A saturated solution of silver bromide is made in  ${10}^{-7}M AgN{O}_3$ solution 
Given:  $K_{sp}ofAgBr=3\times {10}^{-13}$
 $$\begin{gathered} {\lambda }_m^{\infty }\left(A{g}^{+}\right)=6\times {10}^{-3}S{m}^{2}mo{l}^{-1} \\ {\lambda }_m^{\infty }\left(N{O}_3^{-}\right)=7\times {10}^{-3}S{m}^{2}mo{l}^{-1} \\ {\lambda }_m^{\infty }\left(B{r}^{-}\right)=8\times {10}^{-3}S{m}^{2}mo{l}^{-1} \end{gathered}$$
For water,  $k=7.5\times {10}^{-6}S{m}^{-1}$
The conductivity of solution is  $x\times {10}^{-7}S{m}^{-1}$. The value of  $x/10$ is…..

Let solubility of  $\text{AgBr}\left(\text{s}\right){\text{\hspace{0.17em}in\hspace{0.17em}AgNO}}_{\text{3}}$ solution be s(M)

$\text{AgBr}\left(\text{s}\right)\rightleftharpoons {\text{\hspace{0.17em}Ag}}^{+}+{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Br}}^{-}$

$$\begin{gathered} {\left({\text{s+10}}^{\text{7}}\right)}_{\text{s}} \\ \left({\text{s+10}}^{\text{7}}\right)=3\times {10}^{-13} or {\text{s}}^2{\text{+10}}^{\text{-7}}s-30\times {10}^{-14}=0 \\ or {\text{s}}^2\text{+6}\times {\text{10}}^{\text{-7}}s-5\times {10}^{-7}({\text{s}}^2\text{+6}\times {\text{10}}^{\text{-7}})=0 \\ s(\text{s+6}\times {\text{10}}^{\text{-7}})-5\times {10}^{-7}(\text{s+6}\times {\text{10}}^{\text{-7}})=0 \\ or s(\text{s-5}\times {\text{10}}^{\text{-7}})(\text{s+6}\times {\text{10}}^{\text{-7}})=0 \\ {\text{s=5×10}}^{\text{-7}}\left(\text{M}\right) \end{gathered}$$

$$\begin{gathered} {\pi }_m^{\infty }\left(AgN{O}_3\right)={\lambda }_m^{\infty }\left(A{g}^{+}\right)+{\lambda }_m^{\infty }\left(N{O}_3^{-}\right)=13\times {10}^{-3}s{m}^{2}mo{l}^{-1} \\ {\pi }_m^{\infty }\left(AgBr\right)={\lambda }_m^{\infty }\left(A{g}^{+}\right)+{\lambda }_m^{\infty }\left(B{r}^{-}\right)=14\times {10}^{-3}s{m}^{2}mo{l}^{-1} \\ k\left(AgBr\right)=\left[{\pi }_m^{\infty }\left(AgBr\right)\right]C=70\times {10}^{-7}s{m}^{-1} \\ {k}_{solution}=k\left(AgN{O}_3\right)+k\left(AgBr\right)+k\left(H_{2}O\right)=(13\times {10}^{-7}+70\times {10}^{-7}+75\times {10}^{-7}) \\ =158\times {10}^{-7}s{m}^{-1})\Rightarrow \frac{x}{10}=15.8 \end{gathered}$$