Calculate the mole fraction of ethanoic acid (\( CH_3COOH \)) in an aqueous solution containing 20% of \( CH_3COOH \) by mass.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.0697
The mole fraction of a component is its number of moles divided by the total number of moles in the solution:
\( \displaystyle x_{\text{acid}} = \frac{n_{\text{acid}}}{n_{\text{acid}} + n_{\text{water}}} \)
A 20% solution by mass contains 20 g of \( \ce{CH3COOH} \) in 100 g of solution, so the water is 100 − 20 = 80 g. Molar masses: \( \ce{CH3COOH} \) = 2(12) + 4(1) + 2(16) = 60 g mol−1, \( \ce{H2O} \) = 18 g mol−1.
\( \displaystyle n_{\text{acid}} = \frac{20\ \mathrm{g}}{60\ \mathrm{g\,mol^{-1}}} = 0.3333\ \mathrm{mol} \)
\( \displaystyle n_{\text{water}} = \frac{80\ \mathrm{g}}{18\ \mathrm{g\,mol^{-1}}} = 4.444\ \mathrm{mol} \)
\( \displaystyle x_{\text{acid}} = \frac{0.3333}{0.3333 + 4.444} = \frac{0.3333}{4.777} \approx 0.0697 \)
So the mole fraction of ethanoic acid is about 0.0697.
Shortcut: A mole fraction can never exceed 1, which rules out Options (3) and (4) at once.
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 – “Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass.”