Question

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

Answer 1

`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

Answer 2

See.

Explanation:

`22400ml` `O_2` contains `16×N_A` electrons
`1ml` `O_2` contains `(16×N_A)/22400`electrons
Hence,`448ml` contains `(16×N_A×448)/22400` electrons=`0.32×N_A`
Here, `N_A` is Avogadro number.

Answer 3

About `1.93 xx 10^23` electrons.

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Assuming your definition of STP is `0^@ "C"` and `"1 atm"`, the molar volume of `"O"_2`, assuming it is an ideal gas, is based on the ideal gas law:

`PV = nRT`

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

Rearranging, we get...

`V/n = (RT)/P`

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

`=` `"22.414 L/mol"`

So, with this molar volume, we compare ratios:

`"0.448 L"/("x mols") = "22.414 L"/"mol"`

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

Each `"O"` atom contains `8` electrons, so each `"O"_2` molecule contains... well, `16`.

`"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)`

`= ulcolor(blue)(1.93 xx 10^(23)color(white)(.)"e"^(-))`