JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Topic Test - Electrochemistry

Equivalent conductance of saturated \[BaS{{O}_{4}}\] is \[400\,{{\Omega }^{-1}}\,c{{m}^{2}}eq{{u}^{-1}}\] and specific conductance is\[8\times {{10}^{-5}}\,{{\Omega }^{-1}}\,c{{m}^{-1}}\]. Hence, \[{{K}_{sp}}\] of \[BaS{{O}_{4}}\] is

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 \[=\frac{1000\times 8\times {{10}^{-5}}}{400}\]

\[\text{Molarity}=\frac{\text{Normality}}{2}={{10}^{-4}}M\]

\[\therefore \] \[{{K}_{sp}}={{S}^{2}}={{({{10}^{-4}})}^{2}}={{10}^{-8}}{{M}^{2}}\]