Crystals are homogeneous solids, having regular and
definite geometrical shape with faces and sharp edges.
Pure crystals of compounds have sharp melting points.
When a super saturated solution of a solid is prepared
at high temperature and allowed to cool down, then at
lower temperature it cannot hold more solute in
dissolved state. Some of these dissolved solute
particles come out of solution, in solid form having
regular and definite geometric shapes. They are called
crystals. The process in which dissolved solute comes
out of solution and forms crystals is called
Following are two practical examples of super saturated
solution preparation and crystallization.
Preparation of Crystals of Copper
Sulphate (Blue Vitriol CuSO4.5H2O)
Prepare a saturated solution of copper sulphate, in
water at room temperature using a beaker. Heat the
saturated solution and try to dissolve some more
quantity of copper sulphate while stirring the solution
with glass rod. Allow the super saturated solution of
copper sulphate to cool down at room temperature. Upon
cooling and standing, crystals of CuSO4.5H2O,
i.e., blue vitriol will form. Filter out the crystals
and observe the shape of crystals under a light
Preparation of Crystals of Potassium
Take 100 ml water in a beaker. Prepare a saturated
solution of KN03 by dissolving 37g of solute at room
temperature by means of stirring with glass rod. Now
heat this saturated solution to 50°C and dissolve 20g of
additional KNO3 while stirring the solution.
Filter the hot super saturated solution quickly and
collect the filterate in another beaker. Cool the
filterate to room temperature. Upon cooling crystals of
KNO3 are formed.
Filter out the crystals and observe their shape under
Purification of Solids by
An impure solid generally contains two types of
impurities. An insoluble impurity and a soluble
impurity. Insoluble impurity is totally insoluble in the
solvent used for crystallization even at boiling
temperature. While the soluble impurity remains in
soluble form at room temperature. Therefore, a compound
containing such two types of impurities could be easily
removed by means of crystallization technique.
For example, a 42g impure sample of KNO3
contains a small quantity of sand and NaCl. To obtain
pure crystalline KNO3, we perform the
crystallization technique as follows:
Take 50ml of water in a beaker and add the impure sample
(40g) of KNO3 to it while stirring with glass
rod. Supply heat gently till the temperature of the
solution is above 50°C. Stirr the solution at this
temperature till most of the solid is dissolved. Filter
the hot solution and collect the filtrate in a beaker.
Sand being in soluble in water will be removed and
collected on the filter paper. Upon cooling of the
filtrate, crystals of KNO3 will start
appearing. When no further crystals are formed, filter
it again and collect the filtrate in a beaker. Purified
crystals of KNO3 are obtained on the filter
paper. The filtrate will contain some quantity of the
dissolved KNO3 along with the NaCI, being a
7.5 STRENGTHS OF A SOLUTION
The strength (concentration) of a solution means the
mass or volume of the solute present in known amount of
solvent or solution.
The following are the common methods of expressing the
strength (concentration) of a solution.
It is not used in these days.
1. Molality (M):
It is defined as the number of moles of solute dissolved
in 1 liter (1dm3) of a solution; it is
denoted by (M).
Thus, 1 mole of NaOH (i.e. its gram formula mass) 40g
dissolved in 1 liter (1dm3) of solution is
said to be 1 molar (1M) solution. If only half of the
mole i.e. 20g of NaOH is dissolved in one liter (1dm3)
of solution, the solution is said to be one-half molar
The molarity of any solution is found by dividing the
number of moles of solute by the number of liters of
Molarity (M) = Moles of Solute / Liters of Solution or
dm3 of solution
We know that
Number of moles = Given mass of solute / Gram formula
mass of solute
Liters of solution= mls of solution / 1000
Molarity = (M) moles of solute / Gram formula mass
of solute x 100cm3 (ml) / ml(cm3)
Calculate the molarity of a solution containing 4g of
NaOH in 500 cm3 (ml) of solution?
M= Mass of solute x 1000 cm3 / Gram formula
mass x cm3 of solution
Mass of solute NaOH
Gram formula mass of solute =
cm3 of solution
M=4g.1000cm3 / 40g x 500cm3 = 1/5
2. Molality (m):
It is defined as the number of moles of solute dissolved
per 1000g (1kg) of solvent; it is denoted by (m).
