1.00 g of a non-electrolyte solute when dissolved in 50 g of organic solvent, lowered the freezing point of organic solvent by 0.50 K. If the freezing point depression constant of organic solvent is \( 5\, K\, kg\, mol^{-1} \), the molar mass of the solute will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(200\, g\, mol^{-1}\)
Depression in freezing point is a colligative property: it depends on the molality of the solute. For a non-electrolyte (i = 1):
\( \displaystyle \Delta T_f = K_f\,m = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1} \)
Rearranging for the molar mass of the solute:
\( \displaystyle M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} \)
\( \displaystyle M_2 = \frac{5 \times 1.00 \times 1000}{0.50 \times 50} = \frac{5000}{25} = 200\ \mathrm{g\,mol^{-1}} \)
So the molar mass of the solute is 200 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):
Depression of Freezing Point – “1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute.”