Friday, Feb. 21st: “A” Day
Monday, Feb. 24th: “B” Day
Agenda
Collect chromatography labs
Begin Section 13.2: “Concentration and Molarity”
Demos:
Preparing 250 mL of a 0.5000 M CuSO4 solution
Do solvents always add up?
Homework:
Practice pg. 461: #2, 3, 5, 6
Practice pg. 465: #1-7
Concentration
In a solution, the solute is distributed evenly
throughout the solvent. This means that any
part of a solution has the same ratio of solute
to solvent as any other part of the solution.
This ratio is the concentration of the solution.
Concentration: the amount of a particular
substance in a given quantity of a solution
Calculating Concentration
Concentrations can be expressed in many forms.
Calculating Concentrations
One unit of concentration used in pollution
measurements that involve very low
concentrations is parts per million, or ppm.
Parts per million is the number of grams of
solute in 1 million grams of solution.
Sample Problem A, Pg. 461
A chemical analysis shows that there are 2.2 mg of
lead in exactly 500 g of a water sample. Convert
this measurement to parts per million.
First, change mg g: 2.2 mg Pb X 1 g = .0022 g Pb
1,000 mg
Divide by 500 g to get the amount of Pb in 1 g of H2O.
Then multiply by 1,000,000 to get the amount of Pb
in 1,000,000 g H2O:
.0022 g Pb X 1,000,000 parts = 4.4 ppm Pb
500 g
1 million
(2 sig figs)
Additional Example
How many parts per million of mercury are there in a
sample of tap water with a mass of 750 g containing
2.2 mg of Hg?
First, change mg g: 2.2 mg Hg X 1 g = .0022 g Hg
1,000 mg
Divide by 750 g to get the amount of Hg in 1 g of H2O.
Then multiply by 1,000,000 to get the amount of Hg
in 1,000,000 g H2O:
.0022 g X 1,000,000 parts = 2.9 ppm Hg
750 g
1 million
(2 sig figs)
Molarity
Since the mole is the unit chemists use to
measure the number of particles, they often
specify concentrations using molarity.
Molarity (M): a concentration unit of a
solution expressed as moles of solute dissolved
per liter of solution.
Molarity (M) = moles of solute
L of solution
Molarity Example
Suppose that 0.30 moles of KBr are present in
0.40 L of solution.
The molarity of the solution is calculated as
follows:
0.30 mol KBr
= 0.75 M KBr
0.40 L solution
This is called a 0.75 molar solution of KBr.
Preparing a Solution of a Specified Molarity
Note that molarity describes concentration in
terms of volume of solution, NOT volume of
solvent.
If you simply added 1.000 mol solute to
1.000 L solvent, the solution would not be
1.000 M.
The added solute would increase the
volume, so the solution would not have a
concentration of 1.000 M.
The solution must be made to have exactly the
specified volume of solution.
Demo: Preparing 250 mL of a
0.5000 M CuSO4 Solution (pg. 463)
Calculating Molarity
In working with solutions in chemistry, you will
find that numerical calculations often involve
molarity.
The key to all such calculations is the
definition of molarity…
Molarity (M) = moles of solute
L of solution
Calculating Molarity Given Mass of
Solute and Volume of Solution
Sample Problem B, Pg. 465
What is the molarity of a potassium chloride
solution that has a volume of 400.00 mL and
contains 85.0 g KCl?
Molarity = moles of solute
L of solution
First, use molar mass to change g of KCl → moles KCl:
85.0 g KCl X 1 mol KCl = 1.14 mol KCl
74.6 g KCl
1.14 mol KCl = 2.85 M KCl
.400 L
(3 sig figs)
Additional Example
Determine the molarity of a solution prepared by
dissolving 16.9 g of NaOH in enough water to
make 250.00 mL of solution.
Molarity = moles of solute
L of solution
First, use molar mass to change g NaOH mol NaOH
16.9 g NaOH X 1 mol NaOH = 0.423 mol NaOH
40 g NaOH
0.423 mol NaOH = 1.69 M NaOH
0.250 L
(3 sig figs)
Calculating Mass of Solute Given
Molarity and Volume of Solution
Additional Example
How many grams of NaOH are needed to prepare
250.0 mL of a 1.69 M NaOH solution?
First, change mL L: 250 mL X 1 L = 0.2500 L
1,000 mL
Then, multiply by molarity to find moles of solute:
0.2500 L X 1.69 moles NaOH = 0.423 mol NaOH
1L
Finally, use molar mass to find mass of solute:
0.423 mol NaOH X 40 g NaOH = 16.9 g NaOH
1 mole NaOH
(3 sig figs)
Additional Example
How many grams of glucose, C6H12O6, are in
255 mL of a 3.55 M solution?
First, change mL L: 255 mL X 1 L = 0.255 L
1,000 mL
Then, multiply by molarity to find moles of solute:
0.255 L X 3.55 moles glucose = 0.905 mol
1L
glucose
Finally, use molar mass to find mass of solute:
0.905 mol glucose X 180 g glucose = 163 g glucose
1 mol glucose
Demo: Do Solvents Always Add Up?
50.0 mL H2O + 50.0 mL ethanol =
________?___ mL solution
+
Homework
Practice pg. 461: # 2, 3, 5, 6
Practice pg. 465: # 1-7
We will finish section 13.2 next time…