At `25^(@)C` the molar conductances at infinite dillution for the strong electrolytes NaOh, NaCl and `BaCl_(2)` are `248xx10^(-4),126xx10^(-4)` and `280xx10^(-4)Sm^(2)mol^(-1)` respectively `lamda^(@)` of `Ba(OH)_(2)` is `Sm^(2) mol^(-1)` is

Options

A. `52.4 xx 10^(-4)`
B. `524 xx 10^(-4)`
C. `402 xx 10^(-4)`
D. `262 xx 10^(-4)`

Answer

B

Explanation

`Lamda_(m)^(@) Ba(OH)_(2)= Lamda_(m)^(@)BaCl_(2) + 2Lamda_(m)^(@) NaOH - 2Lamdad_(m)^(@)NaCl`
`=280 xx 10^(-4) + 2xx 248 xx 10^(-4)-2 xx 126 xx 10^(-4)= 524 xx 10^(-4) S m^(2) mol^(-1)`