Question

How many molecules of $O_2$ are present in 1 L air containing 80% volume of $O_2$ at STP?

Answer 1

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 $=1000\times {\displaystyle \frac{80}{100}}=800\text{ ml}$
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 $={\displaystyle \frac{6.022\times {10}^{23}}{22400}}$ molecules of oxygen.
800 ml oxygen gas in STP contains
$$={\displaystyle \frac{6.022\times {10}^{23}}{22400}}\times 800$$ molecules of oxygen
$$={\displaystyle \frac{6.022\times {10}^{23}}{\text{224}}}\times 8$$ molecules of oxygen
$$={\displaystyle \frac{6.022\times {10}^{23}}{\text{28}}}$$ 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.