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The solubility product of `AgCl` is `1.56xx10^(-10)`find solubility in g/ltrA. `143.5`B. 108C. `1.57xx10^(-8)`D. `1.79xx10^(-3)` |
Answer» Correct Answer - D `AgClhArrunderset(s)(Ag)^(+)+underset(s)(Cl^(-))` `s^(2)=1.5625xx10^(-10)` `s=1.25xx10^(-5)"mol L"^(-1)` Solubility in g `L^(-1)` = Molar mass `xx` s `=143.5xx1.25xx10^(-5)=1.79xx10^(-3)"g L"^(-1)` |
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