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Solubility product of AgCl is `1 xx 10^(-6)` at 298 K. Its solubility in mole `"litre"^(-1)` would beA. `1 xx 10^(-6)` mol/ litreB. `1 xx 10^(-3)` mol/litreC. `1 xx 10^(-12)` mol/litreD. None of these |
Answer» Correct Answer - B `{:("For",AgClrarr,Ag^(+)+,Cl^(-)),(,,x,x):}` `K_(sp) = x^(2), x = sqrt(K_(sp)), sqrt(1 xx 10^(-6)) = 1 xx 10^(-3)` mole/litre. |
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