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)`)

Options

A. `118.88xx10^(-4)`
B. `154.66xx10^(-4)`
C. `273.54xx10^(-4)`
D. `196.21xx10^(-4)`

Answer

C

Explanation

`Lambda_(m)^(@) CaCl_(2) =lambda^(@)Ca^(2+)+2lambda^(@)Cl^(-)`
`=(118.88xx10^(-4))+2(77.33xx10^(-4))`
`=273.54xx10^(4) m^(2)` mho `mol^(-1)`