Question

An ethylene glycol solution contains 27.6 g of ethylene glycol (C2H6O2) in 92.0 mL of water. (Assume a density of 1.00 g/mL for water.) Determine the freezing point of the solution. Determine the boiling point of the solution.

Answer 1

first calculate the molality of the solution

molality = (W/MW) (1000 / mass of solvent in g)

molality = (27.6 / 62.07) (1000 / 92.0)

molality = 4.83 m

$\Delta$ Tf = Kf x m

$\Delta$ Tf = 1.86 x 4.83

$\Delta$ Tf = 8.98

freezing point of solution = 0 - 8.98

freezing point of solution = - 8.98⁰C

$\Delta$ Tb = Kb x m

$\Delta$ Tb = 0.512 x 4.83

$\Delta$ Tb = 2.47

boiling point of the solution = 100 + 2.47

boiling point of thes solution = 102.47 ⁰C

freezing point will decrease so we are substracting

boiling point will increase so we are adding