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

  • question_answer
    The limiting equivalent conductivity of \[NaCl,~KCl\] and \[KBr\] are 126.5, 150.0 and\[151.5\text{ }S\text{ }c{{m}^{2}}e{{q}^{-1}}\], respectively. The limiting equivalent ionic conductivity for \[B{{r}^{-}}\] is\[78\,Sc{{m}^{2}}e{{q}^{-1}}\]. The limiting equivalent ionic conductivity for \[N{{a}^{+}}\] ions would be:

    A) 128                  

    B) 125  

    C) 49                                

    D) 50

    Correct Answer: D

    Solution :

    [d] \[\Lambda _{m}^{\infty }(NaBr)=\Lambda _{m}^{\infty }(NaCl)+\Lambda _{m}^{\infty }KBr-\Lambda _{m}^{\infty }(KCl)\] \[\lambda _{m}^{\infty }(N{{a}^{+}})+\lambda _{m}^{\infty }(B{{r}^{-}})=126.5+151.5-150\] \[\lambda _{m}^{\infty }(N{{a}^{+}})=50\,Sc{{m}^{2}}e{{q}^{-1}}.\]


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