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

  • question_answer
    Equivalent conductance at infinite dilution, \[{{\lambda }^{{}^\circ }}\] of \[N{{H}_{4}}Cl,\,NaOH\] and \[NaCl\] are 128.0,217.8 and  \[109.3\text{ }oh{{m}^{-1}}c{{m}^{2}}\text{ }e{{q}^{-1}}\] respectively. The equivalent conductance of \[0.01\text{ }N\text{ }N{{H}_{2}}OH\] is  \[9.30\text{ }oh{{m}^{-1}}\,c{{m}^{2}}\text{ }e{{q}^{-1}}\] then the degree of ionization of \[N{{H}_{4}}OH\] at this temperature would be

    A) 0.04                 

    B) 0.1    

    C) 0.39                 

    D) 0.62

    Correct Answer: A

    Solution :

    [a]  \[\Lambda _{eq}^{\infty }(N{{H}_{4}}OH)=\Lambda _{eq}^{\infty }(N{{H}_{4}}Cl)+\Lambda _{eq}^{\infty }(NaOH)-\Lambda _{eq}^{\infty }(NaCl)\] \[=129.8+217.8-109.3=238.3\text{ }oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}\] \[\alpha =\frac{{{\Lambda }_{eq}}}{\Lambda _{eq}^{\infty }}=\frac{9.30}{238.3}=0.04\]


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