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Chapter 12: Solutions
I) Mixtures & Solutions: A Brief Review
1. Mixtures are combinations of two or more pure substances. Mixtures are
classified as being heterogeneous or homogeneous, depending on their
appearance.
2. Heterogeneous mixtures characteristically have more than one phase;
therefore, they are nonuniform. Homogeneous mixtures have only one phase;
meaning that the mixing of components is uniform.
3. Solutions
[see Tables 12.1 & 12.3]
A) A solution is a homogeneous mixture composed of solute & solvent.
B) The solute is the substance being dissolved and the solvent is the
dissolving environment. In general chemistry, the solvent is almost always
water, but others do exist. More about this in future courses.
C) Aqueous solutions refer to solutions that contain water as the solvent.
II) Expressing Solution Concentration
[ see Table 12.5]
1. Molarity (M)
Molarity (M) = moles solute / L solution
2. Mole Fraction
Mole fraction (X) = no. moles of a component / total moles of solution
When liquids are mixed, the liquid present in largest amount is
called the solvent.
3. Mass (Weight) Percent
Mass Percent = (mass of solute / mass solution ) * 100 %
Two other units commonly seen in biology & environmental science
Parts per million (ppm) = (mass component / total mass) x 106
Parts per billion (ppb) = (mass component / total mass) x 109
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4. Molality (m)
Molality (m) = moles of solute / kg of solvent
In very dilute solutions, the magnitude of the molality and the
molarity are almost the same.
Molality is temperature independent, unlike molarity.
Example Problem
An alcohol solution is prepared by mixing 1.00 g of ethanol with 100.0 g water.
Calculate the molarity, mass % ethanol, mole fraction ethanol, and the molality of the
solution.
Given:
Let EtOH = Ethanol (C2H5OH)
Solution composition 1.00 g EtOH + 100.0 g water
Final Mass (Volume) 101.0 g solution = 101 mL Density (water) = 1.00 g/mL
Unknown:
? M EtOH
? mass % EtOH
? mole fraction EtOH
? m EtOH
Rel. Info & Solution to 4 part question:
Molar Mass C2H5OH = (2 * 12.01 g/mole) + (6 * 1.008 g/mole) + 16.00 g/mole
Molar Mass C2H5OH = 46.07 g/mole
M = molarity = moles solute / L solution
moles solute = 1.00 g C2H5OH * (1.00 mole C2H5OH / 46.07 g C2H5OH)
moles solute = 2.17 x 10-2 moles C2H5OH
Volume = 101 mL * ( 1 L / 1000 mL) = 0.101 L
M = 2.17 x 10-2 moles C2H5OH / 0.101 L = 0.215 M = Answer
(check)
mass % = (mass solute / mass solution) * 100 %
mass % = ( 1.00 g C2H5OH / 1.00 g C2H5OH + 100.0 g H2O) * 100 %
mass % = 0.990 % = Answer
(check)
mole fraction () = # moles of component X / total # moles = nX / ntotal
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n C2H5OH = 2.17 x 10-2 moles C2H5OH
n H2O = 100.0 g H2O * ( 1 mole H2O / 18.02 g H2O) = 5.55 moles H2O
n total = (2.17 x 10-2 moles C2H5OH + 5.55 moles H2O) = 5.57 moles
X = 2.17 x 10-2 moles C2H5OH / 5.57 moles = 0.00390 = Answer (check)
molality = moles of solute / kg of H2O
m = 2.17 x 10-2 moles C2H5OH / [ (100.0 g H2O * (1 kg H2O / 1000 g H2O) ]
m = 2.17 x 10-2 moles C2H5OH / 0.1000 kg H2O = 0.217 m = Answer
(check)
-----------------------------------------------------------------------------------------------------------III) Energetics of Solution Formation
1) Like Dissolves Like
A) Three types of interactions go into forming a solution:
a. solute-solute
b. solvent-solvent
c. solute-solvent
B) Polar solvents dissolve polar solutes.
C) Nonpolar solvents dissolve nonpolar solutes.
2) How does a Solution Form ?
[See Table 12.2 & Figure 12.6]
A) Breaking up the solute into individual components (expanding the
solute). This requires energy (endothermic).
B) Overcoming the intermolecular forces in the solvent to make room for
the solute (expanding the solvent). This is also an endothermic step.
C) Allowing the solute and solvent to interact to form the solution. This
step is exothermic.
