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The solubility product of AgCl is 1.56xx...

The solubility product of `AgCl` is `1.56xx10^(-10)`find solubility in g/ltr

A

`143.5`

B

108

C

`1.57xx10^(-8)`

D

`1.79xx10^(-3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`AgClhArrunderset(s)(Ag)^(+)+underset(s)(Cl^(-))`
`s^(2)=1.5625xx10^(-10)`
`s=1.25xx10^(-5)"mol L"^(-1)`
Solubility in g `L^(-1)` = Molar mass `xx` s
`=143.5xx1.25xx10^(-5)=1.79xx10^(-3)"g L"^(-1)`
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