1. Solutions (Chapter 12)
Taylor, Tommy, Kayla

2. Homogeneous mixture - mixture of two or more substances that has a uniform appearance throughout ex: salt water
Heterogeneous mixture - can be separated physically and the different components are visibly distinguishable from one another
ex: chocolate chip cookie

3. Solute - substance being dissolved
ex: Kool-Aid mix
Solvent - dissolves another substance ex: water
Solution - a homogeneous mixture of two or more substances.
ex: Kool-Aid

4. Visual
Salt Water/ Solution
Salt/ Solute
Water/ Solvent
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5. Suspension - a mixture in which large particles can be evenly distributed and the components will settle out
ex: muddy water
Colloid - a mixture in which small particles can be mixed so they remain evenly distributed without settling out ex: milk

6. Dilute - a solution containing a relatively small quantity of solute as compared with the amount of solvent ex: bleach
Concentrated - a solution that contains a large amount of solute relative to the amount that could dissolve ex: alcohol

7. Supersaturated - increase of concentration beyond saturation
Unsaturated - the solute concentration is lower than its equilibrium solubility
Saturated - a point of maximum concentration in which no more solute can be dissolved
Supersaturated - increase of concentration beyond saturation

8. Solubility Curves Solute below the line indicates the solution
is unsaturated at a certain temperature
Solute above the line shows all of the solute
has dissolved and is supersaturated
If the amount of solute is on line then the
solution is saturated
Ex: At 500C and 100g, NaNo3 is saturated
At 200C and 100g, NaNo3 is supersaturated
At 200C and 50g, NaNo3 is unsaturated

9. Stirring A Solution
Stirring a solution helps disperse the solute particles and bring fresh solvent into contact with the solute surface. This increases the contact between the solvent and solute.
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10. Surface Area
The dissolution process occurs at the surface of the solute, so the larger the surface area of the solute the faster it dissolves.
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11. Temperature of a Solution
As the temp. of a solvent increases the solvent molecules move faster. This causes collisions between the solvent and solute molecules. This helps separate solute molecules and disperse them among the solvent molecules.
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12. Concentration
When you increase the concentration of the solute, the rate of solution formation will also increase. They are directly related.

13. Molarity
The concentration of a solution expressed as the number of moles of solute dissolved in each liter of solution
moles of solute or moles
Molarity= Liters of solution L

14. Molarity Calculations
A saline (salt water) solution contains 0.70 g of NaCl per 100 ml of solution. What is its molarity?
First calculate the number of moles NaCl in the solution. You must use dimensional analysis to convert from grams to moles.
0.70 g NaCl x 1 mole NaCl = 0.01 mole
58.44 g NaCl

15. Molarity Calculations
Where do the numbers come from?
0.70 g NaCl x 1 mole NaCl = 0.01 mol
58.44 g NaCl
Formula mass correlates with grams in one mole
Formula mass of NaCl from periodic table
Grams of solute given in problem

16. Molarity Calculations
Now that you have obtained the number of moles solute, use this number in the molarity equation:
M= 0.01 moles NaCl
0.1L
To find this value, convert from ml (given in the problem) to liters.
M= 0.1

17. Molarity Practice
1. How many grams of CaCl2 are needed to make 125 ml of a 1.5 M solution of CaCl2?
2. How many moles of solute are present in 750 ml of a 0.10 M Na2SO4 solution?

18. Molarity Answers
1. How many grams of CaCl2 are needed to make 125 ml of a 1.5 M solution of CaCl2?
1.5M = moles of solute
.125 L
.188 mol x g CaCl2 = 20.9 g
1 mole CaCl2

19. Molarity Answers
2. How many moles of solute are present in 750 ml of a 0.10 M Na2SO4 solution?
.10 M = moles of solute = .075 moles
.750 L Na2SO4

20. Molality
The number of moles of solute dissolved in each kilogram of solvent
moles of solute or moles
Molality= Kilograms of solvent Kg
Molality is represented by "m "

21. Example Molality Problem:
What is the molality of a solution if 100 g of NaCl is dissolved into 250 g of water?
First calculate the number of moles NaCl in the solution. You must use dimensional analysis to convert from grams to moles.
100g NaCl x 1 mole NaCl = 1.71 moles NaCl
58.44g NaCl

22. Molality Calculations
Where do the numbers come from?
100g NaCl x 1 mole NaCl = 1.71 mol NaCl
58.44g NaCl
Formula mass correlates with grams in one mole
Formula mass of NaCl from the periodic table
Grams of solute given in problem

23. Molality Calculations
Now that you have obtained the number of moles solute, use this number in the molality equation:
m = 1.71 moles NaCl
.25 Kg
m = 6.84
To find this value, convert from g (given in the problem) to Kg.

24. Molality Practice
1. A solution was prepared by dissolving 8.55 g of sucrose (table sugar, C12H22011) in 62.5 g of water. Find the molality of this solution.
2. A solution of iodine, I2, in carbon tetrachloride,CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a m solution of iodine in CCl4 if 50.0 g of CCl4 is used?

25. Molality Answers
1. A solution was prepared by dissolving 8.55 g of sucrose (table sugar, C12H22011) in 62.5 g of water. Find the molality of this solution.
8.55g C12H22011 x 1 mol C12H22011 = mol C12H22011
g C12H22011
m = mol C12H22011 = m C12H22011
kg H2O

26. Molality Answers 50.0 g CCl4 x 1 kg = 0.05 kg CCl4 1000 g CCl4
2. A solution of iodine, I2, in carbon tetrachloride,CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a m solution of iodine in CCl4 if 50.0 g of CCl4 is used?
50.0 g CCl4 x 1 kg = 0.05 kg CCl4
1000 g CCl4
0.240 m = X X= moles I2
0.05 kg CCl4
0.012 moles I2 x g I2 = 3.05 g I2
1 mole I2

27. Dilution
The concentration of a solution can be decreased by adding water.
M1 x V1 = M2 x V2
Dilution Equation:
Molarity after dilution
Volume should always be expressed in liters
Original molarity
Volume after dilution
Original volume

28. Dilution Calculations
How much concentrated 6.0 M hydrochloric acid is needed to prepare 50.0 ml of a 1.0 M solution?
First use the information given in the question in order to substitute into the dilution equation.
6.0 M x V1 = 1.0 M x 0.05 L

29. Dilution Calculations
Now, solve for the missing value algebraically.
6.0 M x V1 = 1.0 M x 0.05 L
6.0V1 = 0.05
V1 = L

30. Dilution Practice
1. To what volume should 8.3 ml of 5.0 M nitric acid be diluted to prepare a 1.0 M solution?
2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?

31. Dilution Answers
1. To what volume should 8.3 ml of 5.0 M nitric acid be diluted to prepare a 1.0 M solution?
5.0 M x L = 1.0 M x V2
= 1.0V2
V2= L

32. Dilution Answers
2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?
3.0 M x L = 1.0 M x V2
= 1.0V2
V2 = L
0.038 L L = L of H2O

33. 3.0 M x 0.0125 L = 1.0 M x V2 Further Explanation
2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?
3.0 M x L = 1.0 M x V2
Molarity of solute given in problem
Volume of solute given in problem
Unknown
Final solute molarity given in problem

34. Further Explanation
2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?
3.0 M x L = 1.0 M x V2
= 1.0V2
V2 = L
Solve algebraically

35. Further Explanation 3.0 M x 0.0125 L = 1.0 M x V2 0.0375 = 1.0V2
V2 = L
0.038 L L = L of H2O
Subtract the final volume and the original volume