Solutions are often called ______________________________; no matter where you sample them, they have the same ______________________ but they are composed of ____________ or more materials.
I. Basic Terms
Substance __________ ___________________
Can be individual _______________, ____________________ or ___________
Canít be ____________________
________________________ light (but may absorb some colors)
Substance ____________________ the _________________
1. _________________: _____________ and those substances
ex. alcohol, ammonia
2. ________________________: __________________ with ____________
ex. oil, paint thinner, mineral spirits, dry cleaning fluids
II. Types of Solutions
SOLUTE SOLVENT EXAMPLES
Solid ______________ salt in water
iodine in alcohol
Solid _____________ alloys such as:
14K gold (Cu/Au)
Liquid Liquid alcohol in water
glycols in water
Solid metal tooth fillings
(Hg in Ag)
__________ Liquid Oxygen in water
soda = CO2 in water
Gas Gas Air
III. Degrees of Solubility
How much solute is dissolved in the solvent?
______________________ - could hold more solute
____________________ - holding as much solute as is possible at this temperature
______________________ - holding more solute than is normal at a certain temperature (do this by heating the saturated solution to get extra solute dissolved then carefully cool it)
IV. Factors that Influence the Rate of Dissolving
- important for ____________ and ____________
ex. sugar: __________ __________ ___________
dissolving: __________--------->__________ -----------> __________
Crushing it exposes more ______________________ for the ____________ to attack.
- affects only ______________ ex. can of soda
Suppose we make five ___________________ solutions using the same solute in 100 mL of water, but heat the water each time. We find that __________ dissolves as the water gets hotter.
From this data we construct a _________________ _______________
Any solution on the curve is _________________
What is the solubility at 60oC? ___________________
A solution at 60oC that contains 80 g is _______________________: any solution ________________ the line is ________________________
A solution at 60oC that contains 2 grams is ____________________
IV. Concentrations of Solutions
Example 1: Suppose you have 1.00 mol of sucrose (or about 342.3 grams) and you mix it into some water. It dissolves and makes sugar water. You keep adding water, dissolving and stirring until all the solid is gone. There is exactly 1 liter of solution.
What is the molarity of this solution?
Example 2: Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What is the molarity?
Example 3: What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?
Example 4: Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What is the molarity of the solution?
Example 5: Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.
Example 6: 80.0 grams of glucose (C6H12O6; molar mass = 180.0 g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?
1) Calculate the molarity of 75.0 grams of MgCl2 is dissolved in 500.0 mL of solution.
2) 100.0 grams of sucrose is dissolved in 1.50 L of solution. What is the molarity?
3) 49.8 grams of KI is dissolved in enough water to make 1.00 L of solution. What is the molarity?
Example 1: Suppose you have 1.00 mole of sucrose (about 342.3 grams) and you mix it with exactly 1.00 liter of water. What is the molality?
Example 2: Suppose you had 2.00 mol of solute dissolved into 1.00 L of solvent. Whatís the molality?
Example 3: What is the molality when 0.75 mol is dissolved in 2.50 L of solvent?
Example 4: Suppose you had 58.4 grams of NaCl and you dissolve it in exactly 2.00 kg of pure water. What would the molality of this solution be?
Example 5: Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL of pure water.
Example 6: 80.0 grams of glucose is dissolved in 1.00 kg of water. Calculate the molality.
1) Calculate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 grams of solvent.
2) 100.0 grams of sucrose is dissolved in 1.50 L of water. What is the
3) 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?
V. Colligative Properties
______________________ - a property of a solution that depends on the number of solute particles
A. Boiling Point Elevation
1. Fact: A solution will boil at a _______________ temperature than a pure solvent.
Why? The vapor pressure of a solution is ____________ than that of the pure solvent at any temperature. A _____________ temperature is needed to raise the solutionís vapor pressure to equal external pressure.
Tb = i Kb m
a. Tb = change in the boiling point (units: oC)
if the Tb is the normal boiling point of the pure solvent
then, Tb(solution) = Tb + Tb
b. i = vanít Hoff factor (no units)
the change in boiling point ( Tb) is directly related to the
number of solute particles that are in the solution
for sugar: C12H22O11 (s) --------> C12H22O11 (aq)
* substances that do not dissociate in solution, like sugar, have
i = _______
for NaCl: NaCl (s) -------> Na+ (aq) + Cl- (aq)
* substances which dissociate into two ions, like NaCl, KCl, and
NaOH, have i = ______
for MgCl2: MgCl2 (s) ------> Mg2+ (aq) + 2 Cl- (aq)
* substances which dissociate into three ions, like MgCl2 and
Mg(OH)2 have i = ______
c. Kb is the molal boiling point elevation constant (units: oC/m or
oC . mol/kg)
every solvent has a unique Kb
ex. Kb (H2O) = 0.52 oC/m
d. m is the molality
1) Which solution has a higher boiling point, 1 mol of Al(NO3)3 in 1000 g of water or 1.00 mol of KCl in 1000 g of water? Explain.
2) What is the boiling point of a 1.50 m NaCl solution?
3) What is the boiling point of each solution?
a) 0.50 mol glucose in 1000 g of water
b) 1.50 mol NaCl in 1000 g of water
B. Freezing Point depression
1. Fact: A solution will freeze at a _____________ temperature than the pure solvent.
Why? The vapor pressure of a solution is ___________ than that of the solvent at any temperature, the solution freezes at a temperature than the solvent.
Tf = i Kf m
Tf = change in freeing point
so... Tf (solution) = Tf (solution) - Tf
i = vanít Hoff factor
Kf = molal freezing point depression constant (units: oC/m)
m = molality
1) What is the freezing point of each solution?
a) 1.40 mol Na2SO4 in 1750 g of water
b) 0.60 mol MgSO4 in 100 g of water
2) Determine the freezing points of each 0.20 m solution: