Question

An ethylene glycol solution contains 24.6 g of ethylene glycol (C2H6O2) in 89.8 mL of water. (Assume a density of 1.00 g/mL for water.)

a. Determine the freezing point of the solution in degrees Celsius.

b. Determine the boiling point of the solution in degrees Celsius.

Answer 1

weight of water = volume of solution * density

                          = 89.8*1   = 89.8g

a. molality = W*1000/G.M.Wt* weight of solent in g

                  = 24.6*1000/(62*89.8)   = 4.4184m

Tb = i*Kb*m

           = 1*0.52*4.4184

           = 2.297568⁰C

boiling point of solution = 100+ 2.297568   = 102.3⁰C

b.

Tf = i*Kf*m

          = 1*1.86*4.4184

          = 8.218224⁰C

The freezing point of solution = -8.212⁰C