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. `1 96.21xx10^(-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. `1 96.21xx10^(-4)`
Answer» Correct Answer - C `^^_(m)^(@)(CaCl_(2))=lambda_(m)^(@)(Ca^(2+))+2lambda_(m)^(@)(Cl^(-))` `=118.88xx10^(-4)+2(77+33xx10^(-4))` `=273.54xx10^(-4)m^(2)"mol"^(-1)`

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