The mole fraction of ethylene glycol (C2H6O2) in aqueous solution containing 20% of C2H6O2 by mass is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.068
The mole fraction of a component is its number of moles divided by the total number of moles in the solution:
\( \displaystyle x_{A}=\frac{n_{A}}{n_{A}+n_{B}} \)
A concentration given as mass percentage means grams of solute in 100 g of solution. So a 20% solution of ethylene glycol contains 20 g of \( \ce{C2H6O2} \) and 100 − 20 = 80 g of water.
\( \displaystyle M(\ce{C2H6O2})=2(12)+6(1)+2(16)=62\ \mathrm{g\,mol^{-1}} \)
\( \displaystyle n(\ce{C2H6O2})=\frac{20}{62}=0.322\ \mathrm{mol} \)
\( \displaystyle n(\ce{H2O})=\frac{80}{18}=4.444\ \mathrm{mol} \)
\( \displaystyle x(\ce{C2H6O2})=\frac{0.322}{0.322+4.444}=\frac{0.322}{4.766} \)
\( \displaystyle x(\ce{C2H6O2})=0.068 \)
So the mole fraction of ethylene glycol is 0.068.
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 – …
