If the density of a solution containing 217 .0 g ethylene glycol ( C₂H₆O₂ ) in 200 g water is 1.10 g cm^-3 the molarity of the solution is

Total mass of solution, m = 417 .0 g

Volume of solution $V=\frac{m}{\rho }=\frac{417.0\mathrm{g}}{1.10\mathrm{g}{\mathrm{cm}}^{-3}}=379.1{\mathrm{cm}}^3$

Amount of C₂H₆ O2$n=\frac{m}{M}=\frac{217.0\mathrm{g}}{62\mathrm{g}{\mathrm{mol}}^{-1}}=3.5\mathrm{mol}$

Molarity of solution $M=\frac{n}{V}=\frac{3.5\mathrm{mol}}{379.1\times {10}^{-3}{\mathrm{dm}}^3}=9.23\mathrm{mol}{\mathrm{dm}}^{-3}$