The order of equivalent conductances at infinite dilution for $LiCl$, $NaCl$, and $KCl$ is:

The resistance of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $1500 \ \Omega$. If the conductivity of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $0.1466 \times 10^{-3} \ S \ cm^{-1}$,the cell constant of the conductivity cell in $cm^{-1}$ is:

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

The conductivity of a solution containing $2.08 \ g$ of anhydrous barium chloride in $200 \ mL$ solution is $6 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$. The molar conductivity of the solution (in $\Omega^{-1} \ cm^2 \ mol^{-1}$) is $\underline{x} \times 10^2$. The value of $\underline{x}$ is (Atomic mass of $Ba=137, Cl=35.5$)

What is the molar conductivity of $0.02 \text{ M}$ $KI$ solution if its conductivity is $4.37 \times 10^{-4} \text{ }\Omega^{-1} \text{cm}^{-1}$?

At $298 \text{ K}$, the specific conductance $(\kappa)$ of a $0.0020 \text{ M } NaCl$ solution is $2.50 \times 10^{-4} \text{ S cm}^{-1}$. Calculate the molar conductivity $(\Lambda_m)$ of the solution in $\text{S cm}^2 \text{mol}^{-1}$.