Engineering Chemistry 2
(AAECH2A)
www.vut.ac.za
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Engineering Chemistry 2 (AAECH2A)
Dr. Z NATE
Dept. of Biotechnology & Chemistry
Vaal University of Technology, Vanderbijlpark Campus
Office: J201B
Email: zondin@vut.ac.za
Telephone: 0169506691
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Learning unit 4: Solution and their
properties
Outlook
 Solutions
 Units of concentration
 Some factors affecting solubility
 Physical behaviour of solutions: colligative properties
 Vapour pressure lowering of solutions: Raoult’s law
 Boiling- point elevation and freezing point depression of solutions
 Osmosis and osmotic pressure
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Physical behavior of solutions: colligative properties
The behaviour of solutions is qualitatively similar to that of pure solvents but is
quantitatively different.
Example: Water boils at 100 ℃ and freeze at 0.0 ℃, but 1.00 m (molal) of NaCl(aq)
boils at 101 ℃ and freeze at -3.7 ℃.
This observations are example of colligative properties.
Colligative properties which depends on the amount of dissolved solute but not on
the solute’s chemical identity.
Colligative means bound together in a collection
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Physical behavior of solutions: colligative properties
Changes in colligative properties depend only on the number of solute particles present,
not on the identity of the solute particles.
Among colligative properties in comparing pure solvent and a solution.
1. Boiling-point of solutions is higher (Boiling-point elevation)
∆T= kb×m×i
(kb= boiling point constant, m=molality, i= van’t Hoff factor)
i= moles of particles in solution/ moles of solute dissolved
2. Freezing or melting Point of the solution is lower (Freezing-Point depression)
∆T= -kf×m×i
(kf= freezing point constant, m=molality, i= van’t Hoff factor)
3. Vapor-pressure of solutions is lower (Vapor-pressure lowering)
Psolution = XsolventPsolvent
4.The solution gives rise to osmosis (Osmotic pressure (π)) (π= M×R×T×i (R= gas
constant 0.08206 L.atm/mol.K, M=molarity, T= temperature in K)
Osmosis is the migration of solvent molecules through a semipermeable membrane.
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Vapour pressure lowering of solutions: Raoult’s law
A liquid in a closed container is in equilibrium with its vapour and the pressure
exerted by it is called vapor pressure.
Pure solvent (vp)
v/s vapor pressure of solution
-If the solute is nonvolatile, then the vapour pressure of the solution is lower than that
of pure solvent.
-If the solute is volatile, then the vapour pressure of the mixture is intermediate
between the vapour pressure of the two pure liquids.
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Vapour pressure lowering of solutions: Raoult’s law
Evaporation of pure solvent v/s that of solution.
-A solution evaporates slower than pure solvent, because its vapour pressure is lower
and its molecules escape less rapidly.
According Raoult’s law, the vapor pressure of a solution containing non-volatile
solute is equal to the vapor pressure of the pure solvent times the mole fraction of the
solvent.
Raoult’s law Psolution = XsolventPsolvent
where
Xsolv is the mole fraction of solvent in solution
Psolv is the vapor pressure of pure solvent at that temperature.
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Vapour pressure lowering of solutions: Raoult’s law
Raoult’s law Psoln = XsolvPsolv
We also have vapor-pressure lowering, which is the difference between vapor pressure
of the solvent and vapor pressure of solution.
Psoln =Psolv – Psoln
Alternatively, vapor pressure lowering can be calculated directly by multiplying the mole
fraction of the solute by vapor pressure of the pure solvent, i.e,
Psoln = XsolutePsolv
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Test yourself
1. How many grams of sucrose must be added to 320 g of water to lower the vapor
pressure by 1.5 mm Hg at 25 ℃? The vapor pressure of water at 25℃ is 23.8 mm Hg,
and molar mass of sucrose is 342.3 g/mol.
2. What is the vapor pressure in mm Hg of a solution prepared by dissolving 5.0 g of
benzoic acid (C7H6O2) in 100 g of ethyl alcohol (C2H6O) at 35℃? The vapor pressure
of pure ethyl alcohol at 35℃ is 100.5 mm Hg.
