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Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    The \[Ka\] of an acid is\[3.2\times {{10}^{-5}}\]. The degree of   dissociation of the acid at concentration of \[0.2\,\,M\]is:

    A) \[6.0\times {{10}^{-2}}\]                              

    B) \[1.26\times {{10}^{-2}}\]

    C) \[40\times {{10}^{-4}}\]               

    D) \[0.04\times {{10}^{-3}}\]

    Correct Answer: B

    Solution :

    From Ostwalds dilution law                 \[\alpha =\sqrt{\frac{Ka}{c}}\]                    \[=\sqrt{\frac{3.2\times {{10}^{-5}}}{0.2}}\]                   \[=1.26\times {{10}^{-2}}\]


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