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
moles solute
M
L solution
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?
moles solute
M
L solution
❍ 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 ( )
moles solute 1
1 
moles 1  moles 2      moles n
❍
Percent by Volume (% w/v) AND Percent by Weight (% w/w)
grams solute
%( w / v) 
100 g solution
%( w / w) 
grams solute
• 100%
g solution
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
4. Find molarity – you know moles (0.361 mol) and grams/
mL (64.6 g/mL or 0.0646 L): divide the two