Chemistry
Solution - I
Session objectives

Introduction

Solubility

Henry’s law

Different concentration terms

Vapour pressure

Raoult’s law and its modification

Relative lowering of vapour pressure

Ideal solutions and non-ideal solutions

Maximum and minimum boiling solutions
Introduction
Solute: Component of solution present in smaller amount.
Solvent: Component of solution present in the larger amount.
Solution: a homogenous mixture of two or more substances.
Solubility
Maximum amount of solute in grams which can be
dissolved in a given amount of solvent (generally 100 g)
to form a saturated solution at that particular
temperature is known as its solubility
For solids
 Solubility of ionic compounds
in water generally increases
with increase in temperature.
For gases

The solubility of gases in
water decreases with
increase in temperature.

Solubility tends to zero at the
boiling point of water.
Effect of pressure on solubility of gases

Increase in pressure of the gas above the solution
increases the solubility of the gas in the solution.
More dilute solution
More concentrated solution
Henry’s law
Solubility of a gas in a liquid is proportional to
the pressure of the gas over the solution.
C = kP
C: molar concentration, P: pressure,
k: temperature-dependent constant
Carbonated cold drink is an
application of Henry’s law.
Different concentration terms
Molarity (M) 
Molality (m) 
moles of solute (mol)
volume of solution (L)
moles of solute
Weight of solvent (inkg)
Molarity of a solution changes with temperature due
to accompanied change in volume of the solution.
ni
Mole fraction(x) 
 ni
x1=mole fraction of solvent
x2=mole fraction of solute
Illustrative Example
Determine the molality of a solution prepared
by dissolving 75 g of Ba(NO3)2(s) in 374 g of
water at 25oC.
Solution:
Molar mass of Ba(NO3)2 = 261
Number of moles of Ba(NO3)2 
 0.287 mole
0.287 mole
Molality =
 0.767 m
0.374 kg
75 g
261 g mol-1
Illustrative Example
Calculate the molality of 1 molar solution of NaOH
given density of solution is 1.04 gram/ml.
Solution:
1 molar solution means 1 mole of solute present per
litre of solution.
Therefore, mass of 1 litre solution = 1000 x 1.04
= 1040 gram
Mass of solute = 1 x 40 = 40g
Therefore, mass of solvent 1040 – 40 = 1000g
m =
1
×1000 = 1 molal solution.
1000
Different concentration terms
w
Mass of solute
% by mass =
×100 =
%
Mass of solution
W
% by volume 
Volume of solute
 100
Volume of solution
Parts per million (ppm) =
v
 %
V
mass of solute
×106
mass of solution
Illustrative Example
Calculate the concentration of 1 molal solution
of NaOH in terms of percentage by mass.
Solution:
1 molal solution means 1 mole (or 40g) NaOH
present in 1000g of solvent.
Total mass of solution = 1000 + 40 = 1040g
Therefore, 1040g solution contains 40g NaOH
Therefore, 100g solution contains 
40
 100
1040
= 3.84% by mass.
Different Concentration terms
Relation between Molarity (M) and molality (m)
M
md
MM2
1
103
M2  Molar mass of solute
d  density of solution
Relation between molality(m) and mol-fraction (x2) of solute
x2 =
m
103
m+
M1
Where M1 = Molar mass of the solvent
x2 
mM1
mM1  103
Illustrative Example
Calculate the molality and mole fraction of the solute
in aqueous solution containing 3.0 g of urea
(molecular mass = 60) in 250 g of water.
Solution:
Molality 

Mass of solute
1000

Molecular mass of solute mass of solvent in gram
3
 1000  0.2
60  250
Moles of urea
3 / 60

 0.00359
Mole fraction of urea =
3
250
Total moles

60 18
Mole fraction of water = 1 – 0.00359 = 0.996
Illustrative Example
Calculate the mol fraction of ethanol and water
in a sample of rectified spirit which contains
95% of ethanol by mass.
Solution:
95% of ethanol by mass means 95 g ethanol
present in 100 g of solution.
Hence, mass of water = 100 – 95 = 5 g
Moles of C2H5OH =
95
= 2.07 moles
46
5
= 0.28mol
18
0.28
= 0.88
Mole fraction of C2H5OH =
0.28 + 2.07
Moles of water(H2O)=
Mole fraction of water = 1 – 0.88 = 0.12
Vapour pressure of solution
Liquid molecules evaporate
from the surface
vapour
pressure of
Vapourised
molecules
condensed to liquid
pure liquid
Both processes reach equilibrium
Po=Pressure exerted by the vapour
above the liquid surface at eqm.
Factors affecting Vapour Pressure
Nature of liquid:
More volatile liquids exert more
pressure on the liquid surface.
Temperature:
Increase in temperature
increases vapour pressure.
Vapour pressure  Temperature
Presence of a solute
Due to presence of volatile and non-volatile solute,
vapour pressure of solution decreases.
Vapor Pressure of Solution
Some of the solute particles
will be near the surface.
Block solvent molecules
from entering the gas phase.
Less no. of molecules per
unit surface area are
involved in equilibrium.
Raoult’s law for non-volatile solute
For highly dilute solutions
ps=x1po
po=vapour pressure of pure
liquid
x1=mol. fraction of solvent
ps=vapour pressure of
solution
Raoult’s law for non-volatile solute
Applicable for ideal solution
Hmix  0 Vmix  0
Here, solute-solute and solvent-solvent
interaction exactly equal in magnitude
with solute-solvent interaction.
Relative lowering of vapour pressure
From Raoult’s law,
ps  xsolvent po ,
xsolute  1 

