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BVP Medical BVP Medical Solved Paper-2014

  • question_answer
    The    half-life    for    the    reaction,\[O=16,Cu=63,N=14\]  is 24h at \[1\times {{10}^{-10}}g\].Starting with 10 g of \[1\times {{10}^{-10}}g\], how many grams of \[II<I<\text{ }III<IV\] will remain after a period of 96 h?

    A)  \[II<\text{ }III<I<IV\]                    

    B)  \[III<IV<I<II\]

    C)  \[BaC{{l}_{2}}(s)\]                          

    D)  \[BaC{{l}_{2}}.2{{H}_{2}}O(s)\]

    Correct Answer: B

    Solution :

                    \[k=\frac{0.693}{24}{{h}^{-1}}=\frac{2.303}{96}\log \frac{10}{a-x}\] or            \[\log \frac{10}{a-x}=1.2036\] or            \[1-\log (a-x)=1.2036\] or            \[\log (a-x)=-0.2036\]                                 \[=+1.7964-2\] or            \[(a-x)=(anti\log )\,\,\,(.7964-2)\]                                 \[=0.6258\]                                 \[=0.63g\]


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