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

2 Answers
Jun 12, 2017

Answer:

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!