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)`]

Options

A. 39
B. 55
C. 15
D. 41

Answer

A

Explanation

`ksp(AgBr)=[Ag^+][Br^(-)]`
`12xx10^(-14)=(S+10^(-7)).S`
`S=3xx10^(-7)`
`[Br^(-)]=3xx10^(-7)[Ag^+]=4xx10^(-7)`
`[NO_3^(-)]=10^(-7)`
`K=10^6/1000(lambda_(Ag^+)^@M_(Ag^+)+lambda_(Br^(-))^@M_(Br^(-))+lambda_(NO_3^(-))^@M_(NO_3^(-)))`
`K=10^6/1000(2xx10^(-3)xx4xx10^(-7)+6xx10^(-3)xx3xx10^(-7)+5xx10^(-3)xx10^(-7))`
`K=39xx10^(-13)xx10^6=39xx10^(-7) sm^(-1)`