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?
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