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If the solubility product of lead iodide `(Pbl_(2))` is `3.2 xx 10^(-8)`, then its solubility in moles/litre will beA. `2 xx 10^(-3)`B. `4 xx 10^(-4)`C. `1.6 xx 10^(-5)`D. `1.8 xx 10^(-5)` |
Answer» Correct Answer - A `K_(sp) = 4S^(3)` `4S^(3) = 3.2 xx 10^(-8), S = 2 xx 10^(-3) M`. |
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