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 = mol of NaCL
58g
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 g.
m = molal = moles of solute/mass solvent (kg)
Moles of H2SO4 = 24.4 g H2SO4 x (1 mol/98.09 g) = mol H2SO4
m = mol H2SO4 / kg H2O = 1.26 m

5. 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

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 g
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) = mole NaCl
Need moles of water:
25.0 g H2O x (1 mol H2O/18.02 g H2O) = 1.39 mol H2O
XNaCl = mol NaCl/( ) =
Xwater = 1.39 mol H2O /( ) = 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 g/mL. What is the molality of the solution? The molar mass is 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 / 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) = mol phosphoric acid
2. find kg of solvent
64.6 g = kg water
3. find molality
0.361 mol / kg = 5.59 m
Find molarity – you know moles (0.361 mol) and grams/ mL (64.6 g/mL or L): divide the two