CHAPTER 4
First year
 Solutions
A
By
Dr. Hisham Ezzat
2011- 2012
1
The Dissolution Process

Solutions are homogeneous mixtures of two
or more substances.
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Dissolving medium is called the solvent.
Dissolved species are called the solute.
There are three states of matter (solid, liquid,
and gas) which when mixed two at a time
gives nine different kinds of mixtures.


Seven of the possibilities can be homogeneous.
Two of the possibilities must be heterogeneous.
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The Dissolution Process
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Solute
Solid
Liquid
Gas
Liquid
Solid
Gas
Gas
Solid
Liquid
Seven Homogeneous Possibilities
Solvent
Example
Liquid
salt water
Liquid
mixed drinks
Liquid
carbonated beverages
Solid
dental amalgams
Solid
alloys
Solid
metal pipes
Gas
air
Two Heterogeneous Possibilities
Gas
dust in air
Gas
clouds, fog
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Ways of Expressing Concentration
 Qualitative
Terms:

Dilute Solution – A dilute solution has a relatively
small concentration of solute.

A concentrated solution has a relatively high
concentration of solute.
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Quantitative Terms
Quantitative expressions of concentration require specific
masses,
moles, or liters of the solute,
solvent, or solution
information regarding such quantities as
The solution process: Polar materials dissolve only in polar
solvents (NaCI/H2O), and non - polar substances are soluble
in non - polar solvents. This is the first rule of solubility
"like dissolves like" e.g: benzene in CCI
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4
[A] Weight to Weight expression.
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1. Weight percent (wt %)
Number of grams of solute which present in 100 gram
of solution.
w eightof s olute
mas s %
x 100
totalw eightof s olution
w e ig h to f s o lu t e
w e ig h tf r a c t io n

t o t a lw e ig h to f s o lu t io n
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
e.g. 10% by weight glucose means:

10 gm glucose + 90 gm H2O = 100 gm solution.

% Solute = (10/100) x 100 = 10 %.
% Solvent = (90/100) x 100 = 90 %

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2. Mole fraction (X):
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3. Molality
Molality is a concentration unit
based on the number of moles of
solute per kilogram of solvent.
molesof solute
Molalitym 
kilogramsof solvent
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N.B
W solution = W solute + W solvent
W solvent = W solution - W solute
W solvent =dV solution - W solute
Where d = density of the solution
V = Volume of the solution
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Example 1:
What is the molality of 12.5 % solution of
glucose C6H1206, in water? M.wt. of
glucose is 180.0
Solution:
1) in 12.5 % solution 12.5 gm C6H12O6 is dissolved in l00 gm
solution.
W solvent = 100 - 12.5 = 87.5 g H2O
2) no. of moles glucose = 12.5/180
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Example 2: What are the mole fractions of solute and solvent
in a 1.0m aqueous solution?
Solution: The molecular weight of H2O is 18.0 we find the
number of moles of water in 100 gm of H2O.
1000
 55.6 mol H2O
no of moles of H2O 
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A 1.0 aqueous solution contains

n solute =1.0 mol
n H2O  55.6
n total
The mole fractions are
X solute  n
solute
n total
1.0 mol

 0.018
56.6 mol
 55.6
 0.982
n total  56.6
X water =  n
H2O
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[B] Weight to Volume expression:
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Molarity
moles of solute
Molarity M 
liters of solution
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Example 3:
a) How many grams of concentrated nitric acid solution
should be used to prepare 250 ml of 2.0M HNO3? The
concentrated acid is 70.0 %
Solution: a) 70 gm HNO3  l00 gm solution
M
W
1000
x
M.Wt 250
2
W 1000
x
63 250
2 x 63 x 250
Mass of pure HNO3 
1000
31.5 x 100
 45.0 gm
mass of HNO3 solution =
70
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b) If the density of the concentrated nitric acid solution is
1.42 g/ml. What volume should be used? M.wt. (HNO3)
=63
ml cone. NHO3 = (45/1.42) = 31.7 ml cone. HNO3
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Example 4:
An aqueous solution of acetic acid was prepared by
dissolving 164.2 gm of acetic acid in 800 ml of the solution.
If the density of the solution was 1.026 gm/ml. M. wt of
acetic acid = 60
Calculate:
a) The molar concentration of the solution
b) The molality
c) The mole fraction of both the solute and the solvent
d) The mole %
e) The weight %.
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Solution:
a)
W
1000 164.2x 1000
M
x

 3.4
M.Wtsolute Vml
60 x 800
b) d = 1.026g/ml
V = 800 ml
W solution = V x d = 800 x 1.026 = 820.8 gm
W slvent = 820.8 - 164.2 = 656.6 gm
mass
1000
164.2 x 1000
m
(Solute) x

 4.17 m
M.Wt
W(Solvent)
60 x 656.6
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c) no. of acetic acid moles = 164.2 / 60 = 2.737 mole
no. of H2O moles = 656.6 / 18 = 36.44 mole
2.737
 0.0699
Mole fraction of acetic acid =
2.737  36.44
Mole fraction of H2O =
36.44
 0.9299
2.737  36.44
d) mole % acetic acid = 0.0699 x 100 = 6.99 %
mole % of H2O = 0.9299 x 100 = 92.99 %
164.2 x 100
e) percentage weight of acetic acid = 820.8  20 %
percentage weight of H2O=
656.6 x 100
 80 %
820.8
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Try ?
1. Five grams of NaCl is dissolved in 25.0 g of H2O. What is
the mole fraction of NaCl in the solution? (Answer
=0.0580)
2. What is the mole percent NaCl in the previous problem 1
(Answer = 5.80 mol %)
3. Ten grams of ascorbic acid (vitamin C), C6H8O6, is dissolved
in enough water to make 125 ml of solution. What is the
molarity of the ascorbic acid? (Answer = 5.80 mol %)
4. What is the molality of NaCl in the solution in the previous
problem 1? (Answer = 3.42 m)
5. What is the mass percent of NaCl in the solution in the
previous problem 1? (Answer = 16.7 %)
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Try ?
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Example 1
Five grams of NaCl is dissolved in 25.0 g of
H2O. What is the mole fraction of NaCl in the
solution?
solution:
The formula weight of NaCl is 58.44, so 5.00 g
of NaCl is
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The molecular weight of water is 18.02; so
25.0 g of H2O is
In this solution, then, the mole fraction of
NaCl is 0.0580. (The mole fraction of water is
1.0000 - 0.0580, or 0.9420.)
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Example 2
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What is the mole percent NaCl in the of
Example 1
mol % NaCI = XNaCl X 100 = 5.80 x 10-2 x
100 = 5.80%
The solution is 5.80 mol % NaCI and 94.20
mol % H2O
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