Answer:
Percentage of a sample remains after 60.0 min is 13.03%.
Explanation:
- It is known that the decay of isotopes of C-11 obeys first order kinetics.
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- Half-life time (t1/2) in first order reaction = 0.693/k, where k is the rate constant.
∴ k = 0.693/(t1/2) = 0.693/(20.4 min) = 0.03397 min⁻¹.
- The integrated law for first order reaction is:
<em>kt = ln[A₀]/[A],</em>
where, k is the rate constant (k = 0.03397 min⁻¹).
t is the time of the reaction (t = 60.0 min).
[A₀] is the initial concentration of C-11 ([A₀] = 100.0 %).
[A] is the remaining concentration of C-11 ([A] = ???%).
<em>∵ kt = ln[A₀]/[A]</em>
∴ (0.03397 min⁻¹)(60.0 min) = ln(100%)/[A]
∴ 2.038 = ln(100%)/[A]
∴ 7.677 = (100%)/[A]
<em>∴ [A] </em>= (100%)/(7.677) = <em>13.03%.</em>
<em>So, percentage of a sample remains after 60.0 min is 13.03%.</em>
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