$1 \times 10^{-5} ~\mathrm{M} ~\mathrm{AgNO}_{3}$ is added to $1 \mathrm{~L}$ of saturated solution of $\mathrm{AgBr}$. The conductivity of this solution at $298 \mathrm{~K}$ is _____________ $\times 10^{-8} \mathrm{~S} \mathrm{~m}^{-1}$.

[Given : $\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times 10^{-13}$ at $298 \mathrm{~K}$

$$ \begin{aligned} & \lambda_{\mathrm{Ag}^{+}}^{0}=6 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{Br}^{-}}^{0}=8 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \left.\lambda_{\mathrm{NO}_{3}^{-}}^{0}=7 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\right] \end{aligned} $$

$$ \begin{aligned} & \operatorname{AgBr}(\mathrm{S}) \rightleftharpoons \underset{\left(10^{-5}+\mathrm{x}\right)}{\rightleftharpoons \mathrm{Ag}^{+}}(\mathrm{aq})+\mathrm{Br}_{\mathrm{x}}^{-}(\mathrm{aq}) \\\\ & x\left(x+10^{-5}\right)=4.9 \times 10^{-13} \\\\ & x \simeq 4.9 \times 10^{-8} \mathrm{M} \\\\ & \lambda_{\mathrm{Ag}^{+}}^{\circ}=6 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \lambda_{\mathrm{Br}^{-}}^0=8 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \mathrm{~K}_{\text {solution }}=\mathrm{K}_{\mathrm{Ag}^{+}}+\mathrm{K}_{\mathrm{Br}^{-}}+\mathrm{K}_{\mathrm{NO}_3^{-}} \\\\ & =6 \times 10^{-3} \times 10^{-5} \times 10^3+8 \times 10^{-3} \times 4.9 \times 10^{-8} \times 10^3 \\\\ & +7 \times 10^{-3} \times 10^{-5} \times 10^3 \\\\ & =(6000+39.2+7000) \times 10^{-8} \\\\ & =13039.2 \times 10^{-8} ~\mathrm{Sm}^{-1} \end{aligned} $$