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^^(m((NH(4)OH)))^(o) is equal to...

`^^_(m_((NH_(4)OH)))^(o)` is equal to

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

`Lambda_(m(NH_(4)cl))^(@)=Lambda_(m(NH_(4)^(+)))^(@)+Lambda_(m(Cl^(-)))^(@)`
`Lambda_(m(NaOH))^(@)=Lambda_(m(Na^(+)))^(@)+Lambda_(m(OH^(-)))^(@)`
`Lambda_(m(NaCl))^(@)=Lambda_(m(Na^(+)))^(@)+Lambda_(m(Cl^(-)))^(@)`
- - -
`Lamdba_(m(NH_(4)Cl))^(@)+Lambda_(m(NaOH))^(@)-Lambda_(m(Nacl))^(@)=Lambda_(m(NH_(4)OH))^(@)`
Hence, option (b) is correct choice.
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