At `25^(@)C`, the molar conductances at infinite dilution for the strong electrolytes `NaOH, NaCl " and " BaCl_(2) " are " 248 xx 10^(-4), 126 xx 10^(-4)" and " 280 xx 10^(-4) S m^(2) mol^(-1)` respectively. `Lamda_(m)^(@) Ba (OH)_(2) " in " S m^(2) mol^(-1)`

Answer

`5.232 xx 10^(-2) S m^(2) mol^(-1)`

Explanation

`Lambda^(@) [Ba(OH)_(2)] = Lambda^(@) [BaCl_2] + 2 Lambda^(@) [NaOH] - 2 Lambda^(@) [NaCl]`
`= 2.800 xx 10^(-2) + 2 (2.481 xx 10^(-2) ) - 2 (1.265 xx 10^(-2))`
`= 5.232 xx 10^(-2) S m^(2) mol^(-1).`