We have taken a saturated solution of $AgBr$. $K_{sp}$ of $AgBr$ is $12\times {10}^{-14}$. If ${10}^{-7}$ mole of $AgN{O}_3$ are added to 1 litre of this solution find conductivity (specific conductance) of this solution in terms of ${10}^{-7}$ Sm${}^{-1}$.

Given: ${\lambda }_{A{g}^{+}}^0=6\times {10}^{-3}$ Sm${}^2$ mol${}^{-1}$, ${\lambda }_{B{r}^{-}}^0=8\times {10}^{-3}$ Sm${}^2$ mol${}^{-1}$ and ${\lambda }_{N{O}_3^{-}}^0=7\times {10}^{-3}$ Sm${}^2$ mol${}^{-1}$ [If answer is $X$ then give answer in form of $\frac{X}{11}$]

$AgBr\to \underset{S+{10}^{-7}}{A{g}^{+}}+\underset{S}{B{r}^{-}}$
$K_{sp}=\left[A{g}^{+}\right]\left[B{r}^{-}\right]=[S+{10}^{-7}]\left[S\right]=12\times {10}^{-14}$ ($\because \left[A{g}^{+}\right]={10}^{-7}MfromAgN{O}_3$)
On solving quadratic eqn. for $S$ we get,
$S=3\times {10}^{-7}M$

The solubility of $AgBr$ in presence of ${10}^{-7}$ molar $AgN{O}_3$ is $3\times {10}^{-7}M$

Therefore $\left[B{r}^{-}\right]=3\times {10}^{-4}$ mol/m${}^3$,$\left[A{g}^{+}\right]=4\times {10}^{-4}$ mol/m${}^3$

$\left[N{O}_3^{-}\right]={10}^{-4}$ mol/m${}^3$

We know that specific conductance, $\kappa ={\lambda }_M^o\times M$

Therefore,

${\kappa }_{total}={\kappa }_{B{r}^{-}}+{\kappa }_{A{g}^{+}}+{\kappa }_{N{O}_3^{-}}$ $=55\times {10}^{-7}$ Sm${}^{-1}$
$X=55$

$\therefore$ Value of $\frac{X}{11}=5$
Ans-$5$