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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
`S=(mol)/(Litre)`
Now in `grm //litre = 5xxm.wt`
`Ksp=1.56 25xx10^(-10)`
` s=1.25xx10^(-5)mol//L`
`s=143.5xx1.25xx10^(-5)gm//L`
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