \mathrm{g} \text { as liquid solvent }, $
$i =\text { van’t-Hoff factor }=1 \text { (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 \dfrac{62 \times 1000}{62 \times w_A} \times 1 \\ \Rightarrow \quad & w_A=\dfrac{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} $
BETA
