Two salts $\mathrm{A}_{2} \mathrm{X}$ and $\mathrm{MX}$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of
their molar solubilities i.e. $\frac{\mathrm{S}\left(\mathrm{A}_{2} \mathrm{X}\right)}{\mathrm{S}(\mathrm{MX})}=$_________
(Round off to the Nearest Integer).
For $\mathrm{A}_{2} \mathrm{X}$
$\mathrm{A}_{2} \mathrm{X} \rightarrow 2 \mathrm{~A}^{+}+\mathrm{X}^{2-}$
$2 \mathrm{~S}_{1} \quad \mathrm{~S}_{1}$
$\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}_{1}^{3}=4 \times 10^{-12}$
$\mathrm{S}_{1}=10^{-4}$
for $\mathrm{MX}$
$\mathrm{MX} \rightarrow \mathrm{M}^{+}+\mathrm{X}^{-}$
$\mathrm{S}_{2} \quad \mathrm{~S}_{2}$
$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}_{2}^{2}=4 \times 10^{-12}$
$\mathrm{S}_{2}=2 \times 10^{-6}$
So $\frac{\mathrm{S}_{\mathrm{A}_{2} \mathrm{X}}}{\mathrm{S}_{\mathrm{MX}}}=\frac{10^{-4}}{2 \times 10^{-6}}=50$
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