0.5 L of a saturated solution of Ag_(2)CrO_(4) is mixed with 0.5 L of 2 xx 10^(-6) M AgNO_(3). If concentration of [CrO_(4)^(2-)]after mixing is (10^(-6)/2)M and none of the ions undergo hydrolysis, then calculate a four digit number abcd where, ab = (K_(sp) of Ag_(2)CrO_(4))/(10^(-19) and cd = conductivity of final solution in terms of 10^(-8)Omega^(-1) cm^(-1). Assume temperature of the solution to be 298 K [Given data: Molar conductance at infinite dilution in Omega^(-1) cm^(2) mol^(-1) are lambda_(m)^(infty)(OH^(-)) = 200, lambda_(m)^(infty)NO_(3)^(-) = 75, lambda_(m)^(infty)Ag^(+) = 50 lambda_(m)^(infty)CrO_(4)^(2-) = 100

0.5 L of a saturated solution of Ag_(2)CrO_(4) is mixed with 0.5 L of

1.	0.5 L of a saturated solution of Ag_(2)CrO_(4) is mixed with 0.5 L of 2 xx 10^(-6) M AgNO_(3). If concentration of [CrO_(4)^(2-)]after mixing is (10^(-6)/2)M and none of the ions undergo hydrolysis, then calculate a four digit number abcd where, ab = (K_(sp) of Ag_(2)CrO_(4))/(10^(-19) and cd = conductivity of final solution in terms of 10^(-8)Omega^(-1) cm^(-1). Assume temperature of the solution to be 298 K [Given data: Molar conductance at infinite dilution in Omega^(-1) cm^(2) mol^(-1) are lambda_(m)^(infty)(OH^(-)) = 200, lambda_(m)^(infty)NO_(3)^(-) = 75, lambda_(m)^(infty)Ag^(+) = 50 lambda_(m)^(infty)CrO_(4)^(2-) = 100
Answer» ANSWER :2028