Q49 / 200 · NEET 2007 · Chemistry

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

  • A
    $201 \mathrm{amu}$
  • B
    $202 \mathrm{amu}$
  • C
    $199 \mathrm{amu}$
  • Correct answer
    D
    $200 \mathrm{amu}$

Solution

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

Hence, the correct answer is option: 'D'.

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