Concentration of solutions

A solution has a solute (the component present in smaller amount) dissolved in a solvent (the component present in larger amount). Every concentration scale is a ratio of "how much solute" to "how much solution or solvent," and the scales differ only in what units go into the numerator and denominator.

Mass- and volume-based scales. Mass percentage and volume percentage compare solute to the whole solution:

where: both are dimensionless percentages, useful whenever no density is known and masses (or volumes) are measured directly on a balance or in a measuring cylinder. Parts per million follows the same pattern for trace amounts: $\text{ppm} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$.

Mole-based scale. For a solvent-solute pair with $n_1$ mol solvent and $n_2$ mol solute, the mole fraction is

where: $x_1, x_2$ are dimensionless fractions that must always sum to 1 for a two-component solution. Mole fraction is the natural scale for vapour-pressure work (Raoult's and Henry's laws), since vapour pressure depends on the fraction of molecules of each kind at the surface, not on their mass or volume.

Volume-of-solution scale. Molarity ($M$) is moles of solute per litre of solution:

where: $n_{\text{solute}}$ is in mol and $V_{\text{solution}}$ is the total solution volume in litres, read off a volumetric flask. Because volume expands slightly on heating, molarity is temperature-dependent — the same 1 mol of solute in a slightly larger volume on a hot day reads a slightly lower molarity.

Mass-of-solvent scale. Molality ($m$) is moles of solute per kilogram of solvent:

where: $w_{\text{solvent}}$ is the mass of solvent alone (not the solution) in kilograms. Mass does not change with temperature, so molality is temperature-independent — this is exactly why molality, not molarity, is used later in this chapter for freezing-point depression and boiling-point elevation. Finally, normality ($N$) counts gram-equivalents of solute per litre of solution, $N = M \times n\text{-factor}$, where the $n$-factor is the number of replaceable $H^+$ for an acid, replaceable $OH^-$ for a base, or total charge transferred for a salt (1 for HCl, 2 for $H_2SO_4$). An error in the $n$-factor is the single most common source of wrong answers in normality problems, since it scales the whole result.

Explain, from first principles and unit tracking alone, how the molarity of a solution can be computed directly from its mass percentage and density without ever measuring a volume of solute — a bridge used constantly when a bottle label gives density and mass percentage but a titration calculation needs molarity.

Step 1. Fix a convenient basis. Since molarity is defined per litre of solution, take exactly $V = 1000\ \text{mL} = 1\ \text{L}$ of the solution as the working sample; any consistent basis works, but 1 L makes the final division by volume trivial.

Step 2. Convert that volume to mass using the given density $d$ (in g mL$^{-1}$). Mass of the 1 L sample is $\text{mass}_{\text{solution}} = 1000 \times d$ grams — density is exactly the conversion factor between the volume basis molarity wants and the mass basis percentage gives.

Step 3. Extract the solute's mass using the mass percentage $\%\,\frac{w}{w}$. By definition $\%\,\frac{w}{w} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$, so rearranging, $\text{mass}_{\text{solute}} = \dfrac{\%\,\frac{w}{w}}{100} \times (1000\,d) = \dfrac{1000\,d\,\%\,\frac{w}{w}}{100}$.

Step 4. Convert that mass to moles using the solute's molar mass $M_2$, then divide by the 1 L basis to read off molarity directly:

where: $d$ is in g mL$^{-1}$, $\%\,\frac{w}{w}$ is the mass percentage of solute, and $M_2$ is the solute's molar mass in g mol$^{-1}$; $M$ comes out in mol L$^{-1}$. (Some texts fold the factor of 100 differently and write $M = \tfrac{1000\,d\,\%(w/w)}{M_2}$ with $\%(w/w)$ already meaning the fraction $\times 100$ divided again by 100 inside — the version above, $10\,d\,\%(w/w)/M_2$, is the same formula with the arithmetic pre-simplified.) The reasoning chain is simply volume-to-mass (via density), mass-to-solute-mass (via percentage), solute-mass-to-moles (via molar mass) — three unit conversions stacked into one formula.

Reconsider the aqueous $H_2SO_4$ sample from before: density $1.198\ \text{g mL}^{-1}$, $27\%$ $H_2SO_4$ by mass, $M_2 = 98\ \text{g mol}^{-1}$. Use the bridging formula to find its molarity directly, then find the normality, and finally find the volume to which 250 mL of this stock must be diluted to make it exactly $1.5\ N$.

Step 1. Apply the derived formula: $M = \dfrac{10 \times 1.198 \times 27}{98} = \dfrac{323.46}{98} = 3.30\ \text{mol L}^{-1}$. This matches the value obtained earlier by working through 1 L from scratch — the shortcut formula is exactly that same working, compressed into one line.

Step 2. Convert to normality using the $n$-factor of $H_2SO_4$, which is 2 (two replaceable $H^+$ per molecule): $N = M \times n = 3.30 \times 2 = 6.60\ \text{eq L}^{-1}$.

Step 3. Apply the dilution law. Diluting with solvent changes the volume but not the total gram-equivalents of solute, so $N_1V_1 = N_2V_2$. Taking $V_1 = 250\ \text{mL}$ of the $6.60\ N$ stock and the target $N_2 = 1.5\ N$: $V_2 = \dfrac{N_1V_1}{N_2} = \dfrac{6.60 \times 250}{1.5} = 1100\ \text{mL}$.

Step 4. Interpret: 250 mL of the concentrated stock, made up to a total volume of 1100 mL with water (adding $1100 - 250 = 850\ \text{mL}$ of water), gives the required $1.5\ N$ solution. Molarity $= 3.30\ \text{mol L}^{-1}$, normality $= 6.60\ \text{eq L}^{-1}$, final volume $= 1100\ \text{mL}$.

• Confusing molarity's denominator (volume of the whole solution) with molality's denominator (mass of the solvent alone) — swapping them is the single most frequent numerical slip.
• Forgetting that a computed $x_1 + x_2 \ne 1$ signals an arithmetic error in a mole-fraction problem.
• Plugging percentage as a plain fraction (0.27) instead of the percentage number (27) into the density bridging formula, which throws the answer off by a factor of 100.
• Using the wrong $n$-factor when converting molarity to normality — always check the number of replaceable ions per formula unit before multiplying.

Ask the tutor: stuck on converting between molarity, molality, and normality using density, or want another dilution problem worked through step by step? Ask the AI tutor.