Which one of the following graphs between molar conductivity $$\Lambda_m$$ versus $$\sqrt{C}$$ is correct?

The Debye-Hückel-Onsager Equation

For strong electrolytes, the variation of molar conductivity with concentration is linear and follows the Debye-Hückel-Onsager relation:

$$\Lambda_m = \Lambda_m^\circ - A\sqrt{C}$$

Where:

• $$\Lambda_m$$ is the molar conductivity at concentration $$C$$.
• $$\Lambda_m^\circ$$ is the limiting molar conductivity at infinite dilution.
• $$A$$ is a constant depending on the electrolyte type and solvent.

Since both salts are strong 1:1 electrolytes, plotting $$\Lambda_m$$ against $$\sqrt{C}$$ yields a straight line with a negative slope for both.

Comparing Ionic Mobility and Limiting Molar Conductivity

To identify which line sits higher on the graph, we compare their limiting molar conductivities ($$\Lambda_m^\circ$$) using Kohlrausch's law:

$$\Lambda_m^\circ(\text{KCl}) = \lambda^\circ(\text{K}^+) + \lambda^\circ(\text{Cl}^-)$$

$$\Lambda_m^\circ(\text{NaCl}) = \lambda^\circ(\text{Na}^+) + \lambda^\circ(\text{Cl}^-)$$

Since the chloride anion ($$\text{Cl}^-$$) is common to both salts, the difference depends solely on the alkali metal cations:

• The ionic radius of $$\text{Na}^+$$ is smaller than that of $$\text{K}^+$$, which gives $$\text{Na}^+$$ a higher charge density.
• As a result, $$\text{Na}^+$$ undergoes greater hydration in aqueous solution than $$\text{K}^+$$, giving it a larger hydrated radius.
• A larger hydrated shell moves more slowly through water, meaning $$\text{Na}^+$$ has a lower ionic mobility and lower conductance than $$\text{K}^+$$.

$$\lambda^\circ(\text{K}^+) > \lambda^\circ(\text{Na}^+) \implies \Lambda_m^\circ(\text{KCl}) > \Lambda_m^\circ(\text{NaCl})$$

Conclusion:

Because $$\text{KCl}$$ maintains a higher molar conductivity than $$\text{NaCl}$$ at any given concentration, its linear plot must have a higher vertical intercept and lie entirely above the line for $$\text{NaCl}$$.