The specific conductance (conductivity) of a solution is $0.2 \ \Omega^{-1} cm^{-1}$ and its conductance is $0.04 \ \Omega^{-1}$. The cell constant would be .............. $cm^{-1}$.

The specific conductance of a solution containing $1.0 \ g$ of anhydrous $BaCl_2$ in $200 \ cm^3$ of solution is $0.0058 \ S \ cm^{-1}$. Calculate the molar and equivalent conductivity of the solution. (Molar mass of $BaCl_2 = 208 \ g \ mol^{-1}$)

The limiting molar conductivities $\wedge ^0$ for $NaCl$,$KBr$,and $KCl$ are $126$,$152$,and $150 \ S \ cm^2 \ mol^{-1}$ respectively. The $\wedge ^0$ for $NaBr$ is ............ $S \ cm^2 \ mol^{-1}$.

The equivalent conductance of $1 \, M$ benzoic acid is $12.8 \, \text{mho} \, \text{cm}^2 \, \text{g-eq}^{-1}$. If the molar conductivities of benzoate ion and $H^+$ ion are $42$ and $288.42 \, \Omega^{-1} \, \text{cm}^2$ respectively,then the degree of dissociation is .......... $\%$.

If $X$ is the specific resistance (resistivity) of the solution and $M$ is the molarity of the solution,the molar conductivity of the solution is given by:

What is the conductivity of $0.01 \ M$ $NaCl$ solution if the resistance and cell constant of the $NaCl$ solution are $375 \ \Omega$ and $0.5 \ cm^{-1}$ respectively at $298 \ K$?