• 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.


A homogenous mixture of two or more different chemical substances with uniform chemical and physical properties is called SOLUTION.


 Sugar solution is a mixture of sugar and water. Salt solution is a mixture of salt and water.

A solution which consist of only two substances is called BINARY SOLUTION.


Sugar solution (sugar+water)

Salt solution (salt + water)     

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.


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.


In salt and sugar solutions, water is a solvent.


  • Unsaturated Solution

A solution which can dissolve further amount of solute at a particular temperature is

called an unsaturated solution.


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.

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.

  • If you add 170g of sodium acetate in 100g of water at 0oC, how much will dissolve?

ANS: 119g of Sodium acetate will dissolve in 100g of water at 0oC.

  • Is the solution saturated unsaturated or supersaturated?

ANS: The solution will be saturated

  • If the solution in heated to 100oC, is the solution now saturated, unsaturated or supersaturated.

ANS: The solution will be saturated

  • If the solution is cooled back to 0oC and no crystals appear. Is the solution saturated, unsaturated or supersaturated?

ANS:  the solution will be supersaturated

Q: Descibe various 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.

  • Deep sea divers use a breathing mixture of helium and oxygen.
  • 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.


S.No. Solutions State of Solute State of Solvent Type of Solution
a. Mixture of Helium and Oxygen Helium (g) Oxygen (g) Gas is Gas
b. Brass Zinc(s) Copper(s) Solid in Solid
c. Dental filling Mercury-Hg (l) Silver- Ag (s) Liquid in solid
d. Brine solution Salt (s) Water (l) Solid in liquid
e. Drinking water with chlorine Chlorine (g) Water (l) Gas in liquid
f. Gem stone- Ruby Cr2O3(s) Al2O3(s) Solis in solid
g. H2SO4, 98% 2%, H2O H2SO4, 98%  

Q: Write note on following?

  • Solutions of gases
  • Solutions of liquids
  • Solutions of solids

Solutions of Gases

Gaseous solutions are commonly used by chemical industries to prepare chemical substances.

For Example

  1. 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.
  2. Urea Fertilizer: A gaseous mixture of Ammonia and C02 is used for synthesis of urea.
  3. 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.
  4. Brass is an alloy of copper and zinc
  5. Steel is an alloy of iron with carbon and silicon
  6. 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.

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.


  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?


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.

  1.  Mass by Mass Percentage (m/m)

It is the mass of the solute dissolved per 100 parts by mass of solution.


10% m/m NaCl solution means that 10 g NaCl in 90g water to make 100g of solute

% of solution by m/m =   

  1. 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.

% of solution by m/v=  

  1. 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 =

  1. 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 =

SELF assessment EXERCISE 6.3
  1. 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 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)=


Molarity (M) =   X


Molarity (M) =   X

Units of Molarity

Its units are mol per dm3 (

Q: what these terms refers?

  1. 18 M Sulfuric acid?
  2. 12.1 M Hydrochloric acid


  1.  18M concentrated H2SO4 means that there are 18 moles of H2SO4 in each dm3 of solution.
  2.  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 urea dissolved in 500cm3 of solution. Calculate the molarity of this solution.


Mass of urea = 40g

Molar mass of urea (NH2CONH2=

 =14+1×2 + 12+16+14+1×2 = 60g /mole

Molarity (M ) =

Molarity (M ) =

=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.


                        Moles  of KMnO4 =  0.05 mol

Volume of KMnO4= 600 Cm3

Molarity (M)=

Molarity (M)=

                        = 0.083 mol/dm3

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 that contains. (a) 1.5 moles of this compound dissolved in 250cm3 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.



            Number of moles of KClO3=   1.5 moles

            Volume of solution      =          250 cm3

            Molarity of KClO3        =          ?

Molarity of KClO3        =

                                    =          =6.0 M

(b)        mass of KClO3                           = 75g

            Molar mass of KClO3     = 39+35.5+3×16= 122.5g

            Volume of solution      = 1.25dm3

            Molarity of KClO3               =?

Molarity (M) =   X

                        =   X     

                        = 0.49mol.dm3

 or 0.49 M

Converting the molarity of a solution into its concentration in g/dm3  

            A flask contains 0.25M NaOH solution. What mass of NaOH is present per dm3 of solution?


Molarity: 0.25 M

Volume of solution: 1 dm3

Molar mass od NaOH= 40g/mol

Mass of NaOH= ?

