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At 25^oC, K(sp) for AB2 salt is equal to...

At `25^oC`, `K_(sp)` for `AB_2` salt is equal to `8xx10^-6`. If the salt is 70% dissociated, then the solubility of `AB_2` in mol/litre is

A

`S=(frac{8xx10^-6}{0.98})^frac{1}{3}`

B

`S=(frac{5xx10^-6}{0.76})^frac{1}{3}`

C

`S=(frac{8xx10^-6}{0.98})^frac{1}{2}`

D

`S=(frac{5xx10^-6}{0.76})^frac{1}{2}`

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