When a solution of an electrolyte is heated,the conductance of the solution:

The limiting molar conductivities $\wedge ^0$ for $NaCl$,$KBr$,and $KCl$ are $126$,$152$,and $150 \ S \ cm^2 \ mol^{-1}$ respectively. The $\wedge ^0$ for $NaBr$ is ............ $S \ cm^2 \ mol^{-1}$.

Kohlrausch's law states that at:

If $0.01 \, M$ solution of an electrolyte has a resistance of $40 \, \Omega$ in a cell having a cell constant of $0.4 \, cm^{-1}$,then its molar conductance in $ohm^{-1} \, cm^2 \, mol^{-1}$ will be :-

The equivalent conductances of two strong electrolytes at infinite dilution in $H_2O$ (where ions move freely through a solution) at $25 ^\circ C$ are given below:
$\Lambda _{CH_3COONa}^o = 91.0 \ S \ cm^2 / equiv.$
$\Lambda _{HCl}^o = 426.2 \ S \ cm^2 / equiv.$
What additional information/quantity one needs to calculate $\Lambda ^o$ of an aqueous solution of acetic acid?

What is the conductivity of $0.05 \ M \ BaCl_2$ solution if its molar conductivity is $220 \ \Omega^{-1} \ cm^2 \ mol^{-1}$?