po  ps
po
po  ps
po
ps
po
o
p  ps
o
p
 1  xsolute
 Relative lowering of v.p,
po  ps
o
p
po
n

nN

ps
n
N
n
N
when n  10%
 xsolute
moles of solute
moles of solvent
Modification (two volatile liquids)
According to Raoult’s law,for two
volatile miscible liquids
ps  pA o x A  pBo xB
 pA  pB    (1)
xA  xB  1
pA
xA
Partial vapour pressure of A.
Mol fraction of A in liquid phase.
Modification (two volatile liquids)
Illustrative Example
Vapour pressure of liquids A and B at a particular
temperature are 120 mm and 180 mm of Hg. If 2 moles
of A and 3 moles of B are mixed to form an ideal solution,
what would be the vapour pressure of the solution?
Solution :
pAo  120 mmHg pBo  180 mmHg
nA  2 mol nB  3mol.
xA 
2
5
xB 
3
5
pS  pA o x A  pBo xB
 120 
2
3
 180 
5
5
 48  108  156
Illustrative Problem
At 40oC, the vapour pressure in torr of methyl
alcohol-ethyl alcohol solution is represented by
P = 119Xm + 135 where Xm is the mole fraction
of methyl alcohol. What are the vapour pressures
of pure methyl alcohol & ethyl alcohol ?
Solution
o
P = pm
xm + pEoxE
o
= pm
xm + pEo 1 - xm 


o
= pm
- pEo xm + poE
Comparing it with
p  119xm  135
pEo  135
o
pm
 pEo  119
pom  119  135  245 torr
Illustrative Problem
6g of urea is disolved in 90g water at 25oC ?
What is vapour pressure of sol. If vapour
pressure of water is 40mmHg.
Solution
ps = po x
solvent
Xsolvent =
nsolvent
ntotal
=
90/18
90/18 + 6/60

5
 0.980
5  .1
ps = 0.980 x 40 = 39.2 mm Hg
Modification (two volatile liquids)
From Dalton’s law of partial pressure
pA = yAps - - - (2)
yA=mol. fraction of A in vapour phase
ps=vapour pressure of solution.
From (2)
yA 

pA
pS
pA o x A
pA o x A  pBo xB
Modification (two volatile liquids)
pA o x A  pBo xB
1

yA
pA o x A
1
1
pBo (1  x A )
pA x A
pBo
o
o
pA x A

pBo
pA o
pBo
pBo
1

1 o
 o
yA
pA x A pA
x A  xB  1
Illustrative Problem
An unknown compound is immiscible
with water. It is steam distilled at 98.0oC
and P = 737 Torr.poH20 = 707 torr at
98.0oC. This distillate was 75% by weight of water.
Calculate the molecular weight of the unknown
Solution
Using Dalton’s law of partial pressure
Ptotal = 737 torr PoH2O = 707 torr
Pounknown = 737 – 707 = 30 torr.
If water = 100 g
the unknown = 75.0 g
Pounknown nunknown
75


 18
o
nH2O
M  100
P H2O
30
75  18

707 m  100
m  318.15 gmol
Non-ideal solution
Solute-solvent interaction are different than solute-solute and
solvent solvent in non ideal solutions.
These do not obey Raoult’s Law.
For non ideal solutions
H  0 and V  0
Non ideal solution

For solution showing positive
deviation from Raoult's law.

For solution showing negative
deviation from Raoult's law.
H  0 and V  0
H  0 and V  0
Ps >Pideal
Ps  Pideal
Azeotropic mixtures
Liquid mixtures which distil without any
change in composition are called Azeotropes
or Azeotropic mixtures.
Solution showing positive deviation from
Raoult’s form minimum boiling azeotrope
Interaction between A–B < interaction
between A–A or B–B
Azeotropic mixtures
Solution showing negative deviation from
Raoult’s law form maximum boiling azeotropes
Interaction between A – B > interaction between A – A or B – B
Thank you