Final answer:
After 3 complete half-lives of cobalt-60, 12.5% of the initial amount remains. Therefore, of a 199 g sample, approximately 24.875 g would remain. As slightly more than 3 half-lives have passed after 20 years, the closest answer is 12.5 g, which would be less than the amount after 3 complete half-lives.
Step-by-step explanation:
The student's question is about the half-life of cobalt-60, which is a concept in nuclear chemistry dealing with radioactive decay. To determine the amount of a 199 g sample of cobalt-60 that will remain after 20 years, considering that the half-life is 5.27 years, we need to calculate the number of half-lives that have passed and then apply the concept that the amount of substance remaining is halved with each half-life.
First, we divide 20 years by the half-life of cobalt-60:
20 years ÷ 5.27 years/half-life = 3.797 half-lives
Since we cannot have a fraction of a half-life, we consider 3 complete half-lives to have passed, and a portion of the fourth.
After 3 half-lives (15.81 years), 12.5% remains, and we can calculate the amount of cobalt-60 left after 3 half-lives:
199 g × 0.125 = 24.875 g
For the remaining time (4.19 years), we do not have a complete half-life, so we cannot directly halve the amount again. However, we can infer that it will be less than 24.875 g but more than if a fourth half-life were to complete (which would be half of 24.875 g). Based on the choices given and this calculation, the closest answer would be option B, 12.5 g, which corresponds to the amount that would remain after slightly more than 3 half-lives.