Why is Avogadro constant used to calculate the number density? My book says:

The number density of particles is $nN/V$, where $n$ is the total amount of molecules in the container of volume $V$ and $N$ is Avogadro's constant.

I can do something with the concentration $n/V$, it tells me how many moles of a particle I have in a certain volume, but why times $N$?
And another thing, where is the difference between number density (according to Wikipedia $n/V$) and molar concentration $n/V$?
 A: If n is the number of moles then n/V will be the number of moles per volume.
If you want to know the number of molecules per volume, you need to multiply this by Avogadro's constant.
An example:
If you measure a container's volume is 2 liters. And you know there is 2 moles of gas in this volume. Then you know there is 2moles/2liters = 1 mole per liter.
If you want to know the molecules per liter you can see it as first converting the 2 moles of gas to number of molecules in those 2 moles of gas which is 2N. So in the volume of 2 liters we have 2N molecules.
To get the particle density we have then 2N molecules / 2 liters = N molecules per liter.
Now it doesnt matter if we do "2N molecules/2 liters" or
"N molecules/mole*2moles/liter", they arrive at the same thing.
If that is not clear to you I suggest checking out dimensional analysis.
Hope this helped.
Also, I think number density is usually reserved for gases and molar concentration for solutions.
Number density, as far as I know, usually describes amount of particles per volume.
Molar concentration describes amount of moles per volume.
Indeed they convey a similar concept. Theres a big difference in that with a gas the number density literally describes how many gas particles are in a given volume.
The molar concentration describes how many moles of a substance are dissolved in a given volume of solution.
A: You are converting between absolute number of particles per volume and mol of particles per volume. The total number of particles is obtained by multiplying the moles with Avogardro's constant. Lets say I have 1 mol per volume. How many particles in absolute numbers do I have per volume? To answer that question you multiply with Avogadro's number.
An analogy would be converting between dozens per volume and absolute number per volume. Lets define the constant $d=\frac{12}{\text{[dozen]}}$. To convert  a quantity specified in dozens per volume, you would simply multiply with constant $d$ to obtain the absolute number of particles per volume.
Avogadro's number plays the same role, $N_A=\frac{6.02214076\cdot 10^{23}}{[mol]}$
A: tl;dr–  The alleged-excerpt's wrong.  It's combining several different problems, including messed-up unit-logic and a dated unit-convention.

Two conventions: Normal math and old-fashion Chemistry-math.
There're two relevant conventions here:

*

*Normal math, that everyone uses for everything.


*Old-fashioned silliness.
The book's referring to Convention-(2).  So, let's look at the right way to do this, then we can try to untangle the book's distorted logic.

1: Using normal math.
If you want an amount-of-molecules, $n ,$ then:

*

*Let people plug in whatever value they want, using whatever units they want.


*Let people do their own unit-conversions if they want to.
To fix the textbook's mistakes:

The particle-density is $n/V$, where $n$ is the amount of particles in the container of volume $V .$

Example calculations:
$$
{\newcommand{\Row}[3]{ #1 & #2 & #3 & \frac{#2}{#3} \\ \hline }}
{\newcommand{\Sci}[2]{ #1 \cdot {10}^{#2} }}
{\newcommand{\Stack}[2]{ {\small \begin{array}{c} #1 \\[-25px] #2 \end{array}} }}
\begin{array}{|c|c|c|} \hline
\Row{i}{n}{V}
\Row{1}{\Sci{2.429}{25} \,\text{particles}}{1\,\rm{m}^3}
\Row{2}{\Sci{1.2147}{25} \, \Stack{\text{pairs of}}{\text{particles}} }{1\,\rm{m}^3}
\Row{3}{\Sci{2.0244}{24} \, \Stack{\text{dozens of}}{\text{particles}} }{1\,\rm{m}^3}
\Row{4}{\Sci{1.8687}{24} \, \Stack{\text{baker's dozens}}{\text{of particles}}}{1\,\rm{m}^3}
\Row{5}{40.34 \, \Stack{\text{moles of}}{\text{particles}} }{1\,\rm{m}^3}
\end{array}
$$
Then if you don't like the units of $\frac{n}{V} ,$ you could just do whatever unit-conversions you'd like to arrive at whatever units you'd like.

