A solution is made by dissolving 0609 mol of nonelectrolyte

A solution is made by dissolving 0.609 mol of nonelectrolyte solute in 827 g of benzene. Calculate the freezing point, Tt, and boiling point. Tb, of the solution. Constants may be found here. Number Number

Solution

dTf = i*Kf*m

DTf = Tf-TS c

Tf = freezingpoint of solvent(benzene) = 5.5 c

Ts = freezingpoint of solution = x c

i= vanthoff factor of solute = 1

Kf of benzene = 4.9 c/m

m = molality = (n)*(1000/W)

             = 0.609*(1000/827)

             = 0.74 m

(5.5-x) = 1*4.9*0.74

x = 1.874 c

Tf = freezingpoint of solution = x = 1.874 C

DTb = i*kb*m

DTb = Tb-T0

T0 = boiling point of solvent(benzene) = 80.10 C

Tb = boiling point of solution = x C

m = molality of solution = 0.74 m

i = vanthaff factor of non-electrolyte = 1

kb of benzene = 2.530 °C/m

(x-80.1) = 1*2.53*0.74

x = boiling point of solution = 82.0 C

Tb = boiling point of solution = 82.0 C