 Solution mass and volume bookkeeping

The denominator of molar concentration is the final volume of the chemical solution, including the dissolved solute. It is not the volume of solvent initially poured in. A volumetric flask fixes the final total volume; weighing the solute supplies its mass, and a specified molar mass converts that mass to amount of substance.^[ref]

Consider an original illustrative account, not reported measurements. Suppose $2.922\mathrm{g}$ of pure dry sodium chloride, with stipulated molar mass $58.44\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, is completely dissolved and quantitatively transferred into a chemical solution whose final volume is $250.0\mathrm{mL}$. Assume no losses, no additional chemical reactions, and complete dissociation of an ionic compound. ThenHere 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 $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$. 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 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 $2.922/252.922\approx 0.011553$ expresses the solute share of total mass; the separate mass density would be $252.922/250.0=1.011688\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$ 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 or ideal chemical solution assumption is needed when $V$ is the measured final volume. 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.