$$AgBr(s)$$ $$\rightleftharpoons$$ $$A{g^ + }(aq) + B{r^ - }(aq)$$ ..... [1]

$$AgN{O_3}(aq) \to A{g^ + }(aq) + NO_3^ - (aq)$$ ...... [2]

Suppose the solubility of AgBr in 10^$$-$$7 M AgNO₃ is s mol L^$$-$$1. Substituting in equation (1) and (2), we get,

$$\therefore$$ Total $$[A{g^ + }] = (s + {10^{ - 7}})M$$, $${K_{sp}}(AgBr) = [A{g^ + }][B{r^ - }]$$

or, $$12 \times {10^{ - 14}} = (s + {10^{ - 7}})s$$ or, $${s^2} + {10^{ - 7}}s = 12 \times {10^{ - 14}}$$

or, $${s^2} + {10^{ - 7}}s - 12 \times {10^{ - 14}} = 0$$ or, $$s = 3 \times {10^{ - 7}}M$$

$$\therefore$$ $$[B{r^ - }] = 3 \times {10^{ - 7}}M = 3 \times {10^{ - 7}} \times {10^3}{m^3} = 3 \times {10^{ - 4}}{m^3}$$

$$[A{g^ + }] = 3 \times {10^{ - 7}} + {10^{ - 7}} = 4 \times {10^{ - 7}}M$$

$$ = 4 \times {10^{ - 7}} \times {10^3}{m^3} = 4 \times {10^{ - 4}}{m^3}$$

$$[NO_3^ - ] = {10^{ - 7}}M = {10^{ - 7}} \times {10^3}{m^3} = 1 \times {10^{ - 4}}{m^3}$$

We know, $$\lambda = {\kappa \over C}$$ or, $$\kappa = \lambda \times C$$.

$$\therefore$$ $${\kappa _{B{r^ - }}} = 3 \times {10^{ - 4}} \times 8 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 24 \times {10^{ - 7}}S\,{m^{ - 1}}$$

$$\therefore$$ $${\kappa _{A{g^ + }}} = 4 \times {10^{ - 4}} \times 6 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 24 \times {10^{ - 7}}S\,{m^{ - 1}}$$

$$\therefore$$ $$\kappa _{NO_3^ - }^{} = 1 \times {10^{ - 4}} \times 7 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 7 \times {10^{ - 7}}S\,{m^{ - 1}}$$

$$\therefore$$ $${\kappa _{total}} = {\kappa _{B{r^ - }}} + {\kappa _{A{g^ + }}} + \kappa _{NO_3^ - }^{}$$

$$ = 24 \times {10^{ - 7}} \times 24 \times {10^{ - 7}} + 7 \times {10^{ - 7}} = 55 \times {10^{ - 7}}S\,{m^{ - 1}}$$