Question

An ethylene glycol solution contains 24.6 g of ethylene glycol (C2H6O2) in 96.6 mL of water. (Assume a density of 1.00 g/mL for water.) Part ADetermine the freezing point of the solution. Part B Determine the boiling point of the solution.

Answer 1

weight of water = volue *density

                      = 96.6*1 = 96.6gm

molality = mass of solute*1000/gram molar mass of solute *weight of solvent

          = 24.6*1000/62*96.6

        = 24600/5969.2 = 4.12 m

T_f = m*K_f

kf of water =-1.86g/mole

T_f   = 4.12*-1.86 = -7.66C⁰

freezin point = -7.66C⁰

T_b   = m*K_b

          = 4.12*0.512

         = 0.495C⁰

boiling point = 100+0.495 = 100.495C⁰