# How many molecules are contained in 134.5 L of O_2 at STP?

Jan 21, 2016

$3.6138 \times {10}^{24}$

#### Explanation:

At STP, 1 mole of all gases occupies a volume of $22.4 {\mathrm{dm}}^{3}$.

Therefore, since $1 L = 1000 c {m}^{3}$, we may use ratio and proportion to show that $134.5 L = 134500 c {m}^{3} = 0.1345 {m}^{3} = 134.5 {\mathrm{dm}}^{3}$.

Therefore this volume contains $\frac{134.5}{22.4} = 6 m o l$.

But 1 mole contains an Avagadro number of particles, ie $6.023 \times {10}^{23}$.

Therefore by ration and proportion, 6 moles of oxygen will contain $6 \times 6.023 \times {10}^{23} = 3.6138 \times {10}^{24}$ oxygen molecules.