JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Mock Test - Electrochemistry

  • question_answer
    Given\[l\text{/}a=0.5\,c{{m}^{-1}}\], R = 50 ohm, N = 1.0. The equivalent conductance of the electrolytic cell is

    A) \[100\,oh{{m}^{-1}}c{{m}^{2}}g\,E{{q}^{-1}}\]

    B) \[20\,oh{{m}^{-1}}c{{m}^{2}}g\,E{{q}^{-1}}\]

    C) \[300\,oh{{m}^{-1}}c{{m}^{2}}g\,E{{q}^{-1}}\]

    D) \[10\,oh{{m}^{-1}}c{{m}^{2}}g\,E{{q}^{-1}}\]

    Correct Answer: D

    Solution :

    [d] \[l/a=0.5\,c{{m}^{-1}},R=50\,\Omega \] \[\rho =\frac{Ra}{l}=\frac{50}{0.5}=100\] \[A=k\times \frac{1000}{N}=\frac{1}{\rho }\times \frac{1000}{N}=\frac{1}{100}\times \frac{1000}{1}\] \[=10\,oh{{m}^{-1}}c{{m}^{2}}g\,E{{q}^{-1}}\]


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