1.

Solubility product of AgCl is 1 xx 10^(-6) at 298 K. Its solubility in mole "litre"^(-1) would be

Answer»

`1 xx 10^(-6)` mol/ litre
`1 xx 10^(-3)` mol/litre
`1 xx 10^(-12)` mol/litre
None of these

Solution :`{:("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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