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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1999

  • question_answer
    The ionization percentage of \[0.2M\] \[HCN\] the solution is \[0.02%\]. What will be the value of ionization constant:

    A)  \[8\times {{10}^{-5}}\]                 

    B)                         \[8\times {{10}^{-9}}\]

    C)                         \[8\times {{10}^{-7}}\]                 

    D)                         \[8\times {{10}^{-3}}\]

    Correct Answer: A

    Solution :

    \[{{K}_{h}}=C{{\alpha }^{2}}=0.2\times {{(0.02)}^{4}}\] \[=0.2\times 0.0004\] \[=8\times {{10}^{-5}}\]


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