Solutions
Definitions
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A solution is a homogeneous mixture
Solvent is the liquid in which the solute is dissolved
an aqueous solution has water as solvent
A saturated solution is one where the concentration is
at a maximum - no more solute is able to dissolve.
A solution is composed of a solvent which is the
dissolving medium and a solute which is the
substance dissolved.
In a solution there is an even distribution of the
molecules or ions of the solute throughout the
solvent.
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The concentration of a solution can be
expressed in a variety of ways
(qualitatively and quantitatively).
Formerly, the concentration of a solution
can be expressed in four ways:
– Molarity(M): moles solute / Liter solution
– Mass percent: (mass solute / mass of
solution) * 100
– Molality (m) - moles solute / Kg solvent
– Mole Fraction(cA) - moles solute / total moles
solution
Qualitative Expressions of
Concentration
A solution can be qualitatively described as
 dilute: a solution that contains a small
proportion of solute relative to solvent, or
 concentrated: a solution that contains a
large proportion of solute relative to
solvent.
Semi-Quantitative Expressions of Concentration
 A solution can be semi-quantitatively described as
 Unsaturated: a solution in which more solute will
dissolve, or
 Saturated: a solution in which no more solute will
dissolve.
 The solubility of a solute is the amount of solute
that will dissolve in a given amount of solvent to
produce a saturated solution.
Percent Composition (by mass)
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We can consider percent by mass (or weight percent,
as it is sometimes called)
We need two pieces of information to calculate the
percent by mass of a solute in a solution:
The mass of the solute in the solution.
The mass of the solution.
Use the following equation to calculate percent by
mass:
Molality
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Molality, m, tells us the number of moles of
solute dissolved in exactly one kilogram of
solvent. (Note that molality is spelled with two
"l"'s and represented by a lower case m.)
We need two pieces of information to
calculate the molality of a solute in a solution:
The moles of solute present in the solution.
The mass of solvent (in kilograms) in the
solution.
To calculate molality we use the equation:
Molarity
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Molarity tells us the number of moles of solute in
exactly one liter of a solution. (Note that molarity is
spelled with an "r" and is represented by a capital M.)
We need two pieces of information to calculate the
molarity of a solute in a solution:
The moles of solute present in the solution.
The volume of solution (in liters) containing the
solute.
To calculate molarity we use the equation:
Mole per liter solutions
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A mole per litre (mol/l) solution contains
one mole of a solute dissolved in, and
made up to 1 litre with solvent.
Mole:
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A counting unit
Similar to a dozen, except instead of 12, it’s 602
billion trillion 602,000,000,000,000,000,000,000
6.02 X 1023 (in scientific notation)
A Mole of Particles Contains 6.02 x 1023 particles
1 mole C= 6.02 x 1023 C atoms
1 mole H2O
= 6.02 x 1023 H2O molecules
1 mole NaCl = 6.02 x 1023 NaCl “molecules”
The mole is the SI base unit that measures an
amount of substance.
One mole contains Avogadro's number
(approximately 6.023×1023) (number of atoms or
molecules).
Preparation of mol/l solution
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To preparation a mol/l solution, use the
following formula:
Number of grams to be dissolved in 1 litre of
solution= Required mol/l solution X Molecular
mass of substance
Example
Make a solution of 50 ml of sodium chloride, 0.15
mol/l:
 Required mol/l concentration= 0.15
 Molecular mass of NaCl= 58.44
 Therefore 50 ml NaCl, 0.15 mol/l contains:
0.15X 58.44 X 50 = 0.438 g of the chemical
.
1000
substance dissolved in 50 ml of solvent.
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Make liter of sodium chloride (NaCl),
1 mol/l
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To make 1 litre of sodium chloride (NaCl), 1
mol/l:
Required mol/l concentration= 1
Molecular mass of NaCl= 58.44
Therefore 1 lit NaCl, 1 mol/l contains:
1 X 58.44= 58.44 g of the chemical dissolved
in 1 litre of solvent.
