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States of MatterStates of Matter
VIKASANA ‐ VIGNANA PATHADEDEGE NIMMA NADIGEBridge Course Program for SSLC Students who want to take up Science in I PUC in 2012
States of MatterMatter: It is something that
i i t ioccupies space possessing certainmass. Hence all the Substancesmass. Hence all the Substancespresent in this universe arematters.
Vikasana - CET 2012
Classification of matter on the basis oftheir physical state:On the basis of physical state, matterp y ,can be classified as
solidsolid,liquid,gas andgas andplasma.
Solid:Substances hich possess definite mass Substances which possess definite mass , volume and shape are called solids.In this
l h ld l hstate particles are held very close to each other. These particles have no freedom of pmovement.Ex: Wood, stone, copper etc.Ex: Wood, stone, copper etc.
LiquidSubstances which possess definitemass and volume but indefinite shapemass and volume but indefinite shapeare called liquid. In this state particlesare close to each other and they canare close to each other and they canmove within their volume.Ex: water, Benzene, Carbon
Tetrachloride, Kerosene.Vikasana - CET 2012
Tetrachloride, Kerosene.
GasSubstances which possess definite massbut indefinite volume and shape arepcalled gases. In gases particles are farapart as compared to solid and liquidapart as compared to solid and liquid.The particles can move freely from one
hpoint to another.Ex: Nitrogen, Oxygen, Carbon dioxide
Vikasana - CET 2012Ex: Nitrogen, Oxygen, Carbon dioxideetc.
Classification of matter on the basis of composition:On the basis of the chemical O t e bas s o t e c e cacomposition matter can be classified as fallowsfallows.
Heterogeneous matterHomogeneous matter
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Heterogeneous matterA substance is heterogeneous if itexhibits different properties at itsexhibits different properties at itsdifferent position. Different typesf h t ttof heterogeneous matter area)Suspension b) Colloida)Suspension b) Colloidc) Hetero mixture
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Heterogeneous matterMatter
Heterogeneous matter Homogeneous matter
Suspension colloid Hetero mixture Suspension colloid Hetero mixture
Homogeneous mixture Pure substance
Element Compound
Metal Non‐metal metalloid
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SuspensionIt is a heterogeneous matter inwhich the solute particles do notwhich the solute particles do notdissolve but remain suspendedh h h b lk f hthroughout the bulk of themedium.ed u .Ex: Dirt particles in water, Butter
lk Vikasana - CET 2012milk.
ColloidColloids are the heterogeneousmixture of two components withmixture of two components withthe size of the particle is 1nm to100nm(or 10 A0 to 1000A0) These100nm(or 10 A to 1000A ).Theseparticles of colloid are uniformly
d th h t th l tispread throughout the solution.Ex: Blood,Milk,Butter,Cloud.
Vikasana - CET 2012, , ,
HeteromixtureIt is obtained by mixing two or moresubstance in any ratio. These aresubstance in any ratio. These arepossessing the mixed properties of thecombined substance These can becombined substance. These can beseparated by physical method.
Ex: A mixture of sand and common salt
Vikasana - CET 2012salt.
Homogeneous matterA substance is homogeneous matter ifthe smallest part of it exhibits thethe smallest part of it exhibits thesame chemical and physicalpropertiesproperties.Ex: Air , Solution of sugar with water,an intimate mixture of two or morethan two metals (alloys).
Vikasana - CET 2012( y )
Homogeneous matter can be l ifi d i t t tclassified into two types
(a)Homogeneous mixturesA mixture of two components appears in a single phase is called homogeneous mixtures. These are called as solutionssolutions.A homogeneous mixture of solute and solvent are called solutions. These solutions are also called as true solutions or crystalloids Ex: A homogeneous mixture of sugar and water gives
Vikasana - CET 2012true solutions
Homogeneous matter can be l ifi d i (C d )classified into two types: (Contd.)
(b)Pure substancesThese are made of only one type of particles These are made of only one type of particles such as atoms or molecules. Further these are classified as elements and compounds.classified as elements and compounds.
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ElementSimple forms of matter which cannot bedecomposed into further simple substancesdecomposed into further simple substancesare called elements. 118 elements arediscovered till today out of which 92 arediscovered till today, out of which 92 arenaturally occurring elements and remainingare artificially prepared elementsare artificially prepared elements.Ex: Hydrogen, mercury, gold, iron etc.
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Elements are further classified as metals, non-metals and metalloids
M t lMetalThe elements with the electro-positive nature of losing one or more electrons readily to give positively charged cations are called metals.
M M+ + e-
Ex :Copper, Zinc, IronVikasana - CET 2012
Ex :Copper, Zinc, Iron
Non metal:Non-metal:The electronegative elements whichThe electronegative elements which have a tendency to gain one or more l ll d lelectrons are called non-metals..
