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 ?

Explanation

Given : Mass of the solute, `C_2H_6O_2 = 222.6 g`
Molar mass of `C_2H_6O_2 = 62 g "mol"^(-1)`
Number of moles of the solute = `(222.6 g)/(62g "mol"^(-1)) = 3.59`
Mass of the solvent = 200 g = 0.200 kg
Molality of the solution = 3.59 moles/0.200 kg = 17.95 m
Total mass of the solution = 222.6 + 200 = 422.6 g
Volume of the solution = Mass of solution / Density of solution
` = (422.6 g)/(1.072 g mL^(-1)) = 394.2 mL = 0.3942 L`
Molarity of the solution = `(3.59"moles")/(0.3942 L) = 9.11 M`