04. basic concepts session-4.ppt

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<ul><li><p>SOME BASIC CONCEPT OF CHEMISTRYSESSION -IV</p></li><li><p>Session Opener</p></li><li><p>Session ObjectivesEquivalent mass</p><p>Normality</p><p>Molarity</p><p>Molality</p><p>Strength of solution</p><p>Percentage concentration</p></li><li><p>Equivalent MassEquivalent massBaseSaltAcid</p></li><li><p>Equivalent Mass of AcidEquivalent mass of acid = Example:Equivalent mass of HCl and H2SO4</p></li><li><p>Equivalent Mass of BaseEquivalent mass of base = Example:Equivalent mass of NaOH and Ca(OH)2</p></li><li><p>Equivalent mass of saltEquivalent mass of salt = Example:Equivalent mass of NaCl and MgCl2</p></li><li><p>Concentration of solutions(1) NormalityNumber of equivalents of solutepresent in one litre of solution.Equivalents = N x V (in litre)Milli equivalents = N x V (in ml)</p></li><li><p>Illustrative exampleFind the normality of H2SO4 having49g of H2SO4 present in 500 ml ofsolution.Solution:= 2N</p></li><li><p>Most important point about equivalentsEquivalent and milliequivalents ofreactants reacts in equal number togive same number of equivalents ormilliequivalents of products separately.Example:</p></li><li><p>Illustrative ProblemSolution:By equivalent method, no need of balancing theequation. Because equivalents of reactants andproducts are same.Equivalents of BaCl2 = = 2 x 103= 6 x 103Equivalents of Al2(SO4)3 =</p></li><li><p>Solution contd-Since equivalents of Al2(SO4)3 is in excess, henceequivalents of BaSO4= equivalents of BaCl2= equivalents of AlCl3= 2 x 103Hence, mass of BaSO4 =Equivalents x equivalent mass</p></li><li><p>MolarityNumber of moles of solute present inone litre of solution.Moles = Molarity x volume (in litre)Milli moles = Molarity x volume (in ml)</p></li><li><p>Illustrative exampleCalculate the molarity of a solution ofNaOH in which 0.40g NaOH dissolvedin 500 ml solution.Solution:= 0.02 M</p></li><li><p>Relation between normality and molarityN = M x n factorFor HCl, n = 1 H2SO4, n = 2 H3PO4, n = 3 NaOH, n = 1 Ca(OH)2, n = 2</p></li><li><p>Illustrative ProblemCalculate molarity of 0.6 N AlCl3 solution.Solution:n = 3</p></li><li><p>MolalityNumber of moles of solute present in 1 Kg(or 1000 gram) of solvent. It is representedby m (small letter).</p></li><li><p>Illustrative ProblemCalculate the molality of 1 molar solutionof NaOH given density of solution is 1.04gram/ml.Solution:1 molar solution means 1 mole of solute present perlitre of solution.Therefore, mass of 1 litre solution = 1000 x 1.04= 1040 gramMass of solute = 1 x 40 = 40gTherefore, mass of solvent 1040 40 = 1000g</p></li><li><p>Strength of solutionAmount of solute present in one litresolution.</p></li><li><p>Illustrative ProblemCalculate strength of 0.01 N of NaOHsolution.Solution:Strength = Normality x equivalent mass= 0.01 x 40 = 0.4 gram/litre</p></li><li><p>Concentration in terms of percentage</p></li><li><p>Illustrative ProblemCalculate the concentration of 1 molalsolution of NaOH in terms of percentageby mass.Solution:1 molal solution means 1 mole (or 40g) NaOH presentin 1000g of solvent.Total mass of solution = 1000 + 40 = 1040gTherefore, 1040g solution contains 40g NaOHTherefore, 100g solution contains= 3.84% by mass.</p></li><li><p>Class exercise 10.115 g of pure sodium metal was dissolved in 500 ml distilled water. The molarity of the solution would be (Na = 23)(a) 0.010 M(b) 0.00115 M(c) 0.023 M(d) 0.046 MHence, answer is (a)Solution:</p></li><li><p>Class exercise 2The number of moles of oxygen in one litre of air containing 21% oxygen by volume, in standard conditions, is(a) 0.186 mole(b) 0.21 mole</p><p>(c) 2.10 mole(d) 0.0093 mole21% oxygen by volume means 21 ml oxygen is present in 100 ml of solution.1,000 ml of solution will contain 210 ml.Since at STP 22,400 ml of gas = 1 mole,Hence, answer is (d)Solution:</p></li><li><p>Class exercise 3The vapour density of a gas is 11.2. The volume occupied by 11.2 g of the gas at STP will be(a) 11.2 L(b) 22.4 L(c) 1 L(d) 44.8 LMolecular mass = 2 Vapour density = 2 11.2 = 22.4Since 22.4 g contains 22.4 L of gas at STP,Hence, answer is (a)Solution:</p></li><li><p>Class exercise 4</p><p>The number of water molecules in one litre of water is(a) 18(b) 18 1000</p><p>(c) NA(d) 55.55 NA</p><p>For water d = 1 g/mlSince, One litre water = 1,000 g of waterNumber of water molecules = 55.55 NAHence, answer is (d)Solution:</p></li><li><p>Class exercise 5</p><p>Which is not affected by temperature?(a) Normality(b) Molarity and molality(c) Molarity(d) MolalityMolality involves mass of solute and solvent which are not affected by temperature.Solution:</p></li><li><p>Class exercise 6</p><p>Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea (molecular mass = 60) in 250 g of water.Mole fraction of urea =Mole fraction of water = 1 0.00359 = 0.996Solution:</p></li><li><p>Class exercise 7</p><p>Calculate the molarity and normality of a solution containing 0.5 g ofNaOH dissolved in 500 ml.Or for monovalent compound like NaOH normality and molarity are same.Solution:</p></li><li><p>Class exercise 8</p><p>Calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass.95% of ethanol by mass means 95 g ethanol present in 100 g of solution.Hence, mass of water = 100 95 = 5 g= 2.07 molesSolution:</p></li><li><p>Solution</p><p>Mole fraction of C2H5OH = Mole fraction of water = 1 0.88 = 0.12</p></li><li><p>Class exercise 9</p><p>A solution contains 25% of water, 25% of ethanol and 50% of acetic acid by mass. Calculate the mole fraction of each component.25x + 25x + 50x = 100x = 1Mass of water = 25 gMass of ethanol = 25 gMass of acetic acid = 50 gMoles of water =Solution:</p></li><li><p>SolutionMoles of ethanol =Moles of acetic acid =Mole fraction of ethanol =Mole fraction of acetic acid = 1 0.503 0.196 = 0.301</p></li><li><p>Class exercise 10</p><p>20 ml of 10 N HCl are diluted with distilled water to form one litre of the solution. What is the normality of thediluted solution?N1V1 = N2V2 N2 = 0.2 NSolution:</p></li></ul>