# What is the number of molecules in 1.00 mol of iodine crystals?

## Is the answer 6.022 x 10^25 molec. of I2 ? or do I have double this answer, given that I2 is a diatomic molecule??

Jan 21, 2017

$6.02 \cdot {10}^{23}$

#### Explanation:

The problem wants to make sure that you are familiar with the definition of one mole of a given substance.

In this case, you're dealing with a covalent compound, so you're looking for the number of molecules of iodine, ${\text{I}}_{2}$, present in exactly $1.00$ moles of iodine crystals.

Now, a mole is defined as very, very large collection of particles. In order to have $1.00$ moles of iodine crystals, you need to have exactly $6.022 \cdot {10}^{23}$ molecules of iodine.

color(blue)(ul(color(black)("1 mole I"_2 = 6.022 * 10^(23)"molecules I"_2 $\to$ Avogadro's constant

In your case, you will have

${\text{1.00 moles I"_2 = 6.02 * 10^(23)"molecules I}}_{2}$

The answer must be rounded to three sig figs, the number of significant figures you have for the number of moles of iodine crystals.

Notice that each iodine molecule contains $2$ atoms of iodine, $2 \times \text{I}$, which means that $1$ mole of iodine molecules contains $2$ moles of iodine atoms.

In $1.00$ moles of iodine molecules you have $2$ moles of iodine atoms, or

overbrace(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2))))^(color(blue)(= "1.00 moles I"_2)) * "2 atoms of I"/(1color(red)(cancel(color(black)("molecule I"_2)))) = overbrace(1.20 * 10^(24)"atoms of I")^(color(purple)(="2.00 moles of I"))

Once again, the answer must be rounded to three sig figs.