The molality of 2.4 g of acetic acid in 60 g of benzene is

The correct answer is Option (4) → 0.667 mol kg⁻¹

Molality (m) is the number of moles of solute per kilogram of solvent, so its unit is mol kg⁻¹:

    \( \displaystyle m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)

Here the solute is acetic acid, \( \ce{CH3COOH} \), and the solvent is benzene.

    \( \displaystyle M(\ce{CH3COOH})=2(12)+4(1)+2(16)=60\ \mathrm{g\,mol^{-1}} \)

    \( \displaystyle n=\frac{2.4\ \mathrm{g}}{60\ \mathrm{g\,mol^{-1}}}=0.04\ \mathrm{mol} \)

    \( \displaystyle \text{mass of benzene}=60\ \mathrm{g}=0.060\ \mathrm{kg} \)

    \( \displaystyle m=\frac{0.04\ \mathrm{mol}}{0.060\ \mathrm{kg}}=0.667\ \mathrm{mol\,kg^{-1}} \)

So the molality is 0.667 mol kg⁻¹.

Shortcut: Molality is always in mol kg⁻¹, which rules out Options (2) and (3). The numbers 2.4 and 60 give 0.04 mol, and 0.04/0.06 = 2/3 = 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 – “… Example 1.3: Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.”