CHE1000 module
IMF and Solutions
Intermolecular forces are forces that exist between molecules. It is important to distinguish
between intermolecular forces and intramolecular forces. Intramolecular forces are the forces
that hold atoms together within a molecule. e.g O-H bond in H2O. Intermolecular forces are
much weaker than the intramolecular forces of attraction but are important because they
determine the physical properties of molecules like their boiling point, melting point, density,
and enthalpies of fusion and vaporization. In covalent bonds electrons are being shared. If
electronegativity difference is appreciable, eg. HCl, electron density is more shifted towards Cl giving rise to dipole moment.
Types of intermolecular forces that exist between molecules
1. Ion-ion interaction: It occurs in ionic compounds like NaCl, MgF2 etc. Ion interations are
strong because of electrostatic attraction. Force of attraction depends on the charge of the ions.
Na+1 Cl-
= +1x-1
Mg2 + Cl2-
(+2)x2x(-1)
2. Dipole-dipole interactions: These forces occur when the partially positively charged part of a
molecule interacts with the partially negatively charged part of the neighboring molecule.
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The prerequisite for this type of attraction to exist is partially charged ions like in polar covalent
bonds in hydrogen chloride, HCl. This kind of D-D occurs with molecules with permanent dipole
moment. This is a relatively weak force and becomes weaker as the distance increase. Molecules
like H2O, NH3, R-OH, PCl3 (tetrahedral shape) exhibit dipole-dipole interaction
Figure: Positive and negative ends aligned so that they can interact
Hydrogen bonding: This is a special kind of interaction that occurs specifically between a
hydrogen atom bonded to an oxygen, nitrogen, or fluorine atom of another molecule. Hydrogen
bonding is a relatively strong force of attraction between molecules, and considerable energy is
required to break hydrogen bonds. This explains the exceptionally high boiling points and
melting points of compounds like H2O or HF and biological macromolecules. H-bonds are strong
forces, depends on the polarity of molecule and closest approach. Glycerol has very high
capacity to form H-bonding, which resists its motion also. So glycerol is viscous. Viscosity
reduces with temperature. Another example is grease.
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The most powerful intermolecular force influencing neutral (uncharged) molecules is the
hydrogen bond. If we compare the boiling points of methane (CH4) -161ºC, ammonia (NH3) 33ºC, water (H2O) 100ºC and hydrogen fluoride (HF) 19ºC, we see a greater variation for these
similar sized molecules than expected from the data presented above for polar compounds. This
is shown graphically in the following chart. Most of the simple hydrides of group IV, V, VI &
VII elements display the expected rise in boiling point with molecular mass, but the hydrides of
the most electronegative elements (nitrogen, oxygen and fluorine) have abnormally high boiling
points for their mass.
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When carboxylic acids like acetic acid are dissolved in benzene dimerization of acid occurs due
to H bonding as shown in figure. Because of this association, solute particles appear to have a
relative molecular mass twice as large as expected.
Figure:
London dispersion forces: These are the weakest of the intermolecular forces and exist between
all types of molecules, whether ionic or covalent, polar or nonpolar. The more electrons a
molecule has, the stronger the London dispersion forces are. For example, bromine, Br2 has more
electrons than chlorine, Cl2 so bromine will have stronger London dispersion forces than
chlorine, resulting in a higher boiling point for bromine. Also, the breaking of London dispersion
forces doesn’t require that much energy, which explains why nonpolar covalent compounds like
CH4, oxygen, and nitrogen, which only have London dispersion forces (Vander Wall forces) of
attraction between the molecules freeze at very low temperatures. Straight-chain alkanes are able
to pack and layer each other better than their branched counterparts. This allows greater
intermolecular forces, which raises the melting point since it will take more energy to disperse
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the molecules into a liquid. Similarly, even-numbered alkanes stack better than odd-numbered
alkanes, and will therefore have higher melting points.
Although we assume that electrons in a non-polar molecule are uniformly distributed, a
momentary nonsymmetrical electron distribution can induce a temporary dipole. This
instantaneous polarization can induce a dipole on the neighboring molecule. (Figure)
Figure:
The order of strength of intermolecular forces is; London dispersion forces < dipole-dipole <
Hydrogen bonding < ion-ion interaction.
So a substance that contains Hydrogen bonding will have a far greater boiling point than one
which contains London dispersion forces.
Problem:
1. Identify the different kinds of intermolecular forces present in the following molecules.
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H2S:
London dispersion force-by default every compound will have this force of
attraction between molecules. Since there are polar bonds, dipole-dipole attraction
also is there.
CH3OH :
London dispersion force, dipole-dipole attraction and hydrogen bonding
C2H6:
London dispersion forces: it’s a nonpolar covalent compound-and no other
intermolecular attractions
2. Why are dipole-dipole forces stronger than London dispersion forces?
London dispersion forces are between molecules that for a short time have a very small dipole in
one direction, and in the next moment it may be in a different direction. Thus, they are weaker
than dipole-dipole forces.
Effect of intermolecular forces on physical properties
Boiling point is the temperature at which a liquid boil at 1 atmospheric pressure. A liquid boil
when its vapor pressure equals the atmospheric pressure. Vapor pressure is the pressure of gas
phase of a substance that exists as a liquid or a solid at 25oC and 1 atm. The stronger an
intermolecular force, the higher the boiling point of the substance will be. This is because
stronger intermolecular bonds require more energy to break. As this energy is supplied in the
form of heat when boiling, substances with stronger bonds will have a higher boiling point. A
substance with a high vapor pressure at normal temperatures is often referred to as volatile.
Water has a B.P of 100oC. However, the value is not a constant. The boiling point of water
depends on the atmospheric pressure, which changes according to elevation. water boils at a
lower temperature as you gain altitude (e.g., on a mountain) and boils at a higher temperature if
you increase atmospheric pressure (below sea level).
Capillary action is exhibited by polar liquids which is the spontaneous rising of a liquid in a
narrow glass tube. It occurs because of intermolecular forces between the liquid and surrounding
solid surfaces. If the diameter of the tube is sufficiently small, then the combination of surface
tension (which is caused by cohesion within the liquid) and adhesive forces between the liquid
and container wall act to propel the liquid. Adhesion of water to the walls of a vessel will cause
an upward force on the liquid at the edges and result in a meniscus which is concave. Capillary
action occurs when the adhesion to the walls is stronger than the cohesive forces between the
liquid molecules. Non-polar mercury shows a convex meniscus.
