Section 13.2 Day 1 - Morrison Community Unit District #6
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Transcript Section 13.2 Day 1 - Morrison Community Unit District #6
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…