Question

Boiling point of chloroform is 61°C. After addition of 5.0 g of a non-volatile solute to 20 g chloroform boils at 64.63°C. If kb = 3.63 K kg mol^-1, what is the molecular weight of the solute?

(a) 320 g/mol

(b) 100 g/mol

(c) 400 g/mol

(d) 250 g/mol

This question was posed to me during an internship interview.

My question is from Colligative Properties and Determination of Molar Mass in division Solutions of Chemistry – Class 12

Answer 1

Right answer is (d) 250 g/mol

To explain: Given,

b.p. of chloroform = 61°C

New b.p. after addition = 64.63°C

Mass of solute, w2 = 5.0 g

Mass of solvent, w1 = 20 g

Kb = 3.63 K kg mol^-1

From these, ∆Tb = 64.63 – 61 = 3.63°C

Using ΔTb = (kb x 1000 x w2)/(M2 x w1)

M2 = (kb x 1000 x w2)/(ΔTb x w1)

M2 = (3.63 x 1000 x 5)/(3.63 x 20) = 250 g/mol.