Problem
1. List the following molecules in order of increasing surface tension: C3H8, CH4, CH3COOH,
C2H6
Answer
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Higher surface tension corresponds to stronger intermolecular forces. One of these (CH3COOH)
has the ability to hydrogen-bond. It will probably have the strongest intermolecular forces.
CH3COOH is the only one of these molecules to have a dipole, and we already decided it has
the strongest intermolecular forces. Of the molecules that are left, the largest one (C 3H8) is likely
to have the strongest London dispersion forces. The smallest (CH4) is likely to have the weakest
intermolecular forces.
The answer is: CH4, C2H6, C3H8, CH3COOH
2. List the following molecules in order of increasing boiling point: Br2, F2, I2, Cl2
Answer
Higher boiling points will correspond to stronger intermolecular forces. None of these have
hydrogen bonding. None of these have dipoles. Bigger molecules will have stronger London
dispersion forces. So I2 has the strongest forces, and F2 will have the weakest. Correspondingly,
I2 will have the highest boiling point and F2 will have the lowest boiling point.
Answer: F2, Cl2, Br2, I2
3. Which substance has the higher vapor pressure: C20H42 or C30H62?
oASS
TU
Answer
C20H42 has the higher vapour pressure, since the molecules stick together less.
Solute-Solvent interaction
Highly polar NaCl dissolve in polar solvent like water not in hexane which is non-polar.
Polar organic substance such as sucrose dissolves in water not in hexane because of H-bonding.
Petrol and diesel are miscible, since both are non-polar. Water and ethanol are miscible, because
IMF operating is the same. (Like dissolves like).
Hexane and water forms two separate layers, hexane is unable to disrupt the water structure.
A solution is a mixture of a solvent and a solute. A solvent is a liquid in most cases, and the
solute is the powder or dissolving material. e.g a cup of water with sugar.
Solvent+Solute = Solution
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Phase Diagram for Water
Water is a unique substance in many ways. One of these special properties is the fact that solid
water (ice) is less dense than liquid water just above the freezing point. The phase diagram for
water is shown in the Figure
Phase diagrams plot pressure versus temperature. The labels on the graph represent the stable
states of a system in equilibrium. The lines represent the combinations of pressures and
temperatures at which two phases can exist in equilibrium. The line A-D represents the fusion
curve (solid to liquid), line A-C represents vaporization (liquid to gas). The third line which
curves towards y axis is sublimation curve (solid to gas). The triple point represents the
combination of pressure and temperature that facilitates all phases of matter at equilibrium. The
critical point terminates the liquid/gas phase line and relates to the critical pressure, the pressure
above which a supercritical fluid forms (gas and a liquid can coexist at equilibrium). Beyond the
temperature of the critical point, the merged single phase is known as a supercritical fluid.
The slope of the line between the solid and liquid states is negative rather than positive. The
reason is that water is an unusual substance in that its solid state is less dense than the liquid
state. Ice floats in liquid water. Therefore, a pressure change has the opposite effect on those two
phases. If ice is relatively near its melting point, it can be changed into liquid water by the
application of pressure. The water molecules are actually closer together in the liquid phase than
they are in the solid phase. In the solid forms of water and some other substances, the molecules
crystallize in a lattice with greater average space between molecules, thus resulting in a solid
with a lower density than the liquid. Because of this phenomenon, one is able to melt ice simply
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by applying pressure and not by adding heat. Water can melt at temperatures near its freezing
point when subjected to pressure. The ease with which ice skaters glide across a frozen pond can
be explained by the fact that the pressure exerted by their skates melts a small portion of the ice
that lies beneath the blades.
Normal melting and boiling points
The normal melting and boiling points are those when the pressure is 1 atmosphere. These can be
found from the phase diagram by drawing a line across at 1 atmosphere pressure.
Figure:
Problem
1. What happens to solid ice under high pressure near the freezing point?
It melts. Dry snow can be tightly pressed and will form snowballs because the higher
pressure causes the snowflakes to melt somewhat. However, when you release the pressure,
the snow goes back to a more solid form and the flakes no longer stick together.
2. What happens to this solid if we increase temperature at constant pressure? What happens
if you repeat this at a higher fixed pressure.?
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Solution:
It melts. If you repeat this at a higher fixed pressure, the melting temperature would be
higher because the line between the solid and liquid areas slopes slightly forward. As the
pressure increases, so does the boiling point.
2. What happens if you increase the pressure on a gas (vapour) at a temperature lower than the
critical temperature?
Solution:
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You will eventually cross the liquid-vapour equilibrium line and the vapour will condense to
give a liquid as represented in the diagram.
3. What is the physical change represented by the arrow in the phase diagram given below.
Solution:
Sublimation: You could read the sublimation temperature off the diagram. It will be the temperature at
which the line is crossed.
Problem
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Imagine a substance with the following points on the phase diagram: a triple point at .5 atm and 5°C; a normal melting point at 20°C; a normal boiling point at 150°C; and a critical point at 5
atm and 1000°C. The solid liquid line is "normal" (meaning positive sloping). For this, complete
the following:
1. Describe what one would see at pressures and temperatures above 5 atm and 1000°C.
2. Describe what will happen to the substance when it begins in a vaccum at -15°C and is
slowly pressurized.
3. Describe the phase changes from -80°C to 500°C at 2 atm.
4. What exists in a system that is at 1 atm and 150°?
Solution
1. One would see a super-critical fluid, when approaching the point, one would see the
meniscus between the liquid and gas disappears.
2. The substance would begin as a gas and as the pressure increases; it would compress and
eventually solidify without liquefying as the temperature is below the triple point
temperature.
3. The substance would melt at somewhere around, but above 20°C and then boil at
somewhere around, but above 150°C. It would not form a super-critical fluid as the
neither the pressure nor temperature reach the critical pressure or temperature.
