Chemistry - Electrochemistry Question with Solution | TestHub

$\left[{\mathrm{Ag}}^{+}\right]=10^{-5}$

$\left[{\mathrm{NO}}_3^{-}\right]=10^{-5}$

$\left[{\mathrm{Br}}^{-}\right]=\frac{\mathrm{Ksp}}{\left[{\mathrm{Ag}}^{+}\right]}=4.9\times 10^{-8}$

${\Lambda }_{\mathrm{m}}=\frac{\mathrm{k}}{1000\times \mathrm{M}}$

For ${\mathrm{Ag}}^{+}$

$6\times 10^{-3}=\frac{{\kappa }_{{\mathrm{Ag}}^{+}}}{1000\times 10^{-5}}$

${\kappa }_{Ag+}=6\times 10^{-5}\Rightarrow 6000\times 10^{-8}$

for ${\mathrm{Br}}^{-}$

$8\times 10^{-3}=\frac{{\kappa }_{B{r}^{-}}}{1000\times 4.9\times 10^{-8}}$

${\kappa }_{B{r}^{-}}=39.2\times 10^{-8}$

for ${\mathrm{NO}}_3^{-}$

$7\times 10^{-3}=\frac{{\kappa }_{N{O}_3^{-}}}{1000\times 10^{-5}}$

${\kappa }_{N{O}_3^{-}}=7\times 10^{-5}$

$=7000\times 10^{-8}$

Conductivity of solution

$\Rightarrow (6000+7000+39.2)\times 10^{-8}$

$\Rightarrow 13039.2\times 10^{-8}S{m}^{-1}$