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?

Explanation

Step 1. Calculation of molality of the solution
Mass of ethylene glycol `= 222.6 g`
Molar mass of ethylene glycol `= 2 xx 12+6 xx 1+2xx16 = 62 "g mol"^(-1)`
Mass of water `= 200 g = 0.2 kg`
Molality of solution `(m)=("Mass of ethylene glycol/Molar mass")/("Mass of solvent in kg")`
`=((222.6g)//(62 "g mol"^(-1)))/(0.2 g)=17.95"mol kg"^(-1) =17.95 m`
Step II. Calculation 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 mL"^(-1)`
Volume of solution `= ("Mass of solution")/("Density of solution")=((422.6g))/((1.072"g mL"^(-1)))`
`=394.2 mL=0.3942 L`
Molarity of the solution `(M)=("Mass of ethylene glycol/Molar mass")/("Volume of solution in litres")`
`=((222.6g)//(62"g mol"^(-1)))/((0.3942L))=9.10 "mol L"^(-1)=9.10M`.