Transcript Chapter 18 Solutions

Chapter 16
Solutions
Killarney High School
Section 16.1
Properties of Solutions
 OBJECTIVES:
– Identify
the factors that
determine the rate at which a
solute dissolves.
Section 16.1
Properties of Solutions
 OBJECTIVES:
– Calculate
the solubility of a
gas in a liquid under various
pressure conditions.
Solution formation
 Nature
of the solute and the solvent
– Whether a substance will dissolve
– How much will dissolve
 Factors determining rate of solution...
– stirred or shaken (agitation)
– particles are made smaller
– temperature is increased
 Why?
Making solutions
 In
order to dissolve, the solvent
molecules must come in contact with
the solute.
 Stirring moves fresh solvent next to
the solute.
 The solvent touches the surface of
the solute.
 Smaller pieces increase the amount
of surface area of the solute.
Solution Formation Flash
Temperature and Solutions
 Higher
temperature makes the
molecules of the solvent move
around faster and contact the
solute harder and more often.
– Speeds up dissolving.
 Usually increases the amount
that will dissolve (exception is
gases)
How Much?
 Solubility-
The maximum amount of
substance that will dissolve at a specific
temperature (g solute/100 g solvent)
 Saturated solution- Contains the maximum
amount of solute dissolved
 Unsaturated solution- Can still dissolve more
solute
 Supersaturated- solution that is holding more
than it theoretically can; seed crystal will
make it come out
Liquids
 Miscible
means that two liquids
can dissolve in each other
– water and antifreeze, water
and ethanol
 Partially miscible- slightly
– water and ether
 Immiscible means they can’t
– oil and vinegar
Solubility?
 For
solids in liquids, as the
temperature goes up-the solubility
usually goes up (Fig. 16.5, p.373)
 For gases in a liquid, as the
temperature goes up-the solubility
goes down (Fig. 16.6, p.374)
 For gases in a liquid, as the
pressure goes up-the solubility
goes up (Fig. 16.6, p.373)
Gases in liquids...
 Henry’s
Law - says the solubility
of a gas in a liquid is directly
proportional to the pressure of
the gas above the liquid
– think of a bottle of soda pop,
removing the lid releases pres.
 Equation:
S1
S2
=
P1
P2
Cloud seeding
 Ever
heard of seeding the clouds
to make them produce rain?
 Clouds- mass of air
supersaturated with water vapor
 Silver Iodide (AgI) crystals are
dusted into the cloud
 The AgI attracts the water,
forming droplets to attract others
Section 16.2
Concentration of Solutions
 OBJECTIVES:
– Solve
problems involving the
molarity of a solution.
Section 16.2
Concentration of Solutions
 OBJECTIVES:
– Describe
how to prepare dilute
solutions from more
concentrated solutions of
known molarity.
Section 16.2
Concentration of Solutions
 OBJECTIVES:
– Explain
what is meant by
percent by volume [ % (v/v) ],
and percent by mass [ % (m/v) ]
solutions.
Concentration is...
a
measure of the amount of solute
dissolved in a given quantity of solvent
 A concentrated solution has a large
amount of solute
 A dilute solution has a small amount of
solute
– thus, only qualitative descriptions
 But, there are ways to express solution
concentration quantitatively
Molarity - most important
 The
number of moles of solute in 1
Liter of the solution.
 M = moles/Liter; such as 6.0 molar
 What is the molarity of a solution
with 2.0 moles of NaCl in 250 mL of
solution?
 Sample 16-2, page 378
Making solutions
 Pour
in a small amount of solvent
 Then add the solute (to dissolve it)
 Carefully fill to final volume.
– Fig. 18-10, page 509
 Also: M x L = moles
 How many moles of NaCl are
needed to make 6.0 L of a 0.75 M
NaCl solution?
Making solutions
 10.3
g of NaCl are dissolved in a
small amount of water, then
diluted to 250 mL. What is the
concentration?
 How many grams of sugar are
needed to make 125 mL of a
0.50 M C6H12O6 solution?
Dilution
Adding water to a solution
 The
Dilution
number of moles of solute doesn’t
change if you add more solvent!
 The # moles before = the # moles after
 M1 x V 1 = M 2 x V 2
 M1 and V1 are the starting
concentration and volume.
 M2 and V2 are the final concentration
and volume.
 Stock solutions are pre-made to known
Molarity
Practice
 2.0
L of a 0.88 M solution are diluted
to 3.8 L. What is the new molarity?
 You have 150 mL of 6.0 M HCl. What
volume of 1.3 M HCl can you make?
