chapter 13 solutions. basic definitions solution solution – a homogeneous mixture of 2 or more...
TRANSCRIPT
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CHAPTER 13
SOLUTIONS
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BASIC DEFINITIONS
Solution – a homogeneous mixture of 2 or more substances in a single phase
Solute – The dissolved substance usually in smaller amount
Solvent – the dissolving medium usually in larger amount
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Visual Concepts
Solutes, Solvents, and Solutions
Chapter 13
Solvent Solvent - present in - present in greater amountgreater amount
Solute Solute - substance - substance being dissolvedbeing dissolved
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Visual Concepts
Solutions
Chapter 13
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MORE DEFINITIONS
Saturated – The maximum amount of solute is dissolved in the solventUnsaturated – Less than the maximum amount of solute is dissolved in the solventDilute – The amount of solvent is much greater than the amount of solute dissolved in it.
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MORE DEFINITIONS
Electrolyte – A substance that dissolves in water to give a solution that will conduct electricity
Nonelectrolyte – A substance that dissolves in water to give a solution that will not conduct electricity
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TYPES OF SOLUTIONS
Gases (Air is a mixture of O2, N2, CO2,
H2O, etc.)
Solids (White gold is a mixture of gold & palladium)
Liquids (salt dissolved in water; carbon dioxide in soda)
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Particle Model for a Suspension
Chapter 13 Section 1 What Is a Solution?
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SUSPENSION
A heterogeneous mixture in which particles in a solvent are so large, they will settle out unless it is constantly stirred
Particles can be filtered out due to their size (>1000 nm in diameter)
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Visual Concepts
Suspensions
Chapter 13
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COLLOIDS
A heterogeneous mixture in which particles in a solvent remain suspended by the movement of surrounding molecules Think of fruit suspended in jelloParticles are not easily filtered due to their size (<1000 nm in diameter)
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Visual Concepts
Factors Affecting the Rate of Dissolution
Chapter 13
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FACTORS AFFECTING SOLUBILITY
1. Surface Area – increasing the surface area gives molecules more places to interact causing the solute to dissolve faster
2. Agitation/Stirring – stirring gives molecules more opportunities to interact – faster dissolving
3. Temperature – Adding heat adds energy to molecules = faster dissolving
4. “Like dissolves like” – polar solutes dissolve in polar solvents or Nonpolar solutes dissolve in Nonpolar solvents
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Visual Concepts
Like Dissolves Like
Chapter 13
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CONCENTRATION CALCULATIONS
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Visual Concepts
Concentration
Chapter 13
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Concentration, continuedCalculating Concentration, continued
• Concentrations can be expressed in many forms.
Section 2 Concentration and MolarityChapter 13
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MOLARITY (M)
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MOLARITY (m)
Moles of solute per Liter of solution
molarity (M)
amount of solute (mol)
volume of solution (L)
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SAMPLE PROBLEM A
You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?
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SAMPLE PROBLEM A
You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?
Step 1: Outline what you know.
M = ? Mol / L
mol = 90.0 g
L = 3.50 L
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SAMPLE PROBLEM A
You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?
Step 2: Convert any units necessary.
mol = 90.0 g NaCl
90.0 g NaCl x1 mol NaCl58 g NaCl
= 1.55 mol NaCl
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SAMPLE PROBLEM A
You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?
Step 3: Plug into the equation and solve.M = ? Mol / Lmol = 1.55 molL = 3.50 L
molarity (M)
amount of solute (mol)
volume of solution (L)
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SAMPLE PROBLEM A
You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?
Step 3: Plug into the equation and solve.
molarity (M)
amount of solute (mol)
volume of solution (L)
M=
1.55 mol3.50 L
= 0.44 mol / L
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SAMPLE PROBLEM B
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?
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SAMPLE PROBLEM B
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?
Step 1: Outline what you know.
M = 0.5 Mol / L
mol = ? mol
L = 0.8 L
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SAMPLE PROBLEM B
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?
Step 2: Convert any units necessary.
All units correct! Moving on…
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SAMPLE PROBLEM B
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?
Step 3: Plug into the equation and solve.M = 0.5 Mol / Lmol = ? molL = 0.8 L
molarity (M)
amount of solute (mol)
volume of solution (L)
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SAMPLE PROBLEM B
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?
Step 3: Plug into the equation and solve.
molarity (M)
amount of solute (mol)
volume of solution (L)
0.5 M=
mol
0.8 L
(0.8 L)(0.8 L)
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SAMPLE PROBLEM B
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?
Step 3: Plug into the equation and solve.
mol = 0.4
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MOLALITY (m)
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MOLALITY (m)
Moles of solute per kilogram of solvent
molality (m)
amount of solute (mol)
mass of solvent (kg)
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SAMPLE PROBLEM C
A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.
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SAMPLE PROBLEM C
A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.
Step 1: Outline what you know.
m = ? Mol / kg
mol = 17.1 g
Kg = 125 g
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342.34 g C12H22O11
SAMPLE PROBLEM C
A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.
Step 2a: Convert any units necessary.
mol = 17.1 g C12H22O11
17.1 g C12H22O11 x1 mol C12H22O11
= 0.05 mol C12H22O11
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1000 g
SAMPLE PROBLEM C
A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.
Step 2b: Convert any units necessary.
kg = 125 g H2O
125 g H2O x1 kg
= 0.125 kg H2O
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SAMPLE PROBLEM C
A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.
Step 3: Plug into the equation and solve.m = ? Mol / kgmol = 0.05 molkg = 0.125 kg
molality (m)
amount of solute (mol)
mass of solvent (kg)
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SAMPLE PROBLEM C
A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.
Step 3: Plug into the equation and solve.
m=
0.05 mol
0.125 kg= 0.400 mol /
kg
molality (m)
amount of solute (mol)
mass of solvent (kg)
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SAMPLE PROBLEM D
How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
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SAMPLE PROBLEM D
How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
Step 1: Outline what you know.
m = 0.480 m
mol = ? mol
kg = 100.0 g
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SAMPLE PROBLEM D
How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
Step 2: Convert any units necessary.
kg = 100.0 g
Kg = 0.100 kg
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SAMPLE PROBLEM D
How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
Step 3: Plug into the equation and solve.m = 0.480 Mol / kgmol = ? molkg = 0.100 kg
molality (m)
amount of solute (mol)
mass of solvent (kg)
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SAMPLE PROBLEM D
How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
Step 3: Plug into the equation and solve.
0.480 m =
mol
0.100 kg
(0.100 kg)(0.100
kg)
molality (m)
amount of solute (mol)
mass of solvent (kg)
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SAMPLE PROBLEM D
How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
Step 3: Plug into the equation and solve.
mol = 0.048
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PARTS PER MILLION (ppm)
ppm = grams of soluteLiters of solution
X 1000
Grams of solute in 1 million grams of solution.
Useful in very dilute samples, but for the sake of simplicity, your worksheet will not work with very small exponents.