Solutions suspensions and colloids

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Chapter# 10
Solution Suspensions and Colloids

Course Overview

Solution
A homogenous mixture of solute and solvent
Solute
Lesser quantity compared to solvent
Solvent
More in quantity compared to solvent
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Solutions
Na+
Na+Cl-
Cl-
Na+
Cl-
Solvent :
component in greater amount
(The dissolver)
•Water
•AlcoholSolute:
component in lesser amount
(The dissolved)
•NaCl

• A herterogenous mixture two or more components
• Generally the components appear in distinct phases
• Eg
• A mixture of salt and sand
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Suspension

Separation of mixture components

A heterogeneous mixture having suspended tiny
particles
• Particle size b/w mixture and solution
• Can pass through flters but not through membranes
Particle Size
Solution < Colloids < Suspension
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Colloids

Hydrophilic molecules
• Hydrophilic molecules are the molecules which
having a tendency to mix with, dissolve in, or be
wetted by water. Ex: Salt, Sugar etc

Hydrophobic molecules
• Hydrophobic molecules are the molecules tending to
repel or fail to mix with water. Ex: alkanes, oils, fats ,
greasy substances in general

Nature of Solutions in Liquid Phase
• A solute only dissolve in a solvent according to Like
dissolve Like Principle
• Polar solvent dissolve polar solutes
• Nonpolar solvents dissolve nonpolar solutes
• As NaCl is dissolved in water not in Benzene
• Fats and oils dissolve in Benzene not in water

Chemistry of Dissolution
• Polar solvents solvate (surround) the polar solute by
electrostaic force as a result crystal lattice is broken
and the solute is dissolved.
• Similarly nonpolar solute can develop forces of
attraction with non polar solvent.

Vapour Pressure solvent Vs Solution

Raoult’s Law when Solute is nonvolatile in a
Volatile Solvent
• Statement: Vapour pressure of a solvent above a
solution is directly proportional to mole fraction of
solvent in the solution.
• Mathematically,
• p∞ X1
p=p0 X1 --------------------- (I)
Where

Raoult’s Law
• p= vapour pressure of solvent in solution form
• P0= vapour pressure of solvent in pure form
• X1 = mole Fraction of solvent in the solution
• We know X1 + X2 = 1 OR X1 = 1- X2

Raoult’s Law
• p=p0 X1 --------------------- (I)
• p=p0 (1-X2)
p=p0-p0X2
p-p0 = -p0X2
-p0 +p= -p0X2
p0 -p= p0X2

The lowering of vapour pressure ∆p and
Relative lowering of Vapour pressure ∆p/ p0
• ∆p= p0X2 ---------------------(ii)
• The lowering of vapour pressure ∆p is proportional to
X2
• ∆p/ p0 = X2 ------------------(iii)
∆p/ p0 = Relative lowering of Vapour pressure
• Relative lowering of Vapour pressure ∆p/ p0 is equal
to X2 (Another definition of Raoult’s Law)

Significance of ∆p/ p0
• It’s independent of temperature
• Depends on concentration of solute
• Its constants for equimolar solutes in the same
solvent

Raoult’s Law When Both Components are Volatile
1st Year Chemistry Chap#10 LEC#5

Raoult’s Law When Both Components are
Volatile
• Raoult’s Law can be applied to a solution having both
components volatile
• Consider two liquids A and B
• Their original pressures p0
A and p0
B in the pure state at a
given temperature
• After mixing their vapour pressures changes
• Let the v.p of A becomes pA and pB with
mole fractions XA and XB

Volatile
• Applying Raoult’s Law to both components
• p A=p0
A XA
• p B=p0
B XB
• Total pressure Pt will be
• Pt= pA+pB
• Pt=p0
A XA+p0
B XB

Volatile
• Since
XA + XB = 1
XB = 1 – XA
Putting in Pt Equation
Pt = p0
A XA + p0
B (1 – XA )
Pt = p0
A XA + p0
B – p0
B XA

Volatile
• Pt = p0
A XA – p0
B XA + p0
B
• Pt = (p0
A – p0
B ) XA + p0
B ------------- (B)
• The component A is low boiling point compared to B
• Equation B is an equation of straight line if a graph is
plotted between XB or mole % of B on x-axis and Pt
on y-axis, a straight line will be obtained

