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?
Since, ethylene glycol is in excess, it is a solvent and water is a solute.
Molar mass of water is 18 g/mol.
Number of moles of solute = 200/18 = 11.11 moles
Molality is the number of moles of solute in 1 kg of solvent.
222.6 g of ethylene glycol corresponds to 0.2226 kg.
Molality = 11.11/0.2226 = 49.9 m
Total mass of solution = 222.6 + 200 = 422.6 g
Density of solution = 1.072 g/mL
Volume of solution = Mass/Density = 422.6/1.072 = 394 ml or 0.394 L
Molarity is the number of moles of solute in 1 L of solution.
Molarity = 11.11/0.394 = 28.2 M