An element, X has the following isotopic composition: 200X : 90\% 199X : 8.0\% 202X : 2.0\% The weighted average atomic mass of the naturally occurring element X is closest to
- 201 amu
- 202 amu
- 199 amu
- 200 amu
Answer
(4) 200 amu
Solution
Average atomic mass = (mass \% abundance)100
= 200 90 + 199 8.0 + 202 2.0100
= 18000 + 1592 + 404100
= 19996100 = 199.96 amu
∴ Average atomic mass 200 amu
Answer: (4)