Question

If I have `9.71xx10^22` `"platimum atoms"`, what molar quantity, and what mass of metal are present?

Answer 1

You have approx `0.16*mol` of platinum atoms, with a mass of approx. `30*g`.

Explanation:

`"Moles of platinum"` `=` `(9.71xx10^22*"platinum atoms")/(6.022xx10^23*"platinum atoms"*mol^-1)` `=` `??"moles"`?

Now `1` `mol` of `Pt` atoms has a mass of `195.08*g`.

So you should have a mass of approx. `30*g`.

I appreciate that it is hard to work with such unfeasibly large numbers. But remember that the mole is just another number, like a score, or a dozen, or 1 gross - it's just that a mole is *&&??! LARGE. `"Avogadro's number"` of platinum atoms specifies `1` `mol` and has a mass of `195.08*g`. Likewise `"Avogadro's number"` of `""^1H` atoms has a mass of `1*g` precisely.

This idea of molar and mass equivalence is something that is fundamental to chemistry, and you should take some effort to appreciate it.