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 conductivity of $0.012 \ M$ $NaBr$ solution is $2.67 \times 10^{-4} \ S \ cm^{-1}$. What is its molar conductivity?

Conductivity of a solution is $1.26 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$. Calculate molar conductivity for $0.01 \ M$ solution.

If the equivalent conductance of $1 \ M$ benzoic acid is $12.8 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ and the equivalent conductances of benzoate ion and $H^{+}$ ion are $42$ and $288.42 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively,its degree of dissociation is ............. $\%$

Calculate $\Lambda _{HOAc}^{\infty }$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $H_2O$ at $25\ ^oC$.

Electrolyte	$\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$
$KCl$	$149.9$
$KNO_3$	$145.0$
$HCl$	$426.2$
$NaOAc$	$91.0$
$NaCl$	$126.5$

If the conductivity of $0.08 \ M$ $KCl$ solution is $2 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$,what is the molar conductivity of the solution?