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Chemistry · Ch 9 — Solutions

Expressing Concentration of Solutions

9.3

Expressing Concentration of Solutions

Everyday products come labelled with a concentration -- a chlorhexidine mouthwash might say 0.2% (w/v), commercial hydrogen peroxide is typically 3% (w/v), tap-water purity is reported in ppm (parts per million), and laboratory reagents are labelled molar or normal. In every case, concentration expresses the amount of solute present in a given quantity of solvent (or solution).

Different situations call for different concentration units, because each unit is convenient for a different kind of calculation:

  • Molar (M) solutions are used where reactions proceed in a known, fixed mole ratio -- e.g. complexometric titrations with EDTA, where EDTA and the metal ion react 1:1.
  • Normality (N) is used in redox and acid-base (neutralisation) titrations, where the "reacting unit" (equivalent) matters more than the whole molecule.
  • Mole fraction (x) is used to calculate the partial pressure of a gas in a mixture, and the vapour pressure of a solution (sections 9.5-9.7).
  • Percentage units (w/w, w/v, v/v) are used to state the active-ingredient strength of a therapeutic or commercial product.
  • ppm is reserved for solutes present in very small (trace) amounts, such as dissolved solids in drinking water.

The eight concentration terms (Table 9.2), with a worked illustration for each:

