# How many electrons are found in a sample of dioxygen gas contained in a "0.448 L" container at STP?

Aug 19, 2017

$1.20$ x ${10}^{22}$ electrons

#### Explanation:

The no of electrons present in448 ml of O2 at STP
At STP conditions 1 mole of O2 gas = $6.023$ x ${10}^{23}$ electrons

$1$ mole = $22.4$ L
So, $22.4$ L of ${O}_{2}$ = ($6.023$ x ${10}^{23}$ electrons)
$0.448$ L of ${O}_{2}$= ($6.023$ x ${10}^{23}$ electrons)x ($0.448$)]/$22.4$L
= $1.20$ x ${10}^{22}$ electrons

Aug 19, 2017

See.

#### Explanation:

$22400 m l$ ${O}_{2}$ contains 16×N_A electrons
$1 m l$ ${O}_{2}$ contains (16×N_A)/22400electrons
Hence,$448 m l$ contains (16×N_A×448)/22400 electrons=0.32×N_A
Here, ${N}_{A}$ is Avogadro number.

Aug 19, 2017

About $1.93 \times {10}^{23}$ electrons.

Assuming your definition of STP is ${0}^{\circ} \text{C}$ and $\text{1 atm}$, the molar volume of ${\text{O}}_{2}$, assuming it is an ideal gas, is based on the ideal gas law:

$P V = n R T$

for pressure $P$, volume $V$, mols $n$, temperature $T$, and universal gas constant $R$.

Rearranging, we get...

$\frac{V}{n} = \frac{R T}{P}$

= (("0.082057 L"cdot"atm/mol"cdot"K")("273.15 K"))/"1 atm"

$=$ $\text{22.414 L/mol}$

So, with this molar volume, we compare ratios:

$\text{0.448 L"/("x mols") = "22.414 L"/"mol}$

=> n_("O"_2) = "0.0200 mols"

Each $\text{O}$ atom contains $8$ electrons, so each ${\text{O}}_{2}$ molecule contains... well, $16$.

${\text{0.0200 mols O"_2 harr "0.320 mols e}}^{-}$

And thus, the number of electrons is given by

color(blue)("Number of electrons") = "0.320 mols e"^(-) xx 6.0221413 xx 10^(23) "mol"^(-1)

$= \underline{\textcolor{b l u e}{1.93 \times {10}^{23} \textcolor{w h i t e}{.} {\text{e}}^{-}}}$