In my understanding,

$$1 \ mol \equiv N_A \ (atoms)$$

(Don't say it might not be atoms but molecules and also don't say what's it's not even a unit)

Then the unit should be $N_A = 1 \ mol/atoms = 1 \ mol$, not the reciprocal mole. My first question is what's wrong.

My second question is, we know that defining $atoms$ as being unit is absurd, because it's purely a number. And same applies to mole. There's not difference between them in the sense that they are dimensionless. But why is the mole a SI unit?

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    $\begingroup$ The unit is $\pu{1 mol} = N_\mathrm{A} \text{ particles}$ . What is strange on that ? Note that there is the unitless Avogadro number and the Avogadro constant with dimension $\pu{mol^-1}$. Then $n [mol] \cdot N_\mathrm{A} [1/mol] = N$. The molar amount with 1 mol as the unit is useful concept, as many quantities relate rather to the number of particles than to their total mass. $\endgroup$ – Poutnik Nov 6 '20 at 8:00
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    $\begingroup$ You may take any non-unit and call it a unit, and as long as it doesn't contradict anything, you'll be fine. That's why mole is a unit. $\endgroup$ – Ivan Neretin Nov 6 '20 at 8:01
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    $\begingroup$ "we know that defining atoms as being unit is absurd, because it's purely a number." the same applies to dollars or what ever currency. It seems you are fine with more abstract units (length for instance) but have problem with countable hardware. $\endgroup$ – Alchimista Nov 6 '20 at 8:43
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    $\begingroup$ Does this answer your question? What is the dimension of Avogadro's constant? $\endgroup$ – Mithoron Nov 6 '20 at 17:31

I realised where my confusion lies - so I'm leaving an answer so that other people learning chemistry for the first time are able to fully understand what's actually going on about Avogadro's number.

We have to make a clear distinction between Avogadro's number and Avogadro's constant.

The former is a unitless real number, defined as $6.02214076\times 10^{23}$.

The latter is a physical (or shall I call chemical?) constant, and thus has a unit. As @Poutnik and other people have commented on my question, the fact that $mol$ does not have a dimension doesn't mean it can't be a unit (the same applies to radian as a measure of angles); and that means there should exist a constant called the Avogadro's constant with the dimension $mol^{-1}$ so that the product of the two is dimensionless, i.e.

$$n \ [mol] \ \times N_A \ [mol^{-1}] = N \ [particles]$$

Here it's trivial that [particles] is not a unit, simply because we did not feel the need to (and thus chose not to) assign any unit, dimension, and whatever onto that thing called number of particles.

Again, although $mol$ is dimensionless it should still be treated as a unit so in order to cancel that out to produce a unitless number $N$ we can conclude that we need a number with value $6.02214076\times 10^{23}$ and unit $mol^{-1}$, or namely Avogadro's constant.

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    $\begingroup$ dimension of the mole is mole, as of the kilogram is kilogram and of the metre is metre. Where is the problem ? $\endgroup$ – Poutnik Nov 6 '20 at 11:54
  • $\begingroup$ You have solved your confusion but please see Poutnik comment. $\endgroup$ – Alchimista Nov 6 '20 at 15:47
  • $\begingroup$ @Poutnik Yes you are right. I agree with that, and in my answer I was just emphasising that. I was originally confused about that. $\endgroup$ – curious Nov 6 '20 at 16:26

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