The molar conductivities of HCl , NaCl , CH ₃ COOH , and CH ₃ COONa at infinite dilution follow the order

Step-by-step solution

_ CH ₃ COO ⁻=40.9 10⁻⁴ Sm ² ~mol ⁻¹ _ Na ⁺=50.10 10⁻⁴ Sm ² ~mol ⁻¹ _ Cl ⁻ = 76.35 10⁻⁴ Sm ² ~mol ⁻¹ _ H ⁺=349 10⁻⁴ Sm ² ~mol ⁻¹ _ x HCl = _ u^ t + _ c ⁻ / _ Cl , coon = _ al , coo ⁻ + _ u ⁺ _ Ncl = _ N + _ c r ⁻ / _ CH ₃ com N = _ CH ₃ coo ⁻ + _ N ⁺ _ HC =425.35 sm ² / mol _ cH ₃ coon =389.9 sm ² / mol _ Nacl =126.45 sm ² / mol _ Cl , coON =91 sm ² / mol