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

  • question_answer
    \[{{\lambda }_{ClC{{H}_{2}}COONa}}=224\,oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\],                 \[{{\lambda }_{NaCl}}=38.2\,oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\],                  \[{{\lambda }_{HCl}}=203\,oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\],                         What is the value of \[{{\lambda }_{ClC{{H}_{2}}COOH}}\]                           [JEE Orissa 2004]

    A)                 \[288.5\ oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\]

    B)                 \[289.5\ oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\]

    C)                 \[388.5\ oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\]               

    D)                 \[59.5\ oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\]

    Correct Answer: C

    Solution :

               \[Cl\underset{{{\lambda }_{ClC{{H}_{2}}COONa}}+}{\mathop{C{{H}_{2}}COONa}}\,+\underset{{{\lambda }_{HCl}}}{\mathop{HCl}}\,\underset{=}{\mathop{\to }}\,\underset{{{\lambda }_{ClC{{H}_{2}}COOH}}+}{\mathop{ClC{{H}_{2}}COOH}}\,+\underset{{{\lambda }_{NaCl}}}{\mathop{NaCl}}\,\]                    \[224+203={{\lambda }_{ClC{{H}_{2}}COOH}}+38.2\]                                 \[{{\lambda }_{ClC{{H}_{2}}COOH}}=427-38.2=388.8\ oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\].


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