# How many atoms in 1m^3 of N_2O at "STP"?

Apr 28, 2016

How many atoms in $1$ ${m}^{3}$ of ${N}_{2} O$ at $\text{STP}$?
In $22.4$ $L$ at $\text{STP}$, there are ${N}_{A}$ gaseous particles, where ${N}_{A} = \text{Avogadro's Number } , 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.
So in $1 \cdot {m}^{3}$, there are $\frac{1000 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1}$ $=$ 44.6*mol" *N_2O
So $\text{number of atoms}$ $=$ $44.6 \cdot m o l \times {N}_{A} \times 3 \cdot \text{atoms}$ $=$ $\text{ A lot of atoms}$.