`1 xx 10^(–5) M AgNO_3` is added to 1 L of saturated solution of `AgBr`. The conductivity of this solution at `298 K is ________ xx 10^(–8) S m^(–1)`
Given : `K_(SP)(AgBr) = 4.9 xx 10^(–13)` at `298 K`
`lambda_(Ag^+)^0=6 times10^(-3)S m^2 mol^(-1)`,
`lambda_(Br^-)^0=8times10^(-3)S m^2 mol^(-1)`,
`lambda_(NO_3^(-))^(0)=7times10^(-3)S m^2 mol^(-1)`

We have taken a saturated solution of AgBr,K_(sp) of AgBr is 12xx10^(-14) . If 10^(-7) "mole" of AgNO_(3) are added to 1 litre of this solution then the conductivity of this solution in terms of 10^(-7) Sm^(-1) units will be: [Given: lambda_((Ag^(+)))^(@)=4xx10^(-3)Sm^(2)"mol"^(-1),lambda_((Br^(-)))^(@)=6xx10^(-3) S m^(2) "mol"^(-1),lambda_((NO_(3)^(-)))^(@)=5xx10^(-3)Sm^(2) "mol"^(-1) ]

We have taken a saturated solution of AgBr , whose K_(sp) is 12xx10^(-14). If 10^(-7)M of AgNO_(3) are added to 1L of this solutino, find the conductivity ( specific conductance ) of the solution in terms of 10^(-7)S m^(-1) units. Given : lambda^(@)._((Ag^(o+)))=6xx10^(-3)S m^(2) mol^(-1) lambda^(@)._((Br^(c-)))=8xx10^(-3)S m^(2)mol^(-1) lambda^(@)._((NO_(3)^(C-)))=7XX10^(-3)S m^(2) mol^(-1)

In a saturated aqueous solution of AgBr, concentration of Ag^(+) ion is 1xx10^(-6) mol L^(-1) it K_(sp) for AgBr is 4xx10^(-13) , then concentration of Br^(-) in solution is

At 298K the conductivity of a saturated solution of AgCl in water is 2.6times10^(-6)Scm^(-1) .Its solubility product at 298K (given : lambda^(oo) ( Ag^(+) )=63.0 Scm^(2)mol^(-1) lambda^(oo) (Cl^(-)) =67.0 Scm^(2)mol^(-1) )

The conductivity of a saturated solution of AgCl at 25^(@)C after subtracting the conductivity of water is 2.28 xx 10^(-6) S cm^(-1) . Calculate the solubility product of AgCl at 25^(@)C if Lambda_(m)^(0) (AgCl) 138.3 S cm^(2) mol^(-1)