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 mitre of this solutio then the conductivity of this solution in terms of `10^(-7)Sm^(-1)` units will be [given `lamda((Ag^(+)))^(@)=4xx10^(-3)Sm^(2)mol^(-1)lamda_((Br^(-)))^(@)=6xx10^(-3)Sm^(2)mol^(-1). lamda_((NO_(3)^(-)))^(@)=5xx10^(-3)Smmol^(-1))`
(A). 39
(B). 55
(C). 15
(D). 41

We have taken a saturated solultion of AgBr, K_(SP) is 12 xx 10^(-4) . If 10^(-7) M of AgNO_(3) are added to 1 L of this solution, find conductivity (specific conductance) of this solution in term of 10^(-7) Sm^(-1) units. Given, lambda_((Ag^(+)))^(@) = 6 xx 10^(-3) Sm^(2) mol^(-1) , lambda_((Br^(-)))^(@) = 8 xx 10^(-3) Sm^(2) mol^(-1) , lambda_((NO_(3)^(-)))^@ = 7 xx 10^(-3) Sm^(2) mol^(-1) . Neglect the contribution of solvent.

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)

K_(sp) of AgCl in water at 25^(@)C is 1.8xx10^(-10) . If 10^(-5) mole of Ag^(+) ions are added to this solution. K_(sp) will be:

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

What is the molarity of F^(-) in a saturated solution of In F_(3) ? (K_(sp)=7.9xx10^(-10)

In 1L saturated solution of AgCl[K_(SP)(AgCl)= 1.6xx10^(-10)], 0.1 mole of CuCI [K_(SP)(CuCl)=1.0xx10^(-6)] is added. The resulrant concentration of Ag^(+) in the solution is 1.6xx10^(-x) . The value of "x" is:

If conductivity of water used to make saturated of AgCl is found to be 3.1xx10^(-5)Omega^(-1) cm^(-1) and conductance of the solution of AgCl=4.5xx10^(-5)Omega^(-1)cm^(-1) if lamda_(M)^(0)AgNO_(3)=200Omega^(-1)cm^(2) "mole"^(-1) lamda_(M)^(0)NaNO_(3)=310Omega^(-1)cm^(2) "mole"^(-1) calculate K_(SP) of AgCl

Conductivity of an aqueous solution of 0.1M HX (a weak mono-protic acid) is 5 xx 10^(-4) Sm^(-1) Given: ^^_(m)^(oo) [H^(+)] = 0.04 Sm^(2) mol^(-1): ^^_(m)^(oo) [X^(-)] = 0.01 Sm^(2)mol^(-1) Find pK_(a)[HX]

The solubility product constant of Ag_(2)CrO_(4) and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarities of their saturated solutions.

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: