Colligative Properties
Solutions, Part II, the effect of the number of particles
Colligative Properties-number of particles is key, not the nature of the particles
Properties that only depend on the amount of solute dissolved and not on the identity of the solute are called colligative properties. The properties affected in this way include boiling point, melting point and osmotic pressure. The mechanism that causes the change is the lowering of the vapor pressure of the solvent.
Therefore, the magnitude of the effect will depend on the nature of the solvent but not on the nature of the solute. Solute particles interact with solvent particles. By the way, the solute has to be nonvolatile (does not become a gas easily) as this would seriously mess up the situation since it is vapor pressure lowering of the solvent we are talking about here. If the solute is interacting with the solvent, then fewer solvent particles are available to escape into the vapor phase and the vapor pressure is lowered at any given temperature.
Freezing Point Depression
∆tf = Kf m n
∆tf = the CHANGE in the freezing point not the new freezing point
Kf = the molal freezing point constant and is different for each solvent (i.e., look it up in a table)
m = molality (mol solute/Kg solvent)
n = number of solute particles
Boiling Point Elevation
∆tb = Kb m n
∆tf = the CHANGE in the boiling point not the new boiling point
Kf = the molal boiling point constant and is different for each solvent (i.e., look it up in a table)
m = molality (mol solute/Kg solvent)
n = number of solute particles
Osmotic Pressure
The best way to understand osmosis is remembering that nature does not like two different substances in contact with one another to remain unmixed especially if the substances are liquids (including solutions) and gases. Because solid particles are held in fixed positions, random mixing usually does not occur. Osmosis is a more biological phenomenon as all cell membranes are "semi-permeable" (allowing some particles to pass through while blocking others) and, therefore, osmosis can occur in biological systems and does. If you drink an excess of water, you can "swamp" your cells with too much "solvent" and dilute out your electrolytes enough to even cause death. This is an example of osmosis. Have you ever tried the experiment where you put a carrot in really salty water? What happens to the carrot is that because the solution in the cells of the carrot is more dilute than the salt water, the cells expel water to try to bring the salt concentration on both sides of the cell wall to about the same level. The result is a shriveled carrot.
The pressure required to stop osmosis, known as osmotic pressure, π, is
π = MRT where M is the molarity of the solution
parts per million (ppm)
parts per billion (ppb)
Practice Problems
1. What is the mass percentage of iodine (I2) in a solution containing 0.035 mol I2 in 125 g of CCl4? (b) Seawater contains 0.0079 g Sr2+ per kilogram of water. What is the concentration of Sr2+ in ppm?
2. (a) What is the molality of a solution formed by dissolving 1.12 mol of KCl in 16.0 mol of water? (b) How many grams of sulfur (S8) must be dissolved in 100 g of naphthalene (C10H8) to make a 0.12 m solution?
3. Arrange the following aqueous solutions, each 10% by mass in solute, in order of increasing boiling point: glucose (C6H12O6), sucrose (C12H22O11), and sodium nitrate (NaNO3).
4. Seawater contains 3.4 g of salts for every liter of solution. Assuming that the solute consists entirely of NaCl (over 90% is), calculate the osmotic pressure of seawater at 20°C.
5. Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1°C. What is the approximate molar mass of lauryl alcohol?
Solution Stoichiometry
Most chemistry is done using solutions. Therefore, most stoichiometry involving chemical reactions involves concentrations of solutions. There is really no difference in how you work these other than how you get to moles and the conversion at the end (instead of g, it may be volume or M).
rewrite the big M! Any time you see M in a problem, rewrite it to help you get into moles. For example, 2.5 M NaCl is 2.5 mol NaCl/L of solution. You are already in moles, multiply by volume given and you are in moles.
Practice
6. What is the molarity of ZnCl2 (aq) that forms when 15.0 g of zinc completely reacts with CuCl2 (aq) according to the following reaction? (Assume a final volume of 175 mL).
Zn (s) + CuCl2 (aq) --> ZnCl2 (aq) + Cu (s)
7. Given the following equation:
Ca(OH)2(s) + 2 HCl(aq)---> CaCl2(aq) + 2 H2O(l)
a) What type of chemical reaction is taking place?
b) How many liters of 0.100 M HCl would be required to react completely
with 5.00 grams of calcium hydroxide?
c) If I combined 15.0 grams of calcium hydroxide with 75.0 mL of 0.500 M
HCl, what is the limiting reagent from the reaction?
d) How many grams of calcium chloride would be formed in (c)?
e) How many grams of the excess reagent will be left over after the reaction
in (c) is complete?
8. How many milliliters of 0.35 M calcium hydroxide are needed to neutralize 75 mL of 0.20 M nitric acid?
The video below consists of 12 additional practice problems found below the video link.
Practice from video above
1. How many milliliters of 0.75 M sodium hydroxide are needed to neutralize 275 mL of 0.50 M sulfuric acid?
2. How many grams of lead (II) nitrate are needed to fully react 23.5 mL of 0.55 M sodium chloride in the precipitation of lead (II) chloride?
3. How many mL of 0.40 M barium chloride are needed to react 175 mL of 1.50 M ammonium phosphate in the precipitation of barium phosphate?
4. When 53 mL of 0.75 M cobalt (III) nitrate are added to a sodium sulfate solution, how many grams of cobalt (III) sulfate can be precipitated?
5. How many grams of aluminum hydroxide will be neutralized by 45.3 mL of 0.55 M HCl ?
6. Write the balanced equation and the net ionic equation for each of the following solution mixtures. If no precipitate forms, write “NR”
a. Ammonium Sulfate and Barium Nitrate
b. Lead(II) Nitrate and sodium chloride
c. Sodium Phosphate and Potassium nitrate
d. Sodium bromide and rubidium chloride
e. Copper(II) chloride and Sodium Hydroxide
7. Write a balanced chemical equation for each of the following & determine the type of each reaction.
a. Al(OH)3 + H2SO4 →
b. C2H4 + O2 →
c. mercury(II) oxide → mercury + oxygen
d. potassium chloride + silver nitrate →
e. Ca + HCl → hydrogen + calcium chloride
f. C2H5OH + O2 →
g. hydrogen gas + Fe2O3 → iron + steam
h. carbon + oxygen → carbon monoxide
i. Al2(SO4)3 + Ca(OH)2 →
8. When 9.00g of Al react with an excess of H3PO4, 30.0g of AlPO4 are produced. What is the percentage yield of this reaction?
2Al(s) + 2H3PO4(aq) → 2AlPO4(aq) + 3H2(g)
9. Referring to the problem above, what is the limiting reagent when 100g of H3PO4 react with 25.0 g Al? How much of the excess reactant will be left over?
10. When octane (C8H18) is burned in oxygen, carbon dioxide and water are produced. If 320 g of octane is burned and 392 g of water is recovered, what is the percent yield of the experiment? The equation for the reaction is
2 C8H18 + 25 O2→16 CO2 + 18 H20.
11. Determine the percent yield for the reaction between 2.80 g Al(NO3)3 and excess NaOH if 0.996g Al(OH)3 is recovered.
12. A student places an iron nail with a mass of 2.32g into a flask of CuSO4. The nail reacts completely, leaving a quantity of copper metal in the bottom of the flask. The student finds the mass of the recovered copper to be 2.51g. The equation for this reaction is
Fe + CuSO4→FeSO4 + Cu.
(a) What is the expected yield? (b) What is the percent yield?
What can you do to help your understanding?
Dr. Melinda Oliver
Email: melinda.oliver@ung.edu
Location: Gainesville, GA, USA
Phone: (678)769-5429