Question

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

Answer 1

Correct Answer - 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)`