The properties which depend on the number of solute particles irrespective of their nature relative to the total numbers of particles in the solution are called colligative properties. These are relative lowering of vapour pressure of the solvent, depression of freezing point of the solvent, elevation of boiling point of the solvent and osmotic pressure of the solution. All of these properties help in calculating the observed molar mass of the solute. Determination of osmotic pressure is considered best for the determination of molecular mass of the solute.
A solution of 12.5 g of an unknown solute in 170 g water gave a boiling point elevation of 0.63 K. Kb for water is 0.53 K kg mol−1. The molar mass of the solute is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 61.85 g mol−1
The elevation of boiling point of a dilute solution of a non-volatile solute is proportional to its molality:
\( \displaystyle \Delta T_b = K_b\,m = K_b\times\frac{1000\times w_2}{M_2\times w_1} \)
\( \displaystyle M_2 = \frac{1000\times w_2\times K_b}{\Delta T_b\times w_1} \)
Here w2 = 12.5 g, w1 = 170 g, Kb = 0.53 K kg mol−1, ΔTb = 0.63 K:
\( \displaystyle M_2 = \frac{1000\times12.5\times0.53}{0.63\times170} \)
\( \displaystyle M_2 = \frac{6625}{107.1} = 61.86\ \mathrm{g\,mol^{-1}} \approx 61.85\ \mathrm{g\,mol^{-1}} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Elevation of Boiling Point – “… Example 1.8 The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K kg mol−1
Solution
The elevation (ΔTb) in the boiling point = 354.11 K − 353.23 K = 0.88 K
…
\( \displaystyle M_2 = \frac{2.53\ \mathrm{K\,kg\,mol^{-1}}\times1.8\ \mathrm{g}\times1000\ \mathrm{g\,kg^{-1}}}{0.88\ \mathrm{K}\times90\ \mathrm{g}} \) = 58 g mol−1”