Molarity (M) =   X

Mass of NaOH=  

                        = 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.


            Mass of KOH dissolved in one dm3 of solution = 5.6g

Molar mass of KOH = 39+16+1

                                    = 56g/mole

Molarity (M) =   X

=   X    

= 1 M

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?


Mass of NaOH = 25 g       

Volume of solution: 1 dm3

Molar mass of NaOH= 40g/mol

Molarity  of NaOH solution= ?

Molarity (M) =   X

=   x   = 0.625 M

  • A solution of NaOH has concentration 1.2M. Calculate the mass of NaOH in g/dm3 in this solution.


Molarity: 1.2 M

Volume of solution: 1 dm3

Molar mass od NaOH= 40g/mol

Mass of NaOH= ?

Molarity (M) =   X

Mass of NaOH=  

                  = 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 6.51×104g/mole.


Mass of haemolobin= 10 g          

Volume of solution: 1 dm3

Molar mass of haemoglobin= 6.51x 10 4 g/mol

Molarity  of NaOH solution= ?

Molarity (M) =   X

=   x

                  = 1.536x 10 -4M

Q: How you can Prepare  a solution of given molarity  

Prepare 0.2M KMnO4 solution.


  • Volumetric flasks of capacity 1 dm3, 500cm3, 250cm3, 100cm3, 50cm3 can be used to prepare a solution.
  • Suppose you use a 100cm3 volumetric flask. First find the mass of KMnO4 to give 100 cm3 of 0.2 M KMnO4 solution.


Required volume of solution = 100cm3  

                        Molarity of solution= 0.2 M

Molar mass of KMnO4            = 39+55+16×4

                                                                        = 39+55+64


Molarity (M) =   X

Mass of KMnO4 =  


Mass of KMnO4        =3.16g

To prepare 0.2M KMnO4 in 100cm3 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.


Required volume of solution = 500cm3  

Molarity of solution= 0.2 M

Molar mass of KMnO4            = 39+55+16×4

                                                = 39+55+64


Molarity (M) =   X

Mass of KMnO4 =  


Mass of KMnO4        =15.8g

To prepare 0.2M KMnO4 in 500cm3 volumetric flask, you will add 15.8g of KMnO4.

  • How can you prepare 25cm3 of 0.25M solution of CuSO4.5H2O (Blue vitriol).


Required volume of solution = 25cm3  

Molarity of solution= 0.25 M

Molar mass of CuSO4.5H2O    = 63.5+32+16×4+18X5 =249.5 g/mole

Molarity (M) =   X

Mass of CuSO4.5H2O =  


Mass of CuSO4.5H2O      =1.56g

  • To prepare25 cm3 and  0.25M CuSO4.5H2O , you will add 1.56 g of CuSO4.5H2O in 25 cm3 of solvent.
Preparing a solution of given molarity by diluting a solution of known molarity  

            Concentrated sulphuric acid is 18M H2SO4. How many cm3 of this acid is needed to produce 250cm3 of 0.1M H2SO4?


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                              Desired H2SO4

     M1V1                       =           M2V2

            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. A stock solution of hydrochloric acid is 12.1 M. How many cm3 of this solution should you use to prepare 500cm3 of 0.1 M HCl.


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 cm3

Given HCl                                Desired HCl

                   M1V1            =                 M2V2

V1 =     

                                                      =      = 4.13 cm3

  • Potassium dichromate (K2Cr2O7) is a red-orange compound. It is a strong oxidizing agent and is used in the estimation of iron content in ores. A stock solution is 2.5M K2Cr2O7. How many cm3 of this solution you need to dilute to make 50 cm3 of 0.05 M K2Cr2O7.


M1 = molarity of given K2Cr2O7= 2.5 M

V1 = volume of HCl needed to dilute = ?

M2 = molarity of required HCl solution = 0.05 M

V2 = volume of required H2SO4 = 50 cm3

Given K2Cr2O7              Desired K2Cr2O7

M1V1                       =                        M2V2

V1           =       

        = 1 cm3

  • Commercial acetic acid is 17.8 molar. How can you convert this into 0.1 M acetic acid.


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

V2 = volume of required H2SO4 = 1000 cm3

M1V1    =                   M2V2

V1               =   

            = 5.62 cm3

Q: What do you know about solubility of solution, define and give example?


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.


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?


