Question

the number of electrons present in 448 ml of O2 at STP is

Answer 1

1.92 X 10^23 is the answer

Answer 2

Answer : The number of electrons present in 448 ml of $O_2$ at STP is $1.92\times 10^{23}$.

Solution : Given,

Volume of $O_2$ = 448 ml = 0.448 L         (1L = 1000ml)

At STP, 1 mole contains 22.4 L volume of gas.

As, 22.4 L volume of gas have 1 mole of gas

0.448 L volume of gas have $\frac{1}{22.4}\times 0.448=0.02$ moles of gas

Number of moles of $O_2$ gas = 0.02 moles

1 mole of $O_2$ has $6.022\times 10^{23}$ molecule of $O_2$

0.02 mole of $O_2$ has $(6.022\times 10^{23})\times (0.02)=0.12\times 10^{23}$ molecule of $O_2$

The number of electrons present in 1 molecule of $O_2$ is,

$O_2$ = 2(8) = 16 electrons

Number of electrons present in 1 molecule of $O_2$ = 16 electrons

Number of electrons present in $0.12\times 10^{23}$  molecule of $O_2$ = $(16)\times 0.12\times 10^{23}=1.92\times 10^{23}$ electrons

Therefore, the number of electrons present in 448 ml of $O_2$ at STP is $1.92\times 10^{23}$.