The unit for molar conductivity is

The correct answer is Option (1) → \(S\, m^2\, mol^{-1}\)

Molar conductivity is the conductivity of a solution divided by its molar concentration:

    \( \displaystyle \Lambda_m = \frac{\kappa}{c} \)

In SI units the conductivity \( \kappa \) is in S m⁻¹ and the concentration \( c \) in mol m⁻³:

    \( \displaystyle \text{unit of } \Lambda_m = \frac{\mathrm{S\,m^{-1}}}{\mathrm{mol\,m^{-3}}} = \mathrm{S\,m^{-1+3}\,mol^{-1}} = \mathrm{S\,m^{2}\,mol^{-1}} \)

So the unit of molar conductivity is S m² mol⁻¹ (in the older units, S cm² mol⁻¹).

Why the other options are incorrect

• Option (2) \(S\, m^{-1}\) — Incorrect. This is the unit of conductivity \( \kappa \), before it is divided by the concentration.

• Option (3) \(S\, m\, mol^{-1}\) — Incorrect. The metre comes out with power 2, not 1, because the concentration carries m⁻³.

• Option (4) \(S^{-1}\, m^2\, mol^{-1}\) — Incorrect. \( \mathrm{S^{-1}} \) is the ohm, a unit of resistance; conductivity terms carry siemens to the power +1.

NCERT (2026-27 edition):
Measurement of the Conductivity of Ionic Solutions – “… It, therefore, becomes necessary to define a physically more meaningful quantity called molar conductivity denoted by the symbol Λ_m (Greek, lambda). It is related to the conductivity of the solution by the equation:

    \( \displaystyle \text{Molar conductivity} = \Lambda_m = \frac{\kappa}{c} \)

In the above equation, if κ is expressed in S m^–1 and the concentration, c in mol m^–3 then the units of Λ_m are in S m² mol^–1.”

Note: The same/similar question has been asked in different years by NTA as per the details given below: