The molar conductivities λ m 0 at infinite dilution of KBr ,   HBr and KNH 2 are 120 . 5 , 420 . 6…

For $\mathrm{KBr}$ :-

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{KBr}}= {\mathrm{\lambda }}_{{\mathrm{K}}^{+}} + {\mathrm{\lambda }}_{{\mathrm{Br}}^{-}} \\ 120.5={\mathrm{\lambda }}_{{\mathrm{K}}^{+}} + {\mathrm{\lambda }}_{{\mathrm{Br}}^{-}} \end{gathered}$$

For $\mathrm{HBr}$ :-

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{HBr}}= {\mathrm{\lambda }}_{{\mathrm{H}}^{+}} + {\mathrm{\lambda }}_{{\mathrm{Br}}^{-}} \\ 420.6= {\mathrm{\lambda }}_{{\mathrm{H}}^{+}} + {\mathrm{\lambda }}_{{\mathrm{Br}}^{-}} \end{gathered}$$

For ${\mathrm{KNH}}_2$ :-

$$\begin{gathered} {\mathrm{\lambda }}_{{\mathrm{KNH}}_2}= {\mathrm{\lambda }}_{{\mathrm{K}}^{+}} + {\mathrm{\lambda }}_{{{\mathrm{NH}}_2}^{-}} \\ 90.48={\mathrm{\lambda }}_{{\mathrm{K}}^{+}} + {\mathrm{\lambda }}_{{{\mathrm{NH}}_2}^{-}} \end{gathered}$$

For, ${\mathrm{NaNO}}_3:-$

${\mathrm{\lambda }}_{{\mathrm{NH}}_3}= {\mathrm{\lambda }}_{{\mathrm{H}}^{+}} + {\mathrm{\lambda }}_{{{\mathrm{NH}}_2}^{-}}={\mathrm{\lambda }}_{{\mathrm{KNH}}_2}+{\mathrm{\lambda }}_{\mathrm{HBr}}-{\mathrm{\lambda }}_{\mathrm{KBr}}=90.48+420.6-120.5=390.5$

Hence, molar conductivity of ${\mathrm{NH}}_3$ would be  $390.5 \mathrm{S}.{\mathrm{cm}}^2·{\mathrm{mol}}^{-1}$.