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) "mol" L^(-1)`
B. `3.731xx10^(-3) "mol" L^(-1)`
C. `3.731xx10^(-5) "mol" L^(-1)`
D. `2.6xx10^(-3)g L^(-1)`

Answer

A

Explanation

`lambda_(AgCl)=)62.3+67.7)Omega^(-) "mol"^(-1)`
`=130 Omega^(-)"mol"^(-1)`
`s=(kxx1000)/(lambda_(AgCl))=(3.4xx10^(-6)xx1000)/(130)=2.6xx10^(-5) "mol" L^(-1)`