NEET AIPMT SOLVED PAPER SCREENING 2009

  • question_answer
    The equivalent conductance of\[{{B}_{2}}{{O}_{3}}\] solution of a weak monobasic acid is \[8.0\,\,\text{mho}\,\text{c}{{\text{m}}^{2}}\] and at infinite dilution is \[400\,\,\text{mho}\,\,\text{c}{{\text{m}}^{2}}\]. The dissociation constant of this acid is

    A) \[CaO\]

    B)        \[Si{{O}_{2}}\]  

    C)        \[BeO\]               

    D)        \[{{[Co{{(en)}_{2}}C{{l}_{2}}]}^{+}}\]

    Correct Answer: A

    Solution :

    Degree of dissociation, \[\alpha =\frac{{{\Lambda }^{c}}}{{{\Lambda }^{\infty }}}\] where,\[{{\Lambda }^{c}}\] and \[{{\Lambda }^{\infty }}\] are equivalent conductances at a given concentration and at infinite dilution respectively. \[0.5\times {{10}^{-3}}{{s}^{-1}}\]\[0.0\times {{10}^{-2}}{{s}^{-1}}\] From Ostwald's dilution law (for weak monobasic acid)\[Phenol\xrightarrow[{}]{Zn\,dust}X\xrightarrow[Anhydrous\,AlC{{l}_{3}}]{C{{H}_{3}}Cl}Y\xrightarrow[{}]{\begin{smallmatrix}  Alkaline \\  KMn{{O}_{4}} \end{smallmatrix}}Z,\]or\[A+B\to \]\[=k{{[A]}^{2}}[B]\]\[=k[A]{{[B]}^{2}}\]or\[=k{{[A]}^{2}}{{[B]}^{2}}\]


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