An antifreeze solution is prepared from 222.6 gram of ethylene glycol C2H6O2 and 200 gram 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?

Mass of the solute Molarity of the solution=number of moles of solutevolume of solution in litreC2H6O2=222.6 gramMolar mass of C2H6O2=62 gram per moleNumber of moles of solute 222.6/62 = 3.59Mass of solvent 200 g = 0.200 kgMolality of the solution(m)  : number of moles/weight of solvent in kg m=3.59/0.2=17.95Total mass of the solution : 222.6+200 = 422.6 gramVolume of the solution = mass of solution/density of solution =422.6/1.072=394.2mL=0.3942L=3.59/0.3942=9.11M