  1. Molality (m) =moles of solutemass of solvent in kg=\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}. Example: dissolving 45 g glucose (molar mass 180 g/mol, so 0.25 mol) in 2 kg water gives m=0.252=0.125m=\dfrac{0.25}{2}=0.125 mol/kg.
  2. Molarity (M) =moles of solutevolume of solution in L=\dfrac{\text{moles of solute}}{\text{volume of solution in L}}. Example: 5.845 g NaCl (molar mass 58.45, so 0.1 mol) made up to 500 mL gives M=0.10.5=0.2M=\dfrac{0.1}{0.5}=0.2 M.
  3. Normality (N) =gram equivalents of solutevolume of solution in L=\dfrac{\text{gram equivalents of solute}}{\text{volume of solution in L}}. Example: 3.15 g oxalic acid dihydrate (equivalent mass 63, so 0.05 equivalents) made up to 100 mL gives N=0.050.1=0.5N=\dfrac{0.05}{0.1}=0.5 N.
  4. Formality (F) =formula weights of solutevolume of solution in L=\dfrac{\text{formula weights of solute}}{\text{volume of solution in L}} -- used for ionic compounds that do not exist as discrete molecules. Example: 5.85 g NaCl (formula weight 58.5) made up to 500 mL gives F=5.8558.5×0.5=0.2F=\dfrac{5.85}{58.5\times0.5}=0.2 F.
  5. Mole fraction (x) =moles of one componenttotal moles of all components=\dfrac{\text{moles of one component}}{\text{total moles of all components}}. For a two-component solution of A and B with nAn_A, nBn_B moles respectively, xA=nAnA+nBx_A=\dfrac{n_A}{n_A+n_B} and xB=nBnA+nBx_B=\dfrac{n_B}{n_A+n_B}, and always xA+xB=1x_A+x_B=1. Example: mixing 0.5 mol ethanol with 1.5 mol water gives xethanol=0.52.0=0.25x_{ethanol}=\dfrac{0.5}{2.0}=0.25 and xwater=1−0.25=0.75x_{water}=1-0.25=0.75.
  6. Mass percentage (% w/w) =mass of solute (g)mass of solution (g)×100=\dfrac{\text{mass of solute (g)}}{\text{mass of solution (g)}}\times100. Example: 300 mg (0.3 g) neomycin sulphate in 30 g of ointment gives 0.330×100=1%\dfrac{0.3}{30}\times100=1\% w/w.
  7. Volume percentage (% v/v) =volume of solute (mL)volume of solution (mL)×100=\dfrac{\text{volume of solute (mL)}}{\text{volume of solution (mL)}}\times100. Example: 10 mL benzoin in 50 mL tincture of benzoin gives 1050×100=20%\dfrac{10}{50}\times100=20\% v/v.
  8. Mass by volume percentage (% w/v) =mass of solute (g)volume of solution (mL)×100=\dfrac{\text{mass of solute (g)}}{\text{volume of solution (mL)}}\times100. Example: 3 g paracetamol in 60 mL suspension gives 360×100=5%\dfrac{3}{60}\times100=5\% w/v. …
Table 9.2Concentration units, their expressions and worked illustrations
TermExpressionWorked illustration
Molality (m)Number of moles of soluteMass of solvent in kg\dfrac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}45 g glucose in 2 kg water: m=45/1802=0.252=0.125 mm=\dfrac{45/180}{2}=\dfrac{0.25}{2}=0.125\ m
Molarity (M)Number of moles of soluteVolume of solution in L\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in L}}5.845 g NaCl made up to 500 mL: M=5.845/58.450.5=0.10.5=0.2 MM=\dfrac{5.845/58.45}{0.5}=\dfrac{0.1}{0.5}=0.2\ M
Normality (N)Number of gram equivalents of soluteVolume of solution in L\dfrac{\text{Number of gram equivalents of solute}}{\text{Volume of solution in L}}3.15 g oxalic acid dihydrate (equivalent mass 63) made up to 100 mL: N=3.15/630.1=0.050.1=0.5 NN=\dfrac{3.15/63}{0.1}=\dfrac{0.05}{0.1}=0.5\ N
Formality (F)Number of formula weights of soluteVolume of solution in L\dfrac{\text{Number of formula weights of solute}}{\text{Volume of solution in L}}5.85 g NaCl made up to 500 mL: F=5.8558.5×0.5=0.2 FF=\dfrac{5.85}{58.5\times0.5}=0.2\ F
Mole fraction (x)moles of one componenttotal moles of all components\dfrac{\text{moles of one component}}{\text{total moles of all components}}0.5 mol ethanol + 1.5 mol water: xethanol=0.52.0=0.25x_{ethanol}=\dfrac{0.5}{2.0}=0.25, xwater=0.75x_{water}=0.75
Mass percentage (% w/w)Mass of solute (g)Mass of solution (g)×100\dfrac{\text{Mass of solute (g)}}{\text{Mass of solution (g)}}\times100300 mg neomycin sulphate in 30 g ointment: 0.330×100=1% w/w\dfrac{0.3}{30}\times100=1\%\ w/w
Volume percentage (% v/v)Volume of solute (mL)Volume of solution (mL)×100\dfrac{\text{Volume of solute (mL)}}{\text{Volume of solution (mL)}}\times10010 mL benzoin in 50 mL tincture: 1050×100=20% v/v\dfrac{10}{50}\times100=20\%\ v/v
Misc Evaluate Yourself 1Molarity of KOH from two different volumes

Worked out. An in-text practice box: if 5.6 g of KOH is present in (a) 500 mL and (b) 1 litre of solution, calculate the molarity of each of these solutions. …

Misc Evaluate Yourself 2Mole fraction of glucose and water

Worked out. An in-text practice box: 2.82 g of glucose (molar mass 180 g/mol) is dissolved in 30 g of water (molar mass 18 g/mol). Calculate the mole fraction of glucose and water. Moles of glucose = 2.82/180 = 0.0157 mol; moles of water = 30/18 = 1.667 mol; total = 1.6823 mol, so x(glucose) = 0.0157/1.6823 = 0.0093 and x(water) = 1.667/1.6823 = 0.9907, and the two mole fractions sum to 1 a …

Misc Evaluate Yourself 3Amount of iodopovidone in a 1.5 mL antiseptic dose

Worked out. An in-text practice box: the antiseptic solution of iodopovidone for external application contains 10% w/v of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of 1.5 mL. …

Misc Evaluate Yourself 4Concentration of dissolved oxygen in sea water, in ppm

Worked out. An in-text practice box: a litre of sea water weighing about 1.05 kg contains 5 mg of dissolved oxygen (O2_2). Express the concentration of dissolved oxygen in ppm. …