CHAPTER:6
SOLUTIONS
CEMISTRY :9th
Q: WHY SOLUTIONS ARE OMPORTANT FOR US?
 Brass, steel, german silver, air, soft drinks, juices, shampoo, crude oil, cough syrup and many others are all
solutions. These solutions are widely used for making cooking utensils, surgical tools, cutlery, musical
instruments, health tonics, motor fuel and many other objects.
 Gases when dissolved in liquid also produce some important solutions.
For example, water dissolves small amount of air giving a solution whose oxygen content (Solute) is important
for the survival of fish and other aquatic animals. Carbon dioxide gas dissolves readily in water, for this reason,
it is used in making carbonated drinks.
 In food industries, vegetable oil is converted into vegetable ghee by passing H2 gas through the oil. Nickel
(used as catalyst) accelerates this process by absorbing H2 gas on its surface, producing a solution, which is gas
in solid solution.
Q; DEFINE FOLLOWING WITH EXAMPLE?
 Solution
A homogenous mixture of two or more different chemical substances with uniform chemical and physical properties is called
SOLUTION.
Example:
Sugar solution is a mixture of sugar and water. Salt solution is a mixture of salt and water.
 Binary Solution
A solution which consist of only two substances is called BINARY SOLUTION.
Example
Sugar solution (sugar+water)
Salt solution (salt + water)
 Aqueous Solution
The word aqueous is derived from the Latin word Aqua which means water. A solution in which water is used as a solvent is
called AQUEOUS SOLUTION.
Example
Sugar solution and salt solution. In both solutions, water is used as a solvent
 Solute
The substance or component present in lesser amount in solution is called SOLUTE. Example
In salt sugar solutions, salt and sugar act as a solute.
 Solvent
The substance or component of solution present in relatively large amount in solution
is called SOLVENT.
Example
In salt and sugar solutions, water is a solvent.
Q; DEFINE FOLLOWING WITH SUITABLE EXAMPLES?
 Unsaturated Solution
A solution which can dissolve further amount of solute at a particular temperature is
called an unsaturated solution.
Example
Take a beaker half filled with water. Add a spoon of sugar in it. It will dissolve. Such a solution is unsaturated because it can still
dissolve more amount of solute in it.
 Saturated Solution
A solution which contains the maximum amount, of a solute at a particular temperature and which is unable to dissolve further
amount of solute in it is saturated solution.
Example: Go on adding sugar in the above solution till it starts settling down at the bottom of the beaker at a particular
temperature, it will be saturated solution.
 Supersaturated Solution
A solution which contains more amount of a solute in a particular amount of solvent then the saturated solution is called
supersaturated solution.
For Example
Add more sugar in the above solution, stir it, it will dissolve. Go on a adding more sugar and stir it. A stage will come when no
more sugar will dissolve and will start settling down at the bottom of the beaker. This solution is called saturated solution. Now
heat the above solution, some more solute will dissolves. This solution will be called supper, saturated solution.
1|Page
BY: M. ZIA DOGAR
CHAPTER:6
CEMISTRY :9th
SOLUTIONS
Q: How to know whether a solution is unsaturated, saturated or supersaturated?
A supersaturated solution is not stable in the presence of crystals of solute. If we add a crystal of sodium thiosulphate to its
saturated solution, it will simply drop to the bottom, without dissolving but if we add a crystal of sodium thiosulphate to a
Supersaturated solution of sodium thiosulphate, crystallization will starts while in unsaturated solution the crystal will be
dissolved.
When crystallization has finished, we will have a saturated solution in presence of sodium Thiosulphate.
SELF ASSESSMENT EXERCISE 6.1
The maximum amount of sodium acetate that dissolves in 100g of water at 0oC is 119g and 170g at 100oC.
o
(a) If you add 170g of sodium acetate in 100g of water at 0 C, how much will dissolve?
o
ANS: 119g of Sodium acetate will dissolve in 100g of water at 0 C.
(b) Is the solution saturated unsaturated or supersaturated?
ANS: The solution will be saturated
(c) If the solution in heated to 100oC, is the solution now saturated, unsaturated or supersaturated.
ANS: The solution will be saturated
o
(d) If the solution is cooled back to 0 C and no crystals appear. Is the solution saturated, unsaturated or supersaturated?
