# Which sample of fluorine gas, F2, contains the greatest number of fluorine molecules? (multiple choice)

## A. 1.9 x ${10}^{23}$ F2 molecules. B. 1.9 ${\mathrm{dm}}^{3}$ of F2 at 273 K and 1.00 x ${10}^{5}$ Pa pressure. C. 1.9 g F2 D. 1.9 mol F2 The answer is D. 1.9 mol F2, why? How do you calculate it?

Feb 16, 2016

Let's use the symbol ${N}_{A}$, where ${N}_{A}$ $=$ $6.022 \times {10}^{23}$.

#### Explanation:

A. there are approx. $\frac{1}{3} {N}_{A}$, INDIVIDUAL ${F}_{2}$ molecules.

B. there are approx $\frac{1}{12} {N}_{A}$, INDIVIDUAL ${F}_{2}$ molecules.

C. there are approx. $\frac{1}{19} {N}_{A}$, INDIVIDUAL ${F}_{2}$ molecules.

D. there are approx. $1.9 \times {N}_{A}$, INDIVIDUAL ${F}_{2}$ molecules.

By using the symbol ${N}_{A}$ (where ${N}_{A}$ is Avogadro's number) it is clear that option D is BY FAR THE MOST NUMEROUS.

Option A. was simply a number. To assess B, I know that ${N}_{A}$ ${F}_{2}$ molecules has a volume of $22.4 \cdot L$. To assess C, I know that ${N}_{A}$ ${F}_{2}$ molecules has a mass of $38$ $g$, so in $1.9$ $g$ ${F}_{2}$ there are approx. $\frac{1}{19} \times {N}_{A}$ such molecules. But D. specifies approx. $2 {N}_{A}$ ${F}_{2}$ miolecules.

So, when we use Avogadro's number as a NUMBER, and indeed it is a number, option D CLEARLY contains the most molecules.

I could have replaced the "mole" term with a dozen (because that's the number the mole is equivalent to!) and got nearly the same answer. If you are unconvinced with this treatment, state your objections, and we will try again. The point is that ${N}_{A}$ is simply a number, that has the property that ${N}_{A}$ ${F}_{2}$ molecules has a mass of $38.0$ $g$.