An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) 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?

Molecular mass of glycol C2H4(OH)2 =62

Number of moles of C2H4(OH)2 = mass of glycol / molar mass of glycol

= 222.6/62=3.59  moles

Molality of solution = number of moles of solute / mass of solvent in Kg  

Mass of solvent = 200g = 200/1000 Kg = 0.2 kg

Plug the values we get

Molality = 3.59/0.2

Molality = 17.95

Formula of molarity of solution = number of moles of solute / volume of solution in Kg

Formula of volume = mass / density

Volume = 422.6g/(1.072 g/ml)

Volume = 394.21 ml

Convert in liter   

Volume in liter = 394.21 ml /1000 liter  =0.394 liter

Molarity =9.1M