1.

The solubility product of chalk is `9.3xx10^(-8)`. Calculate its solubility in gram per litreA. `0.3040 gram//litre`B. `0.0304 gram//litre`C. `2.0304 gram//litre`D. `4.0304 gram//litre`

Answer» Correct Answer - B
`CaCO_(3)hArrCa^(2+)+CO_(3)^(2-)`
Let the solubility of `CaCO_(3)` be `s` mole per litre
`:. K_(SP)=[Ca^(2+)][CO_(3)^(2-)]=s.S`
`:. S=sqrt(K_(sp))=sqrt(9.3xx10^(-8))=0.000304 "mole"//litre`
Solubility in `g//L="mole"//litrexx` Molecular weight of `CaCO_(3)=0.000304xx100=0.0304 gram//litre`


Discussion

No Comment Found

Related InterviewSolutions