# An antifreeze solution is prepared from 222.6 g of ethylene glycol [C_(2)H_(4)(OH)_(2)] and 200 g of water. Calculate the molality of the soluti

28 views

closed
An antifreeze solution is prepared from 222.6 g of ethylene glycol [C_(2)H_(4)(OH)_(2)] and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072g mL^(-1) then what shall be the molarity of the solution?

by (90.8k points)
selected

Calculation of molality of the solution.
Mass of ethylene glycol = 222.6 g
Molar mass of ethylene glycol = 2xx12+6xx1+2xx16=62h mol^(-1)
"Mass of solution (m)" = (" Mass of ethylene glycol / Molar mass")/(" Mass of solvent in kg")
= =((222.6g)//(62 g mol^(-1)))/(0.2 kg)=17.95 mol kg^(-1)=17.95 m
Calculatrion of molarity of the solution.
Total mass of solution = Mass of solute + Mass of solvent
=222.6 + 200 422.6 g
Density of solution= 1.072 g mol^(-1)
"Volume of solutin" = (" Mass of solution")/("Density of solution")=((422.6 g))/((1.072 g mol^(-1)))=394.2 mL
=0.3942 L.
"Molarity of solution (M)"=("Mass of ethylene glycon/Molar mass")/("Volume of solution in litres")
=((22.6g)//(62 g mol^(-1)))/((0.3942 L))=9.10 mol L^(-1)=9.10 M