Graphically

Raoult’s Law When and Ideal solutions
• Only those pair of liquids give straight lines which
form ideal solutions
• Rauolts law Determines whether a solution is ideal or
non ideal

Colligative Properties
• Those properties which depend only on the number of
solute and solvent particles/ ions
• Following are the colligative properties of dilute solutions
1. Lowering of vapour pressure
2.Elevation of boiling point
3. Depression of freezing point
4.Osmotic pressure

Why do we study Colligative Properties
1. Helpful in molecular mass determination
2. Contributed in the development of Solution Theory

Conditions to observe colligative properties
1. Solutions must be dilute
2. Solute must be nonvolatile
3. Solute must be non-electrolyte

Lowering of Vapour pressure

Mathematically,
• According to Raoutl’s Law Relative Lowering of V.p
∆p/ p0 = X2 ------------------(iii)
n2
Where X2 = n2 + n1
n2= moles of solute
n1= moles of solvent

Calculation of mass of solute M2
• Putting the value of in equation (iii)
• ∆p/ p0 = n2/ n2 + n1
• For a dilute solution n2 can be ignored
• ∆p/ p0 = n2/ n1 -----------(iv)
• we know
• n2= W2/ M2 and n1= W1/ M1

Calculation of mass of solute M2
• Putting the values of n1=and n2 in equation (iv)
• ∆p/ p0 = W2/ M2
• W1/ M1
• ∆p/ p0 = W2/ M2 X M1/ W1
Or
M2 = p0 / ∆p X W2 M1/ W1 (The molecular mass of
nonvolatile solute can be calculated)

Colligative Properties (Elevation of BP and depression of FP)
1st Year Chemistry Chap#10 LEC#7

Lowering of Vapour pressure
Higher the concentration of solute, lower the v.p accordingly

Lowering of Vapour pressure Results in Elevation
of B.P
50C 55C
Higher the concentration of solute, greater will be elevation of BP

Elevation of Boiling Point, Graphically

Mathematically,
• As we have discussed elevation of BP is directly
proportional to molality of solution (concentration of
solute)
• ∆Tb ∞ m
• ∆Tb = Kb m ----- (I) ( ∆Tb = Elevation of BP)
( ∆Kb = molal bp constant)
(m= molality of solution)

Elevation of Boiling Point
• Molality,
m= mass of solute/Molecular mass of solute X 1/mass of solvent in kg
m= W2/M2 X 1/W1/1000
m= 1000W2/M2W1 -------------------------- (ii)
putting value of m in eq. (i)

Elevation of Boiling Point
∆Tb = Kb m -------(i)
∆Tb = Kb x 1000W2/M2W1 ----------(iii)
Rearranging eq.(iii)
M2 = Kb/ ∆Tb X W2/W1 X1000
Where M2 is the Molecular mass of solute
The above eq. can be used to determine the molecular
mass of an unknown nonvolatile and non- electrolyte
solute

Determination of B.P Elevation
Landberger’s method

Landberger’s method

Colligative Properties
• Those properties which depend only on the number of
solute and solvent particles/ ions
• Following are the colligative properties of dilute solutions
1. Lowering of vapor pressure
2.Elevation of boiling point
3. Depression of freezing point
4.Osmotic pressure( Not included in reduced syllabus)

Depression of Freezing Point
• Freezing point is the temperature at which the solid
and liquid phases of the substance coexist.
• OR
• The temperature at which solid and liquid phases
have same vapour pressure
• We know that the addition of non volatile solute
decreases vp of solvent

Why F.P of a solution decreases
• At freezing point,
• There are two things in the vessel
Liquid solution and (due to lower vp)
Solid solvent
The solution will freeze at that temperature at which vp of
both liquid solution and solid solvent are same
As a result solution will freeze at lower temperature than
pure solvent

Mathematically,
• Depression in freezing point= freezing point of pure
solvent - freezing point of pure solution
• ∆Tf = T1 - T2
• We know
• ∆Tf ∞ m
• ∆Tf = Kf m---------(iv)
• Kf is called molal boiling point consant (cryoscopic)

Mathematically
• Putting value of molality in equation (iv)

Molecular mass of solute M2
The molar mass of nonvolatile solute can be determined by the above eq.

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Determination of F.P Depression

Beckmann’s Freezing point apparatus

Osmotic Pressure

Osmotic pressure

Vant Hoff Equation

Solutions suspensions and colloids

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