ANS: the solution will be supersaturated
Q: Descibe various types of solutions?
TYPES OF SOLUTIONS
Solution exists in any one of the three states of matter i.e. solid, liquid or gas. Physical state of solution is same as that for
solvent. In fact, nine different types of solutions can be prepared by mixing together substances. These substances, in any
physical state can serve as a solute or solvent.
Common types of solutions
S. No.
Solute
Solvent
State of resulting
solution
Examples
1
Gas
Gas
Gas
Air (a mixture of O2, N2, CO2,
etc.), water gas
(CO + Hz)
2
Gas
Liquid
Liquid
Soda water (carbonated drink)
Hydrochloric Acid
3
Gas
Solid
Solid
H2 absorbed on Ni, Pt, Pd
4
Liquid
Gas
Gas
Mist, fog, clouds
5
Liquid
Liquid
Liquid
Alcohol in water
6
Liquid
Solid
Solid
Amalgams Hg in Ag, wet NaCl in
rainy season
7
Solid
Gas
Gas
Carbon particles in air (smoke)
8
Solid
Liquid
Liquid
Sugar in water, salt in water
9
Solid
Solid
Solid
Alloys such as Solder (Pb + Zn),
Bronze (Cu + Zn)
SELF ASSESSMENT EXERCISE 6.2
What are the physical states of solute and solvent in each of the following solutions. Also identify the type of solution.
(a) Deep sea divers use a breathing mixture of helium and oxygen.
2|Page
BY: M. ZIA DOGAR
CHAPTER:6
(b)
(c)
(d)
(e)
(f)
(g)
CEMISTRY :9th
SOLUTIONS
Brass contains 80% copper and 20% zinc.
Dental filling.
Brine (salt in water).
Drinking water containing chlorine as disinfectant.
Gemstone, Ruby contains Cr2O3 and Al2O3.
Conc. H2SO4, we use in the laboratory is 98% H2SO4 and contains only 2% H2O.
Solution:
S.No.
a.
b.
c.
d.
e.
f.
g.
Solutions
Mixture of Helium and
Oxygen
Brass
Dental filling
Brine solution
Drinking water with
chlorine
Gem stone- Ruby
H2SO4, 98%
State of Solute
Helium (g)
State of Solvent
Oxygen (g)
Type of Solution
Gas is Gas
Zinc(s)
Mercury-Hg (l)
Salt (s)
Chlorine (g)
Copper(s)
Silver- Ag (s)
Water (l)
Water (l)
Solid in Solid
Liquid in solid
Solid in liquid
Gas in liquid
Cr2O3(s)
2%, H2O
Al2O3(s)
H2SO4, 98%
Solis in solid
Q: Write note on following?
(i)
Solutions of gases
(ii)
Solutions of liquids
(iii)
Solutions of solids

ANSWER
Solutions of Gases
Gaseous solutions are commonly used by chemical industries to prepare chemical substances.
For Example
i)
Ammonia synthesis: A gaseous mixture of nitrogen and Hydrogen is used with ratio of N2: H2 is 1:3 which is strictly
maintained under varying reaction conditions.
ii)
Urea Fertilizer: A gaseous mixture of Ammonia and C02 is used for synthesis of urea.
iii)
Nitric Acid: A gaseous mixture of NH3 and oxygen is used for the preparation of nitric acid.
In all these cases, gaseous mixture or solution of gases is used. In these solutions, solute and solvent both are gases.
Solutions of Liquids
Solution of liquid in gas, liquid or solid solvents are also very common.
For Example
1) Fog, clouds or mist: In this liquid in gas mixture, water vapours are dissolved in air (solvent).
2) When a person feels weakness. So 0.85% m/m NaCl solution is used in intravenous solution to persons suffering from
dehydration.
3) Amalgam is mixture of mercury as solute in solid like Ag or tin as solvent. It is widely used for dental cavity filling.
Solution of Solids
1. Smoke: It spreads in air forming solution that contains solid carbon particles. In this solution solid particles are solute
and air is solvent. We call such a solution as solid in gas.
2. Saline solution: it contains 0.85% m/m NaCl solution is used in intravenous solution that is given to persons suffering
from dehydration.
