Question

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

Answer 1

`"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.