The molality of solution obtained by dissolving 1.5 g of ethanoic acid \( (CH_3COOH) \) in 25g of benzene is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1.0\, mol\, kg^{-1}\)
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:
\( \displaystyle M(\ce{CH3COOH}) = 2(12) + 4(1) + 2(16) = 60\ \mathrm{g\,mol^{-1}} \)
\( \displaystyle n = \frac{1.5\ \mathrm{g}}{60\ \mathrm{g\,mol^{-1}}} = 0.025\ \mathrm{mol} \)
Mass of benzene = 25 g = 0.025 kg.
\( \displaystyle m = \frac{0.025\ \mathrm{mol}}{0.025\ \mathrm{kg}} = 1.0\ \mathrm{mol\,kg^{-1}} \)
The molality is 1.0 mol kg−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):
Expressing Concentration of Solutions – …
