Question

Calculate the freezing point and boiling point of a solution containing 7.80 g of ethylene glycol (C2H6O2) in 93.4 mL of ethanol. Ethanol has a density of 0.789 g/cm3.

Answer 1

Weight of solute (ethylene glycol) = 7.8 g

molar mass of solute = 62g/mol

weight of solvent = volume x density = 93.4mL x 0.789g/mL = 73.6926g

molality of solution = no. of moles (solute)/W(solvent in Kg)

= (7.80x 1000) / (62x 73.6926) = 1.7071 m

Elevation in boiling point Delta Tb = Kb. molality

Kb of ethanol = 1.19K/Kgmol and boiling point is 78.4C

Delta Tb = 1.19x 1.7071 = 2.03144

Hence boiling point of the solution = 78.4 + 2.03 = 80.431 ^oC

Kf of ethanol = -1.99 and frezzing point is -114.6C

delta Kf (depression in freezing point = Kf x m

= -1.99 x 1.7071 = - 3.3972

Thus freezing point of the solution is -114.6 +(-3.3972) = -117.9972^oC