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Limiting molar conductivity of NH(4)OH [...

Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to -

A

`Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_(m)^(@)(HCl)`

B

`Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`

C

`Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaCl)-Lambda_(m)^(@)(NaOH)`

D

`Lambda_(m)^(@)(NaOH)+Lambda_(m)^(@)(NaCl)-Lambda_(m)^(@)(NH_(4)Cl)`

Text Solution

Verified by Experts

The correct Answer is:
B
`Lambda_(m)^(@)(NH_(4)OH)`
`=lambda_(m)^(@)(NH_(4)^(+))+lambda_(m)^(@)(OH^(-))`
`=lambda_(m)^(@)(NH_(4)^(+))+lambda_(m)^(@)(Cl^(-))+lambda_(m)^(@)(Na^(+))+lambda_(m)^(@)(OH^(-))-lambda_(m)^(0)(Na^(+))-lambda_(m)^(@)(Cl^(-))`
`=Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`
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