The solubility product of chalk is `9.3xx10^(-8)`. Calculate its solubility in gram per litre

Explanation

`CaCO_(3) Leftrightarrow Ca^(2+)+CO_(3)^(2-)`
Let the solublity of `CaCO_3` be S moles per litre.
`K_(sp)=[Ca^(2+)] [CO_(3)^(2-)]=S.S`
`S=sqrt(K_(sp))=sqrt(9.3 xx 10^(-8))=0.000304" mole/litre"`
Solubility in g/L=mole/litre `xx` mol. w.t. of `CaCO_3`
`=0.000304 xx 100`
=0.03004 gram/litre.