# How many oxygen molecules are present in 207 liters of oxygen gas at STP?

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Oct 25, 2017

$5.56 \times {10}^{24} \textcolor{w h i t e}{\text{x}}$molecules$\textcolor{w h i t e}{\text{x}} {O}_{2}$

#### Explanation:

At STP conditions for ideal gases, we use the conversion factor

$1 \textcolor{w h i t e}{\text{x" mol ~~ 22.4 color(white)"x}} L$

Given $207 L$ of oxygen (or any ideal gas for that matter) at STP

$207 \textcolor{w h i t e}{\text{x" cancel(L color(white)"x" O_2) xx (1 color(white)"x" mol color(white)"x" O_2)/(22.4 color(white)"x" cancel(L color(white)"x" O_2))=9.24 color(white)"x" mol color(white)"x}} {O}_{2}$

Then we convert from $m o l$ to number of molecules $\left(m l c\right)$ using Avogadro's number:

$9.24 \textcolor{w h i t e}{\text{x" cancel(mol color(white)"x" O_2)xx(6.02xx10^23 color(white)"x" mlc color(white)"x" O_2)/(1 color(white)"x" cancel(mol color(white)"x" O_2)) = 5.56 xx 10^24 color(white)"x" mlc color(white)"x}} {O}_{2}$

(make sure to use unrounded intermediate answers and round to correct number of significant figures at the end)

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Oct 25, 2017

$5.56$ x ${10}^{24}$ $\text{molecules}$

#### Explanation:

1 mole of any ideal gas at STP occupies volume = $22.414 {\mathrm{dm}}^{3}$

Thus, at STP, 207 liters of oxygen contains

$\text{1 mol"/"22.414 L" xx "207 L" = "9.24 moles}$

$\text{Molecules of oxygen = Number of moles}$ $\times$ $\text{Avogadro's number}$ (${N}_{A}$)

$= {\text{9.24 mols" xx 6.02 xx 10^23 "mol}}^{- 1}$

$= \underline{5.56 \times {10}^{24} \text{molecules}}$

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