4. Depending on how much energy is in the system, there will be different amounts of liquid
and gas at equilibrium. If just enough energy was added to raise the temperature of the
liquid to 150°C, there will just be liquid. If more was added, there will be some liquid
and some gas. If just enough energy was added to change the state of all of the liquid
without raising the temperature of the gas, there will just be gas.
Phase diagram for CO2
Here the solid liquid line has a positive slope and solid CO2 is denser than liquid CO2 (reverse of
water). The only thing special about this phase diagram is the position of the triple point which is
well above atmospheric pressure. It is impossible to get any liquid carbon dioxide at pressures
less than 5.11 atmospheres. That means that at 1 atmosphere pressure, carbon dioxide will
sublime at a temperature of -78°C. This is the reason that solid carbon dioxide is often known as
"dry ice". You can't get liquid carbon dioxide under normal conditions - only the solid or the
vapour.
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Figure:
Application:
In a fire extinguisher, CO2 exists as a liquid at 25oC under high pressure. When we release CO2
into the atmosphere at 1 atm, the liquid changes to vapor. CO2 is heavier than air and its vapor
smothers the fire by keeping O2 away from the flame. The liquid-vapor transition is endothermic,
so cooling occurs (similar to perfumes) which also helps in putting out the fire.
Enthalpy of vaporization
Molar enthalpy of vaporization is the energy needed to vaporize one mole ofa liquid.
Greater the strength of IMF greater the ∆Hvap. Stronger the IMF higher the ∆Hvap and
lower the vapor pressure. Equilibrium processes like fusion, vaporisation, sublimation
can be represented using the Clausius-Clapeyron equation. It relates a solution's vapor
pressures at different temperatures to the heat of vaporisation. The Clausius-Clapeyron
equation is expressed by
ln
p
H x 1 S x


o
p
R T
R
Equation is of the form, y = bx+a
Where H is enthalpy and S is entropy.
x can be vaporization or sublimation
Po =1 atm =101235 Pa = 1.01325 bar = 760 Torr = 760 mmHg
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Gas constant, R = 8.314 J/mol/K
Slope = b = -
H x
R
Intercept = a =
S x
R
Normal B.P is the temperature at which liquid boil at 1 atm, so ln
H vap
RTbp

Svap
⸫
R
Tbp =
1
1
H vap
Svap
The Clausius-Clapeyron equation at two temperatures
T2 >T1
P2 > P1
T1 and T2 are the absolute temperature of the solution in Kelvin
P1 and P2 is the vapor pressure of the solution at temperature T1 and T2
Problem 1:
Determine ΔHvap for a compound that has a measured vapor pressure of 24.3 torr at 273 K and
135 torr at 325 K.
Solution:
P1 = 24.3 torr T1 = 273 K
P2 = 135 torr T2 = 325 K
135
x  1
1 

ln
=


24.3 8.314  273 325 
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1.7148 =
x
0.00058608
8.3144
1.7148 = 0.000070489x
∆Hvap = 24327 J/mol = 24.3 kJ/mol
Problem 2: A certain liquid has a vapor pressure of 6.91 mmHg at 0 °C. If this liquid has a
normal boiling point of 105 °C, what is the liquid's heat of vaporization in kJ/mol?
Solution:
Let us use the Clausius-Clapeyron Equation:
use the following values:
P1 = 6.91 mmHg T1 = 0 °C = 273.15 K
P2 = 760.0 mmHg T2 = 68.73 °C = 378.15 K
-4.70035 = (x / 8.31447) (-0.0010165)
4.70035 = 0.00012226x
∆Hvap = 38445 J/mol = 38.4 kJ/mol
Problem 3: Chloroform, CHCl3 has a vapor pressure of 197 mmHg at 23.0 °C, and 448 mmHg
at 45.0 °C. Estimate its heat of vaporization and normal boiling point.
Solution:
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P1 = 197 mmHg
P2 = 448 mmHg
ln
T1 = 296 K
T2 = 318 K
x  1
1 
197

=


449 8.314  296 318 
∆Hvap = 29227.66 J = 29.2 kJ
P1 = 760 mmHg
P2 = 448 mmHg
T1 = x
T2 = 318 K
Set up equation to solve for the normal boiling point:
ln
29227.66 1
1
760
= ( 
)
8.3144
x 318
448
10.5258048 =
3515.276
x
B.P = 334 K
Problem 4: A 5.00 L flask contains 3.00 g of mercury. The system is at room temperature of
25.0 °C. By how many degrees should we increase the temperature of the flask to triple the
mercury vapor pressure. The enthalpy of vaporization for mercury is 59.11 kJ/mol?
Solution:
Let us use the Clausius-Clapeyrone equation with the following values:
P1 = 1 T1 = 298 K
P2 = 3 T2 = x
ln
3 59110  1
1

 

1 8.3144  298 x 
∆Hvap = 312.4 K = 39.4 °C
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Problem 5: The molar enthalpy of vaporization of hexane (C6H14) is 28.9 kJ/mol, and its normal
boiling point is 68.73 °C. What is the vapor pressure of hexane at 25.00 °C?
Solution:
1) Let us use the Clausius-Clapeyron Equation:
ln (P1 / P2) = - (ΔH / R) (1/T1 - 1/T2)
with the following values:
P1 = x
T1 = 25.00 °C = 298.15 K
P2 = 760.0 mmHg T2 = 68.73 °C = 341.88 K
ln (x / 760) = - (28900 / 8.31447) (1/298.15 - 1/341.88)
ln (x / 760) = - 1.4912
x /760 = 0.2251
x = 171 torr
Comment: note that no pressure is given with the normal boiling point. This is because, by
definition, the vapor pressure of a substance at its normal boiling point is 760 mmHg.
Solutions and their properties
Solutions are homogenous mixture of two or more substances, major component is called solvent
and minor component is solute. e.g NaCl solution, acids, bases etc. Solutions can be non
electrolytes like alcohol in water, soda water, sugar solution. In general formation of liquid
solution takes place in three distinct steps.



Separation of solute into individual components:
∆H1 (endothermic)
Overcoming IMF in the solvent to make room for the solute: ∆H2 (endothermic)
Allowing the solute and solvent to interact to form the solution: ∆H3 (exothermic)
⸫ Enthalpy of solution, ∆Hsoln = ∆H1 + ∆H2 + ∆H3
If ∆Hsoln is negative (exothermic) solution forms, if ∆Hsoln is positive no solution is
formed.
