# How many molecules are there in an 11.2*L volume of gas at "NTP"?

Jul 17, 2017

Approx. $3 \times {10}^{23} \cdot \text{molecules}$

#### Explanation:

As far as I know, and you will have to check your syllabus (do so, because this is important!), $\text{NTP}$ specifies a temperature of $293.15 \cdot K$, and a pressure of $1 \cdot a t m$. Unfortunately there seems to be several standards in use, $\text{STP}$, $\text{NTP}$, and $\text{SATP}$, and several $\text{standard}$ pressures of $1 \cdot a t m$, and $100 \cdot k P a$; and this is bound to cause confusion, and error, and of course it does so. The given standards SHOULD be listed as supplementary material on each exam paper. You still have to be able to use the standards effectively.

So we use the Ideal Gas equation.......$n = \frac{P V}{R T}$

$= \frac{1 \cdot a t m \times 11.2 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 293.15 \cdot K} = 0.465 \cdot m o l$.

We are still not done, because we were axed for THE NUMBER of molecules. And the molar quantity is by definition a NUMBER of particles.

$1 \cdot \text{mole} \equiv 6.022 \times {10}^{23}$ $\text{individual particles}$

And so......$\text{Number of atoms} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times 0.465 \cdot m o l$

$=$ $\text{How many atoms?}$

Capisce?