The specific conductivity of a saturated solution of AgCl is `3.40xx10^(-6) ohm^(-1) cm^(-1)` at `25^(@)C`. If `lambda_(Ag^(+)=62.3 ohm^(-1) cm^(2) "mol"^(-1)` and `lambda_(Cl^(-))=67.7 ohm^(-1) cm^(2) "mol"^(-1)`, the solubility of AgC at `25^(@)C` is:

Options

A. `2.6xx10^(-5)M`
B. `4.5xx10^(-3)M`
C. `3.6xx10^(-5)M`
D. `3.6xx10^(-3)M`

Answer

A

Explanation

`lamda_(Ag^(+))=62.3Scm^(2)mol^(-1),lamda_(cl^(-))=67.7Scm^(2)"mole"^(-1)`
`K_(Agd)=3.4xx10^(-6)Scm^(-1)`
`^^_(AgCl)^(infty)=(62.3+67.5)=(1000xx3.4xx10^(-6))/(S)`
`S=(3.4xx10^(-3))/("normality")`
Normality`=2xx10^(-3)M`
molarity`=(2xx10^(-3))/(2)=10^(-3)M`