3. Alloys: It is mixture of solids in solids and is used on commercial basis.
 Brass is an alloy of copper and zinc
 Steel is an alloy of iron with carbon and silicon
 Gemstones are also solutions e.g Ruby, Opal etc
Q: Define concentration of solution with example?
Concentration of Solution
Def: It is the amount of solute present in a given amount of solvent or solution is called concentration of solution.
Since both parts of the ratio (solute and solvent) can be given in terms of mass, volume or moles, therefore chemists
use a variety of concentration terms e.g.: 5g of solute required to dissolve in 95g of solute to make 100 g of m/m solution.
3|Page
BY: M. ZIA DOGAR
CHAPTER:6
SOLUTIONS
CEMISTRY :9th
Q: Differentiate between dilute & conc. Solution?
Dilute Solution and Concentrated Solution
 A solution which contains small amount of solute in a solvent is called a dilute solution
 while a solution which contains large amount of solute in a solvent is called concentrated solution.
Example
1. 5g of NaCl in 95g of solvent in beaker A to make 100 g solution of NaCl
2. while in beaker B, 10 g of NaCl in 90g of solvent (water) to make 100 gm solution of NaCl
The beaker-A solution will be called dilute solution and beaker-B solution will be called concentrated solution.
Q: What are different units of concentration?
Ans:
Units of Concentration
Chemists uses many concentration units in routine work which are:
1) Percentage composition
2) Molarity
3) Normality
4) Molality
5) Mole fraction
6) Parts per million (ppm)
Q: What is Percentage Composition?
Percentage Composition
The percentage composition is the unit of concentration that specifies the quantity of solute in 100 parts of solution
Quantity of solute and solvent can be expressed by mass in grams or volume in cm3. Therefore, by the percentage of a solution
we mean the mass or volume of solute dissolved in 100g or 100cm3 of solution.
There are four different ways of expression for percentage composition.
i.
Mass by Mass Percentage (m/m)
It is the mass of the solute dissolved per 100 parts by mass of solution.
Example
10% m/m NaCl solution means that 10 g NaCl in 90g water to make 100g of solute
x100
% of solution by m/m =
ii.
Mass by Volume percent (m/v)
It is the mass of the solute dissolved per 100 parts by volume of solution.
For Example
10% m/v NaCl solutions means that 10g NaCl in 90cm3 water to make 100cm3 of solution.
x100
% of solution by m/v=
iii.
Volume by mass Percent (v/m)
It is the volume of solute dissolved per 100 parts by mass of solution.
For Example
10% v/m Alcohol solution and means that 10cm3 of Alcohol in 90g of water to make 100g of solution
% of solution by v/m =
x100
iv.
Volume by volume percent (v/v)
It is the volume of solute dissolved per 100 parts by volume of solution.
For Example
10% v/v Alcohol solution means that 10cm3of Alcohol in 90 cm3 of water to make 100 cm3 of solution.
% of solution by v/v =
x100
SELF ASSESSMENT EXERCISE 6.3
4|Page
BY: M. ZIA DOGAR
CHAPTER:6
1.
2.
SOLUTIONS
CEMISTRY :9th
Write four ways to express percentage of solutions.
Ans:See above topic (percentage composition)
A saline solution is administered intravenously to a person suffering from severe dehydration. This is labeled as 0.85%
m/v of NaCl. What does it mean?
Ans: It means that 0.85g of NaCl in 99.15 cm3 of water to make 100cm3 of solution.
Q Define molarity. What is the formula for finding molarity? What are the units of molarity?
Molarity
Molarity is the concentration unit in which the number of moles of solute dissolved per dm3 of solution. It is denoted by “M”.
Mathematically, the formula for finding Molarity can be written as:
Molarity (M)=
Number of moles of solute
3
Volume of Solution (dm )
or
Molarity (M) =
mass of solute
molar mass of solute
X
1
Volume of Solution (dm3)
Or
mass of solute
1000
Molarity (M) = molar mass of solute X Volume of Solution (cm3)
Units of Molarity
3
-3
Its units are mol per dm (mol.dm )
Q: what these terms refers?
a. 18 M Sulfuric acid?
b. 12.1 M Hydrochloric acid
ANS:
a. 18M concentrated H2SO4 means that there are 18 moles of H2SO4 in each dm3 of solution.
b. 12.1M concentrated HCl means that there are 12.1 moles of HCl in each dm3 of solution.
