Question

An ethylene glycol solution contains 28.8 g of ethylene glycol (C2H6O2) in 98.2 mL of water. (Assume a density of 1.00 g/mL for water.) You may want to reference (Pages 558 - 568) Section 13.6 when completing this problem.

Determine freezing point of this solution.

Determine boiling point of this solution

Answer 1

Ans :

mol ethylene glycol = mass / molar mass

= 28.8 g / 62.07 g/mol = 0.464 mol

Mass of water = density x volume

= 1.00 x 98.2

= 98.2 g or 0.0982 kg

Molality of solution = mol / mass of solvent (kg)

= 0.464 / 0.0982

= 4.72 m

Freezing point depression = kf x m

kf for water = 1.86^oC/m

putting values :

= 1.86 x 4.72

= 8.79^oC

Freezing point of water = 0^oC

freezing point of solution = Freezing point of water - depression

= 0^oC - 8.79^o C

= -8.79^oC

Boiling point elevation = kb x m

kb for water = 0.512^oC/m

putting values :

= 0.512 x 4.72

= 2.42^oC

Boiling point of water = 100^oC

So boiling point of solution = Boiling point of water + elevation

=100^oC + 2.42^oC

= 102.42^oC