Some Basic Concepts of Chemistry - Result Question 15
####15. An element, $X$ has the following isotopic composition :
[2007]
${ }^{200} X: 90 % \quad{ }^{199} X: 8.0 % \quad{ }^{202} X: 2.0 %$
The weighted average atomic mass of the naturally occuring element $X$ is closest to
(a) $201 amu$
(b) $202 amu$
(c) $199 amu$
(d) $200 amu$
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Solution:
- (d) Average isotopic mass of
$ \begin{aligned} & X=\frac{200 \times 90+199 \times 8+202 \times 2}{90+8+2} \\ & =\frac{18000+1592+404}{100} \\ & =\frac{19996}{100}=199.96 \simeq 200 amu \end{aligned} $
Average atomic mass
$=($ Mass of isotope $A \times %$ natural abundance $)+$ (Mass of isotope B $\times %$ natural abundance) + (% Natural abundance of $A+%$ natural abundance of B) + …….