A + e- A-A + e AEx: Chlorine, Bromine, Sulphur etc
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Magnesium metalg
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Sodium metal
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MetalloidsElements possessing of both metalsand non metals properties are calledand non-metals properties are calledmetalloids.Ex: Antimony, arsenic, Germinium
tetc.
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CompoundsThe substances made of two or more than twoelements in a definite ratio by mass are calledycompounds.Ex: (1) In water (H2O) number ofEx: (1) In water (H2O) number ofhydrogen and oxygen atoms are in the ratio2:1 or it contains hydrogen and oxygen in2:1 or it contains hydrogen and oxygen inthe ratio of 1:8 by their mass. i.e. 2g ofhydrogen combines with 16 g of oxygen
Vikasana - CET 2012hydrogen combines with 16 g of oxygen.
CompoundsEx: (2) Sulphuric acid (H2SO4 ):In this compound the ratio of number ofIn this compound the ratio of number ofatoms of H:S:O is 2:1:4The ratio by massf H S O i 1 16 32 I thi d 2of H:S:O is 1:16:32.In this compound 2 g
of hydrogen 32 g of sulpher 64 grams ofoxygen combine with each other to form98 g of H2SO4.
Vikasana - CET 2012g 2 4
SolutionA homogeneous mixtures of solute andsolvent are called solutions. In asolvent are called solutions. In asolution, a substance which is in a lessquantity is solute and other which is inquantity is solute and other which is inmore quantity is solvent. In aqueousl ti f t i l t dsolution of sugar, water is solvent and
sugar is solute.Vikasana - CET 2012
h k Thank You
Concentration of solutionTh t f l t t i it l f• The amount of a solute present in a unit volume ofthe solution is called concentration of the solution.
• Concentration of the solution can be expressed interms such as percentage by mass or by volume,molarity, molality, normality, mole fraction, ppm.
• Out of these terms the most familiar terms toOut of these terms the most familiar terms toexpress the concentration of the solution arepercentage by mass or volume molarity and
Vikasana - CET 2012percentage by mass or volume, molarity andnormality.
Percentage by mass:g yIt is the mass of the solute present in
100 f i100 g of the solution.Percentage by mass =Percentage by mass
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Percentage by volume:It is the mass of the solute present in100 cm3of the solution100 cm3of the solution.
Percentage by volume =
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Problem50 g of glucose is dissolved in 400 g of water then calculate % by mass ofof water, then calculate % by mass of the solute.
Ans: Total mass of the solution = mass of glucose + mass of water =mass of glucose + mass of water = 50 + 400 = 450 g
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Percentage by mass =
11.11
Hence % by mass of glucose in thesolution is 11 11
Vikasana - CET 2012solution is 11.11
Problem300 3 f l ti300 cm3 of solutioncontains 20 g of NaCl dissolved with
h l l % b l fwater, then calculate % by volume ofthe solutionPercentage by volume =
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.
= = 6.67
Hence % by volume of NaCl in the l ti i 6 67solution is 6.67
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MolarityMolarity can be defined as the number of
gram molecular mass of solute dissolvedgram molecular mass of solute dissolved in one dm3 of the solution. It is denoted by Mby M
Molarity = y
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ororMolarity =
WhereWherem= mass of the soluteM M l l f h lM= Molecularmass of the soluteV= volume of the solution
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MolarityMolarity of NaOH solution is 0.5M means 0 5 mole ( or 0 5 gramM means 0.5 mole ( or 0.5 grammolecular mass )of NaOH is)dissolved in one dm3 of the
l tisolution.
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Problem.Calculate the molarity of the
l ti bt i d b di l i 6solution obtained by dissolving 6g of oxalic acid in 200 cm3ofgsolution. Mol. mass of oxalic acidi 126is 126.
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S l tiSolution. Given things m=6 g M= 126 V= 200cm3
Molarity =
= = 0.238M
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NormalityN li b d fi d hNormality can be defined as thenumber of gram equivalent mass ofg qsolute dissolved in one dm3 of thesolution It is represented by Nsolution. It is represented by N
Normality =
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ORNormality =
WhereWherem= mass of the soluteE= Equivalent mass of the soluteE= Equivalent mass of the soluteV= volume of the solution
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NormalityyNormality of H2SO4 solution is0.5 N, It means that 0.5 gramequivalent mass of H2SO4 isequivalent mass of H2SO4 isdissolved in one dm3 of thesolution
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ProblemCalculate the normality of the solutionCalculate the normality of the solutioncontaining 4 g of sodium hydroxide in 500 cm3
of the solution (Equ mass of NaOH is 40)of the solution. (Equ.mass of NaOH is 40)
N liNormality = = 0.2 N
Hence the normality of NaOH solution is 0.2N
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