Problems Involving the Molarity of a Solution
1. Urea (NH2CONH2) is a white solid used as fertilizer and starting material for synthetic plastic. A solution contains 40g
3
urea dissolved in 500cm of solution. Calculate the molarity of this solution.
Solution
Mass of urea = 40g
Molar mass of urea (NH2CONH2=
= 1 4 + 1 x 2 + 1 2 + 1 6 + 1 4 + 1 x2 = 60g /mole
Mass of solute x 1000
Molarity (M ) = molar mass of solute x Volume of Solution (cm3)
Molarity (M ) =
40 x 1000
60 x 500
=1.334 Mol/dm3
Example 6.1
Calculating molarity from moles of solute and volume of solution
Potassium permanganate (KMnO4) is a dark blue-black compound. When it dissolves in water, it forms a bright purple solution. It
is used as disinfectant in water tanks. It is also known as pinky. A solution contains 0.05 moles of KMnO4 in 600cm3 of solution.
Calculate molarity of this solution.
Answer
Moles of KMnO4 = 0.05 mol
3
Volume of KMnO4= 600 Cm
Number of moles of solute x 1000
Molarity (M)=
3
Volume of Solution (cm )
Molarity (M)=
0.05 x 1000
600
3
= 0.083 mol/dm
5|Page
BY: M. ZIA DOGAR
CHAPTER:6
CEMISTRY :9th
SOLUTIONS
SELF ASSESSMENT EXERCISE 6.4
Potassium chlorate (KClO3) is a white solid. It is used in making matches and dyes. Calculate the molarity of solution
3
that contains. (a) 1.5 moles of this compound dissolved in 250cm of solution (b) 75g of this compound dissolved to produce
1.25dm3 of solution.(c) What is the molarity of a 50cm3 sample of potassium chlorate solution that yields 0.25g residue after
evaporation of the water.
Solution:
(a)
Number of moles of KClO3=
1.5 moles
3
Volume of solution
=
250 cm
Molarity of KClO3 =
?
Number of moles of solute x 1000
Molarity of KClO3 = Volume of Solution (cm3)
=
(b)
mass of KClO3
Molar mass of KClO3
Volume of solution
Molarity of KClO3 =?
Molarity (M) =
1.5 x 1000
=
250
6.0 M
= 75g
= 39+35.5+3x16= 122.5g
= 1.25dm3
mass of solute
molar mass of solute
75
X
1
3
Volume of Solution (dm )
1
=
X
122.5
1.25
= 0.49mol.dm3
or 0.49 M
Converting the molarity of a solution into its concentration in g/dm3
3
A flask contains 0.25M NaOH solution. What mass of NaOH is present per dm of solution?
Solution:
Molarity: 0.25 M
Volume of solution: 1 dm3
Molar mass od NaOH= 40g/mol
Mass of NaOH= ?
mass of solute
1
Molarity (M) = molar mass of solute X Volume of Solution (dm3)
Mass of NaOH=
molarity x molar mass x volume of soluton
= 0.25x 40x 1
= 10 g
Converting concentration in g/dm3 into molarity
Potassium hydroxide (KOH) is used in the manufacture of shaving creams, paints and varnish. An analyst makes up a
solution by dissolving 5.8g of KOH in one dm3 of solution. Calculate the molarity of this solution.
Solution:
3
Mass of KOH dissolved in one dm of solution = 5.6g
6|Page
BY: M. ZIA DOGAR
CHAPTER:6
SOLUTIONS
CEMISTRY :9th
Molar mass of KOH = 39+16+1
= 56g/mole
Molarity (M) =
mass of solute
molar mass of solute
5.6
= 56 X
=1M
X
1
Volume of Solution (dm3)
1
SELF ASSESSMENT EXERCISE 6.5
1.
Sodium hydroxide solutions are used to neutralize acids and in the preparation of soaps and rayon. If you dissolve 25g
of NaOH to make 1 dm3 of solution, what is the molarity of this solution?
SOLUTION:
Mass of NaOH = 25 g
3
Volume of solution: 1 dm
Molar mass of NaOH= 40g/mol
Molarity of NaOH solution= ?
