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

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

#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.

About #1.93 xx 10^23# electrons.

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"^(-))#