What is the molality of 2.5 g of ethanoic acid in 75 g of benzene?

The correct answer is Option (1) → 0.556 mol kg⁻¹

Molality (m) is the number of moles of solute per kilogram of solvent:

    \( \displaystyle m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}=\frac{w_2\times1000}{M_2\times w_1} \)

Ethanoic acid (\( \ce{CH3COOH} \)) has a molar mass of 60 g mol⁻¹, and the solvent (benzene) is 75 g = 0.075 kg.

    \( \displaystyle \text{Moles of ethanoic acid}=\frac{2.5}{60}=0.04167\ \mathrm{mol} \)

    \( \displaystyle m=\frac{0.04167}{0.075}=0.556\ \mathrm{mol\,kg^{-1}} \)

Shortcut: Using the second form directly: m = (2.5 × 1000)/(60 × 75) = 2500/4500 = 5/9 ≈ 0.556 mol kg⁻¹.

Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Expressing Concentration of Solutions – …

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: