How many atoms are in 1.45 g of Ag?

1 Answer
anor277
Mar 3, 2016

#"Number of silver atoms"# #=# #0.0134*molxxN_A#. #N_A = "Avogadro's number"#

Explanation:

Silver has a molar mass of #107.87# #g*mol^-1#.

What does this mean? It means that if I have such a mass of silver, there are Avogadro's number, #N_A = 6.022xx10^23#,
individual silver atoms.

So all I have to do is divide the mass by the molar mass, and multiply this number (a molar quantity!) by Avogadro's number.

So, moles of silver #=# #(1.45*cancelg)/(107.87*cancelg*mol^-1)# #=# #0.0134*mol#

And #"number of silver atoms"# #=# #0.0134*cancel(mol)xx6.022xx10^23*" silver atoms "cancel(mol^-1)# #=# #??# #"silver atoms"#

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