DUMET Medical DUMET Medical Solved Paper-2001

  • question_answer
    Degree of dissociation of 0.1 N \[C{{H}_{3}}COOH\]is: (dissociation constant \[=1\times {{10}^{-5}}\])

    A)  \[{{10}^{-5}}\]          

    B)  \[{{10}^{-4}}\]

    C)  \[{{10}^{-3}}\]        

    D)  \[{{10}^{-2}}\]

    Correct Answer: D

    Solution :

    Key Idea: \[\alpha =\sqrt{\frac{K}{C}}\] where, \[\alpha =\] degree of dissociation K = dissociation constant \[=1\times {{10}^{5}}\] C = concentration \[=0.1\,N={{10}^{-1}}\] \[\alpha =\sqrt{\frac{K}{C}}=\sqrt{\frac{1\times {{10}^{-5}}}{{{10}^{-1}}}}=\sqrt{{{10}^{-4}}}\] \[={{10}^{-2}}\]


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