An element, $\mathrm{X}$ has the following isotopic composition: ${ }^{200} \mathrm{X}: 90 \% \quad{ }^{199}…

An element, $\mathrm{X}$ has the following isotopic composition:
${ }^{200} \mathrm{X}: 90 \% \quad{ }^{199} \mathrm{X}: 8.0 \quad{ }^{202} \mathrm{X}: 2.0 \%$
The weighted average atomic mass of the naturally-occurring element $\mathrm{X}$ is closest to
  1. $201 \mathrm{amu}$
  2. $202 \mathrm{amu}$
  3. $199 \mathrm{amu}$
  4. $200 \mathrm{amu}$

Solution

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

Asked in: NEET 2007

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