# Question #18188

May 8, 2017

Let us assume a $100 \cdot g$ mass of gas..........

#### Explanation:

And thus there are:

$\frac{20 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 0.625 \cdot m o l$ with respect to $\text{dioxygen}$.

And $\frac{80 \cdot g}{28.0 \cdot g \cdot m o {l}^{-} 1} = 3.33 \cdot m o l$ with respect to $\text{dinitrogen}$.

And since the molar quantity relates directly to the number of molecules, ${O}_{2} : {N}_{2} \cong 1 : 5$.