# How many molecules are there in 15.7 liters of chlorine gas at STP?

Apr 12, 2016

$\frac{15.7 \cdot L}{22.414 \cdot L \cdot m o {l}^{-} 1}$ $\times$ ${N}_{A}$

#### Explanation:

We idealize the behaviour of this gas. And it is a fact that $S T P$, 1 mole of an ideal gas has a volume of $22.414 \cdot L$.

So we take the quotient, $\frac{15.7 \cdot L}{22.414 \cdot L \cdot m o {l}^{-} 1}$ to get a molar quantity, and multiply this amount by Avogadro's number to get the number of molecules.

$\frac{15.7 \cdot \cancel{L}}{22.414 \cdot \cancel{L} \cdot \cancel{m o {l}^{-} 1}}$ $\times$ $6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1}$ $=$ ??

How many individual chlorine atoms are there in this quantity?