$A$ conductivity cell containing $5 \times 10^{-4} \ M \ NaCl$ solution develops a resistance of $14000 \ \Omega$ at $25^{\circ} C$. Calculate the conductivity of the solution if the cell constant is $0.84 \ cm^{-1}$.

The distance between electrodes of a conductivity cell is $0.98 \ cm$ and area of cross section is $1.96 \ cm^{2}$. What is the cell constant (in $cm^{-1}$)?

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. . . . . .

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 $:-$

At $298 \ K$,the resistance of a $0.01 \ N$ electrolyte solution is $210 \ \Omega$. If the cell constant of the conductivity cell used is $0.66 \ cm^{-1}$,what is the equivalent conductivity of the solution in $mho \ cm^2 \ eq^{-1}$?

Calculate the molar conductivity of a $0.02 \, M$ solution if its conductivity is $2.06 \times 10^{-3} \, S \, cm^{-1}$.