Question

What is the boiling point of a solution made by dissolving 15.4 g of CCl₄ in 80.0 g of benzene? (The boiling point of pure benzene is 80.1°C and K_b = 2.53 °C/m.)

Answer 1

Lets calculate molality first

Molar mass of CCl4,

MM = 1*MM(C) + 4*MM(Cl)

= 1*12.01 + 4*35.45

= 153.81 g/mol

mass(CCl4)= 15.4 g

number of mol of CCl4,

n = mass/molar mass

=(15.4 g)/(153.81 g/mol)

= 0.1001 mol

m(solvent)= 80.0 g

= 0.08 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(0.1001 mol)/(0.08 Kg)

= 1.2515 molal

lets now calculate ΔTb

ΔTb = Kb*m

= 2.53*1.2515

= 3.2 oC

This is increase in boiling point

boiling point of pure liquid = 80.1 oC

So, new boiling point = 80.1 + 3.2

= 83.3 oC

Answer: 83.3 oC