Analyze Statement I

Statement I says: The limiting molar conductivity of $KCl$ (strong electrolyte) is higher than that of $CH_3COOH$ (weak electrolyte).

At infinite dilution, the limiting molar conductivity is $\Lambda_m^\circ = \lambda_+^\circ + \lambda_-^\circ$

For the two electrolytes:

• For $KCl$: $\Lambda_m^\circ(KCl)=\lambda_{K^+}^\circ+\lambda_{Cl^-}^\circ$
• For $CH_3COOH$: $\Lambda_m^\circ(CH_3COOH)=\lambda_{H^+}^\circ+\lambda_{CH_3COO^-}^\circ$

Now, $H^+$ has exceptionally high ionic mobility, so $CH_3COOH$ has a very large limiting molar conductivity.

Typical values: $\Lambda_m^\circ(KCl) \approx 150\ \text{S cm}^2\text{ mol}^{-1}$ $\Lambda_m^\circ(CH_3COOH) \approx 390\ \text{S cm}^2\text{ mol}^{-1}$

Therefore, $\Lambda_m^\circ(CH_3COOH) > \Lambda_m^\circ(KCl)$

So, Statement I is false.

Analyze Statement II

Statement II says: Molar conductivity decreases with decrease in concentration of electrolyte.

But molar conductivity generally increases on dilution, i.e. when concentration decreases.

Reason:

• On decreasing concentration, interionic interactions decrease.
• For weak electrolytes, degree of ionization also increases.

Hence, $\text{as concentration decreases, } \Lambda_m \text{ increases}$

So the given statement is opposite to the correct trend.

Therefore, Statement II is false.

Choose the correct option

• Statement I: False
• Statement II: False

Hence, the correct option is: D. Both Statement I and Statement II are false.

Comparison with stored answer

Stored correct answer: D
Derived answer: D

They match.