The conductivity of $0.001 \ M$ acetic acid at a certain temperature is $5.07 \times 10^{-5} \ S \ cm^{-1}$. If $\wedge_m^0$ of acetic acid at the same temperature is $390 \ S \ cm^2 \ mol^{-1}$,the dissociation constant of acetic acid at that temperature is:

Resistance of a cell containing $0.02 \ M \ KCl$ solution is $164 \ \Omega$. If the cell is filled with $0.05 \ M \ AgNO_3$,the resistance becomes $75.8 \ \Omega$. Calculate the following: [Conductivity of $0.02 \ M \ KCl = 2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$] $(i)$ Conductivity of $0.05 \ M \ AgNO_3$ (ii) Molar conductivity of $AgNO_3$ solution.

The conductivity of $0.3 \ M$ solution of $KCl$ at $298 \ K$ is $0.0627 \ S \ cm^{-1}$. What is its molar conductivity?

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 $.......... \, \times 10^{-3} \, cm^{-1}$.

What happens when the solution of an electrolyte is diluted?

Given $l/a = 0.5 \ cm^{-1}$,$R = 50 \ \Omega$,$N = 1.0$. The equivalent conductance of the electrolytic cell is ................ $\Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.