How many atoms are present in 406.4 g of lead?

1 Answer
May 18, 2017

Answer:

Well in #207.2*g# of lead atoms, there are #"Avogadro's number"# of particles.

Explanation:

And #"Avogadro's Number"-=N_A=6.022xx10^23*mol^-1#........

And thus in a mass of #406.4*g# lead atoms, there are...........

#(406.4*g)/(207.2*g*mol^-1)xx6.022xx10^23*mol^-1#

#=1.18xx10^24# #"individual lead atoms".#

This use of mass and Avogadro's number to define the number of particles, of atoms, and molecules, is fundamental to the study of quantitative chemistry.