Thank you for reporting, we will resolve it shortly
Q.
The solubility product constant of $ Ag_2CrO_4 $ and $ AgBr $ are $1 .1 \times 10^{-12} $ and $5.0 \times 10^{-13} $ respectively. The ratio of the molarities is
Equilibrium
Solution:
$K_{s p}\left( Ag _{2} CrO _{4}\right)>K_{s p}( AgBr )$
$\therefore Ag_2CrO_4 $ is more soluble
$ Ag_2 CrO_4{(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{-2}_4(aq) $,
$K_{sp} =4S^{3}and \therefore S=\left(\frac{1.1\times10^{-12}}{4}\right)^{1 3} = 6.50 \times10^{-5}M $
$AgBr_{\left(s\right)}\rightleftharpoons Ag^{+}_{aq} +Br^{-}_{aq }; K_{sp} =S^{2} $
$\therefore S^{'} =\left(5\times10^{-13}\right)^{12 }=7.07\times10^{-7}M $
The ratio o f the molarities $=\frac{S}{S^{'}} =\frac{6.50\times10^{-5}}{7.07 \times10^{-7}} =91.94$