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 \[\therefore \] \[\frac{0.04}{1.28}={{\left( \frac{1}{2} \right)}^{n}};\frac{1}{32}={{\left( \frac{1}{2} \right)}^{n}};{{\left( \frac{1}{2} \right)}^{5}}={{\left( \frac{1}{2} \right)}^{n}}\] \[n=5\] \[T=5\times 138=690\]You need to login to perform this action.
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