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 beA. `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)`

The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm

1.	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 beA. `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» Correct Answer - A `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)`

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