Question

If there are `2.50xx10^18` lead atoms, what is the mass of this quantity?

Answer 1

Isn't a `"ream"` a quantity of paper, i.e. 480 or 500 individual sheets?

Explanation:

I don't know the mass of a ream of lead, and I have never heard of lead being measured with this quantity.

But the number of atoms of any elemental quantity is `"Number of moles "xx" Molar mass"`, where a `"mole"` specifies `"Avogadro's Number"`, `6.022xx10^23*mol^-1`.

And thus if there were `2.50xx10^18` lead atoms........we have a mass of..........

`(2.50xx10^18*cancel"lead atoms")/(6.022xx10^23*cancel"lead atoms"*cancel(mol^-1))xx207.2*g*cancel(mol^-1)`

`~=1*mg`

Answer 2

I'm not quite sure how much lead you have but this is how to calculate the number of atoms:

`N=n*N_A`

Where:

`N="number of atoms"`

`N_A="Avogadro constant"=6.02*10^23" mol"^-1`

`n="mass"/"molar mass"`

`"molar mass of lead"=207.2" g/mol"`

`N_A` is a constant and is the number of atoms in one mole of any substance. So if you have `n=2" mol"`, you have `2*N_A` atoms.

As an example, if you have `2.50*10^18` grams of lead, you have this many atoms:

`n=(2.50*10^18)/207.2=1.21*10^16" mol"`

`N=n*N_A=1.21*10^16*6.02*10^23=7.28*10^39" atoms"`

This is a lot of lead!