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JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • question_answer
    The specific conductivity of an aqueous solution of a weak monoprotic acid is \[0.00033\,oh{{m}^{-1}}c{{m}^{-1}}\] at a concentration, \[0.02\] M. If at this concentration the degree of dissociation is \[0.043\], calculate the value of \[{{\wedge }_{0}}(oh{{m}^{-1}}c{{m}^{2}}/eqt)\].

    A)  483                             

    B)  438  

    C)  348                             

    D)  384

    Correct Answer: D

    Solution :

    \[a=0.043=\frac{{{\lambda }_{m}}}{{{\lambda }_{\infty }}}\,\] and \[{{I}_{m}}=0.00033\times 50\times {{10}^{3}}\] So \[{{\Lambda }_{0}}\,=\frac{0.33\,\times 50}{0.043}\,=384\]


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