Unit 7
Solution Concentrations
and
Colligative Properties
Molarity
• Most widely used concentration unit
• [HCl] means concentration of HCl in mol/L
• Notice volume is total volume of solution
moles of solute
Molarity (M) =
liters of solution
moles
M
L
Molarity Problems
• Read the problem
– Look for moles
– If you have grams of solute, convert to moles
– Look for the volume of the solution…it must
be in liters
• If volume is in mL or cm3, convert to liters
– Once you have moles and liters, plug those
numbers into the formula to get the big “M”
What is the molarity of a solution that has 84.5 grams of
NaOH dissolved in a total volume of 745 mL?
1 Na
1O
1H
22.99
16.00
1.01
40.00 g
How many grams of KCl are needed to make a
500. mL of solution that is .750 M KCl?
1 K 39.10
1 Cl 35.45
74.55 g
Molality
• Molality (m) expresses concentration in terms
of the mass of the solvent.
• Molality is usually used with formulas revolving
around colligative properties.
moles
moles of solute
Molality (m) =
kilograms of solvent
m
kg
Molarity vs Molality
• Molarity and molality differ in two ways:
– Molarity tells you about moles of solute per
volume of the entire solution (solute & solvent)
– Molality tells you about moles of solute per
mass of the solvent
• Keep in mind that one liter of water masses one
kilogram
• So for a dilute solution, the amount of solution is
about the same as the amount of solvent
• So for a dilute aqueous solution, molarity and
molality are basically the same.
Molality Problems
• Read the problem
– Look for moles
– If you have grams of solute, convert to moles
– Look for the mass of the solvent…it must be
in kilograms
• If volume is in grams, convert to kilograms
– Once you have moles and kilograms, plug
those numbers into the formula to get the
small “m”
What is the molality of a solution that contains 46.8
grams of NaCl dissolved in 545 grams of H2O?
1 Na 22.99
1 Cl 35.45
58.44
How many grams of KNO3 are needed to add to
750. g of water to make .450 m KNO3?
1 K 39.10
1 N 14.01
3 O 48.00
101.11
Mole Fraction
• Mole fraction equals the moles of a solute
divided by total moles of solution
• Remember that we used mole fraction to
complete one of the versions of Dalton’s law
moles of substance a
Xa =
total moles of solution
Determine the mole fraction of MgCl2 in a solution
that contains 94.8 g MgCl2 in 345 g of water.
• first find moles of each
• Total the moles
• Find the mole fraction of each
% Solutions
mass of solute
% solute =
mass of solute + mass of solvent
Mass is usually in grams
Make sure that you understand that the denominator
is the total mass of solution.
Determine the percent of NaCl in a solution that
contains 45.5 g NaCl dissolved in 84.3 g of water.
• first find the total mass of the solution
• Then use the formula to find %
Remember Density
• Density is the measure of mass per unit
volume.
– Since density relates mass and volume, it is
useful if you need to convert between
molarity, which deals with volume, and
molality, which deals with mass.
• D = m/v
units usually g/mL or g/cm3
What do you know if given 12.5 % NaCl?
12.5 g NaCl added to 87.5 g H2O
100 g solution
87.5 g solvent H2O
What can you find? (must be given density)
Moles solute; liters of solution; kg solvent
Molarity
Molality
Mole fraction
What do you know if given 3.0 M KCl?
3.0 moles KCl in 1 liter solution
What can you find? (need density again)
find grams KCl
find grams solvent
Molality
Mole fraction
% solution
What do you know if given 2.7 m NaOH?
2.7 moles NaOH in 1 kg water
What can you find? (need density)
grams NaOH
mole fraction
% solution
Molarity
Effect of temperature on solubility
• Most solids increase solubility when
solution is heated…some do not
• All gases decrease solubility when solution
is heated.
A Solubility Curve
Notice a few things:
• The solubility of NH3 (a gas) decreases as
temperature increases.
• Most of the solid substances increase in
solubility as temperature increases.
• However, one solid, Ce2(SO4)3, decreases
in solubility as the temperature increases.
• Looking at the y-axis, you can see that
solubility units are usually grams of the
substance dissolved in 100 grams of water.
Like Dissolves Like
• Polar or ionic solutes dissolve in polar
solvents
– When ionic solutes dissolve they break up
into ions
• Nonpolar solutes dissolve in nonpolar
solvents.
• Polar and nonpolar do not mix
Colligative Properties
• Colligative properties are properties of a
solution that depend on the number of
solute particles in solution.
