Equivalent conductance of $BaC{{l}_{2}}$ and HCl are ${{x}_{1}},\,{{x}_{2}}$ and ${{x}_{3}}\,S\,c{{m}^{2}}\,equi{{v}^{-1}}$ at infinite dilution. If specific conductance of saturated $BaS{{O}_{4}}$ solution is of y$S\,c{{m}^{-1}}$, then ${{K}_{sp}}$ of $BaS{{O}_{4}}$ is

Question

Accepted Answer

$\frac{{{10}^{6}}{{y}^{2}}}{{{({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}^{2}}}$

Answer

$\frac{{{10}^{3}}y}{2({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}$

Answer

$\frac{{{10}^{6}}{{y}^{2}}}{4{{({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}^{2}}}$

Answer

$\frac{{{x}_{1}}+{{x}_{2}}-2{{x}_{3}}}{{{10}^{6}}{{y}^{2}}}$

electrochemistry

Question: Equivalent conductance of $BaC{{l}_{2}}$ and HCl are ${{x}_{1}},\,{{x}_{2}}$ and ${{x}_{3}}\,S\,c{{m}^{2}}\,equi{{v}^{-1}}$ at infinite dilution. If specific conductance of saturated $BaS{{O}_{4}}$ solution is of y$S\,c{{m}^{-1}}$, then ${{K}_{sp}}$ of $BaS{{O}_{4}}$ is

electrochemistry

Question: Equivalent conductance of $BaC{{l}_{2}}$ and HCl are ${{x}_{1}},\,{{x}_{2}}$ and ${{x}_{3}}\,S\,c{{m}^{2}}\,equi{{v}^{-1}}$ at infinite dilution. If specific conductance of saturated $BaS{{O}_{4}}$ solution is of y$S\,c{{m}^{-1}}$, then ${{K}_{sp}}$ of $BaS{{O}_{4}}$ is

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