II III I II. Concentration Ch. 13 & 14 - Solutions.

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II III I II. Concentration Ch. 13 & 14 - Solutions

Transcript of II III I II. Concentration Ch. 13 & 14 - Solutions.

Page 1: II III I II. Concentration Ch. 13 & 14 - Solutions.

II

III

I II. ConcentrationII. Concentration

Ch. 13 & 14 - SolutionsCh. 13 & 14 - Solutions

Page 2: II III I II. Concentration Ch. 13 & 14 - Solutions.

What is different between the glasses of Kool-aid?

What is different between the glasses of Kool-aid?

Page 3: II III I II. Concentration Ch. 13 & 14 - Solutions.

Solution concentration can be described generally

Solution concentration can be described generally

Dilute - reduced in strength, weak, watered down.

Concentrated – stronger, pure. Has less water.

Page 4: II III I II. Concentration Ch. 13 & 14 - Solutions.

What is the problem with just using dilute and concentrated as descriptions of the solution concentration?

What is the problem with just using dilute and concentrated as descriptions of the solution concentration?

Page 5: II III I II. Concentration Ch. 13 & 14 - Solutions.

Is solution B dilute or concentrated?

The terms dilute and concentrated are relative.

Scientists need a more precise way of referring to the concentration of a solution.

Concentrated Dilute

Solution A Solution B Solution C

Page 6: II III I II. Concentration Ch. 13 & 14 - Solutions.

Solution concentration can be described specifically

Solution concentration can be described specifically

Do you remember the “mole” from Stoichiometry?

What is a mole?

How might you use it to describe the concentration of a solution?

Page 7: II III I II. Concentration Ch. 13 & 14 - Solutions.

A. ConcentrationA. Concentration

The amount of solute in a solution.

Describing Concentration

• % by mass - medicated creams

• % by volume - rubbing alcohol

• ppm, ppb - water contaminants

• molarity - used by chemists

• molality - used by chemists

Page 8: II III I II. Concentration Ch. 13 & 14 - Solutions.

MolarityMolarity

The ratio of the moles of solute to the volume of solution in liters.

Molarity (M) =

Moles of solute

Volume in Liters of Solution

Page 9: II III I II. Concentration Ch. 13 & 14 - Solutions.

How to read MolarityHow to read Molarity

6.0 M NaClRead: “6 molar solution of NaCl”Can be abbreviated 6M solution You must be careful to label the

molarity with a capital M so that it is not confused with m for molality.

Page 10: II III I II. Concentration Ch. 13 & 14 - Solutions.

Types of Calculations with Molarity:

Types of Calculations with Molarity:

1. Finding concentration of a solution.

2. Finding the mass of solute needed.

3. Finding the volume of solution made.

Page 11: II III I II. Concentration Ch. 13 & 14 - Solutions.

Finding ConcentrationFinding Concentration

Antifreeze is a solution of ethylene glycol, C2H6O2 in water. If 4.50 L of antifreeze contains 27.5 g of ethylene glycol, what is the concentration of the solution?

27.5 g C2H6O2

62.08 g C2H6O2

1 mol C2H6O2

4.5 L

= 0.0984 mol/L or

0.0984 M C2H6O2

Page 12: II III I II. Concentration Ch. 13 & 14 - Solutions.

Finding Mass Finding Mass

What mass of sodium carbonate, Na2CO3,

is present in 50 ml of a 0.750M solution?

50 ml 1 L

1000 mL

0.750 mol

1 L 1 mol Na2CO3

105.99 g Na2CO3

= 3.97 g Na2CO3

Conversion Factor

Page 13: II III I II. Concentration Ch. 13 & 14 - Solutions.

Finding VolumeFinding Volume

What volume of 1.50 mol/L HCl solution contains 10.0 g of hydrogen chloride?

10.0 g HCl 1 mol HCl

36.46 g HCl 1.50 mol

1 L

= 0.183 L or 183 mL

Conversion Factor

Page 14: II III I II. Concentration Ch. 13 & 14 - Solutions.

Practice Problems

Page 15: II III I II. Concentration Ch. 13 & 14 - Solutions.

Practice ProblemsPractice Problems

1. A 0.750 L aqueous solution contains 90.0 g of ethanol, C2H5OH. Calculate the molar

concentration of the solution in mol/L.

