Which of the following $M$ $KCl$ solutions has the lowest value of equivalent conductance?

Given that $\Lambda_{m}^{\infty} = 133.4 \, S \, cm^{2} \, mol^{-1} (AgNO_{3})$; $\Lambda_{m}^{\infty} = 149.9 \, S \, cm^{2} \, mol^{-1} (KCl)$ and $\Lambda_{m}^{\infty} = 144.9 \, S \, cm^{2} \, mol^{-1} (KNO_{3})$,the molar conductivity at infinite dilution for $AgCl$ is $....... \, S \, cm^{2} \, mol^{-1}$.

Calculate the cell constant of a conductivity cell containing $0.01 \ M \ AgNO_3$ solution having a resistance of $1440 \ \Omega$ and a conductivity of $0.001262 \ \Omega^{-1} \ cm^{-1}$. (in $cm^{-1}$)

Which of the following is the correct equation representing the change in molar conductivity with respect to concentration for a weak electrolyte,where the symbols carry their usual meaning?

If the values of $\Lambda_{\infty}$ of $NH_4Cl$,$NaOH$ and $NaCl$ are $130$,$217$ and $109 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ respectively,the $\Lambda_{\infty}$ of $NH_4OH$ in $ohm^{-1} \ cm^2 \ equiv^{-1}$ is:

If the resistance of $0.1 \ M \ KCl$ solution in a conductance cell is $300 \ \Omega$ and conductivity is $0.013 \ S \ cm^{-1}$,then the value of cell constant is