The specific conductivity of a standard solution of AgCl is `1.40 xx 10^(-6) "ohm"^(-1) cm^(-1)` at `25^(@)C`. If `lambda_(Ag^(+))^(@) = 62.3 "ohm"^(-1) cm^(2) mol^(-1)` & `lambda_(Cl^(-1)) = 67.7 "ohm"^(-1) cm^(2) "mol"^(-1)`, the solubility of AgCl at `25^(@) C` is:

Options

A. `2.6 xx 10^(-5) M`
B. `4.5 xx 10^(-3) M`
C. `3.6 xx 10^(-5) M`
D. `3.6 xx 10^(-1) M`

Answer

A

Explanation

`lambda_(Ag^(+))= 62.3 S cm^(2) mol^(-1), lambda_(Cl^(-)) = 67.7 S cm^(2)` mole
`K_(AgCl) = 3.4 xx 10^(-6) S cm^(-1)` `wedge_(AgCl)^(infty) = (62.3 + 67.5) =(1000 xx 3.4 xx 10^(-4))/S`
`S = (3.4 xx 10^(-3))/(62.3 + 67.5) =2.6 xx 10^(-5) M`