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The solubility of `BaSO_(4)` in water is `2.33 xx 10^(-3)` gm/litre. Its solubility product will be (molecular weight of `BaSO_(4) = 233`)A. `1 xx 10^(-5)`B. `1 xx 10^(-10)`C. `1 xx 10^(-15)`D. `1 xx 10^(-20)` |
Answer» Correct Answer - B The solubility of `BaSO_(4)` in g/litre is given `2.33 xx 10^(-3)` `because` in mole/litre. `n = (W)/(m.wt) = 1 xx 10^(-5) = (2.33 xx 10^(-3))/(233)` Because `BaSO_(4)` is a compound `K_(sp) = S^(2) = [1 xx 10^(-5)]^(2) = 1 xx 10^(-10)`. |
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