# Question 7a5f8

Dec 17, 2016

Well, once you see that you can convert from $\text{g}$ to $\text{mol}$, and once you identify the units of molar volume, this is fairly straightforward:

$\overline{V} = \frac{V}{n} = 22.4$ is in $\text{L/mol}$, if the standard pressure is $\text{1 atm}$ and the standard temperature is ${0}^{\circ} \text{C}$.

To get the $\text{mol}$s:

11.5 cancel"g Ne" xx "1 mol Ne"/(20.180 cancel"g Ne")#

$=$ $\text{0.570 mols Ne}$

And the only thing left to do is to get rid of the $\text{mol}$s.

$\text{22.4 L"/cancel"mol" xx 0.570 cancel"mols Ne} =$ $\textcolor{b l u e}{\text{12.8 L Ne}}$

So then, what is the predicted density of $\text{Ne}$ in $\text{g/L}$ if assuming ideality? If the true density is $\text{0.9002 g/L}$, is $\text{Ne}$ pretty ideal, or not?