NEET Sample Paper NEET Sample Test Paper-64

  • question_answer
    A monoprotic acid in 1.00 M solution is \[0.01\,%\] ionised. The dissociation constant of this acid is

    A) \[1\,\,\times \,\,{{10}^{-}}^{8}\]                     

    B) \[1\,\,\times \,\,{{10}^{-}}^{4}\]

    C) \[1\,\,\times \,\,{{10}^{-}}^{6}\]                     

    D) \[{{10}^{-}}^{5}\]

    Correct Answer: A

    Solution :

    \[K=\frac{{{a}^{2}}C}{1-\alpha };\,\,\alpha \,\,=\,\,\frac{0.01}{100}\,\,\approx \,\,1\] \[\therefore \,\,\,\,\,K={{a}^{2}}C=\,\left[ \frac{0.01}{100} \right]{{\,}^{2}}\times \,\,1=1\times {{10}^{-\,8}}\]


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