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Some Basic Concepts of Chemistry

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Some Basic Concepts of Chemistry 1 Some Basic Concepts of Chemistry Mole Concept Gram Atomic, Gram Molecular Weight (M) It is 23 the weight of 1.0 mole (Avogadro’s numbers) of atoms, G One Mole Avogadro’s Number ()N = 6.023 ´ 10 . It A moleculesorionsingramunit. is the number of atoms present in exactly 12 g of ()C12 isotope. MA= amu ´ Avogadronumber = A gram Hence, gram molecular weight (M) is numerically equal to G Atomic Weight (A) Atomic weight is the relative weight of one atom of an element with respect to a standard weight. the atomic weight or (molecular weight) in gram unit because Weight of one atom of an element 1.0moleofamuis1.0g. A = 1 th part by weight of an atom of (C12 ) isotope Empirical and Molecular Formula Empirical 12 formula is the simplest formula of a compound with the molecular G amu (atomic mass unit) Weight elements in the simple whole number ratio, and a formula issameoramultipleoftheempiricalformula. 1 12 1amu = thpartbyweightofanatomof ()C isotope e.g. 12 Molecularformula Empiricalformula 1 -24 = g=1.66 ´ 10 g CH6 6 (benzene) CH NA CHO6 12 6 (glucose) CH2 O Atomicweight(A) ´ amu=Absoluteatomicweight. HO2 2 HO NOTE HSO2 2 8 (persulphuricacid) HSO4 G Atomic weight is a relative weight that indicates the relative heavinessofoneatomofanelementwithrespecttoamuweight. G Laws of Chemical Combination Elements combine G Atomicweighthasnounitbecauseitistheratioofweights. in a fixed mass ratio, irrespective of their supplied mass ratio, G Onemoleofanamu=1.00g. e.g. 1 G If an amu is H + O ¾® HO Change of Scale for Atomic Weight 2 2 2 2 defined differently as (1/x)th part by weight of an atom of 2 g 16 g 18 g (C12 )isotope rather (1/12)th part then the atomic weight (A¢ ) Here, H2 and O2 combinesinafixedmassratioof1:8. canbederivedas: No matter in what ratio we mixed hydrogen and oxygen, they æ x ö AA¢ = ç ÷ will always combine in 1:8 mass ratio (stoichiometric mass è 12 ø ratio). Where, A = conventionalatomicweight G Limiting Reactant It is the reactant that is consumed Molecular Weight (MW) Like atomic weight, it is the completely during a chemical reaction. If the supplied mass relative weight of a molecule or a compound with respect to ratio of reactants are not stoichiometric ratio, one of the amuweight. reactant is consumed completely leaving parts of others unreacted. One that is consumed completely is known as Molecularweight limitingreactant. Weight of one molecule of a compound = 1 ‘Limiting reactant determine the amount of th part by weight of an atom of C12 isotope 12 product in a given chemical reaction’ 2 SomeBasicConceptsofChemistry ConcentrationUnits Equivalent Concept, Neutralisation G Normality (N) It is the number of gram equivalent of and Redox Titration solutepresentinonelitreofsolution: G Equivalent Weight Equivalent weight of an element is Eq N = that part by weight which combines with 1.0 g of hydrogen or V ()in litres 8.0gofoxygenor35.5gofchlorine. (i) Molarity (M) Itisthemolesofsolutedissolvein (i) Equivalentweightofasalt(EW) onelitreofsolution. Molar mass = n Net positive (or negative) valency M = : n =Numberofmolesofsolute V e.g.Equivalentweight (\V =Volumeofsolutioninlitre) MMM CaCl =, AlCl =, Al (SO ) = Þ Molarity(M) ´ Volume(V)= n (molesofsolute) 22 33 2 4 3 6 IfvolumeisinmL; MV =millimoles Molar mass (ii) Equivalentweightofacids = If d (g/cc) is density of a solution and it contains x % of solute Basicity ofmolarmass M,itsmolaritycanbeworkedoutas e.g.Equivalentweight M 1000 dx 10dx HCl = M (basicity = 1); H SO = (basicity = 2) Molarity= = 2 4 2 100 M M M H3 PO 4 = (basicity=3) (ii) Molality (m) It is the number of moles of solute 3 Molar mass presentin1.0kgofsolvent. (iii) Equivalentweightofbases = Acidity Moles of solute (n ) m = ´ 1000 e.g.Equivalentweight Weight of solvent in gram M M NOTE NaOH = M, Ca(OH)2 = , Al(OH)3 = 2 3 Molality isatrueconcentrationunit,independentoftemperature while molarity dependsontemperature. G The number of gram-equivalents (Eq) Weight of compound w (iii) Normality (N ) It is the number of gram equivalents of Equivalent = = soluteinonelitreofsolution. Equivalent weight Equivalent weight Gram equivalents of solute (Eq) N = G Mole Equivalent Relationship In a given weight Volume of solution in litre (w) of sample, number of moles (n) and number of equivalents(eq)arerelatedas (iv) Mole Fraction (c i) It is the fraction of moles of a w w particularcomponentinamixtureas n = and Eq = n M Equivalent weight c = i i n Eq M åni Q = = n-factor i = 1 n Equivalent weight (v) ppm (parts per million) Strength It is defined as G n-factor For salt, it is valency, for acid it is basicity, for partsofsolutepresentin(106 part)ofsolution. baseitisacidity. G G Dilution Formula If a concentrated solution is diluted, Normality/Molarity Relationship Eq n N Eq MW followingformulawork N = and M = Þ = = = n-factor V V M n EW MVMV1 1= 2 2 G Acid-Base Titration In acid-base titration, at the (M andV are the molarity and volumes before dilution and 1 1 ‘EndPoint’. M 2 andV2 aremolarityandvolumesafterdilution) Gramequivalentofacid = Gramequivalentofbase G Mixing of two or more solutions of different molarities If two or more solutions of molarities G Titration of a Mixture of NaOH/Na2 CO3 (MMM1 , 2 , 3 , ... ) are mixed together, molarity of the The mixture is analysed by titrating against a standard acid in resultingsolutioncanbeworkedoutas: presenceofphenolphthaleinandmethylorangeindicators. MVMVMV+ + K Phenolphthalein end point occur when the following M = 1 1 2 2 3 3 K neutralisationiscomplete: VVV1+ 2 + 3 Some Basic Concepts of Chemistry 3 NaOH + HCl¾® NaCl + H O ü Quick Balancing of a Redox Reaction 2 ý Na2 CO 3 + 2HCl¾® NaHCO 3 + NaCl þ Cross-multiplication by net change in oxidation number per unit formula of oxidising agent and reducing agent will 1millimolof(HCl) = 1millimolof(NaOH+ Na CO ) 2 3 balance the redox reaction in term of OA and RA as: Methyl orange end point occur when the following neutralisationiscomplete: DON = 1 NaOH + HCl¾® NaCl + H2 O ü 2+ 3+ 3+ ý K2 Cr 2 O 7 + Fe¾® 2Cr + Fe Na2 CO 3 + 2HCl¾® 2NaCl + H 2 O + CO 2 þ methylorangeendpointmillimol(HCl) DON = 12 -6 = 6 2+ = millimol(NaOH)+2millimolof(Na2 CO 3) Hence, multiplying Fe by 6 and K2 Cr 2 O 7 by 1 will balance G Titration of a mixture of NaHCO3 /Na 2 CO 3 thereactionintermsofOAandRA. The mixture is analysed by titrating against a standard acid in Disproportionation Reaction It is a special type