# How many moles of xenon do 5.66 xx 10^23 atoms equal?

Jan 27, 2017

A bit more than $\frac{5}{6}$ $\text{moles}$......

#### Explanation:

$\text{Avogadro's number}$ $\equiv$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And so we take the quotient:

(5.66xx10^23*cancel"atoms")/(6.022xx10^23*cancel"atoms"*mol^-1)=??mol

This is dimensionally consistent, because $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o {l}^{-} 1}} = m o l$ as required.........

Jan 27, 2017

$5.66 \times {10}^{23} \text{Xe atoms}$ equals $\text{0.940 mol Xe atoms}$

#### Explanation:

One mole of anything, including xenon atoms is $6.022 \times {10}^{23}$.

This gives us an equality from which we can derive two conversion factors.

$1 \text{mole Xe atoms"=6.022xx10^23color(white)(.)"atoms}$

Now we can write two conversion factors.

$\left(1 \text{mol Xe atoms")/(6.022xx10^23color(white)(.)"Xe atoms}\right)$ and $\left(6.022 \times {10}^{23} \textcolor{w h i t e}{.} \text{Xe atoms")/(1"mol Xe atoms}\right)$

Now mutliply the given number of atoms by the conversion factor that has moles in the numerator. This will result in the Xe atoms canceling.

$5.66 \times {10}^{23} \cancel{\text{Xe atoms"xx(1"mol Xe")/(6.022xx10^23cancel"Xe atoms")="0.940 mol Xe atoms}}$ rounded to three significant figures