The specific conducitvity of a saturated solution of `AgCI` is ` 3.40 xx 10^6 "ohm"^(-1) cm^(-1) ` at ` 25^@C`. If ` lambda_(Ag+) = 62.3 "ohm"^(-1) cm^2 "mol"^(-1)` and `lambda_(CI-) = 67.7 "ohm"^(-1) cm^2 "mol"^(-1)`, the solubility of `AgCI` 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)`