Λ°m values for electrolytes AB, HB and AD are 120.5, 300.9 and \( 80.0\, S\, cm^2\, mol^{-1} \), respectively. Λ°m value for HD is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(260.4\, S\, cm^2\, mol^{-1}\)
By Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion:
\( \displaystyle \Lambda^{\circ}_{m}(\mathrm{AB}) = \lambda^{\circ}(\mathrm{A^{+}}) + \lambda^{\circ}(\mathrm{B^{-}}) \)
So \( \Lambda^{\circ}_{m} \) of HD can be built from electrolytes that contain \( \mathrm{H^{+}} \) and \( \mathrm{D^{-}} \): add HB and AD, then subtract AB to remove the extra \( \mathrm{A^{+}} \) and \( \mathrm{B^{-}} \):
\( \displaystyle \Lambda^{\circ}_{m}(\mathrm{HD}) = \Lambda^{\circ}_{m}(\mathrm{HB}) + \Lambda^{\circ}_{m}(\mathrm{AD}) - \Lambda^{\circ}_{m}(\mathrm{AB}) \)
\( \displaystyle [\lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{B^-}}] + [\lambda^{\circ}_{\mathrm{A^+}} + \lambda^{\circ}_{\mathrm{D^-}}] - [\lambda^{\circ}_{\mathrm{A^+}} + \lambda^{\circ}_{\mathrm{B^-}}] = \lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{D^-}} \)
\( \displaystyle \Lambda^{\circ}_{m}(\mathrm{HD}) = 300.9 + 80.0 - 120.5 = 260.4\ \mathrm{S\,cm^{2}\,mol^{-1}} \)
So \( \Lambda^{\circ}_{m} \) of HD is 260.4 S cm2 mol−1. The same method gives \( \Lambda^{\circ}_{m} \) of a weak electrolyte such as acetic acid from HCl, NaCl and \( \ce{CH3COONa} \).
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):
Variation of Conductivity and Molar Conductivity with Concentration – “Example 2.8: Λ°m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. Calculate Λ° for HAc.
\( \displaystyle \Lambda^{\circ}_{m}(\mathrm{HAc}) = \lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{Ac^-}} = \lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{Cl^-}} + \lambda^{\circ}_{\mathrm{Ac^-}} + \lambda^{\circ}_{\mathrm{Na^+}} - \lambda^{\circ}_{\mathrm{Cl^-}} - \lambda^{\circ}_{\mathrm{Na^+}} \)
\( \displaystyle = \Lambda^{\circ}_{m}(\mathrm{HCl}) + \Lambda^{\circ}_{m}(\mathrm{NaAc}) - \Lambda^{\circ}_{m}(\mathrm{NaCl}) = (425.9 + 91.0 - 126.4)\ \mathrm{S\,cm^{2}\,mol^{-1}} = 390.5\ \mathrm{S\,cm^{2}\,mol^{-1}} \)”