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

Jun 12, 2017

Isn't a $\text{ream}$ 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 $\text{Number of moles "xx" Molar mass}$, where a $\text{mole}$ specifies $\text{Avogadro's Number}$, $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus if there were $2.50 \times {10}^{18}$ lead atoms........we have a mass of..........

$\left(2.50 \times {10}^{18} \cdot \cancel{\text{lead atoms")/(6.022xx10^23*cancel"lead atoms}} \cdot \cancel{m o {l}^{-} 1}\right) \times 207.2 \cdot g \cdot \cancel{m o {l}^{-} 1}$

$\cong 1 \cdot m g$

Jun 12, 2017

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

$N = n \cdot {N}_{A}$

Where:

$N = \text{number of atoms}$

${N}_{A} = {\text{Avogadro constant"=6.02*10^23" mol}}^{-} 1$

$n = \text{mass"/"molar mass}$

$\text{molar mass of lead"=207.2" g/mol}$

${N}_{A}$ is a constant and is the number of atoms in one mole of any substance. So if you have $n = 2 \text{ mol}$, you have $2 \cdot {N}_{A}$ atoms.

As an example, if you have $2.50 \cdot {10}^{18}$ grams of lead, you have this many atoms:

$n = \frac{2.50 \cdot {10}^{18}}{207.2} = 1.21 \cdot {10}^{16} \text{ mol}$

$N = n \cdot {N}_{A} = 1.21 \cdot {10}^{16} \cdot 6.02 \cdot {10}^{23} = 7.28 \cdot {10}^{39} \text{ atoms}$

This is a lot of lead!