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Write Brief Answer · Q31

Q.Define

(i) molality
(ii) Normality
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
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Concept understanding — Expressing the Concentration of a Solution

Concentration expresses the amount of solute present in a given quantity of solvent or solution. Different situations call for different units, chosen for calculation convenience:

  • Molality (m) =moles of solutemass of solvent in kg=\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} -- independent of temperature (defined by mass, not volume).
  • Molarity (M) =moles of solutevolume of solution in L=\dfrac{\text{moles of solute}}{\text{volume of solution in L}} -- used for 1:1 mole-ratio titrations (e.g. EDTA complexometric titrations); temperature-dependent, since solution volume changes slightly with temperature.
  • Normality (N) =gram equivalents of solutevolume of solution in L=\dfrac{\text{gram equivalents of solute}}{\text{volume of solution in L}} -- used in redox and acid-base (neutralisation) titrations.
  • 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 without discrete molecules.
  • Mole fraction (x) =moles of one componenttotal moles of all components=\dfrac{\text{moles of one component}}{\text{total moles of all components}} (always xA+xB=1x_A+x_B=1 for a binary mixture) -- independent of temperature; used to calculate partial pressures of gases and vapour pressures of solutions.
  • Mass percentage (% w/w), volume percentage (% v/v), and mass-by-volume percentage (% w/v) -- used to state the active-ingredient strength of therapeutic and commercial products.
  • Parts per million (ppm) =mass of solutemass of solution×106=\dfrac{\text{mass of solute}}{\text{mass of solution}}\times10^6 -- reserved for solutes present in very small (trace) amounts, e.g. total dissolved solids in drinking water.

Worked illustrations: 45 g glucose in 2 kg water gives m=0.125m=0.125; 5.845 g NaCl made up to 500 mL gives M=0.2M=0.2; 3.15 g oxalic acid dihydrate (equivalent mass 63) made up to 100 mL gives N=0.5N=0.5; 0.5 mol ethanol with 1.5 mol water gives xethanol=0.25x_{ethanol}=0.25; 20 mg dissolved solids in 50 g tap water gives 400 ppm.

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