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)`

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.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

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Specific conductance of 0.1 M NaCl solution is 1.01xx10^(-2) ohm^(-1) cm^(-1) . Its molar conductance in ohm^(-1) cm^(2) mol^(-1) is

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The conductivity of 0.01 mol L^(-1) KCl solution is 1.41xx10^(-3)" S "cm^(-1) . What is the molar conductivity ( S cm^(2) mol^(-1) ) ?