Question

Calculate the freezing point and boiling point of a solution containing 8.50 g of ethylene glycol (C2H6O2) in 90.6 mL of ethanol. Ethanol has a density of 0.789 g/cm3. Calculate the freezing point of the solution. Calculate the boiling point of the solution.

Answer 1

mass of ethanol = density of ethanol*volume = 0.789*90.6 = 71.48 g

molar mass of ethylene glycol = 62 g/mole

therefore moles of ethylene glycol in 8.50 g of it = 8.50/62 = 0.137 mole

molality = moles of ethylene glycol/mass of ethanol in kg = 0.137/0.07148 = 1.92 m

also Kf for ethanol = 1.99 C/m

and Kb  for ethanol = 1.22 C/m

boiling point of ethanol = 78.4 C

freezing point of ethanol = -114 C

A)depression in freezing point = molality*Kf = 1.92*1.99 = 3.82

thus freezing point of solution = -117.82 C

B)elevation in boiling point = molality*Kb = 1.92*1.22 = 2.34

thus freezing point of solution = 80.74C