What is the mass in grams of #7.9xx10^21# uranium atoms?

Show work and explain please

1 Answer
Nov 25, 2017

Well what is the mass of #"Avogadro's Number of uranium atoms?"#

Explanation:

A quick glimpse at the Periodic Table gives the molar mass of uranium as #231.8*g*mol^-1#.

Now, by definition, this mass is equivalent to the mass of #"Avogadro's Number of uranium atoms."# And so to get the mass of the specified number, we take the product of the quotient:

#(7.9xx10^21*"uranium atoms")/(6.022xx10^23*"uranium atoms"*mol^-1)xx231.8*g*mol^-1#

And my calculator says that this represents approx. a #3*g# mass....