Question #94cff

1 Answer
Nov 5, 2017


#"0.352 moles Br"_2#


Since the problem is providing you with the number of molecules of bromine, #"Br"_2#, and asking for the number of moles present in the sample, a good tool to use here would be the number of molecules needed to have exactly #1# mole of bromine.

By definition, you can't have #1# mole of bromine unless you have #6.022 * 10^(23)# molecules of bromine. This is Avogadro's constant and represents the definition of a mole.

So right from the start, the fact that you have

#2.12 * 10^(23)color(white)(.)"molecules Br"_2 " " < " " overbrace(6.022 * 10^(23)color(white)(.)"molecules Br"_2)^(color(blue)("equivalent to 1 mole of Br"_2))#

means that your sample will contain less than #1# mole of bromine.

More specifically, the sample will contain

#2.12 * color(red)(cancel(color(black)(10^(23)))) color(red)(cancel(color(black)("molec. Br"_2))) * "1 mole Br"_2/(6.022 * color(red)(cancel(color(black)(10^(23)))) color(red)(cancel(color(black)("molec. Br"_2)))) = color(darkgreen)(ul(color(black)("0.352 moles Br"_2#

The answer is rounded to three sig figs, the number of sig figs you have for the number of molecules persent in the sample.