\( Λ_m^o \) for \( NaCl, HCl \) and \( CH_3COONa \) are 126.4, 425.9 and \( 91.0\, S\, cm^2 mol^{-1} \) respectively. Calculate \( Λ_m^o \) for \( CH_3COOH \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(390.5\, S\, cm^2 mol^{-1}\)
By Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the contributions of its cation and anion. So Λ°m of a weak electrolyte such as acetic acid, \( \ce{CH3COOH} \), can be found from strong electrolytes that together supply the same ions:
\( \displaystyle \Lambda^{\circ}_m(\mathrm{CH_3COOH})=\Lambda^{\circ}_m(\mathrm{HCl})+\Lambda^{\circ}_m(\mathrm{CH_3COONa})-\Lambda^{\circ}_m(\mathrm{NaCl}) \)
This works because the \( \ce{Na^{+}} \) and \( \ce{Cl^{-}} \) contributions cancel:
\( \displaystyle =[\lambda^{\circ}_{H^+}+\lambda^{\circ}_{Cl^-}]+[\lambda^{\circ}_{Na^+}+\lambda^{\circ}_{CH_3COO^-}]-[\lambda^{\circ}_{Na^+}+\lambda^{\circ}_{Cl^-}] \)
\( \displaystyle =\lambda^{\circ}_{H^+}+\lambda^{\circ}_{CH_3COO^-} \)
Substituting the values:
\( \displaystyle \Lambda^{\circ}_m(\mathrm{CH_3COOH})=425.9+91.0-126.4 \)
\( \displaystyle =390.5\ \mathrm{S\,cm^2\,mol^{-1}} \)
Note 1: 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):
Variation of Conductivity and Molar Conductivity with Concentration – …
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 11-05 S1 | 03-06 S2 28-05 S1 22-05 S1 15-05 S1 | – | – | 20-08 S1 10-08 S1 |