– The identity of the particles is not important
• There are four colligative properties
– Boiling point elevation
– Freezing point depression
– Vapor pressure lowering
– Osmotic pressure
Van’t Hoff factor (i)
• The Van’t Hoff factor needs to be taken into
account when using any of the four colligative
property formulas.
• The Van’t Hoff factor tells how many ions one
unit of a solute will dissociate into when placed
in solution.
–
–
–
–
C6H12O6 does not dissociate, so i = 1
NaCl dissociates into Na+ and Cl-, so i = 2
CaCl2 dissociates into Ca2+, Cl-, and Cl-, so i = 3
HF partially dissociates, so 1 < i < 2
Boiling Point Elevation
• When a solute is added to a solvent, the
boiling point of the solution increases (that
is, bp is higher than that of the solvent)
∆Tb = ikbm
∆Tb = change in boiling point
i = Van’t Hoff factor
kb = boiling point elevation constant
m = molality of solution
Boiling Point Elevation Problems
• You need to be able to determine “i”
• You need to be able to find the boiling
point elevation constant (from a list)
• You might be given the temperatures
(normal bp and new bp) and asked to find
the molality
• You might be given the normal bp and the
molality and asked to find the new bp
What is the boiling point of a 2.00 m solution of
NaCl in water?
• Need to determine “i”
NaCl has two ions, so the i = 2
• Find the ∆tb
kb = 0.52 ºC/m
∆Tb = ikbm
∆Tb = (2)(0.52 ºC/m)(2.00 m) = 2.08 or 2.1ºC
• Look up the normal bp of water…add the
change in temp
normal bp = 100.0 ºC
New bp = 100.0 ºC + 2.1 ºC = 102.1 ºC
Freezing Point Depression
• When a solute is added to a solvent, the
freezing point of the solution decreases
(that is, fp is lower than that of the solvent)
∆Tf = ikfm
∆Tf = change in freezing point
i = Van’t Hoff factor
kf = freezing point depression constant
m = molality of solution
Freezing Point Depression Problems
• You need to be able to determine “i”
• You need to be able to find the freezing
point depression constant (from a list)
• You might be given the temperatures
(normal fp and new fp) and asked to find
the molality
• You might be given the normal fp and the
molality and asked to find the new fp
What is the freezing point of a 2.00 m solution of
CaCl2 in water?
• Need to determine “i”
CaCl2 has three ions, so the i = 3
• Find the ∆tf
kf = 1.86 ºC/m
∆Tf = ikfm
∆Tf = (3)(1.86 ºC/m)(2.00 m) = 11.16 or 11.2ºC
• Look up the normal fp of water…subtract the
change in temp
normal fp = 0.0 ºC
New fp = 0.0 ºC – 11.2 ºC = -11.2 ºC
Vapor Pressure Lowering
• Called Raoult’s Law – when a solute is
added to a solution, the vapor pressure
will decrease.
P = XPº
P = vapor pressure of solution
Pº = vapor pressure of solvent
X = mole fraction of solvent
Vapor Pressure Problems
• You need to determine the mole fraction of the
solvent
– Find moles of solute, find moles of solvent
– Add the moles together to get total moles
– Then use the formula
• Then use the formula P = XPº
– Pº is the normal vapor pressure of the solvent
Calculate the vapor pressure of a solution made by
dissolving 218 g of glucose (C6H12O6) in 460. g of water.
The vapor pressure of pure water at 30ºC is 31.8 mmHg.
• Determine the moles of solute and solvent
218 g C6H12O6
460. g H2O
1 mol C6H12O6 = 1.21 mol C6H12O6
180 g C6H12O6
1 mol H2O
= 25.5 mol H2O
18.02 g H2O
• Determine the mole fraction of solute
total moles = 1.21 mol + 25.5 mol = 26.7 mol
25.5
H O .955
2
26.7
• Plug values into formula
P = XPº = (.955)(31.8 mmHg) = 30.4 mmHg
Osmotic Pressure
• When a pure solvent and a solution are separated by a
membrane that only allows solvent to pass through, the
solvent will try to pass through the membrane to dilute
the solution. The greater the concentration of solute in
the solution, the greater the osmotic pressure.
Ώ =λ = MRTi
Ώ =λ = osmotic pressure (atm)
M = molarity
R = gas constant = .0821 (L·atm)/(mol·K)
T = Kelvin temperature
i = Van’t Hoff factor
What do you need to know for test?
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Like dissolves like
Affect of temperature on solubility
Molarity
Molality
Mole fractions
% solutions
Colligitave properties (math)