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

Options

A. 0.3040 gram/litre
B. 0.304 gram/litre
C. 2.0304 gram/litre
D. 4.0304 gram/litre

Answer

B

Explanation

`CaCO_(3) Leftrightarrow Ca^(2+)+CO_(3)^(2-)`
Let the solubility of `CaCO_3` be s mole per litre
`therefore K_(SP)=[Ca^(2+)] [CO_(3)^(2-)]=s.s`
` therefore s=sqrt(K_(sp))=sqrt(9.3 xx 10^(-8))`
=0.00304 mole/litre
Solubility in g/l= `"mole//itre" xx "Molecular weight of "CaCO_(3)`
`=0.000304 xx 100`
=0.0304 gram/litre