Expressing Concentration of Solutions

A doctor's saline drip, a chemist's standard flask, and a pollution report quoting "12 ppm of lead" are all answering the same question in different currencies: how much solute is packed into how much solvent or solution? Class 12 Chemistry's Solutions unit opens by making you fluent in the half-dozen ways chemists quote this — mass percentage, volume percentage, mole fraction, molarity, molality, and parts per million — because every later idea in the chapter, from Raoult's law to boiling-point elevation, is stated in one of these currencies and you must convert between them without hesitation. The one distinction examiners chase hardest is molarity versus molality: one uses the volume of the whole solution, the other the mass of just the solvent, and that single difference decides which of the two survives a change in temperature.

The core insight

Every concentration term is a ratio of solute to solvent-or-solution, and the choice of denominator — mass or volume, solvent or solution — is what makes some terms temperature-dependent (molarity) and others not (molality, mole fraction, mass percentage).

The core idea

Mass and volume percentages are the simplest: parts of solute per hundred parts of solution, by mass or by volume.

Mass %=mass of solutemass of solution×100,Volume %=volume of solutevolume of solution×100\text{Mass \%} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100, \qquad \text{Volume \%} = \dfrac{\text{volume of solute}}{\text{volume of solution}} \times 100

where: mass % is used for solid or liquid solutes in any solvent; volume % is used specifically for liquid-in-liquid mixtures (for example, ethanol-water). A mass fraction is the same ratio left as a decimal, without the ×100\times 100.

Mole fraction (xx) compares moles of one component to the total moles of every component present. For a binary solution of n1n_1 mol solvent and n2n_2 mol solute:

x1=n1n1+n2,x2=n2n1+n2,x1+x2=1x_1 = \dfrac{n_1}{n_1+n_2}, \qquad x_2 = \dfrac{n_2}{n_1+n_2}, \qquad x_1 + x_2 = 1

where: the subscripts label solvent (1) and solute (2); the sum rule extends to any number of components and is the fastest sanity check after a mole-fraction calculation.

Molarity (MM) and molality (mm) are the two terms every exam leans on, and they differ in exactly one place — the denominator:

M=nsoluteVsolution (L),m=nsolutewsolvent (kg)M = \dfrac{n_{solute}}{V_{solution}\ (\text{L})}, \qquad m = \dfrac{n_{solute}}{w_{solvent}\ (\text{kg})}

where: nsoluten_{solute} is moles of solute, VsolutionV_{solution} is the total volume of the made-up solution in litres, and wsolventw_{solvent} is the mass of solvent alone in kilograms. Because a liquid's volume expands slightly on heating while its mass does not, molarity drifts with temperature and molality does not — this is the single most tested conceptual fact in the whole topic.

Parts per million (ppm) extends the same mass-ratio idea to very dilute solutions, typically trace impurities in water or air:

ppm=mass (or moles) of solutemass (or moles) of solution×106\text{ppm} = \dfrac{\text{mass (or moles) of solute}}{\text{mass (or moles) of solution}} \times 10^{6}
Figure
Two labelled containers comparing molarity which divides by the volume of the whole solution and molality which divides by the mass of the solvent alone
Molarity vs molality — what goes in the denominator Molarity M whole solution volume (solute + solvent, in litres) changes with temperature Molality m solvent mass only (solute mass excluded, in kg) independent of temperature same solution, two denominators

The molarity-molality interconversion formula

Derivation AConverting molarity to molality using density

The goal is to derive a single algebraic bridge between molarity MM (moles per litre of solution) and molality mm (moles per kilogram of solvent) for a solute of molar mass MBM_B dissolved in a solution of density dd, so that either one can be found from the other without repeating the full mass-and-volume bookkeeping each time — a formula worth having ready for any interconversion question.

Step 1. Fix a convenient sample: exactly 1 litre (1000 mL) of solution. By the definition of molarity, this litre contains MM mol of solute, so it contains nsolute=Mn_{solute} = M mol.

Step 2. Find the total mass of this 1 L sample using density: mass of solution =d (g/mL)×1000 mL=1000d= d\ (\text{g/mL}) \times 1000\ \text{mL} = 1000d grams, where dd is the solution's density in g/mL.

Step 3. Subtract the solute's mass to isolate the solvent's mass. The solute's mass is nsolute×MB=M×MBn_{solute} \times M_B = M \times M_B grams, so the solvent mass is wsolvent=(1000d−M MB)w_{solvent} = (1000d - M\,M_B) grams, which must be converted to kilograms for the molality formula: wsolvent=1000d−M MB1000 kgw_{solvent} = \dfrac{1000d - M\,M_B}{1000}\ \text{kg}.