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Figure:
When a pure crystalline compound is heated, or a liquid cooled, the change in sample
temperature with time is roughly uniform. However, if the solid melts, or the liquid
freezes, a discontinuity occurs and the temperature of the sample remains constant until
the phase change is complete. This behavior is shown in the diagram on the right, with
the green segment representing the solid phase, light blue the liquid, and red the
temperature invariant liquid/solid equilibrium. For a given compound, this temperature
represents its melting point (or freezing point), and is a reproducible constant as long as
the external pressure does not change. The length of the horizontal portion depends on
the size of the sample, since a quantity of heat proportional to the heat of fusion must be
added (or removed) before the phase change is complete
Solubility and its temperature dependance
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Saturated Solution: Saturated solution is formed when no more solute will dissolve in the
solvent.
Unsaturated Solution: Unsaturated solution is formed when it can still dissolve more solute.
Supersaturated Solution: Super saturated is formed when you put solute in the solvent in excess.
The solvent will not be able to absorb the whole solute.
Figure:
Most salts are more soluble in hot water than cold. If in the process of dissolution, heat is
released, reaction is exothermic, then solubility decrease with increase of temperature. If heat is
absorbed then solubility increase with temperature.
Solubility of Gases: Effect of Pressure
Henry's Law describes the relationship between the partial pressure of a gas within a liquid and
its concentration within that liquid. Henry's Law states that the concentration of a gas dissolved
within a liquid, that is moles of the gas per unit liquid volume, is dependent the partial pressure
of the gas within the liquid and the unique chemical properties of the gas and liquid.
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Henry's Law: Cgas = k * Pgas
Cgas = Concentration of gas in the liquid (mol/L)
k = Empirically-derived solubility constant dependent on the unique chemical nature of
the gas and liquid
Pgas = partial pressure of gaseous solute
Henry’s law is found experimentally to hold for all dilute solutions in which the
molecular species is the same in the solution as in the gas.
Problem
1. What is Henry's constant for neon dissolved in water given: CNe = 23.5mL/L solution and
STP (22,414 mL/mole gas) and pressure (1 atm)?
SOLUTION
Henrys law states that
C = kPNe
Convert 23.5 mL/L solution to Molarity.
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(23.5 mL/L soln) (1 mole Ne/22,414 mL) = 0.00105M.
C = 0.00105M and we know that PNe = 1 atm
Rearranging the above equation,
k=CPNe
k = 0.00105 M/atm
Solutions of volatile solute and volatile solvent
A volatile substance is one that evaporates or sublimates at room temperature or below. Volatile
substances have higher vapor pressures versus non-volatile substances at the same temperature.
Examples of volatile substances include alcohol, mercury and gasoline. If you have a volatile
substance, it will have a high vapor pressure and a low boiling point. An increase in temperature
will cause an increase in vapor pressure, or the pressure at which the gas phase is in equilibrium
with the liquid or solid phase. Non-volatile substance refers to a substance that does not readily
evaporate into a gas under existing conditions. Non-volatile substances exhibit a low vapor
pressure and a high boiling point. Sugar and salt are examples of non-volatile solutes.
An ideal mixture is a solution of two volatile liquids, (A and B) of closely similar intermolecular
forces. e.g Hexane and Heptane, propan 1-ol and propan 2-ol. Here the tendency of the mixture
of A and B to become vapor is the same as that of pure A and B. i.e solute has no interaction.
The forces of attraction between solvent and solute are exactly the same as between the original
solvent molecules. An ideal solution is defined as one which obeys Raoult's Law. Only very
dilute solutions obey Raoult's Law to a reasonable approximation.
Raoult's Law is a thermodynamic law and it states that the partial vapor pressure of each
component of an ideal mixture of liquids equals the vapor pressure of the pure component
multiplied by the molar fraction of the mixture. The forces of attraction between solvent and
solute are exactly the same as between the original solvent molecules. Raoult's Law is expressed
by the vapor pressure equation:
Psolution = Xsolvent * Posolvent



Psolution is the vapor pressure of the solution
Xsolvent is the molar fraction of the solvent
Posolvent is the vapor pressure of the pure solvent
If you're working with two or more solutions, you can just add the vapor pressure or each
solution to find the total vapor pressure:
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PTotal = Psolution 1 + Psolution 2 +...
Example 1: If 0.340 mol of a nonvolatile nonelectrolyte are dissolved in 3.00 mol of water, what
is the vapor pressure of the resulting solution? (The vapor pressure of pure water is 23.8 torr at
25.0 °C.)
Solution:
Calculate the mole fraction of the solvent (NOT the solute):
χsolvent = 3.00 mol / (0.340 mol + 3.00 mol)
χsolvent = 0.8982
Calculate the vapor pressure:
Psolution = (χsolvent) (P°solvent)
Psolution = (0.8982) (23.8)
Psolution = 21.4 torr
Example 2: 210.0 g of the nonvolatile solute sucrose (C12H22O11) is added to 485.0 g of water at
25.0 °C. What will be the pressure of the water vapor over this solution? (The vapor pressure of
pure water is 23.8 torr at 25.0 °C.)
Solution:
Determine moles of water and sucrose:
water:
sucrose:
485.0 g / 18.0152 g/mol = 26.92171 mol
210.0 g / 342.3014 g/mol = 0.61349 mol
Determine the mole fraction of the solvent:
χsolvent = 26.92171 mol / (26.92171 mol + 0.61349 mol) = 0.9777
Using Raoult's Law, determine the vapor pressure:
Psolution = (χsolvent) (P°solvent)
Psolution = (0.9777) (23.8) torr
Psolution = 23.3 torr
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Example 3: The vapor pressure of water above a solution of water and a nonvolatile solute at
25.0 °C is 19.3 mm Hg. What is the mole fraction of the solute? (The vapor pressure of pure
water is 23.8 torr at 25.0 °C.)
