A solution containing $62 \mathrm{g}$ ethylene glycol in $250 \mathrm{g}$ water is cooled to $-{10}^{\mathrm{o}}\mathrm{C}$ . If ${\mathrm{K}}_{\mathrm{f}}$ for water is $1.86 \mathrm{K}\mathrm{ }\mathrm{kg}\mathrm{ }{\mathrm{mol}}^{-1}$ , the amount of water (in $\mathrm{g}$ ) separated as ice is:

Molecular weight of ethylene glycol

$\left(C_2{H}_6{O}_2\right)=62$

Moles of ethylene glycol $=\frac{Wt}{M.Wt}$

$=\frac{62}{62}=1$

$\mathrm{ }∆T_f=i\times k_f\times m$

$\therefore ∆T_f=0-\left(-10\right)=10$

$i=1$ for ethylene glycol

$∆T_f=1\times 86\times \frac{1}{\frac{250}{1000}}$

$∆T_f=1.86\times 4=7.44$

As $∆T_f=10$ it implies that some amount of water has frozen and this has led to a greater depression of freezing point.

$10=1.86\times \frac{\frac{62}{62}}{\frac{Wt \mathrm{o}\mathrm{f} H_{2}O}{1000}}$

$W_{H_{2}O}=186 gm$

Hence, the weight of water that has frozen $\text{=}\left(250-186\right)gm$

$=64 gm$