An element, X has the following isotopic composition, 200X:90% 199X:8% 202X:2%The weighted average atomic mass of the naturally occurring element ‘X’ is closest to _____.

Average atomic mass ofX=90100×200+8100×199+2100×202=199.96 amu ≈ 200 amu

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An element, X has the following isotopic composition,
$\begin{array}{l} ^{200} X : 90 % \\ ^{199} X : 8 % \\ ^{202} X : 2 % \end{array}$
The weighted average atomic mass of the naturally occurring element ‘X’ is closest to _____.

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Average atomic mass of

$X = [\frac{90}{100} \times 200] + [\frac{8}{100} \times 199] + [\frac{2}{100} \times 202]$

$= 199.96 amu \approx 200 amu$

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