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 occuring element X is closest to

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

Answer 1

Answer: (D)

Average isotopic mass of
$X=\frac{200\times 90+199\times 8+202\times 2}{90+8+2}$
$=\frac{18000+1892+404}{100}$
$=\frac{19996}{100}=199.96amu$