Answer:
A. 2.74g is the correct answer
Explanation:
[tex]average \: mass = \frac{(mass \: pre - 1982 \times abundance) + (mass \: post - 1982 \times abundance}{100 } \\ ave = \frac{ (3.1g \times 40.0) + (2.5g \times 60.0)}{100} \\ ave = \frac{124 + 150}{100} \\ ave = \frac{274}{10} = 2.74g[/tex]
where, the sum of abundance always have to be 100%
Thus, 40.0 +60.0= 100.0