Which is the most soluble from the following
- $\mathrm{Bi}_{2} \mathrm{~S}_{3}\left[\mathrm{~K}_{\mathrm{sp}}=1 \times 10^{-70}ight]$
- $\operatorname{MnS}\left[\mathrm{K}_{\mathrm{sp}}=7 \times 10^{-16}ight]$
- $\operatorname{CuS}\left[\mathrm{K}_{\mathrm{sp}}=8 \times 10^{-37}ight]$
- $\mathrm{Ag}_{2} \mathrm{~S}\left[\mathrm{~K}_{\mathrm{sp}}=6 \times 10^{-51}ight]$
Solution
$\mathrm{MnS}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}=\sqrt{7 \times 10^{-16}}=2.5 \times 10^{-8}$ *
Asked in: JEE-TOPICTESTS-CHEMISTRY