Question

Ethylene glycol is used as an antifreeze agent. Calculate the amount of ethylene glycol to be added to $4kg$ of water to prevent it from freezing at $-6^{\circ }C.(K_f$ for $H_{2}O=1.85Kmol{e}^{-1}kg)$

Answer 1

Depression in freezing point $△T_f=\frac{1000\times K_f\times \omega }{W\times M}$
In the given case $W=4Kg=4000g$.
$K_f=1.85$
$△T_f=0-(-6^{\circ })=6^{\circ }C$
$M$ of ethylene glycol $C{H}_{2}OH-C{H}_{2}OH=62$
$\therefore$ amount of ethylene glycol required,
$\omega =\frac{M\times W\times △T_f}{1000\times K_f}$

$=\frac{62\times 4000\times 6}{1000\times 1.85}=804.32g$