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**Hint:**Given that, 1 L air contains 80% volume of ${O_2}$

Therefore, find the total volume of ${O_2}$ present in the air i.e. 80% of 1 L, i.e. 800 ml. Now, we know that 1 mole of any gas at STP contains $6.022 \times {10^{23}}$ molecules and occupies a volume of 22.4 litres. Hence, find the number of molecules present in 800 ml of oxygen gas at STP.

**Formula used:**

1 mole = $6.022 \times {10^{23}}$

1 mole of any gas at STP, occupies a volume of 22.4 litres (22400 ml).

**Complete step by step answer:**

Given that, 1 L air contains 80% volume of ${O_2}$

Therefore, total amount of oxygen present in 1 L (1000 ml) air is $ = 1000 \times \dfrac{{80}}{{100}} = 800{\text{ ml}}$

Now, we know that 1 mole of any gas at STP, occupies a volume of 22.4 litres (22400 ml).

Hence, 22400 ml of ${O_2}$ gas in STP, contains $6.022 \times {10^{23}}$ molecules of ${O_2}$.

1 ml of oxygen gas in STP contains $ = \dfrac{{6.022 \times {{10}^{23}}}}{{22400}}$ molecules of oxygen.

800 ml oxygen gas in STP contains

$$ = \dfrac{{6.022 \times {{10}^{23}}}}{{22400}} \times 800$$ molecules of oxygen

$$ = \dfrac{{6.022 \times {{10}^{23}}}}{{{\text{224}}}} \times 8$$ molecules of oxygen

$$ = \dfrac{{6.022 \times {{10}^{23}}}}{{{\text{28}}}}$$ molecules of oxygen

$$ = 2.15 \times {10^{23}}$$ molecules of oxygen

Hence, number of ${O_2}$ molecules present in 1 litre air containing 80% volume of ${O_2}$ at STP, is $$2.15 \times {10^{23}}$$. (answer)

**Note:**We know that, 1 mole of any gas at STP occupies a volume of 22.4 litres (22400 ml).

Also, 1 mole of any substance contains $6.022 \times {10^{23}}$ number of molecules, which is called the Avogadro’s Number. Therefore 22.4 litres of oxygen at STP contains $6.022 \times {10^{23}}$ oxygen molecules.

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