Molar ionic conductivities of divalent cation and anion are $$57 \text{ S cm}^2 \text{ mol}^{-1}$$ and $$73 \text{ S cm}^2 \text{ mol}^{-1}$$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be :

We are given the molar ionic conductivities of a divalent cation and a divalent anion, and we need to find the molar conductivity of the electrolyte formed by them.

Identify the electrolyte formula.

A divalent cation has charge +2 (let us call it $$M^{2+}$$) and a divalent anion has charge -2 (let us call it $$X^{2-}$$). To form an electrically neutral compound, we need one cation and one anion: the formula is $$MX$$.

Apply Kohlrausch's Law of Independent Migration of Ions.

Kohlrausch's law states that the molar conductivity at infinite dilution of an electrolyte is the sum of the contributions of its individual ions:

$$ \Lambda_m^\infty = \nu_+ \lambda_+^\infty + \nu_- \lambda_-^\infty $$

where $$\nu_+$$ and $$\nu_-$$ are the number of cations and anions per formula unit, and $$\lambda_+^\infty$$ and $$\lambda_-^\infty$$ are the molar ionic conductivities.

Substitute the values.

For the electrolyte $$MX$$: $$\nu_+ = 1$$ and $$\nu_- = 1$$

$$ \Lambda_m = 1 \times 57 + 1 \times 73 = 57 + 73 = 130 \text{ S cm}^2 \text{ mol}^{-1} $$

Note: The molar ionic conductivities given (57 and 73 S cm$$^2$$ mol$$^{-1}$$) are already the individual ionic contributions. Since the formula unit contains one cation and one anion, we simply add them.

The correct answer is Option (3): 130 S cm$$^2$$ mol$$^{-1}$$.