Question

An ethylene glycol contains 22.2 g of ethylene glycol (C_2H_6O_2) in 82.4 mL of water Calculate the freezing point of the solution. (Assume a density of 1.00 g/mL for water.) Calculate the boiling point of the solution.

Answer 1

Molar mass of ethylene glycol= 62

Moles of ethylene glycol in 22.2 gm =mass/molar mass= 22.2/60=0.37

Mass of solvent water= 82.4 ml*1g/ml= 82.4gm = 82.4/1000 kg=0.0824 kg

Molality of the solution = moles of solute/ kg of water =0.37/0.0824=4.49

Freezing point depression = i*kf*m

Where I =vant hoff factor for ethylene glycol= 1, kf= freezing point constant = 1.86 deg.c/m

And m= molality

Hecne freezing point depression = i*kf*m= 1*1.86*4.49=8.35 deg.c

Freezing ponit of solution = freezing point of water- freezing point depression =0-8.35= -8.35 deg,c

Boiling point elevation = i*kb*m

Kb=0.512deg.c/m molal elevation boiling point constant

Boiling point elevation = 1*0.512*4.49=2.3

Boiling point of solution =100+2.3= 102.3 deg.c