Solubility of calcium phosphate (molecular mass, `M`) in water is `W g` per `100 mL` at `25^(@)C`. Its solubility product at `25^(@)C` will be approximately

Options

A. `10^(9)((W)/(M))^(5)`
B. `10^(7)((W)/(M))^(5)`
C. `10^(5)((W)/(M))^(5)`
D. `10^(3)((W)/(M))^(5)`

Answer

B

Explanation

`S=(10W)/(M)mol//L`
`K_(sp)` of `Ca_(3)(PO_(4))_(2)=108 S^(6)`
`= 108(10W)/(M)=10^(7)((W)/(M))^(5)`(approximately)