2: Old-fashioned silliness.
Apparently some folks used to think that units were too complicated, so they sought to simplify things by embedding units into equations.
The rules weren't entirely consistent, but the logic went sort of like this:

*

*Everyone in a field adopts a standard unit-set and agrees to always use the same units.


*All variables in all expressions are assumed to be unitless, simply being the number from the value when written in the expected units.


*Any unit-conversion-factors needed to make this work are embedded in equations.
So instead of the simple, straight-forward math in the last section, your textbook apparently wrote:

The number density of particles is $nN/V$, where $n$ is the total amount of molecules in the container of volume $V$ and $N$ is Avogadro's constant.

, reasoning:

*

*The "the total amount of molecules" is an amount, which is always in units of moles, per the field's convention.  So, $n$ is expected to be a unitless number.


*They can't call the amount of molecules the "amount of molecules" when they don't want moles because they're requiring that amounts always be in moles.  So they choose the word "number", which is the amount of molecules that's an exception to the use-only-moles rule.

*

*To be clear: The unit of "moles" is a number, like "dozen" or "thousand", so an "amount of molecules" and a "number of molecules" are logically equivalent in this context, except in that they're implying that the latter is an exception to the rule that all amounts are in units of moles.



*Since they've stripped out all of the units and are requiring specific units to be assumed, they need to embed conversion-factors.  In this case, that's Avogadro's constant, $N_{\text{Avogadro}} .$


*Apparently they don't like writing it as $`` N_{\text{Avogadro}} "$ or even just $`` N_{\text{A}} " ,$ so they write it as $`` N " .$  Given that their primarily use-case for Avogadro's constant is to multiply it by a variable they're calling $`` n " ,$ that seems like a weird choice.
Okay, so that might all sound like a lot of needless silliness, but it's actually even worse in practice.  Because in practice, when folks do a lot of science/engineering/medicine/etc. in real-life, we tend to combine stuff from across fields – so it's not just obnoxious, but stuff can get mixed up.
For example, in 1999, it was reported that the Mars-Climate-Orbiter crashed because two different groups were using implicit-units like this, with an error, causing a huge loss.
And that's the funny sort of problem.  Not that it's funny when missions-to-Mars needlessly crash, wasting a small fortune and setting back scientific progress, but at least we're not talking about dosing mistakes hospitals.
Then there's silly stuff, too.  For example, I've seen equations that involved terms like $`` {\left(T-273\right)}^{2} " ,$ where $`` T "$ was meant to be in terms of $\mathrm{K} .$  My guess was that the equation was originally written in terms of $\sideset{^\circ}{}{\mathrm{C}}$ and that someone later switched it to $\mathrm{K} .$  But did the data they give come from before-or-after the conversion?  Or maybe the data was after the conversion, but whoever did the calculations used the full $-273.15 \,?$  It's not fun having to guess stuff like that when small errors matter.
All of that said... technically,

The number density of particles is $nN/V$, where $n$ is the total amount of molecules in the container of volume $V$ and $N$ is Avogadro's constant.

could be used if correctly navigated:

*

*You start off with the number of particles in terms of moles.


*You want the particle-density not in terms of moles.


*You could calculate $\frac{n \times {N}_{\text{Avogadro}}}{V}$ to do that.
It's an incredibly silly, error-prone, convoluted approach to doing something that would otherwise be relatively very simple, but it's not technically wrong aside from those issues.

Conclusion.
The textbook's in error.  The correct approach would be to just calculate $\frac{n}{V} ,$ then do whatever unit-conversions you'd want.
Technically if you follow their logic far enough it can work when appropriately handled, but it's just too absurd to take seriously.
