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VMMC VMMC Medical Solved Paper-2013

  • question_answer
    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?

    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 \[(\alpha )=\frac{{{A}_{c}}}{{{A}^{o}}}=\frac{9.33}{238.3}=0.0392=3.92%\]


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