Solution:
Use Raoult's Law:
Psolution = (χsolvent) (P°solvent)
19.3 = (x) (23.8)
x = 0.811
Example 4: Calculate the vapor pressure of water above a solution at 35.0 °C that is 1.600 m
fructose, C6H12O6. (The vapor pressure of pure water at 35.0 °C is 42.2 mmHg.)
Solution:
1.600 mol of fructose is dissolved in 1.000 kg of water (molality)
Determine moles of water:
1000 g / 18.015 g/mol = 55.5093 mol of H2O
Determine mole fraction of solvent:
55.5093 / (55.5093 + 1.600) = 0.97198
Using Raoult's Law, the vapor pressure above the solution is:
Psolution = (χsolvent) (P°solvent)
(42.2 mmHg) (0.97198) = 41.0 mmHg
Example 5: The vapor pressure of pure methanol, CH3OH, at 30 °C is 160 torr. How many
grams of the nonvolatile solute glycerol, C3H5(OH)3, must be added to 116 g of methanol to
obtain a solution with a vapor pressure of 127 torr?
Solution:
1) Write an expression for the mole fraction of the solvent (methanol):
Moles of methanol = 116 g /32 g/mol = 3.62 mol
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let the mols of glycerol equal to X
χsolvent = 3.6203/(3.6203 + X)
Solve for 'X' with Raoult's Law:
127 torr = (χsolvent) (P°solvent)
127 =
3.62
x 160 torr
3.62  X
X = 0.940708 mols
Convert moles to grams by multiplying by molar mass of glycerol
0.940708 mol x 92.0932 g/mol = 86.6 g
Raoult's Law and melting and boiling points
The effect of Raoult's Law is that the saturated vapour pressure of a solution is going to be lower
than that of the pure solvent at any particular temperature. Fig) That has important effects on the
phase diagram of the solvent. If you draw the saturated vapour pressure curve for a solution of a
non-volatile solute in water, it will always be lower than the curve for the pure water. If you look
closely at the last diagram, you will see that the point at which the liquid-vapour equilibrium
curve meets the solid-vapour curve has moved. That point is the triple point of the system. That
must mean that the phase diagram needs a new melting point line (a solid-liquid equilibrium
line) passing through the new triple point. This shows that boiling point of the solution is higher
than pure solvent and melting point of solution is lower than pure solvent. (Fig). A volatile solute
will contribute to the vapor pressure above a solution and is equal to the sum of vapor pressures
of solvent and each of the volatile solute particles.
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Figure: The vapor pressure of the solvent above the solution is
lower than the vapor pressure above the pure solvent).
Figure:
Colligative properties
Colligative properties are properties of solutions that depend on the ratio of the number of solute
particles to the number of solvent molecules in a solution, and not on the nature of the chemical
species present. The number ratio can be related to the various units for concentration of
solutions. Colligative properties include:




Relative lowering of vapor pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure
Relative lowering of vapor pressure
The vapor pressure of a liquid is the pressure of the vapor which is in equilibrium with that
liquid. The vapor pressure of a solvent is lowered when a non-volatile solute is dissolved owning
to the extra attraction the solvent has for the dissolved particles. As a result, very few solvent
molecules escape from surface and vapor pressure is low. The solute particles displace some
solvent molecules in the liquid phase and therefore reduce the concentration of solvent, so that
25
the colligative properties are independent of the nature of the solute. The word colligative is
derived from the Latin colligatus meaning bound together.
The relationship between the vapor pressure of solution and vapor pressure of solvent depends
on the concentration of the solute in the solution. According to Raoults law, partial vapor
pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the
pure component multiplied by its mole fraction in the mixture.
Psolution = (χsolvent) (P°solvent)
X solvent 
nsolvent
 mole fraction of solvent
nsolvent  nsolvent
X solvent  X solute  1
Relative lowering of vapour pressure is given by
P o solvent  Psolution
n
 X solute 
o
P solvent
n  N solvent
In dilute solutions, Xsolute =
n
N solvent

m Mr
x
mr M
26
Figure:
Problem:
1. Predict the vapor pressure of a solution prepared by dissolving 35 g of Na2SO4 (Mr
142.05g/mol) in 175 g of water at 25oC. Vapor pressure of pure water is 23.76 Torr.
Solution:
By Raoult’s law
p1 = x1 p°
p1 / p° = x1 = n1 / n1 + n2
n1 = w1 / M1 = 175 g / 18 g mol-1 = 9.72 mol
n2 = w2 / M2 = 35 g / 142.05 g mol-1 = 0.246 mol
Na2SO4(s) ⟶ 2 Na+ + SO421mol
⟶ 2 mol + 1 mol = 3mol
0.246 mol ⟶ 3 (0.246) mol = 0.738 mol
x1 = 9.72 mol / (9.72 mol +0.738 mol)
x1 = 0.929
p1 = x1 p°
p1 = 0.929 × 23.76 torr
p1 = 22.07 torr
2. The vapor pressure of pure benzene at certain temperature is 0.850 bar. 0.5g a non-volatile,
non-electrolyte solid when added to 39.0g of benzene (molar mass 78 g mol-1). vapor pressure of
solution, then, is 0.845 bar. What is the molar mass of that solid solute?
p° = 0.850 bar
p1 = 0.845 bar
w1 = 39.0g
M1 = 78 g mol-1
w2 = 0.5g
M2 = ?
(p°- p1) / p° = w2 M1 / M2 w1
(0.850 bar - 0.845 bar) / 0.850 bar = (0.5g× 78 g mol-1) / (M2 × 39 g)
27
0.0059 = 1 / M2
Ans: M2 = 169.49 g mol-1
3.An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at normal
boiling point of the solvent. What is the molar mass of the solute?
we know that vapor pressure of pure water at boiling point is p° = 1.013bar
vapor pressure of solution p1 = 1.004 bar
2% aqueous solution of solute means 2 parts of solute dissolved in 98 parts of water.
weight of solute w2 = 2g
molar mass of solute M2 = ?
weight of solvent w1 = 98g
molar mass of solute M1 = 18
(p°- p1) / p° = w2 M1 / M2 w1
(1.013-1.004) / 1.013 = (2× 18) / (M2 × 98)
0.0088 = 36 / (M2 × 98)
Ans: M2 = 41.74 g mol-1
A liquid liquid solution that obeys Rauolts law is called ideal solution
Figure: Ideal solution
When a non-volatile solute is dissolved in a solvent there will be lowering of vapor pressure (as
shown in the above figure). Examples of ideal solutions are: n-hexane+n-heptane (same kind of
IMF operating)
28
If we have a solution of ethanol (polar) and hexane (non-polar), due to weak solute-solvent
vaporization is easy implying a positive deviation from Raoults law. Another mixture which
shows a positive deviation is CCl4 (non polar) +CCl3(polar).
