The limiting equivalent conductivity of `NaCl, KCl` and `KBr` are `126.5,150.0` and `151.5 S cm^(2) eq^(-1)` , respectively. The limiting equivalent ionic conductance for `Br^(-)` is `78 S cm^(2) eq^(-1)`. The limiting equivalent ionic conductance for `Na^(+)` ions would be `:`

Options

A. 128
B. 125
C. 49
D. 50

Answer

4

Explanation

`Lambda_(m)^(oo)=(NaBr)=Lambda_(m)^(oo)(NaCl)+Lambda_(m)^(oo)(KBr)-Lambda_(m)^(oo)(KCl)`
`Lambda_(m)^(oo)(Na^(+))+Lambda_(m)^(oo)(Br)=126.5+151.5-150`
`Lambda_(m)^(oo)(Na^(+))=50`