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Solutions
Chapter 12
Types of Mixtures
• Solutions– Solvent dissolves a solute
• Suspensions– The solute particles are too large to be
dissolved. They settle out unless the mixture is constantly stirred.
• Colloids– Particles intermediate in size between
solutions and suspensions.
Solutions
• Solvent– The component of a solution that does the
dissolving.
• Solute– The component of a solution that is dissolved.
• Soluble– Capable of being dissolved.
• Insoluble– Incapable of being dissolved.
Solutions
Solute State Solvent State example
Gas Gas Air
Gas Liquid CO2 in water
Liquid Liquid Alcohol in water
Solutions
Solute State Solvent State Example
Liquid Solid Hg in Ag and Sn (fillings)
Solid Liquid Sugar in water
Solid Solid Cu in Ni
Examples of Solutions• Sugar Water
– Sugar in water (aqueous solution)
• Salt Water– Salt in water (aqueous solution)
• Alloys– Brass (zinc and copper)– Sterling Silver (silver and copper)– Stainless Steel (iron, chromium, and
nickel)
• Air– Oxygen in nitrogen
Suspensions
• Oil and vinegar– Nonpolar and polar
• Oil and water– Nonpolar and polar
• Muddy water– Large particles
Colloids
• Paint• Gelatin• Milk• Mayonnaise• Shaving and whipped cream• Fog• Cheese• butter
Tyndall Effect• Light can be scattered by colloidal particles.
• Used to distinguish a solution from a colloid.– That is why it is hard to
see in the fog, because the light is scattering and is not focused in one direction.
Electrolytes
• Substances that dissolve in water that can conduct current.-Ionic Compounds-
NaCl, CaCl2, MgCl2, KCl
We need electrolytes in our bodies to keep our nervous system and muscular systems functioning properly.
Video
The Assault on SaltSalt vs. Sea Salt
Solubility
• Unsaturated solution– A solution in which more solute can still dissolve.
• Saturated Solution– A solution in which no more solute can dissolve.– **A solution can be saturated when at a warmer
temperature and then when cooled, form a supersaturated solution in which crystals form (rock candy)
• Solubility– The amount of substance required to form a saturated
solution with a specific amount of solvent at a specific temperature.
Solubility curves
Solubility curves Tutorial
Y axis- grams of solute/100g water
X axis temperature
Positive slopes- the higher the temperature, the more solute that will dissolve.
Negative slopes- the higher the temperature, the less solute that will dissolve.
What influences strength of a solution?
Sample Volume of water Mass of sucrose Mass of salt
A 50 mL 5 g 0
B 50 mL 0 5g
C 50 mL 10 g 0
D 50 mL 0 10g
Concentrations Of Solutions
• Molarity– The number of moles
of solute in 1 L of solution.
– Mol– L– Represented by M
• Molality– The number of moles
of a solute per kg of solvent.
– Mol– Kg– Represented by m
What influences strength of a solution?
Sample Mass of sucrose
Mass of salt Moles of solute
Molarity of solution
A 5 g 0 0.015 .30 M
B 0 5g 0.086 1.72
C 10 g 0 0.029 .58 M
D 0 10g 0.171 3.42
Practice Molarity
• You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of the solution?
M= mol
L
90.0 g NaCl x 1 mol NaCl = 1.54 mol
58.44 g
1.54 mol NaCl = 0.440 M NaCl
3.50 L solution
Practice Molarity
• You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? How many grams of HCl would that be?
M = mol
L
0.5 = mol
0.8
Mol = 0.4 mol NaCl x 36.46 g/mol = 14.58 g HCl
Practice Molarity• To produce 40.0 g of Ag2CrO4 you will
need at least 23.4 g of K2CrO4 in solution as a reactant. All you have on hand is 5 L of a 6.0 M K2CrO4 solution. What volume of the solution is needed to give you the 23.4 g K2CrO4 needed for the reaction?
1 mol K2CrO4 = 194.2 g You need at least 23.4 g x 1 mol = 0.120 mol K2CrO4
194.2 g
M= mol therefore, 6.0 M = 0.120 L = 0.020 L K2CrO4 solution
L L
Practice molality
• A solution was prepared by dissolving 17.1 g of C12H22O11, sucrose, in 125 g of water. Find the molal concentration of the solution.
M = mol solute
kg of solvent
17.1 g C12H22O11 = 1 mol = 0.0500 mol C12H22O11
342.34 g
0.0500 mol = 0.400m C12H22O11
0.125 kg H2O
Practice molality
• A solution of Iodine, I2, in CCl4 is used for chemical tests. How much iodine must be added to prepare a 0.480m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
100.0 g = .100 kg CCl4
0.480 m = x mol = 0.0480 mol I2
0.1 kg H2O
0.0480 mol I2 x 253.8 g I2 = 12.2 g I2
1 mol