Question

Find out the amount of water (in g) separated as ice, if a solution containing $62 \, g$ ethylene glycol in $250 \, g$ water is cooled to $-10^{^\circ }C$ . $K_{f}$ for water is $1.86 \, K \, kg \, mol^{- 1}$ .

• A. $48$
• B. $64$
• C. $16$
• D. $32$

Answer 1

Answer: (B)

Molecular weight of ethylene glycol
$\left(C_{2} H_{6} O_{2}\right)=62$
Moles of ethylene glycol $=\frac{W t}{M . W t}$
$=\frac{62}{62}=1$
$ \, \Delta T_{f}=i\times k_{f}\times m$
$\therefore \Delta T_{f}=0-\left(- 10\right)=10$
$i=1$ for ethylene glycol
$\Delta T_{f}=1\times 86\times \frac{1}{\frac{250}{1000}}$
$\Delta T_{f}=1.86\times 4=7.44$
As $\Delta T_{f}=10$ it implies that some amount of water has frozen and this has led to a greater depression of freezing point.
$10=1.86\times \frac{\frac{62}{62}}{\frac{W t \, o f \, H_{2} O}{1000}}$
$W_{H_{2} O}=186 \, gm$
Hence, the weight of water that has frozen $\text{=} \left(\right. 250 - 186 \left.\right) g m$ .
$=64 \, gm$ .