Electrical conductance due to all the ions in $1 \ cm^3$ of a given solution is called as

Molar ionic conductivities of a divalent cation and a divalent anion are $57 \ S \ cm^2 \ mol^{-1}$ and $73 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of a solution of an electrolyte containing these ions will be:

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration $0.1 \, M$ is $100 \, \Omega .$ The conductivity of this solution is $1.29 \, S \, m^{-1}.$ Resistance of the same cell when filled with $0.2 \, M$ of the same solution is $520 \, \Omega .$ The molar conductivity of $0.2 \, M$ solution of electrolyte will be..........$\times 10^{-4} \, S \, m^2 \, mol^{-1}$

The equivalent conductivities of $Al^{3+}$ and $SO_4^{2-}$ ions at infinite dilution are $189$ and $160 \ \Omega^{-1} cm^2 eq^{-1}$ respectively. What is the equivalent conductivity of $Al_2(SO_4)_3$ at infinite dilution?

The property which decreases with dilution is:

The conductivity of a $0.05 \, M$ solution of a weak monobasic acid is $10^{-3} \, S \, cm^{-1}$. If $\lambda_{m}^{\infty}$ for the weak acid is $500 \, S \, cm^{2} \, mol^{-1}$,calculate the $K_{a}$ of the weak monobasic acid.