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-occurring element X is closest to

Correct answer: D. 200 amu Average isotopic mass of aligned X & =(200 × 90+199+8+202 × 2)/(90+8+2) & =(18000+1892+404)/(100) & =(19996)/(100)=199.96 amu . aligned

NEET 2007

Exam: NEET
Year: 2007
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

NEET 2007 · Chemistry · Some Basic Concepts of Chemistry

An element, X has the following isotopic composition: ^ 200 X: 90...

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

Options

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

Correct answer

D. $200 \mathrm{amu}$

Step-by-step 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}

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