An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C_2H_6O_2), and 200 g ofwater. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^(-1) , thenwhat shall be the molarity of the solution ?

An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C

1.	An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C_2H_6O_2), and 200 g ofwater. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^(-1) , thenwhat shall be the molarity of the solution ?
Answer» Solution :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)/(1.072g mL^(-1)) = 394.2 mL = 0.3942 L` Molarity of the solution = `(3.59"moles")/(0.3942 L) = 9.11 M`