The resistance of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $1500 \ \Omega$. If the conductivity of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $0.1466 \times 10^{-3} \ S \ cm^{-1}$,the cell constant of the conductivity cell in $cm^{-1}$ is:

Calculate the degree of dissociation of $0.01 \ M$ acetic acid at $25^{\circ} C$ given that the molar conductivity at concentration $c$ is $\Lambda_{c} = 15.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and the limiting molar conductivity is $\Lambda_0 = 300 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

What is the molar conductivity of $0.02 \text{ M}$ $KI$ solution if its conductivity is $4.37 \times 10^{-4} \text{ }\Omega^{-1} \text{cm}^{-1}$?

The electrodes of a conductivity cell are $3 \text{ cm}$ apart and have a cross-sectional area of $4 \text{ cm}^2$. The cell constant of the cell (in $\text{cm}^{-1}$) is

The values of $\lambda^{\infty}_m$ for $NH_4Cl$,$NaOH$,and $NaCl$ are $129.8$,$248.1$,and $126.4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively. Calculate $\lambda^{\infty}_m$ for $NH_4OH$ solution (in $\Omega^{-1} \ cm^2 \ mol^{-1}$).

How can the ionic conductivity of an unknown solution be determined?