# How many atoms are contained in 0.55 moles of Br?

Apr 12, 2017

$0.55 \times {N}_{A}$ $\text{bromine atoms..........}$

#### Explanation:

Where ${N}_{A} = \text{Avogadro's Number,}$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus, $0.55 \cdot \cancel{m o l} \times 6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1} = 3.31 \times {10}^{23}$ $\text{bromine ATOMS}$.

Note that here we use the mole as would use any other collective term, i.e. dozen, score, gross............Of course the mole specifies a much larger quantity of stuff.

Also note that we specify a quantity of $0.55$ $m o l$ $\text{bromine atoms}$. A quantity of $0.55$ $m o l$ $\text{bromine molecules}$ would contain TWICE this number of bromine atoms. Why?