Calculate the molality of 25 g of ethanoic acid (\( CH_3COOH \)) in 250 g benzene
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1.664\, mol\, kg^{-1}\)
Molality is the number of moles of solute per kilogram of solvent:
\( \displaystyle m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
Molar mass of ethanoic acid, \( \ce{C2H4O2} \):
\( \displaystyle M = 2(12) + 4(1) + 2(16) = 60\ \mathrm{g\,mol^{-1}} \)
\( \displaystyle n = \frac{25\ \mathrm{g}}{60\ \mathrm{g\,mol^{-1}}} = 0.4167\ \mathrm{mol} \)
The solvent is 250 g of benzene = 0.250 kg:
\( \displaystyle m = \frac{0.4167\ \mathrm{mol}}{0.250\ \mathrm{kg}} = 1.67\ \mathrm{mol\,kg^{-1}} \)
The nearest option is 1.664 mol kg−1; the small difference comes only from the rounding of the molar mass.
Shortcut: About 0.4 mol in a quarter of a kilogram is about 1.6 mol kg−1. The other options differ from this by powers of ten or are far away, so Option (2) can be picked at sight.
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):
Expressing Concentration of Solutions – …
