Question

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:

• A. $199amu$
• B. $200amu$
• C. $201amu$
• D. $202amu$

Answer 1

The correct option is B $200amu$

Average atomic mass = $\frac{A_1\times X_1+A_2\times X_2+A_3\times X_3}{100}$

Where,

A= Isotopic mass

X= Percentage of occurrence

Average atomic mass = $\frac{200\times 90+199\times 8+202\times 2}{100}$

Average atomic mass = 199.96