BIO 306
Fact Sheet to assist in solving “concentration-type” problems
PERCENT SOLUTION
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A percent solution is an expression of a proportion of solute in a solution
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Units = gm/100ml (grams of solute / 100 ml of solution)
These units are “by definition” and always apply to a “percent solution”. Once the units are changed,
you are no longer dealing with a “percent solution”.
Many times, for convenience when dealing with more dilute solutions the units are switched to
mgm/100ml (milligrams of solute / 100 ml of solution). In this case the solution is referred to as a “mgm%
solution” (milligram percent solution). The units of measure for the solute have changed from grams to milligrams,
but the unit of measurement for the solution is still 100 ml.
MOLE
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A mole is the atomic mass of an element in grams.
Example:
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Atomic Mass of Carbon = 12 g
Thus, 12 grams of Carbon = 1 mole
If dealing with a molecule, it is the weight in grams equivalent to the sum total of the atomic masses
involved in the molecule.
Example:
Calcium Chloride
CaCl2
A.M. of Calcium
A.M. of Chlorine
A.M. of Chlorine
Molecular Weight of CaCl2
= 40.1 g
= 35.4 g
= 35.4 g
= 110.9 g/mole
Thus, 110.9 grams of CaCl2 = 1 mole
- A mole of a substance will always contain the same number of molecules, 6.02 x 1023 g/mole, regardless of
the substance you are dealing with. (6.02 x 1023 is Avagadro’s number)
MOLARITY
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Units = moles / liter
Molarity is a measures of concentration, as it relates the mass of the solutes to the volume of the solution.
It is calculated by first determining the number of moles of solute in a solution, and then dividing by
the volume (in liters) of the solution.
Example: Find the molarity of a solution containing 70.70 grams of CaCl2 in 750 mls.
Step 1: Calculate the number of moles contained in the solution.
moles = grams / molecular weight
= 70.70 gm / 110.9 g/mole
= 0.638 moles
Step 2: Calculate the molarity.
Molarity = moles / liter
= 0.638 moles / 0.75 L
= 0.85 M
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For more dilute solutions, the term “millimolar” (mM) is often used.
To convert from “M” to “mM”, simply multiply by 1000.
Example
0.85 M x 1000 = 850 mM
or
850 mM  1000 = 0.85 M
OSMOLARITY
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Osmolarity refers to the number of particles per liter of solution. Recall that ionically bonded molecules
will dissociate when placed in solution. When CaCl2 is placed in solution it dissociates into a Ca++ ion and
2 Cl- ions. Thus, 1 mole of CaCl2 in solution will yield 3 moles of particles.
Osmolarity = Molarity x dissociation constant
Example:
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What is the osmolarity of a 0.85 M solution of CaCl2?
Osmolarity = 0.85 M x 3
= 2.55 Osm/L
As with molarity, when dealing with more dilute solutions, the term “milliosmoles” (mOsm) is often
used.
To convert from “Osm/L” to “mOsm/L”, simply multiply by 1000.
Example
0.067 Osm/L x 1000 = 67 mOsm/L or 67mOsm/L  1000 = 0.067 Osm/L
MILLIEQUIVALENTS
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Milliequivalents (mEq) is the number of charges in solution. This is different from the number of particles
in a solution, because some particles such as calcium (Ca+2) are divalent, meaning they carry a double
charge. Milliequivalents is a very commonly used expression of concentration for electrolytes in hospitals
and clinical/pathology laboratories. behaves when it is placed in solution. (Does it dissociate? If so, into
how many particles and how many charges does each particle carry?)
- For monovalent ions (Na+)
mEq = millimolarity (mM)
- For divalent ions (Ca+2)
mEq = mM x 2
Simply count the number of charges and then multiply this number by the millimolarity.
Example:
Calculate the milliequivalents in a 0.02 M solution of NaCl.
NaCl  Na+ + Cl(Total number of charges = 2)
0.02 M x 1000 = 20 mM
mEq
= mM x number of charges
40 mEq = 20mM x 2
Example:
Calculate the mEq in a 0.02M solution of CaCl2.
CaCl2  Ca+2 + Cl- Cl0.02 M x 1000 = 20 mM
mEq = 20 mM x 4
= 80 mEq
(Total number of charges = 4)
NOTE:
You should be able to convert from one unit of concentration to any and all of the others. (In either direction.)
You will be given information such as Atomic Masses of atoms involved, but you will be expected to be able to
calculate everything else including the molecular weight of a compound.
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