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Solution Concentration Units

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.

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Solution Concentration Units

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  1. Solution Concentration Units

  2. 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

  3. 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

  4. 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

  5. Molality vs. Molarity • Molality is never equal to molarity • But the difference becomes smaller as solutions become moredilute(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

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

  7. 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

  8. 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!!!!!!!!!!!!

  9. 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

  10. 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

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