Molarity (M) =
=
2.
mass of solute
molar mass of solute
25
x
X
1
3
Volume of Solution (dm )
1
= 0.625 M
3
A solution of NaOH has concentration 1.2M. Calculate the mass of NaOH in g/dm in this solution.
SOLUTION:
Molarity: 1.2 M
Volume of solution: 1 dm3
Molar mass od NaOH= 40g/mol
Mass of NaOH= ?
mass of solute
1
Molarity (M) = molar mass of solute X Volume of Solution (dm3)
Mass of NaOH=
3.
molarity x molar mass x volume of soluton
= 1.2x 40x 1
= 48 g
A solution is prepared by dissolving 10g of haemoglobin in enough water to make up 1dm3 in volume. Calculate
molarity of this solution. Molar mass of haemoglobin is 6.51x104g/mole.
SOLUTION:
Mass of haemolobin= 10 g
3
Volume of solution: 1 dm
4
Molar mass of haemoglobin= 6.51x 10 g/mol
Molarity of NaOH solution= ?
Molarity (M) =
=
mass of solute
molar mass of solute
10
.
~
x
X
1
3
Volume of Solution (dm )
1
-4
= 1.536x 10 M
7|Page
BY: M. ZIA DOGAR
CHAPTER:6
SOLUTIONS
CEMISTRY :9th
Q: How you can Prepare a solution of given molarity
Prepare 0.2M KMnO4 solution.
Hint


Volumetric flasks of capacity 1 dm3, 500cm3, 250cm3, 100cm3, 50cm3 can be used to prepare a solution.
3
3
Suppose you use a 100cm volumetric flask. First find the mass of KMnO4 to give 100 cm of 0.2 M KMnO4 solution.
Solution:
Required volume of solution = 100cm3
Molarity of solution= 0.2 M
Molar mass of KMnO4
= 39+55+16x4
= 39+55+64
=158g/mole
mass of solute
1000
Molarity (M) = molar mass of solute X Volume of Solution (cm3)
Mass of KMnO4 =
molarity x molar mass x volume of soluton
=
Mass of KMnO4
0.2 x158 x100
=3.16g
3
To prepare 0.2M KMnO4 in 100cm volumetric flask, you will add 3.16g of KMnO4.
SELF ASSESSMENT EXERCISE 6.6
1.
How can you prepare 500cm3 of 0.2M KMnO4 solution.
SOLUTION:
Required volume of solution = 500cm
3
Molarity of solution= 0.2 M
Molar mass of KMnO4
= 39+55+16x4
= 39+55+64
=158g/mole
Molarity (M) =
mass of solute
molar mass of solute
Mass of KMnO4 =
X
1000
3
Volume of Solution (cm )
molarity x molar mass x volume of soluton
=
Mass of KMnO4
0.2 x158 x500
=15.8g
To prepare 0.2M KMnO4 in 500cm3 volumetric flask, you will add 15.8g of KMnO4.
2.
3
How can you prepare 25cm of 0.25M solution of CuSO4.5H2O (Blue vitriol).
8|Page
BY: M. ZIA DOGAR
CHAPTER:6
CEMISTRY :9th
SOLUTIONS
SOLUTION:
Required volume of solution = 25cm3
Molarity of solution= 0.25 M
Molar mass of CuSO4.5H2O
= 63.5+32+16x4+18X5
=249.5 g/mole
Molarity (M) =
mass of solute
molar mass of solute
Mass of CuSO4.5H2O =
3.
1000
3
Volume of Solution (cm )
molarity x molar mass x volume of soluton
=
Mass of CuSO4.5H2O
X
0.25 x249.5 x25
=1.56g
3
3
To prepare25 cm and 0.25M CuSO4.5H2O , you will add 1.56 g of CuSO4.5H2O in 25 cm of solvent.
Preparing a solution of given molarity by diluting a solution of known molarity
3
3
Concentrated sulphuric acid is 18M H2SO4. How many cm of this acid is needed to produce 250cm of 0.1M H2SO4?
Solution:
M1 = molarity of given conc H2SO4 = 18M
V1 = volume of conc H2SO4 needed to dilute = ?
M2 = molarity of required H2SO4 solution = 0.1M
V2 = volume of required H2SO4 = 250cm3
Give H2SO4
M1V1
Desired H2SO4
=
M2V2
M1V1 = M 2 V2
V1 =
M 2 V2
M1
0.1 x 250
18
= 1.39cm3
=
Transfer 1.39cm3 of 18M H2SO4 to a 250cm3 volumetric flask and dilute it by adding water up to the mark and mix.
