# Question #35b10

Apr 25, 2017

Approx. $5.5 \times {N}_{A}$............
In $2.74 \cdot m o l$ fluorine gas, i.e. ${F}_{2}$, there are $5.5 \times {N}_{A}$ $\text{fluorine atoms}$, i.e. $5.5 \times 6.022 \times {10}^{23}$ fluorine atoms.
In fact ALL the elemental gases, save the Noble Gases, are bimolecular, i.e. ${F}_{2} , C {l}_{2} , {N}_{2} , {O}_{2} , {H}_{2}$.
As always, ${N}_{A} \equiv \text{Avogadro's number}$, $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$