Dilution and concentration problems
Concentration of solutions can be expressed in several ways (see lab guidesheets).
Do we understand the meanings of:
weight/weight (w/w), weight/volume (w/v), and volume/volume (v/v)?
Do we understand that M means molar = moles of solute per liter of solution?
“Mole” expresses mass, not concentration, of a substance.
Ex. Glucose has mass (molecular weight) = 180 daltons. One mole of glucose weighs 180 grams.
Ex. A 1 M (one molar) glucose solution contains one mole (180 g) of glucose dissolved in enough water to
make exactly one liter of solution. Note that we could prepare exactly 34 mL (or 216 mL or 1309 mL) of a 1
M solution by appropriately scaling down or up the quantities. Remember that molarity is a ratio; it doesn’t
say what total volume of the solution we may have. We could have a teacup containing 1 M glucose solution
or a bathtub full of 1 M glucose solution.
Ex. The concentration expression 5 mg/mL (w/v) means 5 mg of solute per mL of solution.
▪ 100 mL of this solution would contain 500 mg (= 0.5 g) of the solute.
▪ Put 100 mL of this solution into a graduated cylinder. Wait until the solvent evaporates, leaving exactly
50 mL in the cylinder. The concentration now is 10 mg/mL (w/v). The same 500 mg of solute is now in 50
mL of solution.
Problem #1. You are required to prepare exactly 1 liter of a glucose solution at a concentration of
21.6 micrograms per milliliter. The solvent is water. To do this you are given:
▪ water and whatever glassware you need (pipets, graduated cylinders, beakers, e.g.)
▪ a stock glucose solution, concentration = 4.7 mg/mL.
How do you prepare the required solution?
Size up the problem:
You're given a glucose solution at one concentration and asked to:
a. prepare a solution of a lower concentration (that’s a dilution),
b. prepare exactly one liter of that new solution.
Reasoning:
▪ One liter (1000 mL) of the new solution will contain 21,600 μg of glucose:
21.6 μg/mL X 1000 mL = 21,600 μg (of glucose). This is not a concentration expression.
▪ Where do you get that glucose?-- from the solution that you're given, obviously.
▪ So, how many milliliters of the given solution must you take in order to get 21,600 μg of glucose?
21,600 μg X 1 mg
1000 μg
X
1 mL
4.7 mg
= 4.6 mL
▪ To prepare the new solution, add 995.6 mL of water to the 4.6 mL of the given glucose solution.
Problem #2. There is another way to solve such dilution problems, using the formula C1V1 = C2V2, where
C1 is the concentration of the starting solution,
V1 is the volume of the starting solution,
C2 is the concentration of the final solution,
V2 is the volume of the final solution.
The concentration units must be the same.
In the previous problem:
C1 = 4.7 mg/mL. V1 is the unknown. C2 = 0.0216 mg/mL. V2 = 1 L
Since concentration units must be the same, C2 was converted to mg/mL: 21.6 μg/mL = 0.0216 mg/mL.
So, the setup is: 4.7 mg/mL X V1 = 0.0216 mg/mL X 1 L
V1 = 0.0216 mg/mL X 1 L
4.7 mg/mL
V1 = 4.6 X 10-3 L, which equals 4.6 mL Note that mg/mL canceled.
This procedure gives the same value: the volume of starting solution needed to produce the diluted solution.
Always show the units in the problem; never neglect the units.
We could have converted the 4.7 mg/mL to 4700 μg/mL for C1. The answer would be the same; the
concentration units must be the same.
A variation: you may see this C1V1 = C2V2 written as M1V1 = M2V2. Here the M1 and M2 are the
molar concentrations specifically. As long as the concentrations are the same, the formula works.
next page
Problem #3. Which one of the following solutions has the greatest solute concentration? In each one the
solvent is water and the solute is sodium chloride. A. 1.5% (w/v); B. 15 microgram/mL;
C. 150 mg/L;
D. 1.5 g/L;
E. 0.15 mg/dL; F. 0.15 oz./quart; G. 1.5 X 10-3 oz./gallon
How do you approach to problem? Convert everything to the same set of units. Use g/L, for example.
A.. 1.5% (w/v) means 1.5 g solute per 100 mL of solution. So,
B. 15 μg X 1 g X
mL
106 μg
C.. 150 mg X 1 g
L
1000 mg
D. 1.5 g
L
1000 mL =
L
=
1.5 g X 1000 mL = 15 g
100 mL
lL
L
0.015 g
L
0.15 g
L
No conversion is necessary on this one.
1.5 X 10-3 g
L
E. 0.15 mg X
1g
dL
1000 mg
X
10 dL =
L
F. 0.15 oz.
qt.
X
1 lb. X 454 g =
16 oz.
lb.
X 1.057 qt.
L
G. 1.5 X 10-3 oz.
gal.
X 1 gal.
4 qt.
X
1.057 qt.
L
X
4.5 g
L
1 lb.
16 oz.
X
454 g
lb.
=
1.12 X 10-2 g
L
So, which solution has the greatest solute concentration? A.
This problem is nothing more than drill (practice) of the dimensional analysis work that was presented in
lab session #1; the material is in those guidesheets. The preceding problems here required dimensional
analysis too, but they have the added feature of dilution work, which was introduced in lab session #1.
The C1V1 = C2V2 relation is presented here as another way to view dilution problems, but dilution
problems certainly can be solved without that relation, as shown here and in the lab #1 guidesheet about
dilutions.