The document discusses concentration of solutions in chemistry, including definitions of molarity, dilution processes which involve maintaining the same number of moles of solute before and after dilution, and examples of calculating volumes needed of stock solutions to make target solutions of desired concentrations using the dilution equation. Dilution problems presented include making a 0.100 M CuCl2 solution from a 2.00 M stock and preparing 0.2 M HNO3 from a 4.00 M stock solution.

1. Concentration of Solutions
Molarity (M), or molar concentration, is defined as the number of
moles of solute per liter of solution.
moles solute
molarity =
liters solution

2. Worked Example 9.8
Strategy Convert the mass of glucose given to moles, and use the equations for
interconversions of M, liters, and moles to calculate the answers.
moles of glucose =
For an aqueous solution of glucose (C6H12O6), determine (a) the molarity of
2.00 L of a solution that contains 50.0 g of glucose, (b) the volume of this solution
that would contain 0.250 mole of glucose, and (c) the number of moles of glucose
in 0.500 L of this solution.
Solution
(a) molarity = = 0.139 M
(b) volume = = 1.80 L
(c) moles of C6H12O6 in 0.500 L = 0.500 L×0.139 M = 0.0695 mol
50.0 g
180.2 g/mol
0.277 mol C6H12O6
2.00 L solution
0.250 mol C6H12O6
0.139 M solution
= 0.277 mol

3. 4.5

4. Concentration of Solutions
Dilution is the process of preparing a less concentrated solution from
a more concentrated one.
moles of solute before dilution = moles of solute after dilution

5. Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
Moles of solute
after dilution (f)
=
MiVi MfVf
=
4.5
moles of solute before = moles of solute after
n=CcVc n=CdVd
CcVc = CdVd

6. Concentration of Solutions
In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2
solution. A stock solution of 2.00 M CuCl2 is available.
How much of the stock solution is needed?
Solution: Use the relationship that moles of solute before dilution =
moles of solute after dilution.
(2.00 M CuCl2)(Vc) = (0.100 M CuCl2)(0.2500 L)
Vc = 0.0125 L or 12.5 mL
To make the solution:
1) Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask.
2) Carefully dilute to the calibration mark.
Cc × Vc = Cd × Vd

7. How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
CcVc = CdVd
Cc = 4.00M Cd = 0.200M Vd = 0.06 L Vc = ? L
Vc =
CdVd
Cc
=
0.200M x 0.06L
4.00M
= 0.003 L = 3 mL
3 mL of acid + 57 mL of water = 60 mL of solution

8. Concentration of Solutions
Because most volumes measured in the laboratory are in milliliters
rather than liters, it is worth pointing out that the equation can also be
used with mL as the volume as long as both volume measurements
are the SAME unit.

9. Another Dilution Problem
If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL
What would be the resulting concentration?
CcVc = CdVd
(6.5M) (32 mL) = Cd (500 mL)
Cd =
500 mL
6.5 M * 32 mL
Cd = 0.42 M

10. Worked Example 9.9
Strategy Cc = 12.0 M, Cd = 0.125 M, Vd = 250.0 mL
What volume of 12.0 M HCl, a common laboratory stock solution, must be used
to prepare 250.0 mL of 0.125 M HCl?
Solution
12.0 M × Vc = 0.125 M × 250.0 mL
Vc =
0.125 M × 250.0 mL
12.0 M = 2.60 mL

11. Worked Example 9.11
Strategy Use the concentration given in each case and the stoichiometry
indicated in the corresponding chemical formula to determine the concentration of
the specified ion or compound.
Using square-bracket notation, express the concentration of (a) chloride ion in a
solution that is 1.02 M in AlCl3, (b) nitrate ion in a solution that is 0.451 M in
Ca(NO3)2, and (c) Na2CO3 in a solution in which [Na+] = 0.124 M.
Solution (a) There are 3 moles of Cl- ion for every 1 mole of AlCl3,
AlCl3(s) → Al3+(aq) + 3Cl-(aq)
so the concentration of Cl- will be three times the concentration of AlCl3.
[Cl-] = [AlCl3] ×
= ×
= = 3.06 M
3 mol Cl-
1 mol AlCl3
1.02 mol AlCl3
L
3 mol Cl-
1 mol AlCl3
3.06 mol Cl-
L

12. Homework
Read section 8.7
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