Calculating the value of `Lambda_(m)^(@)` for `NH_(4)OH` solution requires the value of `Lambda_(m)^(@)` for the solutions of `NH_(4)Cl, NaOH` and an electrolyte. Name the electrolyte.

Explanation

`Lambda_(m)^(@)(NH_(4)OH)=lambda_(m)^(@)(NH_(4)^(+))+lambda_(m)^(@)(OH^(-))=Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`
Hence apart from the `Lamnda_(m)^(@)` of `NH_(4)Cl and NaOH, Lambda_(m)^(@)` of `NaCl` is required.