A) 1.76%
B) 2.28%
C) 3.92%
D) 4.15%
Correct Answer: C
Solution :
From Kohlrauschs law, \[{{A}^{o}}\]for \[N{{H}_{4}}OH=\] \[A_{NH_{4}^{+}}^{o}+A_{C{{H}^{-}}}^{o}={{A}^{o}}(N{{H}_{4}}Cl)+{{A}^{o}}(NaOH)\] \[-{{A}^{o}}(NaCl)=129.8+217.4-108.9\] \[=238.3\,S\,c{{m}^{2}}\] Degree of dissociation \[(\alpha )=\frac{{{A}_{c}}}{{{A}^{o}}}=\frac{9.33}{238.3}=0.0392=3.92%\]You need to login to perform this action.
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