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:

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 solibility of AgCl at 25^(@)C is:

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.

At 298 K, the conductivity of a saturated solutioni of Agcl in water is 2.6xx10^(-6)ohm^(-1)cm^(-1) givem lamda_(m)^(oo)(Ag^(+))=63ohm^(-1)cm^(2)mol^(-1) and lamda_(m)^(oo)(Cl^(-))=67ohm^(-1)cm^(2)mol^(-1) The solubility product of AgCl is

The specific conductivity of the saturated aqueous solution of AgCl is 1.82xx10^(-6) ohm^(-1) cm^(-1) at 298K. Calculate me solubility of AgCl. [wedge_(m)^(@)(Ag^(+))=61.83 ohm^(-1) cm^(2) mol^(-1)] and wedge_(m)^(@) (CI)^(-)=76.41 ohm^(-1) cm^(2) mol^(-1)

The specific conductance of a saturated solution of AgCl at 298K is found to be 1.386xx10^-6S cm^-1 . Calculate its solubility (lambda^@_(Ag^+)=62.0 S cm^2 mol^-1 and lambda^@_(Cl^-)=76.3 S cm^2 mol^-1 ).

The conductivity of saturated solution of AgCl is found to be 1.86 xx 10^(-6) ohm^(-1)cm^(-1) and that of water is 6xx 10^(-8) ohm^(-1)cm^(-1) . If lamda^0 AgCl is