How many individual atoms are in a sample of xenon that has a mass of 38.54 grams?

Mar 21, 2016

There are approx. $\frac{1}{3} \times {N}_{A}$ xenon atoms, where ${N}_{A}$ $=$ $\text{Avogadro's number}$, $= 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

Explanation:

Elemental xenon has an equivalent mass of $131.29$ $g \cdot m o {l}^{-} 1$. What does this mean? It means that $\text{Avogadro's number}$ of xenon atoms has a mass of $131.29 \cdot g$.

You have a $38.54 \cdot g$ mass. Thus there are:

$\frac{38.54 \cdot g}{131.29 \cdot g \cdot m o {l}^{-} 1} \times {N}_{A}$ $=$

$\frac{38.54 \cdot g}{131.29 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $=$ ?? $\text{xenon atoms?}$