 Need 450 mL of 0.15 M NaOH. All
you have available is a 2.0 M stock
solution of NaOH. How do you make
the required solution?
Percent solutions...
 Percent
means parts per 100, so
 Percent by volume:
= Volume of solute x 100%
Volume of solution
 indicated %(v/v)
 What is the percent solution if 25 mL
of CH3OH is diluted to 150 mL with
water?
 Percent
Percent solutions
by mass:
= Mass of solute(g)
x 100%
Volume of solution(mL)
 Indicated %(m/v)
 More commonly used
 4.8 g of NaCl are dissolved in 82 mL of
solution. What is the percent of the
solution?
 How many grams of salt are there in 52
mL of a 6.3 % solution?
Section 16.3
Colligative Properties of
Solutions
 OBJECTIVES:
– Explain
on a particle basis why
a solution has a lower vapor
pressure than the pure solvent
of that solution.
Section 16.3
Colligative Properties of
Solutions
 OBJECTIVES:
– Explain
on a particle basis why
a solution has an elevated
boiling point, and a depressed
freezing point compared with
the pure solvent.
Colligative Properties
Depend only on the number of
dissolved particles
Not on what kind of particle
Vapor Pressure decreased
 The
bonds between molecules keep
molecules from escaping.
 In a solution, some of the solvent is
busy keeping the solute dissolved.
 Lowers the vapor pressure
 Electrolytes form ions when they are
dissolved = more pieces.
 NaCl  Na+ + Cl- (= 2 pieces)
 More pieces = bigger effect
Boiling Point Elevation
 The
vapor pressure determines
the boiling point.
 Lower vapor pressure = higher
boiling point.
 Salt water boils above 100ºC
 The number of dissolved
particles determines how much,
as well as the solvent itself.
Freezing Point Depression
 Solids
form when molecules make
an orderly pattern.
 The solute molecules break up the
orderly pattern.
 Makes the freezing point lower.
 Salt water freezes below 0ºC
 How much depends on the number
of solute particles dissolved. Simulation
Section 16.4
Calculations Involving
Colligative Properties
 OBJECTIVES:
– Calculate
the molality and
mole fraction of a solution.
Section 16.4
Calculations Involving
Colligative Properties
 OBJECTIVES:
– Calculate
the molar mass of a
molecular compound from the
freezing point depression or
boiling point elevation of a
solution of the compound.
Molality
a
new unit for concentration
 m = Moles of solute
kilogram of solvent
 m = Moles of solute
1000 g of solvent
 What is the molality of a solution
with 9.3 mole of NaCl in 450 g of
water?
Why molality?
 The
size of the change in boiling
point is determined by the molality.
 DTb = Kb x m x n
 DTb is the change in the boiling point
 Kb is a constant determined by the
solvent (Table 16.2, page 387).
 m is the molality of the solution.
 n is the number of pieces it falls into
when it dissolves.
What about Freezing?
 The
size of the change in freezing
point is also determined by molality.
 DTf = -Kf x m x n
 DTf is the change in freezing point
 Kf is a constant determined by the
solvent (Table 16.3, page 388).
 m is the molality of the solution.
 n is the number of pieces it falls into
when it dissolves.
Problems
 What
is the boiling point of a
solution made by dissolving 1.20
moles of NaCl in 750 g of water?
 What is the freezing point?
 What is the boiling point of a
solution made by dissolving 1.20
moles of CaCl2 in 750 g of water?
 What is the freezing point?
Mole fraction
 This
is another way to express
concentration
 It is the ratio of moles of solute to
total number of moles of solute +
solvent (Fig. 16-19, p.386)
na
X=
na + n b
Sample 18-8,
page 521
Molar Mass
 We
can use changes in boiling
and freezing to calculate the
molar mass of a substance
 Find: 1) molality 2) moles, and
then 3) molar mass
 Sample 16-10, page 388
A solution of 7.50g of a nonvolatile compound in 22.60 g of water boils
at 100.78 degrees celsius at 1 atm. What is the molar mass of the solute?
1. Use the BP elevation to calculate the molality of the solution.
Tb = Kb x m x n
thus
m=
tb
Kb
Kb = 0.512 C x kg of water
Mol of solute
0.78 C
m=
0.512 C / m
= 1.5 m
2. Now calculate the moles of solute I the solution
1.5 mol solute
0.0226 Kg
0.034 mol solute
Kg of water
3. Use the # of moles of solute and its mass to get the molar mass
Molar mass =
mass of solute
Moles of solute
Molar mass =
7.50 g
0.0344 mol
Molar mass = 0.022 g/mol
Key Equations
 Note
the key equations to solve
problems in this chapter.