An antifreeze solution is prepared by mixing 217.0 g of ethylene glycol ( C₂H₆O₂ ) with 200 g water. The molality of solution is

Molar mass of C₂H₆O₂ = 62 g mol^-1 

Amount of C₂H₆O₂ =$n=\frac{m}{M}=\frac{217.0\mathrm{g}}{62\mathrm{g}{\mathrm{mol}}^{-1}}=3.5\mathrm{mol}$

Molality of solution, $m=\frac{n}{m_{\mathrm{l}}}=\frac{3.5\mathrm{mol}}{0.200\mathrm{kg}}=17.5\mathrm{mol}{\mathrm{kg}}^{-1}$