A) \[4\times {{10}^{-8}}{{M}^{2}}\]
B) \[1\times {{10}^{-8}}{{M}^{2}}\]
C) \[2\times {{10}^{-4}}{{M}^{2}}\]
D) \[1\times {{10}^{-4}}{{M}^{2}}\]
Correct Answer: B
Solution :
[b] \[{{\lambda }_{0}}(BaS{{O}_{4}})=\frac{1000\times \text{sp}\,\text{.}\,\,\text{conductance}}{\text{conc}\text{.}\,\text{(Normality)}}\] |
\[\therefore \] Normality \[=\frac{1000\times 8\times {{10}^{-5}}}{400}\] |
\[\text{Molarity}=\frac{\text{Normality}}{2}={{10}^{-4}}M\] |
\[\therefore \] \[{{K}_{sp}}={{S}^{2}}={{({{10}^{-4}})}^{2}}={{10}^{-8}}{{M}^{2}}\] |
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