A solution of 2.5 g of ethanoic acid in 75 g of benzene is prepared. What is the molar mass of ethanoic acid, moles of ethanoic acid, mass of benzene and molality of ethanoic acid?
(A) \( 0.556\, mol\, kg^{-1} \)
(B) \( 0.0417\, mol \)
(C) \( 60\, g\, mol^{-1} \)
(D) \( 0.075\, kg \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (B), (D), (A)
Molality is the number of moles of solute dissolved in one kilogram of solvent:
\( \displaystyle m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
The four quantities asked for are the steps of this calculation, in that order.
Molar mass of ethanoic acid, \( \ce{CH3COOH} \) (\( \ce{C2H4O2} \)):
\( \displaystyle M = 2(12) + 4(1) + 2(16) = 60\,\mathrm{g\,mol^{-1}} \quad \text{(C)} \)
Moles of ethanoic acid:
\( \displaystyle n = \frac{2.5\,\mathrm{g}}{60\,\mathrm{g\,mol^{-1}}} = 0.0417\,\mathrm{mol} \quad \text{(B)} \)
Mass of benzene:
\( \displaystyle 75\,\mathrm{g} = \frac{75}{1000}\,\mathrm{kg} = 0.075\,\mathrm{kg} \quad \text{(D)} \)
Molality of ethanoic acid:
\( \displaystyle m = \frac{0.0417\,\mathrm{mol}}{0.075\,\mathrm{kg}} = 0.556\,\mathrm{mol\,kg^{-1}} \quad \text{(A)} \)
Shortcut: The first quantity asked for is the molar mass, which can only be (C), the one value in g mol−1. Option (4) is the only option that starts with (C).
Note: 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 – …
