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Lambda(m)^(@) for NH(4)Cl, NaOH and NaCl...

`Lambda_(m)^(@)` for `NH_(4)Cl, NaOH` and NaCl are 130, 248 and `126.5 "ohm"^(-1)cm^(2)"mol"^(-1)` respectively. The `Lambda_(m)^(@)` of `NH_(4)OH` will be,

A

251.5

B

244.5

C

130

D

504.5

Text Solution

Verified by Experts

The correct Answer is:
A
`Lambda_(m)^(@)(NH_(4)OH)=Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NH_(4)OH)-Lambda_(m)^(@)(NaCl)`
`=130+248-126.5`
`=251.5 "ohm"^(-1) cm^(2) "mol"^(-1)`
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