Concept of Solution
There are several ways by which concentration of solution is quantitatively described: molality, molarity, mole fraction, ppm, mass % etc. Liquid- liquid solutions can be classified as ideal and non-ideal solution on the basis of Raoult's law. Suppose the molecules of two components are represented by A and B respectively. In pure components, the intermolecular attractive interaction will be of types A-A and B-B, whereas in the binary solution, A-B type of interactions will also be present. If intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, ideal solution is formed.
The properties of solution which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution are called colligative properties.
What would be the molality of 1.5 g of ethanoic acid (CH3COOH) in 75 g of benzene?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.33 mol kg−1
Molality (m), one of the concentration terms named in the passage, is the number of moles of solute dissolved in 1 kg of the solvent:
\( \displaystyle m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
Here the solute is ethanoic acid and the solvent is benzene.
Molar mass of ethanoic acid:
\( \displaystyle M(\ce{CH3COOH})=2(12)+4(1)+2(16)=60\ \mathrm{g\,mol^{-1}} \)
Moles of ethanoic acid:
\( \displaystyle n=\frac{1.5\ \mathrm{g}}{60\ \mathrm{g\,mol^{-1}}}=0.025\ \mathrm{mol} \)
Mass of benzene in kg: 75 g = 0.075 kg.
Molality:
\( \displaystyle m=\frac{0.025\ \mathrm{mol}}{0.075\ \mathrm{kg}}=0.33\ \mathrm{mol\,kg^{-1}} \)
So the molality of the solution is 0.33 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 – “Example 1.3 Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene.”