# How many atoms are present in 95.1 L H_2O at STP?

There are 95.1 kg of water at STP; i.e $\frac{95.1 \times {10}^{3} \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $\text{moles}$.
Now the mole represents Avogadro's number of particles, $6.022 \times {10}^{23}$ $m o {l}^{-} 1$, ${N}_{A}$. So multiply the number of moles first by ${N}_{A}$ to give the number water molecules, and then mulitply this quantity by $3$. Why three?
Because this is a large number, you would be justified in using the symbol ${N}_{A}$ in your answer, provided you defined it.