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 .............. $cm^{-1}$.

Limiting molar conductivity of $CH_3COOH$ (i.e.,$\Lambda^o_{m(CH_3COOH)}$) is equal to:

Molar conductivities $\left(\Lambda_{m}^{\circ}\right)$ at infinite dilution of $NaCl$,$HCl$,and $CH_{3}COONa$ are $126.4$,$425.9$,and $91.0 \ S \ cm^{2} \ mol^{-1}$ respectively. $\Lambda_{m}^{\circ}$ for $CH_{3}COOH$ will be $:-$

The molar conductivity of $0.02 \ M$ $KCl$ solution is $410 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ at $25^{\circ} C$. Calculate its conductivity?

The ionic conductances of $Ba^{2+}$ and $Cl^{-}$ at infinite dilution are $127$ and $76 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively. The equivalent conductance of $BaCl_2$ at infinite dilution is ..... (in $.5$)

Equivalent conductances of $BaCl_2$,$H_2SO_4$ and $HCl$ at infinite dilution are $\Lambda_1^\infty$,$\Lambda_2^\infty$ and $\Lambda_3^\infty$ respectively. Then,the equivalent conductance of $BaSO_4$ at infinite dilution will be: