Molarity and dilution calculations are discussed and explained, along with the associated concepts. General Chemistry

1. Molarity and Dilutions
By Shawn P. Shields, Ph.D.
This work is licensed by Shawn P. Shields-Maxwell under a Creative Commons
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2. Molarity
Molarity (M) is a concentration unit that
describes how much of a substance is
dissolved in solution.
Molarity =
moles of solute
Liters of solution
M =
molsolute
Lsoln

3. Molarity Example
Suppose we wanted to make a 1.0 M solution of
NaCl. What should we do?
Calculate the mass of 1 mole of NaCl, then weigh this
amount on a balance.
Molar mass = 22.990g + 35.453g = 58.443 g/mol NaCl
Next, add this amount of NaCl to a flask, and fill it until
the volume of the solution (water AND salt) equals one
liter.

4. Molarity Example 2
What is the molarity of a 250 mL solution containing
0.35 moles NaCl?
First, convert mL to L,
250 mL
1 L
1000 mL
= 0.25 L
Now, calculate the molarity of the solution
M =
molsolute
Lsoln
=
0.35 mol NaCl
0.25 L solution
= 1.4 M NaCl

5. Molarity Example 3
Suppose we have 35 mL of a 0.85 M MgCl2 solution. How
many moles of MgCl2 are present in the solution?
Convert mL to L 35 mL
1 L
1000 mL
= 0.035 L
Now, rearrange the molarity equation and solve for moles.
M =
molsolute
Lsoln
molsolute = Lsoln × M = 0.035 L ×
0.85 mol
Lsoln
= 0.030 mol MgCl2
Sig figs!

6. Molarity Example 4
How many mL of a 1.2 M MgCl2 solution are needed to have 0.35
moles MgCl2?
Rearrange the molarity equation and solve for volume.
M =
molsolute
Lsoln
Lsoln =
molsolute
M
=
0.35 moles MgCl2
1.2 moles MgCl2
Lsoln
= 0.29 Lsoln
Convert L to mL
0.29 L
1000 mL
1 L
= 290 mL MgCl2 solution
Sig figs!

7. Dilution
Dilution is the process of adding solvent to
a more concentrated solution to yield a
solution of lower concentration.
The equation for dilution calculations is
𝐌 𝟏 𝐕𝟏 = 𝐌 𝟐 𝐕𝟐
Where M1 is the initial molarity, V1 is the initial volume,
M2 is the final molarity, and V2 is the final volume.

8. Dilution
Recall that multiplying a volume in Liters and a
molarity gives the moles of the substance.
𝐌 𝟏 𝐕𝟏 = 𝐌 𝟐 𝐕𝟐
Therefore, this relationship is simply
moles = moles!
The moles of solute remains constant, but
the volume and molarity change.

9. Dilution Example
How many mL of a 2.5 M HCl stock
solution would we need to make 100 mL
of 0.80 M HCl?
𝐌 𝟏 𝐕𝟏 = 𝐌 𝟐 𝐕𝟐

10. Dilution Example Solution
How many mL of a 2.5 M HCl stock solution would
we need to make 100 mL of 0.80 M HCl?
𝐌 𝟏 𝐕𝟏 = 𝐌 𝟐 𝐕𝟐
First, let’s identify what we have and what we are solving
for…
M1 is the molarity of the stock solution, but we don’t know
how much we need.
M2 is the final concentration (0.80 M) and V2 is the final
volume (100 mL).

11. Dilution Example Solution
How many mL of a 2.5 M HCl stock solution would
we need to make 100 mL of 0.80 M HCl?
𝐌 𝟏 𝐕𝟏 = 𝐌 𝟐 𝐕𝟐
Plug everything in…
(𝟐. 𝟓 𝐌)𝐕𝟏 = (𝟎. 𝟖𝟎𝐌)(𝟏𝟎𝟎𝐦𝐋)
Solve for V1
𝐕𝟏 =
(𝟎. 𝟖𝟎𝐌)(𝟏𝟎𝟎𝐦𝐋)
(𝟐. 𝟓 𝐌)
= 𝟑𝟐 𝐦𝐋
Watch sig figs… only 1 sf, so the final answer is 30 mL.

12. What You Should Be Able to Do
Calculate the molarity of a solution
Describe how to prepare a solution
with a specific molarity
Use the molarity relationship to solve
for moles and/or volume of solution.
Perform dilution calculations