calculate the boiling point of 500 gram aqueous solution that contains 20% ethylene glycol (C2H6O2, a non volatille liquid) if the solvent is water

Question

helppp, how to get its boiling point

Answer 1

∆T = imK

∆T = change in boiling point = ?

i = van't Hoff factor = 1 for ethylene glycol (EG)

m = molality = mols EG/kg water (see below for calculation of m)

K = boiling constant for water = 0.512º/m

calculation of m (molality):

For 500 g solution, 20% is EG. 0.20 x 500 g = 100 g EG

The remainder is water = 400 g H2O

m = moles EG / kg water. Moles EG = 100 g EG x 1 mol/62 g = 1.61 moles EG

kg water = 400 g x 1 kg/1000 g = 0.400 kg

m = molality = 1.61 mols / 0.400 kg = 4.03 m

∆T = (1)(4.03 m)(0.512º/m)

∆T = 2.06º

Boiling point of solution = 100º + 2.06º = 102.6ºC