Concentration – the size characterizing substance content in unit of mass or volume of mix. It can be expressed in various ways. Distinguish the following concentration: mass fraction, molar share, volume fraction and molar concentration.
Instruction
1. The mass fraction is the relation of mass of substance to the mass of solution or mix: w = m (v)/m (solution) where w is a mass fraction, the mass of substance is m (v), m (solution) there is a mass of solution, or w = m (v)/m (cm) where m (cm) is the mass of mix. Is expressed in shares of unit or percent. Additional formulas which can be necessary for the solution of tasks on a mass fraction of substance: 1) m = V*p where m is weight, V – the volume, p – density.2) m = n*M where m – weight, n – amount of substance, M – molar weight.
2. The molar share is the attitude of number of moles of a substance towards number of moths of all substances: q = n (v)/n (obshch) where q is a molar share, n (v) is amount of a certain substance, n (obshch) – total amount of substances. Additional formulas: 1) n = V/Vm where n is amount of substance, V – the volume, Vm – the molar volume (it is under normal conditions equal to 22.4 l/mol).2) n = N/Na, where n – amount of substance, N – number of molecules, Na – constant Avogadro (is a constant and it is equal 6.02*10 in 23 degrees 1/mol).
3. The volume fraction is the relation of volume of substance to mix volume: q = V (c) / V (cm) where q is a volume fraction, V (c) – the volume of substance, V (cm) – mix volume.
4. Molar concentration – the relation of amount of this substance to mix volume: Cm = n (v)/V (cm) where Cm is molar concentration (mol/l), n is amount of substance (mol), V (cm) – the volume of mix (l). Let's solve a problem on molar concentration. Define molar concentration of the solution received at dissolution of sulfate of sodium weighing 42.6 g in water weighing 300 g if density of the received solution is equal to 1.12 g/ml. We write a formula for calculation of molar concentration: Cm = n(Na2SO4)/V (cm). We see that it is necessary to find amount of substance of sodium and volume of solution. We count: n(Na2SO4) = m(Na2SO4)/M(Na2SO4).M (Na2SO4) = 23*2+32+16*4 = 142 @/mol.n (Na2SO4) = 42.6/142 = 0.3 mol. We look for solution volume: V = m/pm = m(Na2SO4) + m(H2O) = 42.6 + 300 = 342.6 g of V = 342.6/1.12 = 306 ml = 0.306 l. We substitute in the general formula: Cm = 0.3/0.306 = 0.98 mol/l. The task is solved.