This document provides information on various chemistry concepts related to solutions including: 1) Definitions of key terms like solvent, solute, concentration, and activity. 2) Explanations of concentration terms like molarity, molality, and normality. 3) Procedures for preparing solutions of specific concentrations. 4) Calculations for determining molecular weights, equivalent weights, and concentrations in parts per million and billion. 5) A description and procedure for standardizing a potassium permanganate solution through titration with oxalic acid.

2. Assignment
1)Solution
2)Solvent
3)Solute
4)Exothermic reaction
5)Endothermic reaction
6)Concentration
7)Why standardisation is required
8) Why to study the concept of solution, preparation of solution and
concentration
9)Definition of concentration
10)What do you understand by the term activity
11) Explain the concept of concentration and activity with example
12) What are the different types of concentration
13) Define number of mole or mole
14) How will you define the gram molecular weight
15 ) how will you define the elemental weight
16) Explain the concept of vapourisation and condensation with respect to
solution

3. Standardisation
Collective Properties
Solution
Solvent
Solute
Molality
Concentration terms
Concentrated solution
Dilute solution
Evaporation
Condensation
Electrolyte
Non electrolyte
Volatile solvent
Non volatile solvent
Vapour Pressure

4. Non electrolyte: Do not dissociate in to ions.
Soluble or may be insoluble in water
Example: Naphthalene,Cyclohexane,
Sugar,Urea, ether,Benzene, Chloroform ,
Ethanol
Volatile Solvent : Alcohol, acetone,
ether, ammonia. They have
appreciable vapour pressure (higher
than non volatile)

5. Non Volatile Solvent : Not appreciable vapour
pressure, having low rate of evaporation
Non Volatile Solute: Example: Naphthalene,
Sucrose,Urea,potassium nitrate, sodium
chloride
Vapour Pressure: Pressure exerted by
vapour when its surface is saturated
with the same vapour

6. Boiling point: is the temperature at which the
vapour pressure become equal to the
atmospheric pressure (1 atm=760 mm Hg)
Vapour Pressure: Pressure exerted by
vapour when its surface is saturated
with the same vapour

8. Preparation of 0.1 N NaOH
Molecular weight of sodium hydroxide
Atomic weight of Na=23
Atomic weight of O=16
Atomic weight of hydrogen=1
23+16+1=40
Molecular weight of NaOH=40
40 gram of sodium hydroxide dissolved in 1000cm3 of solvent,
it gives 1N solution
40 gram = 1000ml
40 =100 ml
4 gram of sodium hydroxide = 100 ml of solvent =1N solution
0.40 gram of sodium hydroxide = 100 ml of solvent =0.1N

9. Normality
Gram equivalent weight of solute
dissolved in 1000 ml of solvent is called
as Normality
1000 ml of 1N NaOH= 59 gram of succinic
acid
1) Titration of x N NaOH with 0.05 N
succinic acid

10. Assignment
2)How will you prepare250ml of 0.1
N solution of sodium hydroxide
Answer) 40 gram of NaOH when
dissolved in 1000 ml of solution gives
1N Solution

11. Molarity
Gram molecular weight of solute dissolved in 1000cm3 of the
solvent
1) Prepare a 2.0% (W/V) concept of preparation of solution
2) Prepare a 2.0% (V/V) concept of preparation of solution
3) Calculate the molecular weight of acetic acid
(CH3COOH)
Atomic weight of Carbon=12
Atomic weight of Hydrogen=1
12+(1*3)+12+16+16+1
12+3+12+16+16+1
=60

12. Gram molecular weight of solute dissolved in
1000 ml of solvent is called as Molarity
1000 ml of 1N Silver nitrate= 74.55 gram of
potassium chloride
1) Titration of 0.1 N Silver nitrate with 10 ml of
Potassium chloride, find out the amount of
potassium chloride in given 10 ml of unknown
solution
Molarity

13. 2) How will you prepare 50ml of 0.02 M
solution of potassium nitrate. (Molecular
weight of potassium nitrate is 101.10 gram)
Answer) 101.10 gram of potassium nitrate
when dissolved in 1000 ml of solution gives
1M Solution

14. Formality
One formula weight of solute dissolved in 1000
ml of solvent is called as Formality
1000 ml of 1N K2SO4= 36.45 gram of
hydrochloric acid
1) Titration of 0.05 N K2SO4 with 25 ml
hydrochloric acid

15. 2) What will be the formality of solution, when 21
gram of ferric sulphate Fe2 (SO4)3
dissolved in 250 ml of solution
2* Atomic weight of Fe (55.84 ) + 3* (32.06+64)
Molecular weight of ferric sulphate 399.25 g/mol
Answer)
399.25 gram dissolved in 1000 ml gives 1 F
solution

16. What is difference between molarity and
formality. As per formality the answer obtained
in previous slide is concentration of salt solution
but molarity of ions which are involved in the
reaction are different and they are calculated as
follows. Suppose the answer obtained in pervious
numerical is z , then
Fe2 (SO4)3 = 2Fe3+ + 3 (SO4)2-
2Fe3+ = 2*z
3 (SO4)2- = 3*z

17. Mole
Mole= Weight in gram / Molecular weight in gram
Millimole= Weight in milligram / Molecular weight
in milligram
Equivalent= Weight in gram / Equivalent weight in
gram
Milliequivalent= Weight in milligram / Equivalent
weight in Milligram

