Question #3fa99

1 Answer
Aug 20, 2017

1.48xx10^241.48×1024 "molecules NH"_3molecules NH3

Explanation:

We're asked to find the number of molecules of "NH"_3NH3 present in 2.452.45 "mol NH"_3mol NH3.

To do this, we can use the fact that one mole of any substance contains underbrace(6.022xx10^23)_"Avogadro's number" particles of that substance.

Therefore, we have

2.45cancel("mol NH"_3)((6.022xx10^23color(white)(l)"molecules NH"_3)/(1cancel("mol NH"_3)))

= color(red)(ulbar(|stackrel(" ")(" "1.48xx10^24color(white)(l)"molecules NH"_3" ")|)

This goes for any substance as well. That is to say, 2.45 moles of carbon dioxide also contains color(red)(1.48xx10^24color(white)(l)"molecules" of "CO"_2.