Question

Calculate the freezing point and boiling point of a solution containing 13.6 g of naphthalene (C10H8) in 110.0 mL of benzene. Benzene has a density of 0.877 g/cm3. A)Calculate the freezing point of a solution. (Kf(benzene)=5.12∘C/m.) B)Calculate the boiling point of a solution. (Kb(benzene)=2.53∘C/m.)

Answer 1

molar mass of naphtalene = 128 g/mol
number of mole naphtalene = (given mass)/(molar mass)
= 13.6/128
= 0.106 mole
mass of benzene = (volume)*(density)
= 110*0.877
= 96.5 g
= 0.0965 kg
molality = (number of mole)/(mass of solvent in kg)
= 0.106/0.0965
= 1.10 m

melting point oof pure benzene = 5.5 oC
delta Tf = Kf*m
5.5 - Tf = (5.12*1.1)
5.5 - Tf = 5.6
Tf = -0.1 oC

boiling point of pure benzene = 80..1 oC
Tb - 80.1 = Kb*m
Tb - 80.1 = 2.53*1.1
Tb - 80.1 = 2.8
Tb = 82.9 oC