Molar ionic conductivities of divalent cation and anion are 57 S cm² mol⁻¹ and 73 S cm² mol⁻¹ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:

Solution

Chemistry ma’am · 3 steps

A divalent cation and a divalent anion pair up one to one, so just add their two ionic conductivities.

1. Find the formula. The cation carries +2 and the anion carries -2, so one of each makes a neutral unit. The electrolyte is of the MX type, like MgSO₄.
2. Apply Kohlrausch. Kohlrausch's law adds the contribution of each ion, multiplied by how many of that ion one formula unit gives. Here each count is 1.

   Λₘ = ν₊λ₊ + ν₋λ₋ = 1 × 57 + 1 × 73 = 130 S cm² mol⁻¹

3. So. The molar conductivity is 130 S cm² mol⁻¹, option (d).

The idea to keep. In Kohlrausch's law the multipliers are the number of ions per formula unit, not the charges. Write the formula first, then count the ions.

Answer from the NTA final answer key. Solution by PrepWiser.