# How many atoms in a 25.0*g mass of water?

Mar 28, 2016

We calculate the number of moles of water, and then multiply this number by $3 {N}_{A}$

#### Explanation:

$\text{Moles of water}$ $=$ $\frac{25.0 \cdot g}{\text{18.01 } g \cdot m o {l}^{-} 1}$ $=$ $1.39 \cdot m o l .$

In each mole of water there are $3$ $m o l$ atoms.

So $\text{Number of atoms}$ $=$ $1.39 \cdot m o l \times {N}_{A} \times 3 \text{ atoms } m o {l}^{-} 1$ $=$ $4.16 \times {N}_{A} \text{ atoms}$.

${N}_{A} = \text{Avogadro's number}$ $=$ $6.022 \times {10}^{23}$