If the conductance and specific conductance of a solution are both $1$,then its cell constant would be:

The resistance of a conductivity cell containing $0.01 \, M \, KCl$ solution at $298 \, K$ is $1750 \, \Omega$. If the conductivity of $0.01 \, M \, KCl$ solution at $298 \, K$ is $0.152 \times 10^{-3} \, S \, cm^{-1}$,then the cell constant of the conductivity cell is $.......... \, \times 10^{-3} \, cm^{-1}$.

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 $....... \, S \, cm^{2} \, mol^{-1}$.

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:

The equivalent conductances of $CH_3COONa$,$HCl$,and $CH_3COOH$ at infinite dilution are $91$,$426$,and $391 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively. What is the equivalent conductance of $NaCl$ at infinite dilution?

If the specific conductance of $1 \, M \, H_2SO_4$ solution is $26 \times 10^{-2} \, S \, cm^{-1}$,then the equivalent conductivity would be