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. `3.73xx10^(-3)g//L`
C. `3.7xx10^(-5)M`
D. `2.6xx10^(-)g//L`

Answer

A, B

Explanation

`A^(oo)=62.3+67.7=130Scm^(2)mol^(-1)`
`rArrDelta^(oo)=(K1000)/(S)rArrS=(3.4xx10^(-6)xx1000)/(130)`
`=2.6xx10^(-5)"molL"^(-1)`
In `gL^(-1)` unit , solubility =`2.6xx10^(-5)xx143.5`
`=3.7xx10^(-3)gL^(-1)`