A) \[1.62\times 1{{0}^{-}}^{4}\]
B) \[1.62\times 1{{0}^{-}}^{8}\]
C) \[1.62\times 1{{0}^{-}}^{2}\]
D) \[1.62\times 1{{0}^{-}}^{6}\]
Correct Answer: C
Solution :
Let solubility of \[PbC{{l}_{2}}=s\] \[Moles\,\,\,\,\,\underset{s}{\mathop{PbC{{l}_{2}}}}\,\underset{s}{\mathop{P{{b}^{2+}}}}\,+\underset{2s}{\mathop{2C{{l}^{-}}}}\,\] \[{{K}_{sp}}=\left[ P{{b}^{2+}} \right]{{\left[ C{{l}^{-}} \right]}^{2}}\] \[\therefore \,\,\,\,\,\,1.7\,\,\times \,\,10\,\,-\,\,5=\,\,(s)\,{{(2s)}^{2}}\] \[1.7\,\,\times \,\,1{{0}^{-\,5}}\,\,=\,\,4{{s}^{3}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,s{{=}^{3}}\sqrt{\frac{1.7\times {{10}^{-\,5}}}{4}}\,\,=\,\,1.62\,\,\times 1{{0}^{-\,2}}\]
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