Resulting solution is 0.1M H2SO4.
SELF ASSESSMENT EXERCISE 6.7
1.
3
SOLUTION:
9|Page
3
A stock solution of hydrochloric acid is 12.1 M. How many cm of this solution should you use to prepare 500cm of
0.1 M HCl.
BY: M. ZIA DOGAR
CHAPTER:6
CEMISTRY :9th
SOLUTIONS
M1 = molarity of given HCl = 12.1M
V1 = volume of HCl needed to dilute = ?
M2 = molarity of required HCl solution = 0.1M
V2 = volume of required H2SO4 = 500 cm
3
Given HCl
M1V1
Desired HCl
=
M2V2
V1
=
M1
.
=
2.
3
12.1
= 4.13 cm
Potassium dichromate (K2Cr2O7) is a red-orange compound. It is a strong oxidizing agent and is used in the
3
estimation of iron content in ores. A stock solution is 2.5M K2Cr2O7. How many cm of this solution you need to
3
dilute to make 50 cm of 0.05 M K2Cr2O7.
Solution:
M1 = molarity of given K2Cr2O7= 2.5 M
V1 = volume of HCl needed to dilute = ?
M2 = molarity of required HCl solution = 0.05 M
3
V2 = volume of required H2SO4 = 50 cm
Given K2Cr2O7
M1V1
Desired K2Cr2O7
=
V1
M2V2
=
M1
.
=
3.
2.5
= 1 cm3
Commercial acetic acid is 17.8 molar. How can you convert this into 0.1 M acetic acid.
Solution:
M1 = molarity of given acetic acid = 17.8 M
V1 = volume of HCl needed to dilute = ?
M2 = molarity of required HCl solution = 0.1 M
3
V2 = volume of required H2SO4 = 1000 cm
Given acetic acid
M1V1
Desired acetic acid
=
V1
M2V2
=
=
M1
.
3
17.8
= 5.62 cm
Q: What do you know about solubility of solution, define and give example?
Solubility
10 | P a g e
BY: M. ZIA DOGAR
CHAPTER:6
CEMISTRY :9th
SOLUTIONS
Solubility of a substance in a particular solvent at a definite temperature is, the maximum amount of the solute in grams that
can dissolve in 100g of the solvent to form a saturated solution.
Example
Solubility of NaCl in 100g of water at room temperature (25°C) is 35.7g while at 100°C it is 39.12g where as sodium thiosulphate
is 50 per 100g water at 25°C but when we add 60g in 100g water at 20°C, the solubility will decrease.
Q: How does solubility take place in different compounds or
How nature of solute and solvent could effect the solubility?
Ans:
Solubility and Solute-Solvent Interactions
In chemistry analysis, “Like dissolves like” is a general rule for guiding in solubility of substances. It has been observed that nonpolar covalent solutes are soluble in non polar covalent solvents while Ionic and polar solutes dissolve in polar solvent.
Explanation with Examples
1. Methanol and water are miscible with all proportions
 because water molecules are polar and having Hydrogen bonding between partial positive charge H+Ϭ and partial
negative charge on O-Ϭ.
 Similarly methanol molecules are also polar and exhibit hydrogen bonding like water molecules. Thus both have similar
structure and intermolecular force and therefore miscible with each other in all proportions.
2.
Glucose, whose molecule has many -O -H bonds creates Hydrogen bonding with
water so it is very soluble in water.
+
-
3. When we place a crystal of sodium chloride in water, it dissociate into Na and Cl
As a result of which, the negative end of water molecule is attracted to Na+ and the positive end to Cl - ions. Thus sodium
chloride dissolves readily. Figure shows attraction of Na+ and Cl - ions for water molecules called Ions dipole forces. Thus the
interaction of ions of solute with the molecules of solvent (water) is called Hydration or salvation and the ions will be called
Hydrated ions or solvated ions.
Attraction of Na+ and Cl - ions for water molecules
11 | P a g e
BY: M. ZIA DOGAR
CHAPTER:6
SOLUTIONS
CEMISTRY :9th
4.