A pair like H2O + HNO3 will show a negative deviation from Raoults law because of strong H
bonding between them.
Boiling point elevation (ebullioscopy)
The boiling point of a liquid at a given external pressure is the temperature at which the vapor
pressure of the liquid equals the external pressure. The normal boiling point is the boiling point
at a pressure equal to 1 atm. The boiling point of a pure solvent is increased by the addition of a
non-volatile solute. In order to boil the solution, the temperature has to be raised further so as to
make its vapour pressure equal to atmospheric pressure. Magnitude of elevation depends on the
concentration of solute (colligative property). The change in boiling temperature, ∆T, can be
represented by the equation,
∆T = iKbmsolute
∆T = difference between boiling point of pure solvent and solution
Kb= molal boiling point elevation constant (K.mol-1kg)
msolute = molality of solute in the solvent (mol/kg)
i = Vant Hoff factor
The van 't Hoff factor is the ratio between the actual concentration of particles produced
when the substance is dissolved, and the concentration of a substance as calculated from
its mass. For most non-electrolytes dissolved in water, the van' t Hoff factor is essentially
1. For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the
number of discrete ions in a formula unit of the substance. This is true for ideal
solutions only. Sometime ion pairing occurs in solution. At a given instant a small
percentage of the ions are paired and count as a single particle. Ion pairing occurs to
some extent in all electrolyte solutions. This causes deviation from the van 't Hoff factor.
The deviation for the van 't Hoff factor tends to be greatest where the ions have multiple
charges.
for non-electrolytes like sugar, i = 1,
for strong electrolytes like NaCl, i =2
for concentrated solutions there will be reassociation and = i ⁓1
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Figure: Reassociation of NaCl
Figure:
Problem:
1. A 166.5 mg sample of the compound eugenol was dissolved in 1.00 g of chloroform (Kb =
3.63 °C/m), increasing the boiling point of chloroform by 3.68 °C. What is the molar mass of
eugenol?
Solution:
Δt = i Kb m
3.68 °C = (1) (3.63 °C kg mol¯1) (x / 0.00100 kg)
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x = 0.00368 kg °C / 3.63 °C kg mol¯1 = 0.0010137741 mol
0.1665 g / 0.0010137741 mol = 164.2 g/mol
Note: °C / m equals °C kg mol¯1
2.How many grams of fructose (C6H12O6) must be dissolved in 937 g of acetic acid to raise the
boiling point by 9.1 °C? The boiling point constant for acetic acid is 3.08 °C/m
Solution:
Δt = i Kb m
9.1 °C = (1) (3.08 °C kg mol-1) (x / 0.937 kg)
9.1 °C = (3.287 °C mol-1) (x)
x = 2.7684 mol
2.7684 mol times 180.1548 g/mol = 499 g
3: A 5.00 g sample of a large biomolecule was dissolved in 16.0 g of carbon tetrachloride. The
boiling point of this solution was determined to be 77.85 °C. Calculate the molar mass of the
biomolecule. For carbon tetrachloride, the boiling point constant is 5.03 °C/m, and the boiling
point of pure carbon tetrachloride is 76.50 °C.
Solution:
1) Determine moles of the compound dissolved:
Δt = i Kb m
1.35 °C = (1) (5.03 °C kg mol-1) (x / 0.0160 kg)
1.35 °C = (314.375 °C mol-1) (x)
x = 0.00429423 mol
2) Determine the molar mass:
5.00 g / 0.00429423 mol = 1160 g/mol
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Depression of freezing point (cryoscopy)
Freezing-point depression is the decrease of the freezing point of a solvent on addition of a nonvolatile solute. Examples include salt in water, alcohol in water. The freezing point is the
temperature at which the vapor pressures of liquid solvent and solid solvent are equal. When a
non-volatile solute is added to a volatile liquid solvent, the solution vapor pressure will be lower
than that of the pure solvent. As a result the solid will reach equilibrium with the solution at a
lower temperature than with the pure solvent.
Colligative properties depend on the number of particles present, not on the type of particles or
their mass. So, for example, if both calcium chloride (CaCl2) and sodium chloride (NaCl)
completely dissolve in water, the calcium chloride would lower the freezing point more than the
sodium chloride because it would produce three particles (one calcium ion and two chloride
ions), while the sodium chloride would only produce two particles (one sodium and one chloride
ion). Raoult’s law states that the freezing and boiling points of an ideal solution are respectively
depressed and elevated relative to that of the pure solvent by an amount proportional to the mole
fraction of solute. Figure:
Figure:
Freezing-point depression is used by some organisms that live in extreme cold. Such creatures
have evolved means through which they can produce high concentration of various compounds
such as sorbitol and glycerol (antifreeze compound). This elevated concentration of solute
decreases the freezing point of the water inside them, preventing the organism from freezing
solid even as the water around them freezes, or as the air around them becomes very cold. Road
salting takes advantage of this effect to lower the freezing point of the ice it is placed on.
Lowering the freezing point allows the street ice to melt at lower temperatures, preventing the
accumulation of dangerous, slippery ice. Commonly used sodium chloride can depress the
freezing point of water to about −21 °C (−6 °F). The freezing points of solutions are all lower
than that of the pure solvent and is directly proportional to the molality of the solute.
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The freezing point depression equation is:
ΔT = i Kf msolute
where ΔT is the freezing point depression, Kf is the freezing point depression constant, and m is
the molality and i is the van 't Hoff factor. The value of ‘i ’ is the actual number of particles in
solution after dissociation divided by the number of formula units initially dissolved in solution
and means the number of particles per formula unit of the solute when a solution is dilute.
