When 1.5 g of a non-volatile solute is dissolved in 30 g of a solvent, the boiling point of the solvent is raised by 2K. Calculate the molar mass of the solute, given that \(K_b\) for the solvent is 1.85 \(K kg mol^{-1}\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 46.25 \(g mol^{-1}\)
Elevation of boiling point is a colligative property. For a dilute solution of a non-volatile, non-electrolyte solute it is proportional to the molality:
\( \displaystyle \Delta T_b = K_b\, m = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1} \)
so
\( \displaystyle M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1} \)
With w2 = 1.5 g, w1 = 30 g, ΔTb = 2 K and Kb = 1.85 K kg mol−1:
\( \displaystyle M_2 = \frac{1.85 \times 1.5 \times 1000}{2 \times 30} = \frac{2775}{60} = 46.25\ \text{g mol}^{-1} \)
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 – “… 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.”