Question

Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:
A. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(Na_(4)Cl)-Lambda_(m)^(@)(NaOH)`
B. `Lambda_(m)^(@)(NaOH)+Lambda_(m)^(@)(NaCl)-Lambda_(m)^(@)(NH_(4)Cl)`
C. `Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_(m)^(@)(HCl)`
D. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`

Answer 1

Correct Answer - D
According to Kohlrausch law of independent migration of ions:
`Lambda_(m)^(@)(NH_(4)OH) = Lambda_(m)^(@)(NH_(4)^(+))+Lambda_(m)^(@)(OH^(-))`
`= Lambda_(m)^(@)(NH_(4)^(+))+Lambda_(m)^(@)(Cl^(-))+Lambda_(m)^(@)(OH^(-))`
`= Lambda_(m)^(@)(NH_(4)^(+))-Lambda_(m)^(@)(Cl^(-))-Lambda_(m)^(@)(Na^(+))`
`= Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`