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NEET Chemistry Electrochemistry / विद्युत् रसायन Question Bank Self Evaluation Test - Electrochemistry

  • question_answer
    Specific conductance of 0.1 MHA is \[3.75\times {{10}^{-4}}oh{{m}^{-1}}c{{m}^{-1}}\]. If \[{{\lambda }^{\infty }}(HA)=250\,oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}\], the dissociation constant \[{{K}_{a}}\] of HA is:

    A) \[1.0\times {{10}^{-5}}\]         

    B) \[2.25\times {{10}^{-4}}\]

    C) \[2.25\times {{10}^{-5}}\]                    

    D) \[2.25\times {{10}^{-13}}\]

    Correct Answer: C

    Solution :

    [c] \[{{\lambda }_{m}}=\frac{1000\kappa }{0.1}=\frac{1000\times 3.75\times {{10}^{-4}}}{0.1}=3.75;\] \[\alpha =\frac{{{\lambda }_{m}}}{\lambda _{m}^{\infty }}=\frac{3.75}{250}=1.5\times {{10}^{-2}}\]; \[{{K}_{a}}=C{{\alpha }^{2}}=0.1\times {{(1.5\times {{10}^{-2}})}^{2}}=2.25\times {{10}^{-5}}\]


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