How many silver atoms are there in a 0.566*kg mass of silver?

Oct 9, 2016

$0.566 \cdot k g$ of silver $\cong$ $30 \times {10}^{23} \text{ silver atoms}$

Explanation:

We know (or should know) that the formula mass of silver is $107.87 \cdot g \cdot m o {l}^{-} 1$. How did I know this? Did I know it off the top of my head? How will you know this if you are asked the question in an exam?

Now $1$ $\text{mole}$ of silver atoms or in a $107.87 \cdot g$ mass, there are, by definition, $6.022 \times {10}^{23}$ $\text{individual atoms of silver.}$

So all I have to do to get the number of silver atoms, is (i) calculate the molar quantity, and (ii) multiply this molar quantity by $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, i.e. $\text{Avogadro's number}$.

So.......

$\frac{0.566 \times {10}^{3} \cdot g}{107.87 \cdot g \cdot m o {l}^{-} 1}$ $=$ $5.25 \cdot m o l$.

And $5.25 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $\cong$ $30 \times {10}^{23} \text{ silver atoms}$