Problem
1. What is the freezing point of an aqueous solution when enough NaCl has been added to
create a 0.25 m solution? The Kf value for water is 1.858 oC/m.
To solve this, you must remember that NaCl breaks into two ions, Na+ and Cl–, when it
dissolves in water. In simplest terms, this means it has an “i” factor of 2.
This might seem like the end of the problem, but it is not. The value of 0.93 oC is the
change in the freezing point. The new freezing point of water, which is normally 0 oC, is
equal to: 0 – 0.93 = -0.93 oC.
2) A 1.00 m solution of acetic acid (CH3COOH) in benzene has a freezing point depression of
2.6 K. Calculate the value for i and suggest an explanation for its value.
Solution
The freezing point depression equation is: ΔT = i Kf m
The only value we do not know is i:
2.6 K = (x) (5.10 K m¯1 ) (1.00m)
x = 0.51
I'd like you to think about the answer. If i = 1, then one item dissolving yields one item in
solution. If i = 2, then one item dissolving yields two items in solution. What about i = 0.5?
Hint : Recall association of acetic acid
33
3) 2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from
5.53 to 4.90 ∘C. What is the molar mass of the compound?
Solution
2. First we must compute the molality of the benzene solution, which will allow us to find
the number of moles of solute dissolved.
m=ΔTf−Kf
m = (4.90−5.53) oC−5.12OC/m =0.123m
Amount Solute=0.07500 kgbenzene × 0.123m1kgbenzene
=0.00923 m solute
Mass = 2.00 g
Molecular Weight =216.80g/mol
3. 4.1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of benzene. The freezing point of
pure benzene is 5.5 oC, and the freezing point of the mixture is 2.8 oC. What is the molal
freezing point depression constant, Kf of benzene?
Tf = (Freezing point of pure solvent) - (Freezing point of solution)
(5.5 oC) - (2.8 oC) = 2.7 oC
Calculation of molal concentration of the solution.
molality = moles of solute / kg of solvent
moles of naphthalene = (1.60 g) (1 mol / 128 g) = 0.0125 mol naphthalene
molality of solution = (0.0125 mol) / (0.0200 kg) = 0.625 m
∆Tf = (Kf) (m)
(2.7 oC) = (Kf) (0.625 m)
Kf = 4.3 oC/m
4. How many grams of ethylene glycol, C2H4(OH)2, must be added to 400.0 g of water to
yield a solution that will freeze at -8.35 °C?
Solution:
34
Δt = i Kf m
8.35 °C = (1) (1.86 °C kg mol-1) (x / 0.4000 kg)
8.35 °C = (4.65 °C mol-1) (x)
x = 1.7957 mol
1.7957 mol times 62.07 g/mol = 111 g (to three sig figs)
The solution freezes at -1.37 °C.
5. A 33.7 g sample of a nonelectrolyte was dissolved is 750. g of water. The solution's freezing
point was -2.86 °C. What is the molar mass of the compound? Kf = 1.86 °C/m.
Solution:
Δt = i Kf m
2.86 °C = (1) (1.86 °C kg mol-1) (x / 0.750 kg)
2.86 °C = (2.48 °C mol-1) (x)
x = 1.1532 mol
33.7 g / 1.1532 mol = 29.2 g/mol
6. A 1.60 g sample of napthalene (a non-electrolyte with a formula of C10H8) is dissolved in 20.0
g of benzene. The freezing point of benzene is 5.5 °C and Kf = 5.12 kg/mol. What is the freezing
point of the solution?
Solution:
(1.60 g / 128.1732 g/mol) / 0.0200 kg = 0.624155 m
Δt = i Kf m
x = (1) (5.12 °C kg mol-1) (0.624155 mol/kg)
x = 3.2 °C
5.5 - 3.2 = 2.3 °C
7. Camphor (C6H16O) melts at 179.8 °C, and it has a particularly large freezing point depression
constant, Kf = 40.0 °C/m. When 0.186 g of an organic substance of unknown molar mass is
dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C.
What is the molar mass of the solute?
35
Solution:
179.8 - 176.7 = 3.1 °C
Δt = i Kf m
3.1 °C = (1) (40.0 °C kg mol-1) (x / 0.02201 kg)
3.1 °C = (1) (1817.356 °C mol-1) (x)
x = 0.001705775 mol
0.186 g / 0.001705775 mol = 109 g/mol
8. Which sample, when dissolved in 1.0 liter of water, produces a solution with the lowest
freezing point?
(1) 0.1 mol of C2H5OH
(2) 0.1 mol of LiBr
(3) 0.2 mol of C6H12O6
(4) 0.2 mol of CaCl2
9. Which solution has the highest boiling point at standard pressure?
(1) 0.10 M KCl(aq)
(2) 0.10 M K2SO4(aq)
(3) 0.10 M K3PO4(aq)
(4) 0.10 M KNO3(aq)
Osmotic pressure
The osmotic pressure of a solution is the difference in pressure between the solution and the pure
liquid solvent when the two are in equilibrium across a semipermeable membrane, which allows
the passage of solvent molecules but not of solute particles. Osmosis is the diffusion of a fluid
through a semipermeable membrane. Osmosis is the movement of water from an area of low
concentration of solute to an area of higher concentration of solute. When a semipermeable
membrane (animal bladders, skins of fruits and vegetables) separates a solution from a solvent,
then only solvent molecules are able to pass through the membrane. The osmotic pressure of a
solution is the pressure difference needed to stop the flow of solvent across a semipermeable
membrane. If the two phases are at the same initial pressure, there is a net transfer of solvent
across the membrane into the solution known as osmosis. The process stops and equilibrium is
attained when the pressure difference equals the osmotic pressure. The osmotic pressure of a
dilute solution at constant temperature is directly proportional to its molar concentration and is a
colligative property.
36
π = icRT
where π is the osmotic pressure, R is the ideal gas constant (0.0821 L atm / mol K), T is the
temperature in Kelvin, c is then the molar concentration of the solute (moles/L).
