The `K_(sp) " for " Cr(OH)_(3) " is " 1.6xx10^(-30)`. The solubility of this compound in water is :

Options

A. `root(4)(1.6xx10^(-30))`
B. `root(4)(1.6xx10^(-30)//27)`
C. `1.6xx10^(-30//27)`
D. `sqrt(1.6xx10^(-30)`

Answer

B

Explanation

`Cr(OH)_(s)(s) hArr Cr^(3+)(aq)underset(S)+3OH^(-)underset(3S)(aq)`
`(s) (3s)^(3)=K_(sp)`
`27S^(4)=K_(sp)`
`s=((K_(sp))/(27))^(1//4)=((1.6xx10^(-30))/(27))^(1//4)`