A saturated solution in AgA `(K_(sp)=3xx10^(-14))` and AgB `(K_(sp)=1xx10^(-14))` has conductivity of `375xx10^(-10) S cm^(-1)` and limiting molar conductivity of `Ag^(+)` and `A^(-)` are 60S `cm^(2) "mol"^(-1)` and 80 S `cm^(2) "mol"^(-1)` respectively , then what will be the limiting molar conductivity of `B^(-)` (in `S cm^(2) "mol"^(-1)`)?

Options

A. 150
B. 180
C. 190
D. 270

Answer

D

Explanation

`{:(AgA(s)hArr,Ag^(+)+,A^(-)),(,S_(1)+S_(2),S_(1)):}`
`((K_(sp))_(AgA))/((K_(sp))_(AgB))=(S_(1))/(S_(2))=(3)/(1)`
`{:(AgB(s)hArr,Ag^(+)+,B^(-)),(,S_(1)+S_(2),S_(2)):}`
`S_(1)+S_(2)= sqrt((K_(sp))_(AgA)+(K_(sp))_(AgB))=2xx10^(-7)`
On solving `S_(1)=(3)/(2)xx10^(-7)` and `S_(2)=(1)/(2)xx10^(-7)`
`K_("solution")=k_(Ag^(+))+k_(A^(-))+k_(B^(-))`
`375xx10^(-4)=60xx2xx10^(-10)+(3)/(2)xx10^(-10)xx80+(1)/(2)xx10^(-10)xxlambda_(B^(-))^(oo)`
`375=120+120+(lambda_(B^(-))^(oo))/(2)`
`lambda_(B^(-))^(oo)=270` Scm `mol^(-1)` .