Question #8b2ee

Dec 10, 2016

The mass of one atom of iron is $\left(9.2741105 \times {10}^{-} 23\right)$.

Explanation:

The formula we are using is $m = \frac{M}{n}$. Where...
=> $m$ is the mass in grams.
=> $M$ is the molar mass in $\frac{g}{m o l}$.
=> ${n}_{\text{A}}$ is Avogadro's number - $6.02 \times {10}^{23}$.

$m = \frac{M}{n} _ \text{A}$

$= \frac{55.85}{6.02 \times {10}^{23}}$

$= \left(9.2741105 \times {10}^{-} 23\right)$

The mass of one atom of iron is $\left(9.2741105 \times {10}^{-} 23\right)$.

Hope this helps :)