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The EMF of the cell Ag|0.1(N)AgNO(3)||...

The `EMF` of the cell
`Ag|0.1(N)AgNO_(3)||1(N) KBr, AgBr(s)|Ag` was found to be `-0.64V` at `298 K. 0.1 N Ag NO_(3)` is `81.3%` dissociated and` 1N KBr` is `75.5%` dissociated.
Calculate `:`
`a.` Solubility
`b.` Solubility product of `AgBr ` at `298 K`

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