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Chapter 8 Molarity Parts 1 & 2

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Chapter 8. Molarity Parts 1 & 2. Solutions. Solute The substance that dissolves (the minor component of a solution). KMnO 4. Solutions. Solvent The substance that the solute dissolves in (the major component of a solution). Solution A homogeneous mixture of the solute and solvent. - PowerPoint PPT Presentation

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Page 1: Chapter 8

Chapter 8

MolarityParts 1 & 2

Page 2: Chapter 8

Solutions

Solute • The substance that dissolves (the minor

component of a solution).

KMnO4

Page 3: Chapter 8

Solutions

• Solvent

• The substance that the solute dissolves in (the major component of a solution).

Page 4: Chapter 8

KMnO4 solution

Solution

A homogeneous mixture of the solute and solvent.

Page 5: Chapter 8

Solutions

Dilute Concentrated

Page 6: Chapter 8

Concentration of a solution

Molarity(M) is the moles of solute per liter of solution.

Molarity(M) =moles of soluteLiters of solution

Page 7: Chapter 8

Common Volume Units

= 1000 mL= 1000 mL

Page 8: Chapter 8

Common Volume Units

= 1000 mL= 1000 mL

Page 9: Chapter 8

Units of Molarity

Page 10: Chapter 8

Units of Molarity

2.0 M HCl

Page 11: Chapter 8

Units of Molarity

2.0 M HCl = 2.0 moles HCl

L solution

Page 12: Chapter 8

Units of Molarity

2.0 M HCl = 2.0 moles HCl

L solution

6.0 M HCl

Page 13: Chapter 8

Units of Molarity

2.0 M HCl = 2.0 moles HCl

L solution

6.0 M HCl = 6.0 moles HCl

dm3 solution

Page 14: Chapter 8

Molarity

• Calculate the molarity of an aqueous Ba(NO3)2 solution containing 0.283 grams of barium nitrate in 0.931 L of solution.

Page 15: Chapter 8

Molarity

• Calculate the molarity of an aqueous Ba(NO3)2 solution containing 0.283 grams of barium nitrate in 0.931 L of solution.

0.283 g Ba(NO3)2

0.931 L

Page 16: Chapter 8

Molarity

• Calculate the molarity of an aqueous Ba(NO3)2 solution containing 0.283 grams of barium nitrate in 0.931 L of solution.

0.283 g Ba(NO3)2

0.931 L 261.3 g

mol

Page 17: Chapter 8

Molarity

• Calculate the molarity of an aqueous Ba(NO3)2 solution containing 0.283 grams of barium nitrate in 0.931 L of solution.

0.283 g Ba(NO3)2

0.931 L 261.3 g

mol= 0.00116 mol Ba(NO3)2/L

= 0.00116 M Ba(NO3)2

Page 18: Chapter 8

Molarity Calculation

• NaOH is used to open stopped sinks, to treat cellulose in the making of nylon.• It is also used to remove potato peels

commercially.

Page 19: Chapter 8

If 4.0 g NaOH are used to make 500.0 mL of NaOH solution, what is the molarity (M) of the solution?

0.20 mol NaOH/L = 0.20M NaOH

Page 20: Chapter 8

If 4.0 g NaOH are used to make 500.0 mL of NaOH solution, what is the molarity (M) of the solution?

4.0 g NaOH

500.0 mL

Page 21: Chapter 8

If 4.0 g NaOH are used to make 500.0 mL of NaOH solution, what is the molarity (M) of the solution?

4.0 g NaOH

500.0 mL 40.0 g

mol

Page 22: Chapter 8

If 4.0 g NaOH are used to make 500.0 mL of NaOH solution, what is the molarity (M) of the solution?

4.0 g NaOH

500.0 mL 40.0 g

mol

0.001 L

mL

Page 23: Chapter 8

If 4.0 g NaOH are used to make 500.0 mL of NaOH solution, what is the molarity (M) of the solution?

4.0 g NaOH

500.0 mL 40.0 g

mol

0.001 L

mL= 0.20 mol NaOH/L = 0.20M NaOH

Page 24: Chapter 8

Molarity Conversion Factors

An NaOH solution has a molarity of 0.20M.

• Write this molarity in the form of conversion factors (ratios).

Page 25: Chapter 8

Molarity Conversion Factors

0.20 moles NaOH and L solution ____

L solution 0.20 moles NaOH

Page 26: Chapter 8

Diabetes is a serious medical issue facing our country at this time.

Page 27: Chapter 8

• 23.6 million people in the United States (7.8% of the total population) have diabetes. Of these, 5.7 million have undiagnosed diabetes.

• In 2007, about 1.6 million new cases of diabetes were diagnosed in people aged 20 years or older.

• If current trends continue, 1 in 3 Americans will develop diabetes sometime in their lifetime, and those with diabetes will lose, on average, 10–15 years of life.