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 non-polar 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 and partial negative charge on O.
  • 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.
  • Glucose, whose molecule has many -O -H bonds creates Hydrogen bonding with

water so it is very soluble in water.

  • 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 Clions. 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

  • 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.                    


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.

  1. 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..

  • 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.
  1. 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                           


(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 .


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.


A mixture in which solute particles do not dissolve in solvents is called suspension.


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 to 1nm Particles size is greater than 103nm Particles size vary from 1 to 103nm
3. Particles are invisible by naked eye, ordinary microscope as well as electron microscope Particles are visible by naked eye Particles are invisible by naked eye and in ordinary microscope but visible under electron microscope
4. Particles can pass through ordinary as well as ultra filter paper Particles can not pass through ordinary as well as ultra filter paper Particles can pas through ordinary filter paper  but cannot pass through ultra filter paper
5. Cannot scatter light Scatter light 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)

Manymedicine bottles contain an insoluble solid in water. The bottle has to beshaken before use to produce a suspension, so that the solid is spread evenlythroughout the bottle and the patient takes the correct amount of the medicine.

Q.2:        Give short answers

  1. Differentiate between saturated and unsaturated solution?
  2. Give example of a solid solution containing two solids.
  3. Can you call collide a solution?
  1. Gasoline does not dissolve in water, why?
  2. Are gem stones solutions?

Q.3:        A tiny crystal of a solid substance is added to an aqueous solution of the same substance. What would happen if the original solution was:

                (a) supersaturated                               (b) unsaturated    (c)          saturated

Q.4:        Explain why CH3OH is soluble in water but C6H6 is not.

Q.5:        How can you prepare 250cm3 of 0.5M MgSO4 from a stock solution of 2.5M MgSO4?

Q.6:        Copy and complete the following table for aqueous solution of NaOH.

Mass of solute Moles of solute Volume of solution Molarity
20g   500cm3  
  0.25   0.25
    200cm3 0.1

Q.7:        Give examples of the following solutions:

                (a)           a liquid solution of a liquid solvent and gaseous solute

                (b)           a solid solution of two solids

 Q.8:       What is the molarity of a solution prepared by dissolving 1.25g of HCl gas into enough       water to make 30cm3 of solution.

Q.9:        Formalin is an aqueous solution of formaldehyde (HCHO), used as a preservative for          biological specimens. A biologist wants to prepare 1dm3 of 11.5M formalin. What mass                of formaldehyde he requires?

Q.10:      A solution of Ca(OH)2 is prepared by dissolving 5.2 mg of           Ca(OH)2 to a total volume of              1000cm3. Calculate the molarity of this solution.

Q.11:      Calculate the number of moles of solute present in 1.25cm3 of 0.5M H3PO4 solution.

Q.12:      Calculate the new molarity when 100cm3 of water is added to 100cm3 of 0.5M HCl.

Q.13:      How are solutions useful for society? Give three examples.

1:            A 10.0g of solid solute is placed in 100g of water at 20oC and all of it dissolves. Then           another 4.0g of the solute is added at 20oC and all of it dissolves.

                (a)           Is the first solution saturated, unsaturated or supersaturated?

                (b)           Is it possible to tell from this information that the final solution is unsaturated or                                  saturated?

2:            What should you do to change:

                (a)           a  saturated solution to an unsaturated solution.

                (b)           an unsaturated solution to a saturated solution.

3:            Knowing the molarity of a solution is more meaningful than knowing whether a solution    is dilute or concentrated. Explain.

4:            Design an experiment to determine the solubility of table sugar in water at room                temperature.

5:            Design an experiment to prepare 10% mass by volume solution of CuSO4.5H2O    (Nelathota).

 6:           Which solution is more dilute 50cm3 of 0.2M NaOH or 100cm3 of 0.1M NaOH.

7:            Which solution is more concentrated 100cm3 of 0.1 M HCl or 100cm3 of 0.1M NaOH.

8:            Benzene is a common organic solvent. Its use is now restricted because this can cause      cancer. The recommended limit of exposure to benzene is 0.32 mg per dm3 of air.       Calculate the molarity of this solution.

9:            A patient in a hospital is often administered an intravenous (IV) drip containing an             aqueous solution. This aqueous solution contains 0.85% (mass by volume) of sodium   chloride or 5% (mass by volume) of glucose. Calculate the molarity of both these                 solutions.

10:          100cm3 of NaOH solution was heated to complete dryness, 1.5g residue left behind.                         What      was the molarity of the solution.