Resistance of $0.1\, M\, KCl$ solution in a conductance cell is $300\, \Omega$ and conductivity is $0.013\, S\, cm^{-1}$. The value of cell constant is

The resistance of a $0.1 \, N$ solution of a salt is $2.5 \times 10^{3} \, \Omega$. If the cell constant is $1.15 \, cm^{-1}$,what will be the equivalent conductance of the solution in $\Omega^{-1} \, cm^{2} \, eq^{-1}$ (in $.6$)?

$A$ solution of concentration $C \ g \ equiv/L$ has a specific resistance $R$. The equivalent conductance of the solution is

Molar conductivity of $0.02 \ M$ weak acid is $7.92 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and its molar conductivity at infinite dilution is $232.7 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. Calculate the degree of dissociation of the weak acid.

Equivalent conductivity of $Cr_2(SO_4)_3$ is related to molar conductivity by the expression

Which solution of $NaCl$ will show the highest resistance during the passage of current (in $N$)?