$A$ $0.5\,M\,NaOH$ solution offers a resistance of $31.6\,\Omega$ in a conductivity cell at room temperature. What shall be the approximate molar conductance of this $NaOH$ solution if the cell constant of the cell is $0.367\,cm^{-1}$? (in $S\,cm^2\,mol^{-1}$)

The specific conductivity of $N/10$ $KCl$ solution at $20 \, ^oC$ is $0.012 \, \Omega^{-1} \, cm^{-1}$ and the resistance of the solution in the cell at $20 \, ^oC$ is $56 \, \Omega$. The cell constant is ........... $cm^{-1}$.

Resistance of a conductivity cell filled with $0.1 \ M$ $KCl$ solution is $100 \ \Omega$ and conductivity of the solution is $1.29 \ S/m$. What will be the value of the cell constant (in $m^{-1}$)?

At $298 \ K$,the limiting molar conductivity of a weak monobasic acid is $4 \times 10^2 \ S \ cm^2 \ mol^{-1}$. At $298 \ K$,for an aqueous solution of the acid,the degree of dissociation is $\alpha$ and the molar conductivity is $y \times 10^2 \ S \ cm^2 \ mol^{-1}$. At $298 \ K$,upon $20$ times dilution with water,the molar conductivity of the solution becomes $3y \times 10^2 \ S \ cm^2 \ mol^{-1}$. $(1)$ The value of $\alpha$ is. . . . . . $(2)$ The value of $y$ is. . . . . .

The limiting molar conductivities $(\Lambda_0)$ for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ S \ cm^2 \ mol^{-1}$ respectively. What is the $\Lambda_0$ of $NaBr$?

Discuss the construction and uses of a conductivity cell.