# How many neutrons are present in a 54*mL volume of ""^(1)H_2""^(16)O under standard conditions?

Apr 30, 2017

Approx...$24 \cdot {N}_{A}$.........

#### Explanation:

Each individual water molecule, ${\text{^(1)H_2}}^{16} O$ molecule contains $\text{8 neutrons}$ (of course for this problem we ignore isotopic distribution, and these all are contained in the oxygen nucleus.

And so $\text{number of neutrons}$

$= \text{8 neutrons} \times \frac{54 \cdot m L \times 1.00 \cdot g \cdot m {L}^{-} 1}{18.01 \cdot g \cdot m o {l}^{-} 1} \times {N}_{A}$,

where ${N}_{A} = \text{Avocado Number} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

$= 24 \cdot {N}_{A}$.

And here I use ${N}_{A}$ as I would any other collective number, $\text{dozen, score, gross}$ etc. I am certainly free to do so, and any other chemist would immediately appreciate the quantity.