How many calcium atoms are present in a mass of #169*g# of this metal?

1 Answer
Feb 5, 2017

Answer:

#"Number of atoms"# #="Mass"/"Molar mass"xx"Avogadro's number."#

Explanation:

By definition, #40.1*g# of calcium atoms contains #"Avogadro's number of molecules"#. We specify this quantity as #1*mol# of #"calcium atoms"#.

And so we take the quotient, #(169*g)/(40.1*g*mol^-1)#, and multiply this by #N_A, "Avogadro's number of molecules",# where #N_A=6.022xx10^23*mol^-1#.

And thus the product, #(169*cancelg)/(40.1*cancelg*cancel(mol^-1))xx6.022xx10^23*cancel(mol^-1)# gives a DIMENSIONLESS number as required, approx. #4xxN_A#.

If I asked how many eggs were in 3 dozen eggses, I think you would immediately be able to reply. The question here is much the same sort of question, and can be viewed in the same simple light.