Solution Concentration Units

Units of Concentration 1 Molality (m) aka “molal concentration” Solvent= does not include solute Molarity (M) aka “molar concentration” Most common concentration unit in Chemistry Solution=solute+solvent

Example What is the molarity of a solution prepared by dissolving 45.0 grams of NaCl in enough water to give a total volume of 489 mL? Na = 23 a.m.u. Cl = +35 a.m.u. 58 a.m.u.  1mol NaCl = 58g 45g X 1 mol = 0.7759 mol of NaCL 58g 0.7759 mol of NaCl = 1.6 M 0.489 L

Example 12.3 Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g of water. The molar mass of sulfuric acid is 98.08 g. m = molal = moles of solute/mass solvent (kg) Moles of H2SO4 = 24.4 g H2SO4 x (1 mol/98.09 g) = 0.249 mol H2SO4 m = 0.249 mol H2SO4 / 0.198 kg H2O = 1.26 m

Molality vs. Molarity Molality is never equal to molarity But the difference becomes smaller as solutions become more dilute (denominators are very similar) Molarity is more useful when dealing with solution stoichiometry Molality is more appropriate for dealing with physical chemistry Question: which is temperature-dependent? Molarity depends on temperature. Molarity will decrease as temperature increases since the amount of the solution will decrease (from evaporation). Temp  M Molality does not depend on temperature since mass (kg) does not change with temperature

Units of Concentration 2 Mole Fraction ( ) Percent by Volume (% w/v) AND Percent by Weight (% w/w)

Example Calculate the percent by weight NaCl in a solution comprised of 45.0 g NaCl and 457 g of water. % (w/w) = grams of solute x 100% = 45g x 100% = 8.96% grams of solution 502g Solute = 45g of NaCl Solution = 457g of NaCl (solute) + H2O (solvent) = 502g

Mole Fraction Example What is the mole fraction of each component in a solution in which 3.57 g of sodium chloride, NaCl, is dissolved in 25.0 g of water? Need moles of NaCl: 3.57 g NaCl x (1 mol NaCl/58.44 g NaCl) = 0.0611 mole NaCl Need moles of water: 25.0 g H2O x (1 mol H2O/18.02 g H2O) = 1.39 mol H2O XNaCl = 0.0611 mol NaCl/(0.0611 + 1.39) = 0.0421 Xwater = 1.39 mol H2O /(0.0611 + 1.39) = 0.958 Always needs to equal 1!!!!!!!!!!!!

Converting Between Units Density can be a conversion factor between molarity and molality The density of 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL. What is the molality of the solution? The molar mass is 32.04 g. 1. Find the mass of water using density (D = m/V) 1 L soln x (1000 mL / 1 L) x (0.976 g / 1 mL soln) = 976 g 2. Knowing there are 2.45 moles of methanol Mass of H2O = mass of soln – mass of solute Mass = 976 – (2.45 mol CH3OH x (32.04 g CH3OH / 1 mol CH3OH)) = 898 g 3. molal = 2.45 mol CH3OH / 0.898 kg H2O Molal = 2.73 m

Converting Between Units Calculate the molality AND molarity of a 35.4 % (by mass) aqueous solution of phosphoric acid (H3PO4). The molar mass of phosphoric acid is 98.0 g. It is convenient to assume one starts with 100 g, therefore 35.4 g is phosphoric acid and 64.6 g is water. Molal = mol solute / kg solvent 1. find moles of solute 35.4g x (1 mol / 98.0 g) = 0.361 mol phosphoric acid 2. find kg of solvent 64.6 g = 0.0646 kg water 3. find molality 0.361 mol / 0.0646 kg = 5.59 m Find molarity – you know moles (0.361 mol) and grams/ mL (64.6 g/mL or 0.0646 L): divide the two