$1\times {10}^{-5} \mathrm{M} {\mathrm{AgNO}}_3$ is added to 1 L of saturated solution of AgBr. The conductivity of this solution at 298K is _____________$\times {10}^{-8}{\mathrm{Sm}}^{-1}$
Given: ${\mathrm{K}}_{\mathrm{SP}}\left(\mathrm{AgBr}\right)=4.9\times {10}^{-13}\text{ at }298\mathrm{K}$
$$\begin{gathered} {\mathrm{\lambda }}^0{\mathrm{Ag}}^{+}=6\times {10}^{-3}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1} \\ {\mathrm{\lambda }}^0{\mathrm{Br}}^{-}=8\times {10}^{-3}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1} \\ {\mathrm{\lambda }}^0{\mathrm{NO}}_3^{-}=7\times {10}^{-3}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1} \end{gathered}$$

${\mathrm{AgNO}}_3\to {\mathrm{Ag}}^{+}+{\mathrm{NO}}_3^{-}$

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\left[{\mathrm{NO}}_3^{-}\right]={10}^{-5} \\ \left[{\mathrm{Br}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left[{\mathrm{Ag}}^{+}\right]} \\ \left[{\mathrm{Br}}^{-}\right]=\frac{4.9\times {10}^{-13}}{{10}^{-5}}=4.9\times {10}^{-8} \end{gathered}$$

Using the following formula:

$$\begin{gathered} {\wedge }_{\mathrm{m}}=\frac{\mathrm{k}}{1000\times \mathrm{M}} \\ 6\times {10}^{-2}=\frac{{\mathrm{k}}_{{\mathrm{Ag}}^{+}}}{1000\times {10}^{-5}} \\ {\mathrm{k}}_{{\mathrm{Ag}}^{+}}=6\times {10}^{-5}=6000\times {10}^{-8} \\ 8\times {10}^{-3}=\frac{{\mathrm{k}}_{{\mathrm{Br}}^{-}}}{1000\times 4.9\times {10}^{-8}} \\ {\mathrm{k}}_{{\mathrm{Br}}^{-}}=39.2\times {10}^{-8} \\ 7\times {10}^{-3}=\frac{{\mathrm{k}}_{{\mathrm{NO}}_3^{-}}}{1000\times {10}^{-5}} \\ {\mathrm{k}}_{{\mathrm{NO}}_3^{-}}=7\times {10}^{-5}=7000\times {10}^{-8} \end{gathered}$$

Conductivity of solution:
$\left(6000+39.2+7000\right)\times {10}^{-8}=13039.2\times {10}^{-8} {\mathrm{Sm}}^{-1}$