Explanation:
\({\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{2A}}{{\text{g}}^{\text{ + }}}{\text{ + CrO}}_{\text{4}}^{{\text{2 - }}}{\text{;}}{{\text{K}}_{{\text{sp}}}}{\text{ = 1}}{\text{.1 1}}{{\text{0}}^{{\text{ - 12}}}}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,{\text{S}}\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{2S}}\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{S}}\)
(solubility of \(\mathrm {\mathrm{AgBr}}\) is \(\mathrm {\mathrm{S} \mathrm{mol} \mathrm{L}^{-1}}\) )
\(\mathrm {\mathrm{K}_{\mathrm{sp}}=\left[A g^{+}\right]\left[B r^{-}\right]=S . S=S^{2}}\)
or \(\mathrm {S=\sqrt{K_{s p}}=\sqrt{5.0 \times 10^{-13}}=\sqrt{50 \times 10^{-14}}}\)
\(\mathrm {S=7.07 \times 10^{-7} \mathrm{M}}\)
Ratio of their solubilities
\(\mathrm {\dfrac{S\left(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right)}{S(\mathrm{AgBr})}=\dfrac{6.50 \times 10^{-5} \mathrm{M}}{7.07 \times 10^{-7} \mathrm{M}}=91.93 \approx 92}\)
\(\mathrm {\mathrm{Ag}_{2} \mathrm{CrO}_{4}}\) is 92 times more soluble than \(\mathrm {\mathrm{AgBr}}\).