This document discusses units of concentration used to quantify the amount of solute dissolved in a solution. It defines percent by volume and percent by mass as ratios comparing the amount of solute to the total solution. Molarity is introduced as moles of solute per liter of solution, which is a common unit of concentration. Examples are provided for calculating concentration using various units. The document also explains how to calculate the amount of solute needed to make a stock solution of a specific molarity and how to dilute a stock solution to obtain lower concentrations using dilution equations.

1. Units of Concentration
A solution is a homogeneous
mixture of one substance (the
solute) dissolved in another
substance (the solvent).
Concentration is a ratio of the
amount of solute to the amount
of solvent.

2. Units of Concentration
Percent volume
% volume = volume solute (ml) x 100
volume solution (ml)
Percent mass
% mass = mass solute (g) x 100
mass solution (g)
Solution = solvent + solute

3. Units of Concentration
Example 1:
What is the percent by volume concentration of a
solution in which 75.0 ml of ethanol is diluted to a
volume of 250.0 ml?
75.0 ml x 100 = 30.0%
250.0 ml

4. Units of Concentration
Example 2:
What volume of acetic acid is present in a bottle containing
350.0 ml of a solution which measures 5.00%
concentration?
x = 0.05
350.0 ml
x = 17.5 ml

5. Units of Concentration
Example 3:
Find the percent by mass in which 41.0 g of NaCl is
dissolved in 331 grams of water.
41 g x 100 = 11.0%
372 g

6. Molarity (M) is the most common unit of
concentration
Molarity is an expression of moles/Liter of
the solute.
Units of Concentration

7. A mole is the SI unit of number of particles and
can be used as an expression of the molecular
weight of a substance.
Units of Concentration
The formula weight of an
element is expressed as
grams/mole

8. The molar mass of a compound can be
calculated by adding the molar mass of the
individual elements.
Units of Concentration
22.99 + 35.45 = 58.44 g/mol

9. You just calculated the molar mass of sodium
chloride to be 58.44 g/mol.
To determine how to make a stock solution of
sodium chloride, use the formula:
Making Solutions
g = M x L x molar mass

10.  How many grams of NaCl would you need to prepare
200.0 mL of a 5 M solution?
Making Solutions
g = M x L x molar mass
g = (5mol/L) (0.2L) (58.44g/mol)
g = 58.44 g

11. Diluting Solutions
Often once you have made a stock
solution, you need to dilute it to a working
concentration.
To determine how to dilute the stock
solution, use the formula:
C1V1 = C2V2
C1 – concentration of stock
C2 - concentration of diluted solution
V1 – volume needed of stock
V2 – final volume of dilution

12. Example 5:
How many milliliters of a 5 M stock solution of NaCl are
needed to prepare 100 ml of a 0.4 M solution?
Diluting Solutions
C1 V1 = C2 V2
(5) V1 = (0.4)(100)
V1= 8 ml

13. Diluting Solutions
Serial Dilutions are dilutions made in series (for
example, if you needed to make solutions that
were 2M, 1M, 0.5M, and 0.25 M)
The formula for serial dilutions is:
Dilution Factor = (V1 + V2)
V1
V1 – volume of solution being diluted
V2 – volume of solvent

14. Units of Concentration
Example 6:
Propose a method to prepare 100 ml of a 0.5 M glucose
solution from a 5 M glucose solution.
10 = (v1 + 100)
v1
10v1 = v1 + 100
-v1 -v1
9v1 = 100
9 9
v1 = 11.1 ml of 5 M glucose + 100 ml H2O

15. Dilutions tutorial
Diluting Solutions