Question

The coolant usually contains a solution of antifreeze prepared by mixing equal volumes of ethylene glycol $C_2{H}_4(OH{)}_2$ and water. The density of ethylene glycol is $1.113gc{m}^{-3}$. Thus, the freezing point of the mixture is: $(K_{f}of{H}_{2}O=1.86Kkgmo{l}^{-1})$

• A. $-{7.8}^{\circ }C$
• B. $+{26}^{\circ }C$
• C. $-{5.6}^{\circ }C$
• D. $-{33.4}^{\circ }C$

Answer 1

The correct option is D $-{33.4}^{\circ }C$

$\text{Molar mass of ethylene glycol}\left(M_1\right)=62g$

$\text{Density of ethylene glycol is}1.113gc{m}^{-3}$
Thus, mass of ethylene glycol in 1 mL is $1.113g\left(w_1\right)$

Density of water is $1gc{m}^{-3}$
Mass of water in $1mL=1g\left(w_2\right)$

Depreesion in freezing point is given by:
$△T_f=K_f\times m$
where,
$△T_f$ is the depression in freezing point.
m is molality of the solution.

$\text{Molality (m)}=\frac{\frac{w_{\text{solute}}}{M_1}\times 1000}{w_{\text{solvent}}}$

$\Delta T_f=\frac{1000K_f{w}_1}{M_1{w}_2}=\frac{1000\times 1.86\times 1.113}{62\times 1}$

$={33.39}^{\circ }C$

$\therefore \text{Freezing point of the solution mixture}=0-{33.39}^{\circ }=-{33.39}^{\circ }C$