The values of $\lambda^{\infty}_m$ for $NH_4Cl$,$NaOH$,and $NaCl$ are $129.8$,$248.1$,and $126.4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively. Calculate $\lambda^{\infty}_m$ for $NH_4OH$ solution (in $\Omega^{-1} \ cm^2 \ mol^{-1}$).

Molar conductivities $\left(\Lambda_{m}^{\circ}\right)$ at infinite dilution of $NaCl$,$HCl$,and $CH_{3}COONa$ are $126.4$,$425.9$,and $91.0 \ S \ cm^{2} \ mol^{-1}$ respectively. $\Lambda_{m}^{\circ}$ for $CH_{3}COOH$ will be $:-$

Why does the conductivity of a solution decrease with dilution?

If the values of $\Lambda_{\infty}$ of $NH_4Cl$,$NaOH$ and $NaCl$ are $130$,$217$ and $109 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ respectively,the $\Lambda_{\infty}$ of $NH_4OH$ in $ohm^{-1} \ cm^2 \ equiv^{-1}$ is:

The variation of molar conductivity with concentration of an electrolyte $X$ in aqueous solution is shown in the given figure.
The electrolyte $X$ is :

Conductivity of $0.001 \ M$ aqueous solution of $Na_2SO_4$ is found to be $2.6 \times 10^{-3} \ S \ cm^{-1}$ at $25 \ ^oC$. If limiting molar conductance of $Na^+$ is $50 \ S \ cm^2 \ mol^{-1}$,then limiting molar conductance of $SO_4^{2-}$ will be .............. $S \ cm^2 \ mol^{-1}$ (neglect conductivity of water).