Thus mole of Na2CO3g (i.e. its
gram formula mass) 106g dissolved in 1000g of solvent is
said to be 1 molal (lm) solution. If only half of the
mole i.e. 53g of Na2C3 is
dissolved in 1000g of solvent, the solution is said to
be one-half molal (i.e. 0.5m)
The molality of a solution is found by the following
Mass of solute * 1000c No. of moles of solute
Molality (m) =Mass of solute x 1000g / Gram formula mass
x grams of solvent or
No. of moles of solute / Kg of solvent
Calculate the molality of solution containing 5.3g of Na2CO3
in 500g of water.
Molality (m) = Mass of solute x1000g / Gram formula mass
of solute x grams of solvent
= (m) =?
Mass of solute i.e. Na2CO3
Gram formula mass of solute Na2CO3
Grams of solvent (water) =
Molarity =m= Mass of solute x 1000g / Gram formula mass
of solute x gram of solvent
M = 5.3g x 1000g / 106 x 500g = 1/10 = 0.1m
3. Mole fraction (X):
Mole fraction (X) of any component in a solution is the
number of moles of the component divided by total number
of moles making up a solution. It is denoted by X.
Mole fraction (X) = Moles of component
Total number of moles making
For example, a solution is prepared by dissolving 1 mole
of ethyl alcohol C2H5-OH in 3
moles of water (H2O), where NA and
NB the represent the number of moles of ethyl
alcohol and water respectively.
Mole fraction of ethyl alcohol = XA = NA/NA+NB
=1/1+3 = 1/4 = 0.25
Mole fraction of water = XB = NB/NA+NB
= 3/1+3 = 3/4 = 0.75
Results: Mole fraction of ethyl alcohol = XA
Mole fraction of water = XB =
Note, the sum of the mole of fractions is equal to 1.
Mole fraction of ethyl alcohol = 0.25
Mole fraction of water = 0.75
Sum of the mole of fractions = 1.00
The mole fraction is dimensionless quantity that
expresses the ratio of the number of moles of
one component to the number of moles of all
components present. The sum of mole fractions of
all components of a solution must equal 1
4. Percentage (%):
This is based on mass (m) or volume (v) of components of
solution. It is of four types.
in M/M (mass/mass)
in M/V (mass/volume)
in V/M (volume/mass)
in V/V (volume/volume)
solution means solute 5g in 95gm solvent.
solution means solute 10g in solution 100cm3.
solution means solute 5cm3 in solution 100 g.
solution means solute 15 cm3 in 85 cm3 solvent.
Problem of %:
6.5g of NaCI are dissolved in 43.5 g water. Calculate
the percent by mass of NaCI in this solution.
% by mass of solute NaCl = Mass of solutex100/Mass of
solute+Mass of solvent
= 6.5x100/6.5+43.5 = 6.5x100/50.0
Percent by Mass of NaCl =13.0%
If sugar or salt is dissolved in water, the crystals
dissolve into molecules or ions. The resulting
homogenous mixture is called true solution. If fine sand
is stirred in water, the crystals do not dissolve, but
even after several days some of the smallest particles
remain suspended, such mixture is called a suspension.
Suspension is defined as a heterogeneous mixture
consists of visible particles, each of which contains
many thousands or even millions of molecules, surrounded
by molecules of liquid.
Important to note
mixtures can be classified according to size of the
constituent particles, as solution, suspension or
colloids. In solution the dispersed particles are of
molecular size (0.1-lnm). In suspension, the
dispersed particles are much larger than molecules
Between these two
extreme type of dispersion, there is another type of
homogenous mixture in which dispersed particles are
larger than molecules (2.0-1000 nm) but not large
enough to settle out. This intermediate type of
mixture is called colloid.
Difference between Solution and Suspension
6. Components cannot be
separated by filtration. Solution
1. The size of particles
is between 0.1 to lnm.
2. Particles cannot be
seen with low power microscope.
3. It is homogeneous.
4 Particles do not
5. It is transparent.
6. Components cannot be
separated by filtration.
1. The size of
particles is larger than 1000 nm.
2. Particles can
be seen by low power microscope.
3. It is
5. It is not
6. Components can
be separated by filtration.
Examples of Suspensions in Daily Life:
1) Smoke: A suspension of the particles of carbon
in a gas or air.
2) Mud (slime): A suspension of fine particles of
solid in small quantity of liquid.
3) Foam (froth): A suspension of fine particles
of a gas in a liquid.
4) Emulsion: A suspension of droplets of one
liquid into another in which it is not soluble.