3. How many grams of NaBr must be added to 250 g of water to lower the vapor
pressure by 1.30 mm Hg at 40 ℃ assuming complete dissociation? The vapor
pressure of water at 40 ℃ is 55.3 mm Hg
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Solutions with nonvolatile solute
If an ionic substance (NaCl) is the solute rather than molecular substance, we
calculate mole fractions based on total concentration of solute particles (ions) rather
than NaCl formula units.
CHECK Page 492 of your prescribed textbook.
Ionic substances rarely dissociate completely, so a solution of an ionic compound
usually contains fewer particles than the formula of the compound would suggest.
The actual extend of dissociation can be expressed as a van’t Hoff factor (i)
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Solutions with nonvolatile solute
van’t Hoff factor (i) = Moles of particles in solutions
Moles of solute dissolved
Rearranging this equation
Moles of particles in solution= i x moles of solute dissolved
For all/any nonelecrolytes we always assume i= 1.
For Electrolytes (substance that dissociates into ion in solutions) we used
experimentally determined van’t Hoof factor (i)
Look at worked example 13.7 and 13.8 and then attempt practice 13.13 and 13.14
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Solutions with a volatile solute
According to Dalton’s Law of partial pressures, the overall vapor pressure Ptotal of a
mixture of volatile liquids, is the sum of the vapour pressure contribution of the
individual component in the mixture.
Consider volatile liquids A and B
Ptotal =PA + PB
This individual vapour pressures are calculated by Raoult’s law
Ptotal =PA + PB = XA PA + XB PB
 i.e the vapour pressure of A is equal to mole fraction of A times vapour pressure of
pure A, same as that of B.
Look at worked example 13.9 and attempt practice 13.17
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Bp elevation and Fp depression of Sol
The vapour pressure of a liquid rises with increasing temperature and that the liquids
boil when its vapour pressure equals atmospheric pressure.
Boiling point: temperature where a liquid turns into a gas
A solution’s boiling point can be raised by the amount of solute in the solvent.
Since non-volatile solute solution have lower vapour pressure than pure solvent, it
means it must be heated to higher temperature for it to boil.
The actual amount of boiling point elevation for solution of ionic substance depends
on the extent of dissociation, as given by van’t Hoof factor (i).
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Bp elevation and Fp depression of Sol
The change in boiling point is proportional to the molality of the solution:
Tb = Kb ×m × I
where Kb is the molal boiling-point elevation constant, a property of the
solvent, m is molal concentration of the solute and the van’t Hoof factor.
Concentration in molality so that is independent of temperature.
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Bp elevation and Fp depression of Sol
 The freezing point of a solution depends on the concentration of the
solute particles
Freezing point: temperature at which a liquid becomes a solid
The freezing point of a liquid solvent decreases when a solute is dissolved
in it.
The change in freezing point can be found similarly:
Tf = -Kf ×m × i
Here Kf is the molal freezing-point depression constant of the solvent.
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Osmosis and Osmotic Pressure
When a solution and pure solvent or two solutions of different concentration(s)
are separated by the right kind of permeable membrane, solvent molecules pass
through the membrane in a process called osmosis.
Osmosis: diffusion of solvent through the semi permeable membrane from a
solution of lower concentration towards a solution of higher concentration.
Osmosis: The selective passage of solvent molecules through a porous membrane
from a dilute solution to a more concentrated one.
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Osmosis and Osmotic Pressure
The amount of pressure necessary for the passage of molecule through the
membrane is called the solution’s osmotic pressure (π or ∏) (Greek capital pi).
Osmotic pressure (π or ∏): is given by
π = iMRT
where M is molar concentration of solute, T is temperature in kelvins, R = 0.08206
(Latm)/(molK) and i is the van’t Hoof factor.
Look at worked example 13.10 and 13.11 and attempt practice 13.19 and 13.21
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End of learning unit 4
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