To make 1 litre of sodium chloride,
0.15 mol/l (physiological saline)
Conversion a percentage solution into
a mol/l solution:
By the following formula:
 Mol/l solution =
g% (w/v) solution X 10
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Molecular mass of the substance
Examples:
 To convert a 4% w/v NaOH solution into a mol/l
solution:
 Gram % solution = 4
 Molecular mass of NaOH= 40
 Conversion to mol/l= 4X10 = 1
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40
 Therefore 4% w/v NaOH is equivalent to NaOH, 1
mol/l solution.
Convert a 0.9% w/v NaCl solution
into a mol/l solution
Conversion of a normal solution into a
mol/l solution:
By the following formula:
mol/l solution= Normality of solution
Valence of substance
 Examples:
Convert 0.1 N (N/10) HCl into a mol/l solution:
Normality of solution= 0.1
Valence of HCl= 1
Conversion to mol/l= 0.1/1= 0.1
 Therefore 0.1 N HCl is equivalent to HCl, 0.1mol/l
solution.
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To convert 1 N Na2CO3 into a mol/l
solution (valence =2)
How to dilute solutions and body fluids
In the laboratory it is frequently necessary to dilute
solutions and body fluids to reduce its concentrations.
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Diluting solutions:
A weaker solution can be made from a stronger
solution by using the following formula:
Volume (ml) of stronger solution required= R X V
O
Where: R= concentration of solution required
V= volume of solution required.
O= strength of original solution.
Examples:
 To make 500 ml of NaOH, 0.25 mol/l
from a 0.4 mol/l solution:
C= 0.25 mol/l, V= 500 ml, S=0.4 mol/l ml
of stronger solution required:
0.25 x 500 = 312.5 ml
0.4
 Therefore, measure 312.5 ml NaOH,
0.4 mol/l and make up to 500 ml with
distilled water.
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Example 1
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Make 100 ml glucose, 3 mmol/l in 1 g/l
benzoic acid from glucose 100 mmol/l
solution :
C= 3 mmol/l V= 100 ml S= 100 mmol/l
ml of stronger solution required= 3 x 100 = 3
100
Therefore, measure 3 ml of glucose, 100
mmol/l and make up to 100 ml with 1 g/l
benzoic acid.
Example 2
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To make 500 ml H2SO4, 0.33 mol/l from concentrated
H2SO4 which has an approximate concentration of 18
mol/l:
C= 0.33 mol/l, V= 500 ml S= 18 mol/l
ml of stronger solution required= 0.33 x 500 = 9.2
18
Therefore, measure 9.2 ml conc. H2SO4, and slowly add
it to about 250 ml of distilled water in a volumetric flask.
Make up to 500 ml with DW.
Make 1 litre of HCl, 0.01 mol/l from a
1.0 mol/l solution:
Diluting body fluids and calculating
dilutions
To prepare a dilution or series of dilutions of a body
fluid:
Examples: To make 8 ml of 1 in 20 dilution of blood:
 Volume of blood required= 8/20 = 0.4 ml
Therefore, to prepare 8 ml of a 1in 20 dilution, add
0.4 ml of blood to 7.6 ml of diluting fluid.
2- To make 4 ml of a 1 in 2 dilution of serum in
physiological saline:
Volume of serum required= 4/2 = 2 ml
Therefore, to prepare 4 ml of a 1in 2 dilution, add 2 ml
of serum to 2 ml of physiological saline.
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To calculate the dilution of a body fluid
Examples:
1- Calculate the dilution of blood when using 50 µl of
blood and 950 µl of diluting fluid:
 Total volume of body fluid and diluting fluid=
50+950=1000 µl
 Therefore, dilution of blood: 1000/50= 20
 i. e. 1 in 20 dilution.
2- Calculate the dilution of urine using 0.5 ml of urine
and 8.5 ml of diluting fluid (physiological saline):
 Total volume of urine and diluting fluid= 8.5+0.5=9
ml
 Therefore, dilution of urine: 9.0/0.5= 18
 i. e. 1 in 18 dilution.