What is the molality \( (\mathrm{mol\ kg^{-1}}) \) of 5 g of ethanoic acid in 125 g of benzene?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.667
Molality (m) is the number of moles of solute dissolved in 1 kg of solvent:
\( \displaystyle m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
Since it uses the mass of the solvent (not the volume of the solution), molality does not change with temperature.
Here the solute is ethanoic acid, \( \ce{CH3COOH} \), and the solvent is benzene.
\( \displaystyle M(\ce{CH3COOH})=2(12)+4(1)+2(16)=60\ \mathrm{g\,mol^{-1}} \)
\( \displaystyle \text{moles of acid}=\frac{5}{60}=0.0833\ \mathrm{mol} \)
\( \displaystyle \text{mass of benzene}=125\ \mathrm{g}=0.125\ \mathrm{kg} \)
\( \displaystyle m=\frac{0.0833}{0.125}=0.667\ \mathrm{mol\,kg^{-1}} \)
Shortcut: 125 g is 1/8 kg, so m = 8 × (5/60) = 40/60 = 0.667.
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 – …
