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Q. At 291 K the molar conductivities at infinite dilution of $ N{{H}_{4}}Cl,NaOH $ and $ NaCI $ are 129.8, 217.4 and $ 108.9\,Sc{{m}^{2}} $ respectively,. If the molar conductivity of a centinormal solution of $ N{{H}_{4}}OH $ is $ 9.33\,Sc{{m}^{2}}, $ What is the precentage dissociation of $ N{{H}_{4}}OH $ at this dilution?

VMMC MedicalVMMC Medical 2013

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% $