An antifreeze solution is prepared from $222.6 \ g$ of ethylene glycol $(C_2H_6O_2)$ 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?

(N/A) Mass of solute $= 222.6 \ g$
Molar mass of solute $C_2H_6O_2 = (12 \times 2) + (1 \times 6) + (16 \times 2) = 62 \ g \ mol^{-1}$
Moles of solute $= \frac{222.6 \ g}{62 \ g \ mol^{-1}} = 3.59 \ mol$
Mass of solvent $= 200 \ g = 0.2 \ kg$
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3.59 \ mol}{0.2 \ kg} = 17.95 \ mol \ kg^{-1}$
Total mass of solution $= 222.6 \ g + 200 \ g = 422.6 \ g$
Volume of solution $= \frac{\text{mass of solution}}{\text{density}} = \frac{422.6 \ g}{1.072 \ g \ mL^{-1}} = 394.21 \ mL = 0.39421 \ L$
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{3.59 \ mol}{0.39421 \ L} = 9.11 \ mol \ L^{-1}$