Solubility if `M_(2)S` type salt is `3.5xx10^(-6)`, then find out its solubility product

Options

A. `1.7 xx 10^(-6)`
B. `1.7 xx 10^(-16)`
C. `1.7 xx 10^(-18)`
D. `1.7 xx 10^(-12)`

Answer

B

Explanation

`M_(2)S hArr 2M^(+) +S^(2-)`
If S is solubility in mol `L^(-1)`
Then `K_(sp)=4s^(2)=4 xx (3.5 xx 10^(-6))^(3)`
`=1.715 xx 10^(-16)`