of presenceofphenolphthaleinandmethylorangeindicators. redox reaction in which similar species is oxidised as well Phenolphthalein end point occur when the following reduced, e.g. Br2 + NaOH¾® NaBr + NaBrO3.Inthis neutralisationiscomplete: reaction, bromine is reduced to bromide ion and the same is oxidised to bromate ion, hence bromine is undergoing Na CO + HCl¾® NaHCO + NaCl 2 3 3 disproportionationreaction. millimol(HCl)=millimol(Na CO ) 2 3 EquivalentWeightofOA/RA Methyl orange end point occur when the following neutralisationiscomplete: Equivalent weight of OA/RA Na CO +2 HCl¾® 2 NaCl + H O + CO ü Molar mass 2 3 2 2 ý = NaHCO3 + HCl¾® NaCl + H 2 O + CO2 þ Change in ON per formula unit Methylorangeendpointmillimol(HCl) + 2+ æ M ö =millimol (NaHCO3 ) +2millimolof(Na2 CO 3) e.g. KMnO4 + H¾® Mn:ç E = ÷ è 5 ø G Percentage Strength of Oleum It is the mass of H2 SO 4 obtainedonhydrolysisof100gofoleumas: + 3+ æ M ö K2 Cr 2 O 7 + H¾® 2 Cr :ç E = ÷ H2 S 2 O 7 + H 2 O¾® 2H 2 SO 4 è 6 ø Thenetreactionis: 2KI¾® I + 2K+ : SO + H O¾® H SO 2 3 2 2 4 - 80 18 98 [EM= (DON per I = 1 )] + 80 2Na2 S 2 O 3¾® Na 2 S 4 O 6 + 2Na : Þ % offree SO3 inoleum = (% Strength-100) 18 [E = M (D ON per Na2 S 2 O 3 = 1)] G G Redox Reaction and Redox Titration n-Factor and Normality/Molarity Relationship Eq (i) Oxidation Loss of electrons or increase in oxidation N = numberiscalledoxidation. V n (ii) Reduction Gain of electron or decrease in oxidation and M = numberiscalledreduction. V N Eq +6 + 2 + 3 Þ = KCrO2 2 7 +FeSO 4 +HSO 2 4¾® Fe(SO) 2 4 3 M n +3 MW = = n-factor + Cr2 (SO 4 ) 3 EW In the above redox reaction, chromium is reduced from (n-factor=Changeinoxidationnumberperformulaunit). (+6to+3) and iron is oxidised from (+ 2 to + 3). Hence, G Redox Titration Attheendpoint: K2 Cr 2 O 7 is known as oxidising agent (itself reduced) GramequivalentsofOA=GramequivalentofRA. and FeSO4 reducingagent(itselfoxidised). 4 SomeBasicConceptsofChemistry Topic 1 Mole Concept ObjectiveQuestionsI (Onlyonecorrectoption) 10. Thenormalityof0.3Mphosphorusacid(H3PO3)is 1. The molecular formula of a commercial resin used for (a)0.1 (b)0.9 (1999,2M) (c)0.3 (d)0.6 exchanging ions in water softening is C8 H 7 SO 3 Na (molecular weight = 206). What would be the maximum 11. In which mode of expression, the concentration of a solution 2+ uptake of Ca ions by the resin when expressed in mole per remainsindependentoftemperature? (1988,1M) gramresin? (2015JEEMain) (a)Molarity (b)Normality (c)Formality (d) Molality 1 1 2 1 (a) (b) (c) (d) 12. Amolalsolutionisonethatcontainsonemoleofsolutein 103 206 309 412 (a)1000gofsolvent (1986,1M) 2. 3g of activated charcoal was added to 50mL of acetic acid (b)1.0Lofsolvent solution (0.06N) in a flask. After an hour it was filtered and (c)1.0Lofsolution the strength of the filtrate was found to be 0.042 N. The (d)22.4Lofsolution amountofaceticacidadsorbed(pergramofcharcoal)is 13. If 0.50 mole of BaCl is mixed with 0.20 mole of Na PO , (a)18mg (b)36mg (2015JEEMain) 2 3 4 the maximum number of moles of Ba (PO ) that can be (c)42mg (d)54mg 3 4 2 formedis (1981,1M) 3. The ratio mass of oxygen and nitrogen of a particular gaseous (a)0.70 (b)0.50 (c)0.20 (d)0.10 mixtureis1:4.Theratioofnumberoftheirmoleculeis 14.
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