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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    Given\[{{K}_{a}}=6\times {{10}^{-4}},{{K}_{{{a}_{2}}}}=1.5\times {{10}^{-4}}\]The ratio of relative acidic strength is

    A)  4                  

    B)  2

    C)  1                  

    D)  3

    Correct Answer: B

    Solution :

     \[\frac{{{(Acid\text{ }strength)}_{1}}}{{{(Acid\text{ }strength)}_{2}}}=\sqrt{\frac{{{K}_{{{a}_{1}}}}}{{{K}_{{{a}_{2}}}}}}=\sqrt{\frac{6\times {{10}^{-4}}}{1.5\times {{10}^{-4}}}}\] \[=\sqrt{\frac{4}{1}}=2\]


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