# How many molecules of neon are associated with a 75*L volume of neon gas?

##### 1 Answer
May 16, 2016

The question should have specified a temperature. I am going to assume that the gas is at room temperature, $298 \cdot K$.

#### Explanation:

We know that at room temperature and pressure, $1$ $m o l$ of ideal gas occupies a volume of $25.4 \cdot L$.

We assume (reasonably) that $N e$ behaves ideally.

Thus the number of moles of $N e$ $=$ $\frac{75 \cdot L}{25.4 \cdot L \cdot m o {l}^{-} 1}$ $\cong$ $3 \cdot m o l$.

But by definition, $1 \cdot m o l$ of stuff possesses $6.022 \times {10}^{23}$ individual items of that stuff. If you can grasp this concept, it will save you a ton of angst.

Thus $\text{Molecules (atoms) of neon}$ $=$ $\frac{75 \cdot L}{25.4 \cdot L \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \text{ neon atoms per mole}$

$\cong 18 \times {10}^{23} \text{ neon atoms}$.

What is the mass of this quantity of neon gas?

(Neon gas is a monatomic molecule).

See also here.