How do I find the amount of atoms in 1 Liter H20?

Sep 12, 2017

For a start we ASSUME that the temperature is $25$ ""^@C, i.e. we gots LIQUID water......

Explanation:

We then work out the molar quantity....

$\frac{1 \cdot L \times {10}^{3} \cdot g \cdot {L}^{-} 1}{18.01 \cdot g \cdot m o {l}^{-} 1} = 55.52 \cdot m o l$

What is the $m o l$? This is Avogadro's number, and is equal to $6.02214 \times {10}^{23}$; and this quantity of water MOLECULES has a mass of $18.01 \cdot g$.

But clearly there are THREE MOLES OF ATOMS ($2 \times H + O$) per mole of water, and so we got.....

$166.57 \cdot m o l$ atoms......

And finally we multiply this molar quantity by Avogadro's number....

$166.57 \cdot m o l \times 6.02214 \times {10}^{23} \cdot m o {l}^{-} 1 = 1.003 \times {10}^{26} \cdot \text{atoms}$.

Are you happy with this?

Here I have used the mole as I would any other collective number, 10, dozen, gross, etc. As a number, the mole is unfeasibly large.

Sep 12, 2017

Yes Answer is $1.10 \times {10}^{26}$ Atoms

Explanation:

The concept behind the question is " 1 Mole of water contains $6.023 \times {10}^{23}$ water molecules and 1 Mole of water weighs $18 g$ or $18 m l$ "

Now we are provided with 1 liter i.e, $1000 m l$

Hence number of moles in 1 liter of water is $\frac{1000}{18} = 55.55556 m o l s$

So that total number of molecules is given by $m o l s \times 6.023 \times {10}^{23}$= $55.55556 \times 6.023 \times {10}^{23}$

Since every molecule of water ( ${H}_{2} O$) contains 3 atoms(1 oxygen and 2 hydrogen atoms) the total number of atoms in 1 liter
of water is given by $3 \times 55.55556 \times 6.023 \times {10}^{23}$= $1.10 \times {10}^{26}$ Atoms
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