Question #7a5f8

1 Answer
Truong-Son N.
Dec 17, 2016

Well, once you see that you can convert from #"g"# to #"mol"#, and once you identify the units of molar volume, this is fairly straightforward:

#barV = V/n = 22.4# is in #"L/mol"#, if the standard pressure is #"1 atm"# and the standard temperature is #0^@ "C"#.

To get the #"mol"#s:

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

#=# #"0.570 mols Ne"#

And the only thing left to do is to get rid of the #"mol"#s.

#"22.4 L"/cancel"mol" xx 0.570 cancel"mols Ne" =# #color(blue)("12.8 L Ne")#

So then, what is the predicted density of #"Ne"# in #"g/L"# if assuming ideality? If the true density is #"0.9002 g/L"#, is #"Ne"# pretty ideal, or not?

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