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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
$^{200}X$ : 90%, $^{199}X$ : 8.0%, $^{202}X$ : 2.0%
The weighted average atomic mass of the naturally occurring element X is closest to
A
201 amu
B
202 amu
C
199 amu
D
200 amu
Detailed Solution
Average atomic mass $= \dfrac{\Sigma(\text{isotopic mass} \times \text{percentage abundance})}{100}$
Average isotopic mass of X $= \dfrac{200 \times 90 + 199 \times 8 + 202 \times 2}{90 + 8 + 2}$
$= \dfrac{18000 + 1592 + 404}{100}$
$= \dfrac{19996}{100}$
$= 199.96$ amu
This is closest to 200 amu.
Average isotopic mass of X $= \dfrac{200 \times 90 + 199 \times 8 + 202 \times 2}{90 + 8 + 2}$
$= \dfrac{18000 + 1592 + 404}{100}$
$= \dfrac{19996}{100}$
$= 199.96$ amu
This is closest to 200 amu.