3. Units of Concentration
Definition
same
unitless
moles A
total moles
mass A
 100%
total mass
same
Chapter 11
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Advantages/disadvantages of each (read only; no lecture)
Chapter 11
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Example problems
1. Determine the molarity of a 0.258 m solution of glucose
given that the solution’s density is 1.0173 g ml-1 and that
the molar mass of glucose is 180.2 g.
•first, write down definitions: molarity is moles solute per
liter solution; molality is moles solute per kg solvent;
density is mass solution per volume solution
we want: moles solute/liter solution
•assume some amount of solution to start: in this case 0.258
moles glucose and 1 kg water
so we have 0.258 moles solute
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•get the total volume of the solution - glucose plus water
mass sol’n
180.2 g
0.258moles 
 46.49 g glucose
mol
46.49 g glucose  1000 g water  1046.49 g sol' n
volume sol’n
1ml
1046.49 g sol ' n 
 1028.70ml
1.0173g
•solve for the molarity
0.258moles
 0.251M
1.02870 L
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2. Determine the molality of a 0.500 M solution of
acetic acid (molar mass 60.02 g) with a density of
1.0042 g ml-1.
•Again, write down the units and their definitions.
•Start by assuming 1 L of solution and thus 0.500 moles
acetic acid
•determine the mass of the solution from the density
times the volume
•determine the mass of water (in kg) by subtraction
(subtract the mass of the acetic acid from the mass of
the solution)
•solve for the molality
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3. Describe how 1.50 L of a 12.0% KBr solution is made if
the solution density is 1.10 g ml-1. In other words,
determine the mass of KBr and the mass of water to be
mixed together.
•Write out the units and their definitions
•1.50 L or 1500 ml of solution are required; determine the
total mass of solution required (HOW?).
•Of the total mass of solution, how much must be KBr?
•Determine the mass of water necessary (HOW?).
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total mass solution
g
1500ml  1.10
ml
 1650 g
mass KBr
0.12  1650  198 g KBr
mass water
1650 g sol' n  198 g KBr  1452 g water
dissolve 198 g KBr in 1452 g water
Chapter 11
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