18. 1000ml of 1N solution =1 Equivalent
1000ml of 1N solution =1000 Milliequivalent
1ml of 1N solution =1 Milliequivalent
Equivalent/ Litre = Milliequivalent /Millilitre
Normality =Number of Equivalents/volume in litre
Normality =Number of Milliequivalents/volume in
Millilitre

19. 1000ml of 1M solution =1 Mole
1000ml of 1M solution =1000 Milliemoles
1ml of 1M solution =1 Millimole
moles/ Litre = Millimole /Millilitre
Molarity =Number of Moles/volume in litre
Molarity =Number of Millimoles/volume in Millilitre

20. 1) Calculate the molarity of solution , if 0.1 mole of
solute is present in 0.250 litre of solution.
Molarity =Number of Moles/volume in litre
Molarity =Number of Millimoles/volume in Millilitre

21. 2) Calculate the molarity of solution , if 1231 millimole
mole of solute is present in 750 millilitre of solution.
Molarity =Number of Millimoles/volume in Millilitre

22. Parts Per Million (ppm)
1 mg /litre= 1 ppm
Parts Per Billion (ppb):
1 microgram /litre= 1 ppb
1gram=1000mg
1gram=1000000 microgram

23. 1) How will you prepare a 10 ppm
solution of potassium
39.09 gram of K=74.55gram of KCl
39.09 microgram of K=74.55 Milligram of
KCl
10 Milligram of K= ? Milligram of KCl
74.55*10/39.09 = 19.07 Milligram of KCl
or 0.019 gram of KCL

24. 1) How will you prepare a 10 ppb solution
of potassium
39.09 gram of K=74.55gram of KCl
39.09 Microgram of K=74.55 Microgram
of KCl
10 microgram of K= ? microgram of KCl
74.55*10/39.09 = 19.07 Microgram of KCl
or 0.019 milligram of KCL
Or 0.000019 gram of KCl

25. 1) How will you prepare a 20 ppm
solution of sodium from sodium chloride
23.0 gram of Na = 58.44 gram of NaCl

26. 2) How will you prepare a 20 ppb solution
of sodium from sodium chloride
23.0 gram of Na = 58.44 gram of NaCl

27. For HCl, basicity =1
Equivalent weight = Molecular weight
1 ) Calculate the molecular weight and equivalent
weight of hydrochloric acid
For H2SO4 : Equivalent weight= Molecular weight/2
As sulphuric acid has basicity =2
2) Calculate the molecular weight ang equivalent
weight of sulphuric acid
For H3PO4 : Equivalent weight= Molecular weight /3
3) Calculate the molecular weight and equivalent
weight of phosphoric acid

28. For NaOH: Equivalent weight= Molecular
weight as acidity of NaOH is equal to 1
1) Calculate the molecular weight and
equivalent weight of sodium hydroxide
2) For KOH: Equivalent weight= Molecular
weight as acidity of KOH is equal to 1
1) Calculate the molecular weight and
equivalent weight of Potassium
hydroxide

29. For Ca(OH)2
:Equivalent weight= Molecular weight /2
as acidity = 2
2) Calculate the molecular weight and equivalent
weight of Calcium hydroxide
For Ba(OH)2
:Equivalent weight= Molecular weight /2
as acidity = 2
2) Calculate the molecular weight and equivalent
weight of Barium hydroxide

30. Standardisation of potassium permanganate
solution
To find the Normality of given sample of 0.2N Potassium
permanganate.
REQUIREMENTS: Apparatus : Burette, Iodine flask, Pipette,
Reagents : Potassium Permanganate and oxalic acid.
PRINCIPLE: The principle of standardization of potassium
permanganate is based upon redox titration in which strength of
an oxidizing agent is estimated by titrating it with a reducing agent
and viceversa.Potassium permanganate acts as an strong oxidizing
agent in acidic medium that oxidizers oxalic acid in to
carbondioxide. Known strength of oxalic acid is titrated directly
with potassium permanganate. End point can be detected with
appearance of permanent pink colour potassium permanganate
acts as self indicator
2KMnO4+5H2C2O4+3H2SO4→K2SO4+2MnSO4+8H2O +10CO2

31. PROCEDURE: Preparation of 0.2 N Potassium
permanganate solution. Dissolve 6.4g of potassium
permanganate in 1000ml of water, heat on a water bath
for 1 hour, allow to stand for 2 days. Filter the solution
through glass wool.
Preparation of 0.2 N Oxalic acid: 12.6 gm of oxalic acid
dissolve in 1000 ml of distilled water.
Standardisation of 0.1N Potassium permanganate:
Take 25 ml of Oxalic acid solution ,add 5 ml of 1M
sulphuric acid. Warm the mixture to about 70 oC .Titrate
with potassium permanganate solution taken in the
burette. End point permanent pink colour ( persistence
30 second)

32. Burette
reading
I II III
Initial 0.0 cm3 15.0 cm3 15.0cm3
Final 0.0 cm3 15.2cm3 15.2cm3
difference 0.0 cm3 15.0 cm3 15.0cm3
CBR 15.0cm3
N 1V 1=N 2V 2
N 2 = N 1V 1 /V 2
Observation Table
Result : The normality of the provided solution of potassium permanganate
is

33. Thank you