# How many atoms are in 1.45 g of Ag?

Mar 3, 2016

$\text{Number of silver atoms}$ $=$ $0.0134 \cdot m o l \times {N}_{A}$. ${N}_{A} = \text{Avogadro's number}$

#### Explanation:

Silver has a molar mass of $107.87$ $g \cdot m o {l}^{-} 1$.

What does this mean? It means that if I have such a mass of silver, there are Avogadro's number, ${N}_{A} = 6.022 \times {10}^{23}$,
individual silver atoms.

So all I have to do is divide the mass by the molar mass, and multiply this number (a molar quantity!) by Avogadro's number.

So, moles of silver $=$ $\frac{1.45 \cdot \cancel{g}}{107.87 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $0.0134 \cdot m o l$

And $\text{number of silver atoms}$ $=$ $0.0134 \cdot \cancel{m o l} \times 6.022 \times {10}^{23} \cdot \text{ silver atoms } \cancel{m o {l}^{-} 1}$ $=$ ?? $\text{silver atoms}$