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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Cell constant and Electrochemical cells

  • question_answer
    The molar conductances of \[NaCl,\,HCl\] and \[C{{H}_{3}}COONa\] at infinite dilution are 126.45, 426.16 and \[91\,oh{{m}^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] respectively. The molar conductance of \[C{{H}_{3}}COOH\] at infinite dilution is             [CBSE PMT 1997]

    A)                 \[201.28\,oh{{m}^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\]

    B)                 \[390.71\,oh{{m}^{-1}}c{{m}^{2}}\,mo{{l}^{-1}}\]

    C)                 \[698.28\,oh{{m}^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\]

    D)                 \[540.48\,oh{{m}^{-1}}c{{m}^{2}}\,mo{{l}^{-1}}\]

    Correct Answer: B

    Solution :

               \[\wedge _{m}^{o}(C{{H}_{3}}COOH)=\,\,\]                    \[{{\wedge }^{o}}(C{{H}_{3}}COONa)\]\[+{{\wedge }^{o}}(HCl)-\,\,{{\wedge }^{o}}(NaCl)\]                                 \[=91+426.16-126.45=390.71\,\,oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}\].


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