1. Use the relation for conductance of a saturated solution

For a solution placed in the same conductivity cell, $C = \kappa \times \text{cell constant factor}^{-1}$ so comparison of measured conductance is equivalent to comparing conductivity $\kappa$.

For an electrolyte, $\kappa = \sum_i c_i z_i F u_i$ Thus conductivity depends on:

• number of ions in solution / concentration (solubility), and
• ionic mobilities $u_i$.

2. Analyse $\mathrm{NaCl}$ saturated solution

$\mathrm{NaCl}$ is highly soluble in water, so its saturated solution has large ionic concentration. Also, with increase in temperature, ionic mobility increases because viscosity of water decreases. Hence conductivity (and therefore measured conductance) of saturated $\mathrm{NaCl}$ solution increases with temperature.

So: $C_{\mathrm{NaCl}}(T_2) > C_{\mathrm{NaCl}}(T_1) \quad \text{for } T_2>T_1$ Therefore B is true.

3. Analyse $\mathrm{BaSO_4}$ saturated solution

$\mathrm{BaSO_4}$ is sparingly soluble, so its saturated solution contains very few ions. Thus its conductance is much smaller than that of saturated $\mathrm{NaCl}$ at a given temperature.

However, for a saturated solution, conductivity depends not only on mobility but also on solubility. For $\mathrm{BaSO_4}$, dissolution is generally exothermic, so its solubility decreases with increase in temperature. Although ionic mobilities increase with temperature, the ionic concentration in the saturated solution decreases.

Hence the statement $C_{\mathrm{BaSO_4}}(T_2) > C_{\mathrm{BaSO_4}}(T_1)$ for $T_2>T_1$ is not necessarily true; in fact for such sparingly soluble salts it can decrease because solubility decreases.

Therefore C is false.

4. Check option A

Ionic mobility of ions in aqueous solution generally increases with temperature. So for both salts, mobilities of their ions increase with $T$.

Therefore A is true.

5. Check option D

At a given temperature, saturated $\mathrm{NaCl}$ solution has far greater ionic concentration than saturated $\mathrm{BaSO_4}$ solution because:

• $\mathrm{NaCl}$ is highly soluble,
• $\mathrm{BaSO_4}$ is very sparingly soluble.

So, $C_{\mathrm{NaCl}} \gg C_{\mathrm{BaSO_4}}$ Therefore D is true.

6. Conclusion

The false statement is: $\boxed{\text{C}}$

The stored correct answer is D, but that is not correct because D is actually true for saturated solutions at a given temperature.