The resistance of a conductivity cell filled with $0.1 \ M$ $KCl$ solution is $100 \ \Omega$ and the conductivity is $1.70 \times 10^{-4} \ S \ cm^{-1}$. What is the cell constant of the cell (in $cm^{-1}$)?

The molar conductivities $(\lambda_{m}^0)$ at infinite dilution of $KBr$,$HBr$,and $KNH_2$ are $120.5$,$420.6$,and $90.48 \ S \ cm^2 \ mol^{-1}$ respectively. Find the value of $\lambda_{m}^0$ for $NH_3$.

The resistance of a conductivity cell filled with $0.1 \ M$ $KCl$ solution is $100 \ \Omega$. If the resistance of the same cell when filled with $0.02 \ M$ $KCl$ solution is $520 \ \Omega$,the molar conductivity of $0.02 \ M$ solution (in $S \ cm^2 \ mol^{-1}$) is (Given: conductivity of $0.1 \ M$ $KCl$ solution $= 1.29 \ S \ m^{-1}$)

$A$ solution of concentration $C \ g \ equiv/L$ has a specific resistance $R$. The equivalent conductance of the solution is

The correct order of equivalent conductances at infinite dilution in water at room temperature for $H^{+}$, $K^{+}$, $CH_{3}COO^{-}$ and $HO^{-}$ ions is

The equivalent conductance at infinite dilution $\wedge^0$ for electrolytes $BA$ and $CA$ are $140$ and $120 \ S \ cm^2 \ eq^{-1}$,respectively. The equivalent conductance at infinite dilution for $BX$ is $198 \ S \ cm^2 \ eq^{-1}$. The $\wedge^0$ (in $S \ cm^2 \ eq^{-1}$) of $CX$ is: