The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will be

Options

A. `280 S cm^(2) mol^(-1)`
B. `330.98 S cm^(2) mol^(-1)`
C. `90.98 S cm^(2) mol^(-1)`
D. `203.6 S cm^(2) mol^(-1)`

Answer

A

Explanation

`Lambda_(m)^(oo) (BaCl_(2)) = lambda^(oo) (Ba^(2+)) + 2 lambda^(oo) (Cl^(-))`
`= 127.32 + 2 xx 76.34`
`=280 S cm^(2) mol^(-1)`