• Diabetes is the leading cause of new cases of blindness, kidney failure, and nontraumatic lower-extremity amputations among adults.

Page 28: Chapter 8

• Diabetes was the sixth leading cause of death on U.S. death certificates in 2006. Overall, the risk for death among people with diabetes is about twice that of people without diabetes of similar age.

• In 1999–2000, 7% of U.S. adolescents aged 12–19 years had impaired fasting glucose (prediabetes), putting them at increased risk of developing type 2 diabetes, heart disease, and stroke.

Page 29: Chapter 8

Diabetes Is Preventable and Controllable

• Recent studies show that lifestyle changes can prevent or delay the onset of type 2 diabetes among people at high risk.

• For people with prediabetes, lifestyle changes, including a 5%–7% weight loss and at least 150 minutes of physical activity per week, can reduce the rate of onset of type 2 diabetes by 58%.

• Disability and premature death are not inevitable consequences of diabetes. By working with their support network and health care providers, people with diabetes can prevent premature death and disability by controlling their blood glucose, blood pressure, and blood lipids and by receiving other preventive care in a timely manner.

Page 30: Chapter 8

Blood sugar is a term referring to the amount of glucose (C6H12O6) present in the blood. A normal blood glucose level is about 0.083g in 100.0mL. What is the molarity of normal blood glucose?

0.0046M C0.0046M C66HH1212OO66

Page 31: Chapter 8

How many grams of Na2SO4 are required to prepare 1000.0 mL of 1.0000 M Na2SO4 solution?

Page 32: Chapter 8

How many grams of Na2SO4 are required to prepare 1000.0 mL of 1.0000 M Na2SO4 solution?

1000.0 mL1000.0 mL

Page 33: Chapter 8

How many grams of Na2SO4 are required to prepare 1000.0 mL of 1.0000 M Na2SO4 solution?

1000.0 mL 0.001 L1000.0 mL 0.001 L

mLmL

Page 34: Chapter 8

How many grams of Na2SO4 are required to prepare 1000.0 mL of 1.0000 M Na2SO4 solution?

1000.0 mL 0.001 L 1.0000 mol Na1000.0 mL 0.001 L 1.0000 mol Na22SOSO44

mL LmL L

Page 35: Chapter 8

How many grams of Na2SO4 are required to prepare 1000.0 mL of 1.0000 M Na2SO4 solution?

1000.0 mL 0.001 L 1.0000 mol Na1000.0 mL 0.001 L 1.0000 mol Na22SOSO4 4 142.04 g142.04 g

mL L molmL L mol

Page 36: Chapter 8

How many grams of Na2SO4 are required to prepare 1000.0 mL of 1.0000 M Na2SO4 solution?

1000.0 mL 0.001 L 1.0000 mol Na1000.0 mL 0.001 L 1.0000 mol Na22SOSO4 4 142.04 g142.04 g

mL L molmL L mol

= 142.04 g Na= 142.04 g Na22SOSO44

Page 37: Chapter 8

a

a

How to prepare 1000ml (1.000L) How to prepare 1000ml (1.000L) of a 1.0000M solution of Naof a 1.0000M solution of Na22SOSO44..

Page 38: Chapter 8

In a college chemistry class you are asked to prepare 100.0 mL solution of 0.250 M NiCl2 (gfm = 129.7g/mol).

Page 39: Chapter 8

In a college chemistry class you are asked to prepare 100.0 mL solution of 0.250 M NiCl2 (gfm = 129.7g/mol).

How do you prepare the solution?

Page 40: Chapter 8

In a college chemistry class you are asked to prepare 100.0 mL solution of 0.250 M NiCl2 (gfm = 129.7g/mol).

How do you prepare the solution.

3.24 grams of NiCl3.24 grams of NiCl22

Page 41: Chapter 8

• Add 3.24g of NiClAdd 3.24g of NiCl22 to water in a half-filled to water in a half-filled

100.0 ml volumetric flask.100.0 ml volumetric flask.

• Swirl until completely dissolved.Swirl until completely dissolved.

Page 42: Chapter 8

• Add water to the 100.0 ml. mark.Add water to the 100.0 ml. mark.

Page 43: Chapter 8

• Thoroughly mix the solution by capping and inverting the flask several times or stirring the contents.

Page 44: Chapter 8

Information for Upcoming Mole Quiz

• How does a mole relate to particles, mass, atoms? (example: One mole of H2O is 6.02 x 1023 H2O molecules. It has a mass of 18g and contains 2 moles of H and 1 mole of O).

• Be able to use this information in factor-label problems.

• Be able to determine the number of grams/molecule.

• Identify the elements that exist primarily as diatomic gases.

• Do metric conversions.

Page 45: Chapter 8

Homework Molarity Worksheet. (Note on

question number 3 you must put your answers in sentence form in addition to solving the problem).