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

2 Answers
Jun 12, 2017

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

Jun 12, 2017

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!