Figure:
Osmotic pressure is the basis of filtering ("reverse osmosis"), a process commonly used in water
purification. The water to be purified is placed in a chamber and put under an amount of pressure
greater than the osmotic pressure exerted by the water and the solutes dissolved in it. Part of the
chamber opens to a differentially permeable membrane that lets water molecules through, but not
the solute particles. The osmotic pressure of ocean water is about 27 atm. Reverse osmosis
desalinates fresh water from salt water. Osmosis is also impacted by how the solute behaves in
37
water, which is where Van't Hoff's factor comes in. Basically, the Van't Hoff's factor of a solute
is determined by whether or not a solute stays together or breaks apart in water.
Solutions with same osmotic pressure are called isotonic solutions. Isotonicity is the presence of
a solution that produces no change in cell volume. Hypertonicity is the presence of a solution
that causes cells to shrink. Hypotonicity is the presence of a solution that causes cells to swell.
Problem:
1. What is the osmotic pressure of a 1.35 M solution of NaCl at 25 oC?
i = 2 (NaCl breaks into two particles)
M = 1.35 moles/L
R = 0.0821 L×atmK×mol
T = 25 oC + 273 = 298 K
π=2×1.35×0.0821×298
π=66.1 atm
2. Sea water contains dissolved salts at a total ionic concentration of about 1.13 mol L–1.
What pressure must be applied to prevent osmotic flow of pure water into sea water
through a membrane permeable only to water molecules?
Solution:
π = icRT
(1.13mol/L)(0.0821Latmmol–1K–1)(298K)=27.6atm
3. The osmotic pressure of a benzene solution containing 5.0 g of polystyrene per liter was found
to be 7.6 torr at 25°C. Estimate the average molecular weight of the polystyrene in this sample.
Solution:
osmotic pressure : π = (7.6 torr) / (760 torr atm–1) = 0.0100 atm
Using the form of the van't Hoff equation PV = nRT, the number of moles of polystyrene is
n = (0.0100 atm)(1 L) ÷ (0.0821 L atm mol–1 K–1)(298 K) = 4.09 x 10–4 mol
Molar mass of the polystyrene:
38
(5g) ÷ (4.09 x 10–4 mol) = 12200 g mol–1.
4.Calculate osmotic pressure for 0.10 M Na3PO4 at 20°C. Calculate molarity if solution in
water (300 K) has osmotic pressure of 3.00 atm.
Solution:
****
π = icRT
c = (3.00atm)/[0.0821 atm.L/mol.K)(300K)] = 0.122M
Since Na3PO4 ionizes into 4 particles (3 Na+1 +PO4 3- ), the ion concentration is 0.40 M
(0.40)(0.0821)((293) = 9.6 atm
6.51×104g/mol
Van’t Hoff factor and degree of dissociation
The degree of dissociation is the fraction of the original solute molecules that have dissociated
into ions. It is usually indicated by the Greek symbol α. The van 't Hoff factor i is the number of
particles obtained when a solute dissolve. For most ionic compounds dissolved in water, the
maximum van 't Hoff factor is equal to the number of discrete ions in a formula unit of the
substance.
For K₃Fe(CN)₆: K₃Fe(CN)₆(s) → 3K⁺(aq) + Fe(CN)₆³⁻(aq), so i = 4.
These are the maximum values of i. They are found only in ideal solutions- solutions that are
infinitely dilute. In real solutions, some of the ions are paired within the same solvation shell and
count as a single particle. This causes deviations from the maximum van 't Hoff factor. Here are
some observed values of the van't Hoff factor for 0.1 mol/kg aqueous solutions.
39
There is a simple relationship between this parameter and the van 't Hoff factor, i. If a fraction α
of the solute dissociates into n ions, then,
α=
i 1
n 1
Problem:
1. What is the osmotic pressure of a 0.30 M solution of MgSO4 if the MgSO4 is 80%
dissociated at 20.0 °C?
Solution
1) Calculate the van 't Hoff factor from the degree of dissociation:
α = 0.80
n=2
i = 1 + 0.80(2 - 1)
i = 1.80
Solve for the osmotic pressure:
π = iMRT = (1.80)(0.30 mol/L) (0.08206 L-atm/mol-K) (293 K)
π = 12.98 atm
2. 2.00 mols of Ba(ClO4)2 were placed in 1.00 L of solution at 45.0 °C. 15% of the salt was
dissociated at equilibrium. Calculate the osmotic pressure of the solution.
40
Solution
Calculate the van 't Hoff factor from the degree of dissociation:
α=
i 1
n 1
α = 0.15
n=3
i = 1.30
Solve for the osmotic pressure:
π = iMRT = (1.30) (2.00 mol/L) (0.08206 L-atm/mol-K) (318 K)
π = 67.85 atm
3. Find the osmotic pressure of an aqueous solution of BaCl2 at 288 K containing 0.390 g per
60.0 mL of solution. The salt is 60% dissociated.
Solution:
Calculate the van 't Hoff factor from the degree of dissociation:
α = 0.60
n=3
α=
i 1
n 1
i = 2.20
MV = mass / molar mass
(x) (0.0600 L) = 0.390 g / 208.236 g/mol
x = 0.0312146 M
Solve for the osmotic pressure:
π = iMRT = (2.20) (0.0312146 mol/L) (0.08206 L-atm/mol-K) (318 K)
π = 1.623 atm
41
4. 3.58 g of NaCl was dissolved in 120.0 mL of solution at 77.0 °C. The osmotic pressure is
26.31 atm. Calculate the degree of dissociation of NaCl.
Solution
Calculate the molarity of the NaCl:
MV = mass / molar mass
(x) (0.1200 L) = 3.58 g / 58.443 g/mol
x = 0.510469 M
2) Calculate the van 't Hoff factor:
26.31 atm = (i) (0.510469 mol/L) (0.08206 L atm / mol K) (350 K)
i = 1.7945
Use the van 't Hoff factor to determine the percent dissociation:
α=x
n=2
1.7945 = 1 + x(2 - 1)
x = 0.7945 = 79.45% dissociated
5.The freezing point depression of a 0.10 m solution of HF(aq) solution is -0.201 °C. Calculate
the percent dissociation of HF(aq).
Solution
The freezing point depression equation is: ΔT = i Kf m
The only value we do not know is i :
0.201 °C = (x) (1.86 °C m¯1 ) (0.10m)
x = 1.08 The HF is about 8% dissociated.
42
43