# How many hydrogen atoms are present in an 18*mL volume of water?

Jul 1, 2017

We have $2 \times {N}_{A}$ hydrogen atoms in an $18 \cdot m L$ volume of water, where ${N}_{A} = \text{Avogadro's Number}$, $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$. Why?
Given that ${\rho}_{\text{water}} = 1 \cdot g \cdot m {L}^{-} 1$, we have a mass of $1 \cdot \cancel{m L} \times 1 \cdot g \cdot \cancel{m {L}^{-} 1} = 18 \cdot g$.......
And thus a molar quantity of $2 \cdot H \cdot m o {l}^{-} 1 \times {N}_{A} = 1.204 \times {10}^{24}$ $\text{individual hydrogen ATOMS.......}$