# An antifreeze solution is prepared from 222.6 g of ethylene glycol

Question:

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL−1, then what shall be the molarity of the solution?

Solution:

Molar mass of ethylene glycol $\left[\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{OH})_{2}\right]=2 \times 12+6 \times 1+2 \times 16$

= 62 gmol−1

Number of moles of ethylene glycol $=\frac{222.6 \mathrm{~g}}{62 \mathrm{gmol}^{-1}}$

= 3.59 mol

Therefore, molality of the solution $=\frac{3.59 \mathrm{~mol}}{0.200 \mathrm{~kg}}$

= 17.95 m

Total mass of the solution = (222.6 + 200) g

= 422.6 g

Given,

Density of the solution = 1.072 g mL−1

$\therefore$ Volume of the solution $=\frac{422.6 \mathrm{~g}}{1.072 \mathrm{~g} \mathrm{~mL}^{-1}}$

= 394.22 mL

= 0.3942 × 10−3 L

$\Rightarrow$ Molarity of the solution $=\frac{3.59 \mathrm{~mol}}{0.39422 \times 10^{-3} \mathrm{~L}}$

= 9.11 M