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NEET Sample Paper NEET Sample Test Paper-13

  • question_answer
    The half-life of a substance in a certain enzyme- catalysed reaction is 13 8 s. The time required for the concentration of the substance to fall from \[1.28\text{ }mg\text{ }{{L}^{-1}}\]to \[0.04\text{ }mg\text{ }{{L}^{-1}}\] is

    A)  414 s                           

    B)  552 s

    C)  690 s                           

    D)  276 s

    Correct Answer: C

    Solution :

     For a first-order reaction Total time T= no. of half livers \[(n)\times \]half-life \[({{t}_{1/2}})\] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] Where \[n=\]no. of half-lives Given \[{{N}_{0}}\](original amount) = 1.28 mg/L = 0.04 mg/L \[\therefore \]      \[\frac{0.04}{1.28}={{\left( \frac{1}{2} \right)}^{n}};\frac{1}{32}={{\left( \frac{1}{2} \right)}^{n}};{{\left( \frac{1}{2} \right)}^{5}}={{\left( \frac{1}{2} \right)}^{n}}\] \[n=5\] \[T=5\times 138=690\]


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