The resistance of a conductivity cell of $0.1 \ M$ $KCl$ solution is $120 \ \Omega$ and conductivity is $1.64 \times 10^{-4} \ S \ cm^{-1}$. What is the value of cell constant (in $cm^{-1}$)?

The conductivity of $0.20 \, M$ solution of $KCl$ at $298 \, K$ is $0.0248 \, S \, cm^{-1}$. Calculate its molar conductivity.

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:

The molar conductivity of a $0.02 \ M$ solution of an electrolyte is $124 \times 10^{-4} \ S \ m^2 \ mol^{-1}$. What is the resistance of the same solution (in ohms),kept in a cell with a cell constant of $129 \ m^{-1}$?

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 $m^{-1}$)?

Equivalent conductances of $BaCl_2$,$H_2SO_4$ and $HCl$ at infinite dilution are $\Lambda_1^\infty$,$\Lambda_2^\infty$ and $\Lambda_3^\infty$ respectively. Then,the equivalent conductance of $BaSO_4$ at infinite dilution will be: