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. `139.5"ohm"^(-1) "cm"^(2) "eq"^(-1)`
B. `203"ohm"^(-1) "cm"^(2) "eq"^(-1)`
C. `279"ohm"^(-1) "cm"^(2) "eq"^(-1)`
D. `101.5"ohm"^(-1) "cm"^(2) "eq"^(-1)`

Answer

A

Explanation

`BaCl_2 to Ba^(2+) + 2Cl^(-)`
`Lambda_(BaCl_2)^@ = Lambda_(Ba^(2+))^@ + 2Lambda_(Cl^-)^@`=127 + ( 2 x 76 )
`279 "ohm"^(-1) "cm"^(2) eq^(-1)`
Equivalent conductivity =`279/2`=139.5 `"ohm"^(-1) "cm"^(2) eq^(-1)`