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

(i) Calculation of molality .
Mass of ethylene glycol = 222.6 g
Molar mass of ethylene glycol ` = 2 xx ` At. Mass of `C + 6 xx At` mass of `H+2 xx At` , mass of O = `2 xx 12 + 6 xx 1 + 2 xx 16 = 62`
Moles of ethylene glycol ` = (222.6)/(62) = 3.59 ` mol .
Mass of water = 200 g
Molality ` = ("Moles of ethylene glycol")/("Mass of water ")xx100 `
` = (3.59)/(200) xx 1000 = 17.59m`
(ii) Calculation of molarity.
Moles of ethylene glycol = 3.59 mol .
Mass of solution `= 200+222.6 = 422.6 g `
Volume of solution ` = ("Mass")/("Density")`
` = (422.6)/(1.072) = 394.22mL`
Molarity ` = ("Moles of ethylene glycol")/("Vol.of solution") xx 1000`
` = (3.59)/(394.22) xx 100 = 9.11 M`