THE CHEMIST’S TOOLKIT 11 Measures of concentration Let A be the solvent and B the solute. The molar concen- 2. The relation between molality and molar concentration tration (informally: ‘molarity’), cB or [B], is the amount of solute molecules (in moles) divided by the volume, V, of The total mass of a volume V of solution (not solvent) of the solution: mass density ρ is m = ρV. The amount of solute molecules = = n in this volume is nB cBV. The mass of solute present is mB c = B (11.1) B V nBMB = cBVMB. The mass of solvent present is therefore mA = m – mB = ρV − cBVMB = (ρ − cBMB)V. The molality It is commonly reported in moles per cubic decimetre is therefore (mol dm−3) or, equivalently, in moles per litre (mol L−1). It is ⦵ −3 convenient to define its ‘standard’ value as c = 1 mol dm . nB cVB cB bB == = (11.4a) The molality, bB, of a solute is the amount of solute spe- mA ()ρ −cMBBV ρ −cMBB cies (in moles) in a solution divided by the total mass of the The inverse of this relation, the molar concentration in solvent (in kilograms), mA: terms of the molality, is nB bB = (11.2) mA b ρ c = B (11.4b) B 1+ bM Both the molality and mole fraction are independent of BB temperature; in contrast, the molar concentration is not. It 3. The relation between molar concentration and is convenient to define the ‘standard’ value of the molality mole fraction as b⦵ = 1 mol kg−1. By inserting the expression for bB in terms of xB into the 1. The relation between molality and mole fraction expression for cB, the molar concentration of B in terms of Consider a solution with one solute and having a total its mole fraction is amount n of molecules. If the mole fraction of the solute xBρ is xB, the amount of solute molecules is nBB= xn. The mole cB = (11.5) xMAA+xMBB fraction of solvent molecules is xA = 1 − xB, so the amount of solvent molecules is n = x n = (1 − x )n. The mass of sol- A A B with xA = 1 − xB. For a dilute solution in the sense that vent, of molar mass M , present is m ==nM (1−xn) M . A AAABA xMBB<< xMAA, The molality of the solute is therefore ρ nB xnB xB cB ≈ xB (11.6) bB == = (11.3a) xMAA mA (1−xnBA)(M 1)−xMBA The inverse of this relation, the mole fraction in terms of If, moreover, xB <<1, so xA ≈1, then the molality, is ρ c ≈ x (11.7) bMBA B B MA xB = (11.3b) 1+ bMBA.