Answer:
B) 271 g.
Step-by-step explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of As-76 = 26.0 hours.
- For, first order reactions:
k = ln(2)/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(26.0 hours) = 0.02665 hour⁻¹.
- Also, we have the integral law of first order reaction:
kt = ln([A₀]/[A]),
where, k is the rate constant of the reaction (k = 0.02665 hour⁻¹).
t is the time of the reaction (t = 538 min = 8.97 hour).
[A₀] is the initial concentration of (As-76) ([A₀] = 344 g).
[A] is the remaining concentration of (As-76) ([A] = ??? g).
∴ (0.02665 hour⁻¹)(8.97 hour) = ln((344 g)/[A])
∴ 0.239 = ln((344 g)/[A]).
- Taking exponential for both sides:
∴ 1.27 = ((344 g)/[A]).
∴ [A] = (344 g)/(1.27) = 270.88 g ≅ 271 g.
- So, the right choice is: B) 271 g.