# How many molecules of ethane gas, C_2H_6, are in 33.6 L at STP?

Aug 17, 2016

Well, one mole of ideal gas at $\text{STP}$ occupies $22.4 \cdot L$.
So, assuming ethane behaves ideally, $33.6 \cdot L$ of the stuff represents $\frac{33.6 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1}$ $=$ $1.5 \cdot m o l$.
Now, it is a fact that $1$ $m o l$ of any substance contains $\text{Avogadro's Number, } {N}_{A}$ of particles, i.e. $6.022 \times {10}^{23}$ particles.
And thus $\text{number of ethane gas molecules } = 6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1} \times 1.5 \cdot \cancel{m o l}$ $\cong$ $9 \times {10}^{23}$ $\text{ethane formula units}$.