Question

What is the boiling point (in °C) of a solution of 9.06 g of I2 in 74.1 g of toluene, assuming the I2 is nonvolatile? (For toluene, Tb = 110.63°C and Kb = 3.40°C·kg/mol.)

Answer 1

Lets calculate molality first

Molar mass of I2 = 253.8 g/mol

mass(I2)= 9.06 g

number of mol of I2,

n = mass of I2/molar mass of I2

=(9.06 g)/(253.8 g/mol)

= 3.57*10^-2 mol

m(solvent)= 74.1 g

= 0.0741 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(3.57*10^-2 mol)/(0.0741 Kg)

= 0.4817 molal

lets now calculate ΔTb

ΔTb = Kb*m

= 3.4*0.481746

= 1.64 oC

This is increase in boiling point

boiling point of pure liquid = 110.63 oC

So, new boiling point = 110.63 + 1.64

= 112.27 oC

aNSWER: 112.27 oC