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The solubility product of AgCl is 10^(-1...

The solubility product of AgCl is `10^(-10)` `M^(2)`. The minimum volume `(in m^(3))` of water required to dissolve `14.35`mg of AgCl is approximately :

A

`0.01`

B

`0.1`

C

100

D

10

Text Solution

Verified by Experts

The correct Answer is:
A

`[Ag^(+)]or S=sqrt(K_(sp))=10^(-5)M" "rArr" "10^(-5)=(14.35xx10^(-5))/((143.5)/(V("in litre")))`
`V=10` litre or `0.01m^(3)`
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