friday, feb. 21 st : “a” day monday, feb. 24 th : “b” day agenda
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Friday, Feb. 21 st : “A” Day Monday, Feb. 24 th : “B” Day Agenda. Collect chromatography labs Begin Section 13.2: “Concentration and Molarity” Demos: Preparing 250 mL of a 0.5000 M CuSO 4 solution Do solvents always add up? Homework: Practice pg. 461: #2, 3, 5, 6 - PowerPoint PPT PresentationTRANSCRIPT
Friday, Feb. 21st: “A” DayMonday, Feb. 24th: “B” Day
AgendaCollect chromatography labsBegin Section 13.2: “Concentration and Molarity”Demos:
Preparing 250 mL of a 0.5000 M CuSO4 solutionDo solvents always add up?
Homework:Practice pg. 461: #2, 3, 5, 6Practice pg. 465: #1-7
ConcentrationIn 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 ConcentrationConcentrations can be expressed in many forms.
Calculating ConcentrationsOne 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. 461A 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 ExampleHow 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)
MolaritySince 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 ExampleSuppose 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 MolarityNote 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 MolarityIn 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. 465What 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 KCl1.14 mol KCl = 2.85 M KCl
.400 L (3 sig figs)
Additional ExampleDetermine 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 NaOH16.9 g NaOH X 1 mol NaOH = 0.423 mol NaOH
40 g NaOH0.423 mol NaOH = 1.69 M NaOH
0.250 L (3 sig figs)
Calculating Mass of Solute Given Molarity and Volume of Solution
Additional ExampleHow 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 mLThen, multiply by molarity to find moles of solute: 0.2500 L X 1.69 moles NaOH = 0.423 mol NaOH
1 LFinally, 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 ExampleHow 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 mLThen, multiply by molarity to find moles of solute: 0.255 L X 3.55 moles glucose = 0.905 mol
1 L glucoseFinally, 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, 6Practice pg. 465: # 1-7
We will finish section 13.2 next time…