# Question 5af27

Sep 14, 2017

$1.91 \cdot {10}^{22}$

#### Explanation:

Your strategy here will be to go from grams to moles using the molar mass of lithium iodide and from moles to formula units using Avogadro's constant.

Now, lithium iodide has a molar mass equal to ${\text{133.85 g mol}}^{- 1}$, which means that every mole of lithium iodide has a mass of $\text{133.85 g}$.

You can thus say that your sample will contain

4.24 color(red)(cancel(color(black)("g"))) * "1 mole LiI"/(133.85color(red)(cancel(color(black)("g")))) = "0.03168 moles LiI"

As you know, in order to have $1$ mole of lithium iodide, you need to have $6.022 \cdot {10}^{23}$ formula units of lithium iodide $\to$ this is Avogadro's constant.

In your case, the sample will contain

0.03168 color(red)(cancel(color(black)("moles LiI"))) * (6.022 * 10^(23)color(white)(.)"formula units LiI")/(1 color(red)(cancel(color(black)("mole LiI"))))#

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.91 \cdot {10}^{22} \textcolor{w h i t e}{.} \text{formula units LiI}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of lithium iodide.