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

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Answer 1 · 2017-01-27T07:35:25

A bit more than #5/6# #"moles"#......

#"Avogadro's number"# #-=# #6.022xx10^23*mol^-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 #1/(mol^-1)=1/(1/(mol^-1))=mol# as required.........

Answer 2 · 2017-01-27T07:38:09

#5.66xx10^23"Xe atoms"# equals #"0.940 mol Xe atoms"#

One mole of anything, including xenon atoms is #6.022xx10^23#.

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

#1"mole Xe atoms"=6.022xx10^23color(white)(.)"atoms"#

Now we can write two conversion factors.

#(1"mol Xe atoms")/(6.022xx10^23color(white)(.)"Xe atoms")# and #(6.022xx10^23color(white)(.)"Xe atoms")/(1"mol Xe atoms")#

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.

#5.66xx10^23cancel"Xe atoms"xx(1"mol Xe")/(6.022xx10^23cancel"Xe atoms")="0.940 mol Xe atoms"# rounded to three significant figures