Solubility product of silver bromide is 5.0 × 10^ -13 . The...

$\mathrm{Ag}^{+}+\mathrm{Br}^{-} \rightleftharpoons \mathrm{AgBr}$ Precipitation starts when ionic product just exceeds solubility product $$\begin{aligned} \mathrm{K}_{\text{sp }}= \left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]\\ & {\left[\mathrm{Br}^{-}\right]=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{Ag}^{+}\right]}=\frac{5 \times 10^{-13}}{0.05}=10^{-11}} \end{aligned}$$i.e., precipitation just starts when 10^{-11}$moles of$\mathrm{KBr} is added to 1 \mathrm{ L} of \mathrm{AgNO}_3 solution. No. of moles of \mathrm{KBr} to $be added$$\begin{aligned} \therefore\text{ weight\text{ of } } \mathrm{KBr} \text{ to be added } & =10^{-11} \times120 \\ & =1.2 \times10^{-9} \mathrm{ g} \end{aligned}$$