Step 4. Substitute into m=nsolute/wsolventm = n_{solute}/w_{solvent}. With nsolute=Mn_{solute} = M mol from Step 1 and wsolventw_{solvent} from Step 3:

m=M(1000d−M MB1000)=1000 M1000d−M MBm = \dfrac{M}{\left(\dfrac{1000d - M\,M_B}{1000}\right)} = \dfrac{1000\,M}{1000d - M\,M_B}

where: MM is molarity (mol/L), dd is solution density (g/mL), MBM_B is the solute's molar mass (g/mol), and mm is molality (mol/kg). Every quantity on the right is something you either measure or look up, so this formula converts molarity directly to molality (or, rearranged, back again) without redoing the litre-by-litre bookkeeping.

Worked example

Problem: A solution is prepared by dissolving 20 g of KI (molar mass 166 g mol⁻¹) in enough water to make 100 g of solution, which has density 1.202 g mL⁻¹. Calculate (a) the mass percentage, (b) the molality, (c) the molarity by direct calculation, and (d) verify (c) using the Step 4 formula.

Step 1 — Mass percentage. Mass % =20100×100=20%= \dfrac{20}{100} \times 100 = 20\%.

Step 2 — Molality by definition. Moles of KI =20166=0.1205 mol= \dfrac{20}{166} = 0.1205\ \text{mol}. Mass of water =100−20=80 g=0.080 kg= 100 - 20 = 80\ \text{g} = 0.080\ \text{kg}.

m=0.1205 mol0.080 kg=1.506 mol kg−1m = \dfrac{0.1205\ \text{mol}}{0.080\ \text{kg}} = 1.506\ \text{mol kg}^{-1}

Step 3 — Molarity by definition. Volume of solution =massdensity=100 g1.202 g mL−1=83.19 mL=0.08319 L= \dfrac{\text{mass}}{\text{density}} = \dfrac{100\ \text{g}}{1.202\ \text{g mL}^{-1}} = 83.19\ \text{mL} = 0.08319\ \text{L}.

M=0.1205 mol0.08319 L=1.449 mol L−1M = \dfrac{0.1205\ \text{mol}}{0.08319\ \text{L}} = 1.449\ \text{mol L}^{-1}

Step 4 — Check with the interconversion formula. Using m=1000M1000d−MMBm = \dfrac{1000M}{1000d - M M_B} with M=1.449M = 1.449, d=1.202d = 1.202, MB=166M_B = 166:

m=1000×1.4491000(1.202)−1.449(166)=14491202−240.5=1449961.5=1.507 mol kg−1m = \dfrac{1000 \times 1.449}{1000(1.202) - 1.449(166)} = \dfrac{1449}{1202 - 240.5} = \dfrac{1449}{961.5} = 1.507\ \text{mol kg}^{-1}

This matches the 1.506 mol kg⁻¹ found directly in Step 2 (the tiny 0.001 gap is rounding). Final answers: mass % = 20%, molality ≈ 1.506 mol kg⁻¹, molarity ≈ 1.449 mol L⁻¹, and the formula reproduces molality independently.

Common traps

  • Using solvent volume instead of solution volume for molarity. Dissolving a solute changes the total volume — molarity always uses the volume of the finished solution, never the solvent's original volume before mixing.
  • Using solution mass instead of solvent mass for molality. Molality's denominator deliberately excludes the solute; only the solvent's mass counts.
  • Forgetting which term survives a temperature change. Molarity depends on volume, and volume expands with heat, so molarity changes with temperature; molality depends only on masses, which don't change with temperature, so molality does not.
  • Misapplying the interconversion formula's units. The formula needs dd in g/mL and MBM_B in g/mol to output mm in mol/kg directly — mixing in g/L density or kg/mol molar mass silently wrecks the answer.

Ask the tutor: stuck on when a concentration question wants molarity versus molality, or want another density-based interconversion problem worked step by step? Ask the AI tutor.

Why do chemists bother with both molarity and molality instead of picking one?

Because each is convenient for a different job. Molarity is what you actually measure and use in the lab — you fill a volumetric flask to a mark, so titrations and reaction stoichiometry are naturally written in moles per litre. Molality, however, is unaffected by temperature since it is built entirely from masses, so it is the term of choice whenever a calculation must stay valid across a temperature range, such as boiling-point elevation or freezing-point depression later in this unit.

Do I need to memorise the molarity-to-molality formula, or can I always just work it out from scratch?

Either approach earns full marks, so use whichever is faster for you under exam conditions. Working from scratch (assume a convenient volume or mass, find moles, find solvent mass, divide) is safer because it cannot be misremembered, and it is exactly what Steps 1 to 4 of the derivation do. The formula is a shortcut once you trust it, and it is worth using to check your from-scratch answer, as the worked example does in Step 4.

Why does ppm use a factor of 10 to the 6 instead of 100 like percentage?

Because ppm is built for solutions so dilute that a percentage would be an awkward string of zeros after the decimal point. A pollutant at 12 ppm is 12 parts per million, which as a percentage would be 0.0012% — correct, but clumsy to write and read repeatedly. Multiplying by 10610^6 instead of 100 simply rescales the same mass-ratio idea into whole numbers that are easy to state and compare, which is why ppm is the standard unit for trace impurities in drinking water or air quality reports.

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