Question

An element, X has the following isotopic composition; ${}^{200}X:90\%{;}^{199}X:8.0\%{;}^{202}X:2.0\%.$ The weighted average atomic mass of the naturally occurring element X is closest to :

• A. 199 amu
• B. 200 amu
• C. 201 amu
• D. 202 amu

Answer 1

The correct option is B

200 amu

Even without calculating the weighted average atomic mass, we can solve this one. This is basically a question testing the idea of weighted mean or weighted average. The overall weighted mean will be closer to the value that has the highest "weightage," which in this case is ${}^{200}X:90\%$. In other words, if 90 samples out of hundred have 200 as their atomic mass, then the average will be closest to 200!

The weighted average atomic mass of element $\left(X\right)=0.9\left(200\right)+0.08\left(199\right)+0.02\left(202\right)=180+15.92+4.04=199.96\approx 200$