Gasoline and oils are non-polar in nature and do not dissolve in water (Polar). Gasoline and oils as both non polar
molecules, are soluble in one another.
Q: How does the change of temperature effect the solubility of certain compounds? Justify with Example.
Answer
Effect of Temperature on Solubility
We can divide the substance in three groups as far as solubility is concerned.
1. Those substance whose solubility is increased on increasing the temperature e.g. KCl, KBr, NaNO3, NH4Cl.
Heat is a absorbed when solutions of these substances are prepared in water. When these substances are dissolved, the vessel
cools down.
Reason is that during dissolution the heat of solvent and the vessel is taken up in the process, of solution formation. Whenever
temperature of such solution is increased, solubility’s of solutes increase.
2. Those substance whose solubility decreases on increasing the temperature, e.g. Na2SO4. These substances heat
produce when dissolved in water. The vessel in which these substances are dissolved becomes hot and its temperature
increases.
The solubility of gases e.g, air in water decreases with increasing temperature because when water is heated, we will see small
bubbles form at the side of the beaker before the water boils. These bubbles are composed of air and come out of water in the
form of bubbles. This means that solubility of air in water decreases with increasing temperature..
3. Similarly we have observed in a home aquarium, that the fish shows signs of stress on a hot day. This is because less
oxygen from air dissolves in the warm water.
4. The solubility of some substances is least affected by change in temperature e.g. NaCl. The reason is that a very small
amount of heat is absorbed during their solution formation as shown in a graph.
The Graph shows the variation of solubility with temperature.
Q (a) What do you understand by
i) True Solution ii) Suspension
iii) Colloids
(iv)Give the characteristics of each (v)Give the comparison of each
Answer
(i)Solution or True Solution
A homogenous mixture in which solute particles are completely homegnized in the solvent is called solution or True solution,
e.g. solution of NaCl, sugar in water .
(ii)Suspension
A heterogeneous mixture containing particles large enough to be seen with the naked eye and clearly distinct from the
surrounding fluid is called a suspension.
or
A mixture in which solute particles do not dissolve in solvents is called suspension.
(iii)Colloids
A heterogeneous mixture of tiny particles of a substance dispersed through a medium is called a colloid
 Solute particles pass through the filter paper.
 Solute particles do not settle down in the bottom by keeping the solution for some time.
c) Comparison of properties of solutions, suspension and colloids.
S.No.
Solutions
Suspensions
Colloids
1.
Homogeneous
Heterogeneous
Heterogeneous
2.
Particles size vary from 0.1 Particles size is greater
Particles size vary from 1
3
3
to 1nm
than 10 nm
to 10 nm
3.
Particles are invisible by
Particles are visible by
Particles are invisible by
naked eye, ordinary
naked eye
naked eye and in ordinary
microscope as well as
microscope but visible
electron microscope
under electron
microscope
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CHAPTER:6
SOLUTIONS
4.
Particles can pass through
ordinary as well as ultra
filter paper
Particles can not pass
through ordinary as well
as ultra filter paper
5.
Cannot scatter light
Scatter light
CEMISTRY :9th
Particles can pas through
ordinary filter paper but
cannot pass through ultra
filter paper
Scatter light
Importance of solutions in our daily life
Most of the substances we need for our existence are solutions. The air we breathe is a gaseous solution containing N2, O2, CO2
and rare gases. The water we use for drinking, cooking and washing is not pure. It contains dissolved gases and many minerals
that are essential for our health. In fact natural water is a liquid solution.
Beverages, vinegar, soft drinks etc. are liquid solutions. Commercial products such as window cleaners, sanitary cleaners,
shampoo, gasoline, kerosene, diesel, etc. are also liquid solutions. Most medicines are dispensed in solution form. We also use
many solid solutions in our daily life.
Gold is a solid solution of gold containing some copper. Brass and steel are used for making utensils, musical instruments, buses,
cars, trains etc are solid solution of metals. Parts of aero plane are made of solid solution of metals such as Al and Mg. Dental
filling are liquid solutions of metals in mercury.
Science tit bits (suspensions)
Many medicine bottles contain an insoluble solid in water. The bottle has to be shaken before use to produce a suspension, so
that the solid is spread evenly throughout the bottle and the patient takes the correct amount of the medicine.
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BY: M. ZIA DOGAR