= Solution mass and volume bookkeeping
{title2=$c=n/V$}
The denominator of <molar concentration> is the final <volume (thermodynamics)> of the <chemical solution>, including the dissolved <solute>. It is not the <volume (thermodynamics)> of <solvent> initially poured in. A <volumetric flask> fixes the final total <volume (thermodynamics)>; weighing the <solute> supplies its <mass>, and a specified <molar mass> converts that <mass> to <amount of substance>.<Chemistry 2e by Paul Flowers and others/3.3 Molarity>[Chemistry 2e, §3.3, molarity definition and Example 3.16: mass converted to moles and divided by total solution volume]{ref}
Consider an original illustrative account, not reported measurements. Suppose $2.922\,\mathrm g$ of pure dry <sodium chloride>, with stipulated <molar mass> $58.44\,\mathrm{g\,mol^{-1}}$, is completely dissolved and quantitatively transferred into a <chemical solution> whose final <volume (thermodynamics)> is $250.0\,\mathrm{mL}$. Assume no losses, no additional <chemical reactions>, and complete <dissociation of an ionic compound>. Then
$$
n_{\mathrm{NaCl}}=\frac{2.922\,\mathrm g}{58.44\,\mathrm{g\,mol^{-1}}}=0.05000\,\mathrm{mol},\qquad
c_{\mathrm{NaCl,analytical}}=\frac{0.05000\,\mathrm{mol}}{0.2500\,\mathrm L}=0.2000\,\mathrm{mol\,L^{-1}}.
$$
Here <analytical concentration> counts the <sodium chloride> inventory, not intact dissolved $\mathrm{NaCl}$ <molecules>. Under these assumptions, the <amount of substance> of <sodium> <cations> and of <chloride> <anions> is separately $0.05000\,\mathrm{mol}$; each has <molar concentration> $0.2000\,\mathrm{mol\,L^{-1}}$. Their combined ionic <amount of substance> is $0.1000\,\mathrm{mol}$, but this does not double either the salt inventory or the <mass>. Each <sodium> <cation> still contains a <sodium> <atomic nucleus>, and each <chloride> <anion> a <chlorine> <atomic nucleus>. Counting <chemical elements> survives <dissociation of an ionic compound>.
The <mass> ledger is different from the <volume (thermodynamics)> ledger. If the final <chemical solution> had an illustrative net <mass> of $252.922\,\mathrm g$, the <water> contribution would be $252.922-2.922=250.000\,\mathrm g$, assuming nothing entered or escaped except those two components. The ratio $2.922/252.922\approx0.011553$ expresses the solute share of total <mass>; the separate <mass density> would be $252.922/250.0=1.011688\,\mathrm{g\,mL^{-1}}$ before rounding. These stipulated quantities are mutually consistent bookkeeping inputs, not a measured <mass density> for saltwater. Neither establishes that $250.0\,\mathrm{mL}$ of <water> was added.
More generally, with full transfer and conserved <solute>, $n=m/M$ and $c=m/(MV)$. No additive <volume (thermodynamics)> or ideal <chemical solution> assumption is needed when $V$ is the measured final <volume (thermodynamics)>. If only a fraction $f$ of the weighed <solute> reaches the flask, the actual result is $c=fm/(MV)$. This short calculation explains why an undissolved residue or an incomplete rinse creates a systematic deficit even when the final calibration mark is read correctly.
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