Question

a solution is made by dissolving 0.585mol of nonelectrolyte solute in 889g of benzene. calculate the freezing point and boiling point of the solution.

Answer 1

Freezing point depression= m*kf

kf =5.12deg.c/m Freezing point constant

mass of Benzene= 889gm =889/1000 kg= 0.889

m= molality= moles of solute/kg of solvent Benzene= 0.585/0.889=0.591moles of solute/ kg of solvent

Freezing point depression= 5.12*0.591=3.025

Freezing point of Benzene =5.5deg.c

Freezig point of solution = 5.5- 3.025=2.475 deg.c

b) kb=2.53 , Boiling point constant

Boiling point elevation= kb*m= 2.53*0.591=1.5

Boiling point of Benzene =80.1 deg.c

Boiling point = 80.1+1.5=81.6 deg.c