An antifreeze solution is prepared from 222.6 gram of ethylene glycol $\left(C_2{H}_6{O}_2\right)$and 200 gram of water. Calculate the molality of the solution. If the density of the solution is  $1.072gm{L}^{-1}$, then what shall be the molarity of the solution?

Mass of the solute $\text{Molarity of the solution}=\frac{\text{number of moles of solute}}{\text{volume of solution in litre}}$$C_2{H}_6{O}_2=222.6gram$

Molar mass of $C_2{H}_6{O}_2=62$ gram per mole

Number of moles of solute 222.6/62 = 3.59

Mass of solvent 200 g = 0.200 kg

Molality of the solution(m)  : number of moles/weight of solvent in kg 

$m=3.59/0.2=17.95$

Total mass of the solution : 222.6+200 = 422.6 gram

Volume of the solution = mass of solution/density of solution 

$=422.6/1.072=394.2mL=0.3942L$

$=3.59/0.3942=9.11M$