If an electrolyte on dissociation gives $v_+$ cations and $v_-$ anions, then its limiting molar conductivity is given by

$Λ_m^0 = v_+λ^0_+ + v_-λ^0_-$

The correct answer is Option (4) → $Λ_m^0 = v_+λ^0_+ + v_-λ^0_-$
The limiting molar conductivity of an electrolyte is equal to the sum of the limiting ionic conductivities of the cation and anion multiplied by their respective stoichiometric numbers.

Reasoning:

• Λₘ⁰ = limiting molar conductivity of the electrolyte.

• λ₊⁰ = limiting ionic conductivity of the cation.

• λ₋⁰ = limiting ionic conductivity of the anion.

• ν₊ and ν₋ = number of cations and anions produced on dissociation.

Since conductivity is due to both ions, their contributions are added, not subtracted.
This is known as Kohlrausch’s Law of Independent Migration of Ions.