What is the conductivity of $0.02 \text{ M}$ $AgNO_3$ solution having cell constant $1.2 \text{ cm}^{-1}$ and resistance $95.0 \text{ } \Omega$?

If $X$ is the specific resistance (resistivity) of the solution and $M$ is the molarity of the solution,the molar conductivity of the solution is given by:

What is the $SI$ unit of resistivity?

The equivalent conductance of $H_2SO_4$ at infinite dilution is $384 \, \Omega^{-1} \, cm^2 \, eq^{-1}$. If the solution contains $49 \, g$ of $H_2SO_4$ per liter and the specific conductance is $18.4 \, \Omega^{-1} \, cm^{-1}$,calculate the degree of dissociation in percentage.

Calculate the cell constant of a conductivity cell containing $0.1 \ M$ $KCl$ solution having a resistance of $60 \ \Omega$ and a conductivity of $0.014 \ \Omega^{-1} \ cm^{-1}$ at $25^{\circ} \ C$. (in $cm^{-1}$)

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.