Page 16: II III I II. Concentration Ch. 13 & 14 - Solutions.

Practice ProblemPractice Problem

2. What mass of NaCl are dissolved in 152 mL of a solution if the concentration of the solution is 0.364 M?

Page 17: II III I II. Concentration Ch. 13 & 14 - Solutions.

Practice ProblemPractice Problem

3. What mass of dextrose, C6H12O6 is dissolved in 325 mL of 0.258 M solution?

Page 18: II III I II. Concentration Ch. 13 & 14 - Solutions.

Practice ProblemPractice Problem

4. A mass of 98 g of sulfuric acid, H2SO4, is dissolved in water to prepare a 0.500 M solution. What is the volume of the solution?

Page 19: II III I II. Concentration Ch. 13 & 14 - Solutions.

Practice ProblemPractice Problem

5. A solution of sodium carbonate, Na2CO3, contains 53.0 g of solute in 215 mL of solution. What is its molarity?

Page 20: II III I II. Concentration Ch. 13 & 14 - Solutions.

Practice ProblemPractice Problem

6. What is the molarity of a solution of HNO3 that contains 12.6 g of solute in 5.00 L of solution?

Page 21: II III I II. Concentration Ch. 13 & 14 - Solutions.

B. MolalityB. Molality

solvent ofkg

solute of moles(m)molality

mass of solvent only

1 kg water = 1 L waterkg 1

mol0.25 0.25m

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B. MolalityB. Molality

Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.

75 g MgCl2 1 mol MgCl2

95.21 g MgCl2

= 3.2m MgCl2

0.25 kg water

kg

molm

Page 23: II III I II. Concentration Ch. 13 & 14 - Solutions.

B. MolalityB. Molality

How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water?

0.500 kg water 1.54 mol NaCl

1 kg water

= 45.0 g NaCl

58.44 g NaCl

1 mol NaCl

kg 1

mol1.5 1.5m

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Practice ProblemPractice Problem

What is the molality of a solution made

from 2.4 moles of NaCl and 0.80 kg of

water?

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Practice Problem Practice Problem

What is the molality of a solution made

from 63 g of HNO3 in 0.50 kg of water?

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Practice ProblemPractice Problem

How much water is needed to make a 0.50 m solution from 3.2 g of NaCl?

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Practice ProblemsPractice Problems

What mass of CH3OH is needed to add to

1.20 kg of water to make a 3.00 m

solution?

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2211 VMVM

C. DilutionC. Dilution

Preparation of a desired solution by adding water to a concentrate.

Moles of solute remain the same.

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C. DilutionC. Dilution

What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:

M1 = 15.8M

V1 = ?

M2 = 6.0M

V2 = 250 mL

WORK:

M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

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Dilution PracticeDilution Practice

Water is added to 200. mL of a 2.0 M

solution of CaCl2 to increase the volume of

the solution to 400. mL. What is the new

concentration?

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Dilution PracticeDilution Practice

To what volume must 1.0 L of a 6.0 M

solution of HCl be diluted in order to

prepare a 0.2 M solution?

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D. Preparing SolutionsD. Preparing Solutions

500 mL of 1.54M NaCl

500 mLwater

45.0 gNaCl

• mass 45.0 g of NaCl• add water until total

volume is 500 mL• mass 45.0 g of NaCl• add 0.500 kg of water

500 mLmark

500 mLvolumetric

flask

1.54m NaCl in 0.500 kg of water

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D. Preparing SolutionsD. Preparing Solutions

250 mL of 6.0M HNO3 by dilution

• measure 95 mL of 15.8M HNO3

95 mL of15.8M HNO3

water for

safety

250 mL mark

• combine with water until total volume is 250 mL

• Safety: “Do as you oughtta, add the acid to the watta!”

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Solution Preparation LabSolution Preparation Lab

Turn in one paper per team. Complete the following steps:

A) Show the necessary calculations.

B) Write out directions for preparing the solution.

C) Prepare the solution. For each of the following solutions:

1) 100.0 mL of 0.50M NaCl

2) 0.25m NaCl in 100.0 mL of water

3) 100.0 mL of 3.0M HCl from 12.1M concentrate.