Question

Calculate the freezing point and boiling point of a solution containing 10.6 g of naphthalene (C10H8) in 114.0 mL of benzene. Benzene has a density of 0.877 g/cm3.

Calculate the freezing point of a solution. (Kf(benzene)=5.12∘C/m.)

Calculate the boiling point of a solution. (Kb(benzene)=2.53∘C/m.)

Answer 1

mass of benzene   = volume * density

                               = 114*0.877   = 99.978g

molality     = weight of solute*1000/gram molecular weight * weight of solvent in g

               = 10.6*1000/128*99.978    = 0.828m

Tf     = Kf*m

           = 5.12*0.828   = 4.2394C⁰

freezing point of naphthalene    = 80.26+4.2394    = 84.5C⁰

Tb    = Kb*m

            = 2.53*0.828   = 2.09484C⁰

bolilint point of naphthalene   = 218+2.09484   = 220.09C⁰