Question

Calculate the freezing point and boiling point of each of the following solutions: the boiling point of the solution: 60.0 g of glucose, C₆H₁₂O₆, added to 113 g of water (K_b=0.52∘C) Express your answer using one decimal place.

Answer 1

Boiling point depression delta Tb is given as :

Delta Tb= kb.m

where kb is the boiling point constant and m is the molality

m = number of moles / mass of solvent in kg

m = (60/180.1559) / 0.113

m = 2.95m

Delta Tb = 0.512 x 2.95

Delta Tb = 1.51

So the boiling point of solution will be 100 + 1.51

= 101.51^oC

Freezing point depression of the solution Delta Tf = kf.m

putting the values we get :

Delta Tf = 1.86 x 2.95

= 5.5^oC

So the freezing point of the solution will be 0^o - 5.5^o

= -5.5^oC