Let represent molar conductivity, represent equivalent conductivity, represent specific conductance, — Electrochemistry Chemistry Question

💡 Solution & Explanation

The key relation is $\Lambda_m = \Lambda_{eq} \times \text{n-factor (charge } z)$. For $Al_2(SO_4)_3$, total cationic charge is $2 \times (+3) = 6$, so $\Lambda_m = 6 \Lambda_{eq}$. The standard derived formula for equivalent conductivity uses Normality: $\Lambda_{eq} = 1000 \kappa / N$. For 1:1 electrolytes (like $Na^+Cl^-$), the n-factor is 1, so Molarity exactly equals Normality, making $\Lambda_m = \Lambda_{eq}$.