A) \[1\times {{10}^{-5}}\]
B) \[1\times {{10}^{-10}}\]
C) \[1.435\times {{10}^{-5}}\]
D) \[108\times {{10}^{-3}}\]
Correct Answer: B
Solution :
\[AgCl\,A{{g}^{+}}+C{{l}^{-}}\] \[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]\] \[{{K}_{sp}}={{S}^{2}}\] (S = solubility in mol/ L) \[S=\frac{1.435\times {{10}^{-3}}g/L}{143.5}\] \[=1\times {{10}^{-5}}\,mol/L\] \[\therefore \] \[{{K}_{sp}}={{(1\times {{10}^{-6}})}^{2}}\] \[=1\times {{10}^{-10}}\,mo{{l}^{2}}/{{L}^{2}}\]
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