Q.The molality of a solution containing 1.8 g of glucose dissolved in 250 g of water is

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) $=\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$ -- independent of temperature (defined by mass, not volume).
• Molarity (M) $=\dfrac{\text{moles of solute}}{\text{volume of solution in L}}$ -- used for 1:1 mole-ratio titrations (e.g. EDTA complexometric titrations); temperature-dependent, since solution volume changes slightly with temperature.
• Normality (N) $=\dfrac{\text{gram equivalents of solute}}{\text{volume of solution in L}}$ -- used in redox and acid-base (neutralisation) titrations.
• Formality (F) $=\dfrac{\text{formula weights of solute}}{\text{volume of solution in L}}$ -- used for ionic compounds without discrete molecules.
• Mole fraction (x) $=\dfrac{\text{moles of one component}}{\text{total moles of all components}}$ (always $x_A+x_B=1$ 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) $=\dfrac{\text{mass of solute}}{\text{mass of solution}}\times10^6$ -- 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 $m=0.125$; 5.845 g NaCl made up to 500 mL gives $M=0.2$; 3.15 g oxalic acid dihydrate (equivalent mass 63) made up to 100 mL gives $N=0.5$; 0.5 mol ethanol with 1.5 mol water gives $x_{ethanol}=0.25$; 20 mg dissolved solids in 50 g tap water gives 400 ppm.