If the molar conductivity values of `Ca^(2+) and Cl^(-)` at infinite dilution are, respectively, `118.88 xx 10^(-4) and 77.33 xx 10^(-4) S m^(2) mol^(-1)`, then that of `CaCl_(2)` is (`Sm^(2) mol^(-1)`)

Options

A. `118.88 xx 10^(-4)`
B. `154.66 xx 10^(-4)`
C. `273 xx 10^(-4)`
D. `196.21 xx 10^(-4)`

Answer

C

Explanation

`Lamda_(m)^(@) (CaCl_(2))= lamda_(m)^(@) (Ca^(2+)) + 2lamda_(m)^(@) (Cl^(-))` (Kohlrausch.s law)
`=118.88 xx 10^(-4) + 2(77.33 xx 10^(-4)) = 273.54 xx 10^(-4) S m^(2) mol^(-1)`