3) Enthalpy (Heat) of Solution
A) Each of the three processes mentioned above has an associated
enthalpy value (H).
B) Hsoln is the sum of three processes:
Hsoln = H1 + H2 + H3
IV) Factors Affecting Solubility
1. Solubility is defined as the amount of solute per unit solvent needed to form a
saturated solution. A solution is saturated when no more solute can dissolve
into a given quantity of solvent.
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2. The solubility of a solute in a given solvent is a physical property
characteristic of that substance.
3. Some factors that affect solubility of substances.
A) Structure
- hydrophobic - water fearing (nonpolar substance)
- hydrophilic - water loving (polar substances)
B) Temperature
- As the temperature increases, the rate of dissolution increases.
- General Trends (Usually holds true, but not always)
i. As temperature increases, solubility of solids increases.
ii. As temperature increases, solubility of gases decreases.
C) Pressure
- No appreciable effect on liquids or solids; however, pressure
has a profound effect on gases
- Henry’s Law
P = kC
V) Colligative Properties
1. Colligative properties depend ONLY on the number of solute particles that are
present in a solution, not on the identity of the solute particles.
2. Examples of colligative properties are vapor pressure, boiling point elevation,
freezing point depression, and osmotic pressure.
A) Vapor Pressure
- Definition:
Pressure of a vapor over a liquid @ equilibrium.
- Presence of a nonvolatile solute lowers the vapor pressure of
solvent by reducing the tendency of solvent molecules to escape
from solution.
- Raoult’s Law Psoln = Xsolvent Posolvent
Psoln = observed vapor pressure of the solution
Xsolvent = mole fraction of solvent
Posolvent = vapor pressure of pure solvent
- NonIdeal Solutions (Liquid-liquid solutions) one uses
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Ptotal = PA + PB = XAPoA + XBPoB
B) Boiling point elevation
Tb = kb msolute
Tb is change in boiling temperature
kb is molal boiling point elevation constant
m refers to molal concentration (molality NOT molarity)
C) Freezing point depression
Tf= kf msolute
Tf is change in freezing temperature
kf is molal freezing point depression constant
m refers to molal concentration (molality NOT molarity)
Example Problem:
A solution is prepared by dissolving 5.6 g sucrose (C12H22O11) in 185 g water.
Calculate the boiling point of this solution. Sucrose is a nonelectrolyte with a
molar mass of 342 g/mole.
Given: Solution containing 5.6 g sucrose (C12H22O11) in 185 g water.
Unknown: ? b.p. of solution
Rel. Info:
Tb = kb msolute
m = molality = moles solute / kg water
kb = 0.51 oC kg / mole
Solution:
5.6 g C12H22O11 * ( 1 mole C12H22O11 / 342 g C12H22O11) = 0.0163 moles C12H22O11
m = 0.0163 moles C12H22O11 / 0.185 kg water = 0.0885 m
Tb = kb msolute

Tb = (0.51 oC kg / mole * 0.0885 mole / kg ) = 0.045 oC
Now since pure water boils at 100.00 oC (1 atm P), the boiling point of the solution is
100.045 oC which is the answer. Note that the boiling temperature is elevated (it is higher
than the pure solvent water, which is what we expect).
(check looks good )
Review Examples in Text
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D) Osmotic pressure - minimum pressure that stops osmotic flow
from occuring.
- What is osmotic flow (osmosis)?
Process that allows solvent to flow into solution through a
semipermeable membrane is called osmosis.
Solution & pure solvent separated by semipermeable
membrane (allows solvent but not solute molecules to pass
through).
 = MRT
 = osmotic pressure (atm)
M = concentration of solution in molarity
R = gas law constant (0.08206 L atm / mol K)
T = Kelvin temperature
- Applications of Osmosis
1. dialysis of blood
2. desalination (reverse osmosis)
VI) Colligative Properties of Electrolyte Solutions
1. Van’t Hoff Factor (i)
A) Factor used to account for electrolytes dissociating in solution since
colligative properties are dependent upon the number of solute particles in
solution.
B) One need only keep track of the number of ions in solution to find out
the van’t Hoff factor.
C) Once the van’t Hoff factor is known, it can be used in the equations for
bp, fp, or osmotic pressure.
2. Example Problem:
A student prepared several solutions each with a concentration of 0.25 m.
The solutes used were KCl, CaBr2, NaNO3, & K3PO4. Which solution
would yield the highest boiling point elevation?
Answer
K3PO4