If the molar conductance values of `Ca^(2+)` and `Cl^(-)` at infinite dilution are respectively `118.88xx10^(-4) m^(2)` mho `mol^(-1)` and `77.33xx10^(-4) m^(2)` mho `mol^(-1)` then that of `CaCl_(2)` is : (in `m^(2)` mho `mol^(-1)`)A. `118.88xx10^(-4)`B. `154.66xx10^(-4)`C. `273.54xx10^(-4)`D. `196.21xx10^(-4)`

If the molar conductance values of `Ca^(2+)` and `Cl^(-)` at infinite

1.	If the molar conductance values of `Ca^(2+)` and `Cl^(-)` at infinite dilution are respectively `118.88xx10^(-4) m^(2)` mho `mol^(-1)` and `77.33xx10^(-4) m^(2)` mho `mol^(-1)` then that of `CaCl_(2)` is : (in `m^(2)` mho `mol^(-1)`)A. `118.88xx10^(-4)`B. `154.66xx10^(-4)`C. `273.54xx10^(-4)`D. `196.21xx10^(-4)`
Answer» Correct Answer - C `Lambda_(m)^(@) CaCl_(2) =lambda^(@)Ca^(2+)+2lambda^(@)Cl^(-)` `=(118.88xx10^(-4))+2(77.33xx10^(-4))` `=273.54xx10^(4) m^(2)` mho `mol^(-1)`

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