1. Chapter 13

2.  A solution forms when one substance disperses uniformly throughout another.  The reason substances dissolve is due to intermolecular forces.  Ion-Dipole forces between water and ionic solids.  Dispersion forces between C 6 H 14 and CCl 4  The interactions between solute and solvent molecules are know as solvation.  When the solvent is water they are know as hydration.

3.  NaCl dissolves in water because the water molecules have strong enough attractive forces for the Na + and Cl - ions.  The water molecules must also move away from each other to make space for the Na + and Cl - ions.

5.  As a solid dissolves the concentration of solute particles in the solution increases.  This leads to a certain amount of crystallization (the opposite of solvation)  When the solute concentration reaches a certain point the rate of solvation and the rate of crystallization will be equal  At this point the solution is said to be saturated.

6.  The amount of solute needed to form a saturated solution in a given quantity of solvent is known as the solubility of that solute.  The solubility of a compound is the maximum amount of that compound that will dissolve in a given amount of solvent at a specific temperature.  The solubility of NaCl at 0 o C is 35.7 g per 100 mL of water.  If we dissolve less than this amount the solution is said to be unsaturated.  Under certain conditions it is possible for a solution to contain more than the amount of solute needed to form a saturated solution.

8.  Along with the chemical nature of the solvent and solute the solubility of a substance depends on…  Temperature  Pressure (For gases only)

9.  One factor that affects solubility is the chemical nature of both the solute and the solvent.  The stronger the attractions between solute and solvent molecules the higher the solubility.  If we try to dissolve a polar solute into a polar solvent it would work very well.  This is due to favorable dipole-dipole interactions between solute and solvent particles.  If we tried to dissolve a polar solute into a nonpolar solvent it would not work.  This is because there would be little to know attractive forces between solute and solvent particles.

10.  The solubility of solids and liquids is affected so little by ambient pressure that we do not need to account for it.  However the solubility of gases is highly affected by pressure.  The solubility of a gas increases as pressure increases.  Soda  Henry’s Law:  S g = kP g

11.  Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4 atm over the liquid at 25 o C. The Henry’s Law constant for CO 2 in water at this temperature is 3.1 x 10 -2 mol/L-atm.

12.  The solubility of most solid solutes in water increases as the temperature of the solution increase.  The solubility of gases in water decreases with increasing temperature.  Poor fishies…

13.  Mass Percentage : % Mass  Parts Per Million: ppm  Mole Fraction

14.  Molarity: M  Molality: m

15.  A solution is made by dissolving 13.5 g of glucose (C 6 H 6 O 6 ) in 0.100 kg of water. What is the mass percentage of solute in this solution?  11.9 %  A 2.5 g sample of ground water was found to contain 5.4 μg of Zn 2+. What is the concentration of Zn 2+ in parts per million?  2.2 ppm

16.  A solution is made by dissolving 4.35 g of glucose in 25.0 mL of water at 25 o C. Calculate the molality of the glucose in solution.  0.964 m  An aqueous solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in solution and the molality of HCl in solution.  Mole fraction = 0.22  Molality = 15 m

17.  A solution contains 5.0 g of toluene (C 7 H 8 ) dissolved in 225 g of benzene (C 6 H 6 ) and has a density of 0.876 g/mL. Calculate the molarity of toluene in solution.  0.2 M

18.  The physical properties of many solutions differ from those of the pure solvent.  Example:  Pure water freezes at 0 o C but aqueous solutions freeze at lower temperatures.  Pure water boils at 100 o C but aqueous solutions boil at higher temperatures.

19.  We saw in a previous chapter that all liquids have a vapor pressure associated with them.  We find that the vapor pressure of a pure solvent is higher than the vapor pressure of a solution made using that solvent.  Raoult’s Law ◦ P A = X A P o A

20.  Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 o C. Calculate the vapor pressure at 25 o C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of water at 25 o C is 23.8 torr.

21.  The boiling point of a solution is higher than that of the pure solvent.  The change in boiling point is denoted as ΔT b  We can solve for the change in boiling point using the equation…  ΔT b = K b m  Where K b is the molal boiling point elevation constant and m is the molal concentration of the solution.

22.  Automotive antifreeze consists of ethylene glycol (C 2 H 6 O 2 ), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0% by mass solution of ethylene glycol in water.  K b = 0.51 o C/m  K f = 1.86 o C/m

23.  List the following aqueous solutions in order of their expected freezing point: 0.050 m CaCl 2, 0.15 m NaCl, 0.10 m HCl, 0.050 m CH 3 COOH, 0.10 m C 12 H 22 O 11.

24.  Osmosis is the movement of materials across a permeable membrane.  The pressure required to prevent osmosis of a pure solvent is called the osmotic pressure.  Isotonic – Two solutions with identical osmotic pressure.  Hypotonic vs Hypertonic  π = MRT

25.  The average osmotic pressure of blood is 7.7 atm at 25 o C. What molarity of glucose would be isotonic with blood?

26.  A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.25 g of the substance in 40.0 g of CCl 4. The boiling point of the resultant solution was 0.357 o C higher than that of the pure solvent. Calculate the molar mass of the solvent.