A) \[3.75\times {{10}^{-4}}\]
B) \[3.34\times {{10}^{-2}}\]
C) \[3.34\times {{10}^{2}}\]
D) None of these
Correct Answer: B
Solution :
\[\underset{S}{\mathop{PbC{{l}_{2}}}}\,\]⇌\[\underset{S\,\,\,}{\mathop{P{{b}^{++}}}}\,+\underset{{{(2S)}^{2}}}{\mathop{2C{{l}^{-}}}}\,\] \[{{K}_{sp}}=S\times {{(2S)}^{2}}=4{{S}^{3}}\] \[S=\sqrt[3]{\frac{{{K}_{sp}}}{4}}=\sqrt[3]{\frac{1.5\times {{10}^{-4}}}{4}}=3.34\times {{10}^{-2}}\].
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