At infinite dilution, the equivalent conductance of CH3COONa, HCl and CH3COOH are 91, 426 and 391 mho cm2 eq–1 respectively at 25 °C. The equivalent conductance of NaCl at infinite dilution will be :126209391908

The equivalent conductance of a substance at infinite dilution is the sum of the equivalent conductances of its constituent ions.

In this case, we are given the equivalent conductances of CH3COONa, HCl and CH3COOH at infinite dilution and we are asked to find the equivalent conductance of NaCl at infinite dilution.

The equivalent conductance of CH3COONa at infinite dilution is given as 91 mho cm2 eq–1. This is the sum of the equivalent conductances of the CH3COO- ion and the Na+ ion.

The equivalent conductance of HCl at infinite dilution is given as 426 mho cm2 eq–1. This is the sum of the equivalent conductances of the H+ ion and the Cl- ion.

The equivalent conductance of CH3COOH at infinite dilution is given as 391 mho cm2 eq–1. This is the sum of the equivalent conductances of the CH3COO- ion and the H+ ion.

We can use these values to find the equivalent conductance of the Na+ ion and the Cl- ion, and hence the equivalent conductance of NaCl at infinite dilution.

The equivalent conductance of the Na+ ion is the difference between the equivalent conductance of CH3COONa and the equivalent conductance of the CH3COO- ion. The equivalent conductance of the CH3COO- ion is the difference between the equivalent conductance of CH3COOH and the equivalent conductance of the H+ ion.

So, the equivalent conductance of the Na+ ion = 91 - (391 - 426) = 91 - (-35) = 126 mho cm2 eq–1.

The equivalent conductance of the Cl- ion is the difference between the equivalent conductance of HCl and the equivalent conductance of the H+ ion.

So, the equivalent conductance of the Cl- ion = 426 - 391 = 35 mho cm2 eq–1.

Therefore, the equivalent conductance of NaCl at infinite dilution is the sum of the equivalent conductances of the Na+ ion and the Cl- ion = 126 + 35 = 161 mho cm2 eq–1.