1.

Equilibrium constant K_(p) for the reaction: CaCO_(3)(s)

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Solution :`CaCO_(3)(s) `K_(p) = P_(CO_(2)) = 0.82` ATM,
`n_(CO_(2)) =(PV)/(RT) = (0.82 xx 20)/(0.082 xx 1000) = 0.2` mole
Mole of `CaCO_(3)` dissociated `=n_(CO_(2)) =0.2`
AMOUNT dissociated `=0.2 xx 100 = 20g`


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