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  2. Question
  3. Chemistry
  4. An element, X has the following isotopic composition: 200X: 90 % 199X: 8.0 % 202X: 2.0 % The weighted average atomic mass of the naturally occuring element X is closest to

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Q. 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

Structure of Atom

Solution:

Average isotopic mass of
$X = \frac{200 \times90 + 199 \times8+ 202 \times2}{90+8+2} $
$= \frac{18000 + 1892 + 404}{100} $
$= \frac{19996}{100} = 199.96 amu$