A) \[7\times {{10}^{-24}}mo{{l}^{2}}{{L}^{2}}\]
B) \[2.65\times {{10}^{-12}}mol\,{{L}^{-1}}\]
C) \[1.32\times {{10}^{-12}}mol\,{{L}^{-1}}\]
D) \[14\times {{10}^{-24}}mo{{l}^{3}}{{L}^{2}}\]
Correct Answer: B
Solution :
\[MS{{M}^{2+}}+{{S}^{2-}}\] \[{{K}_{sp}}=7\times {{10}^{-24}}mo{{l}^{2}}{{L}^{-2}}\] \[{{K}_{sp}}={{S}^{2}}\] \[=\sqrt{{{K}_{sp}}}\] \[=\sqrt{7\times {{10}^{-24}}}\] \[=2.645\times {{10}^{-12}}\] \[\approx 2.65\times {{10}^{-12}}mol\,{{L}^{-1}}\]
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