Some Basic Concept of Chemistry

Some Basic Concept of Chemistry INTRODUCTION Classification of matter There are a large number of objects around us which we can see and feel. on the basis of Anything that occupies space and has mass is physical behaviour called matter. It was John Dalton who firstly developed a theory Solids on the structure of matter, later on which is known as Dalton’s atomic theory. on the basis of chemical behaviour Liquids Gases Pure substances Mixtures DALTON’S ATOMIC THEORY     Matter is made up of very small indivisible particles called atoms. Element Compound BASIC DEFINITIONS All the atoms of a given element are identical in all Relative atomic mass : respect i.e. mass, shape, size, etc. One of the most important concept come out from Atoms cannot be created or destroyed by any Dalton’s atomic theory was that of relative atomic chemical process. mass or relative atomic weight. This is done by expressing mass of one atom with respect to a fixed Atoms of different elements are different in nature. standard. Dalton used hydrogen as the standard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the reference. Therefore relative atomic mass is given as 1 Some Basic Concept of Chemistry On hydrogen scale : Relative atomic mass (R.A.M) = Mass of one atom of an element mass of one hydrogen atom On oxygen scale : Relative atomic mass (R.A.M) = Mass of one atom of an element 1  mass of one oxygen atom 16 Ex.-1 Find the relative atomic mass of ‘O’ atom and its atomic mass. Sol. The number of nucleons present in ‘O’ atom is 16.  relative atomic mass of ‘O’ atom = 16. Atomic mass = R.A.M × 1 amu = 16 × 1 amu = 16 amu MOLE The Mass / Number Relationship  The present standard unit which was adopted internationally in 1961, is based on the mass of one Mole is a chemical counting S unit and defined as carbon-12 atom. follows : Relative atomic mass (R.A.M) A mole is the amount of a substance that contains as many entities (atoms, molecules or Mass of one atom of an element = other particles) as there are atoms in exactly 1  mass of one C  12 atom 0.012 kg (or 12 gm) of the carbon-12 isotope. 12 From mass spectrometer we found that there are Atomic mass unit (or amu) : 6.023 × 1023 atoms present in 12 gm of C-12 th  1 isotope. The atomic mass unit (amu) is equal to   mass  12  The number of entities in 1 mol is so important that it of one atom of carbon-12 isotope. is given a separate name and symbol known as 1 Avogadro constant denoted by NA.  1 amu = × mass of one C-12 atom 12 i.e. on the whole we can say that 1 mole is the ~ mass of one nucleon in C-12 atom. collection of 6.02 × 1023 entities. Here entities may represent atoms, ions, molecules or even pens, chair, = 1.66 × 10–24 gm or 1.66 × 10–27 kg paper etc also include in this but as this number (NA)  one amu is also called one Dalton (Da). is very large therefore it is used only for very small  Today, amu has been replaced by ‘u’ which is things. known as unified mass Atomic & molecular mass : It is the mass of 1 atom of a substance it is expressed Note : In modern practice gram-atom and grammolecule are termed as mole. in amu.  Atomic mass = R.A.M × 1 amu Relative molecular mass GRAM ATOMIC MASS = mass of one molecule of the subs tan ce 1  mass of one  C 12 atom 12  Molecular mass = Relative molecular mass × 1 amu The atomic mass of an element expressed in gram is called gram atomic mass of the element. or It is also defined as mass of 6.02 × 1023 atoms. Note : Relative atomic mass is nothing but the number of nucleons present in the atom. or It is also defined as the mass of one mole atoms. 2 Some Basic Concept of Chemistry For example for oxygen atom : Atomic mass of ‘O’ atom = mass of one ‘O’ atom gram atomic mass = mass of 6.02 × 1023 ‘O’ atoms Ex.-3 The molecular mass of H2SO4 is 98 amu. Calculate the number of moles of each element in 294 g of H2SO4. = 16 amu × 6.02 × 1023 Sol. Gram molecular mass of H2SO4 = 98 gm, moles of = 16 amu –24 = 16 × 1.66 × 10 23 g × 6.02 ×10 = 16 g ( 1.66 × 10–24 × 6.02 × 1023 ~ 1 ) H2SO4 = H2SO4 294 98 = 3 moles H One molecule 2 atom Ex.-2 How many atoms of oxygen are their in 16 g oxygen. Sol. Let x atoms of oxygen are present So, 16 × 1.66 × 10 –24 S O one atom 4 atom 1 × NA 2 × NA atoms 1 × NA atoms 4 × NA atoms  one mole 2 mole one mole  3 mole 6 mole 3 mole 4 mole 12 mole × x = 16 g GAY-LUSSAC’S LAW OF COMBINING 1 x = 1.66 x 10  24 = NA VOLUME GRAM MOLECULAR MASS The molecular mass of a substance expressed in gram is called the gram-molecular mass of the substance. or It is also defined as mass of 6.02 × 1023 molecules According to him elements combine in a simple ratio of atoms, gases combine in a simple ratio of their volumes provided all measurements should be done at the same temperature and pressure H2 (g) + Cl2 (g)  2HCl 1 vol 1 vol 2 vol Avogadro’s hypothesis : or It is also defined as the mass of one mole molecules. For example for ‘O2’ molecule : Molecular mass of ‘O2’ molecule Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressure condition. S.T.P. (Standard Temperature and Pressure) = mass of one ‘O2’ molecule At S.T.P. condition : temperature= 0°C or 273 K pressure = 1 atm = 760 mm of Hg = 2 × mass of one ‘O’ atom = 2 × 16 amu and volume of one mole of gas at STP is found to be experimentally equal to 22.4 litres which is known as molar volume. = 32 amu gram molecular mass = mass of 6.02 × 1023 ‘O2’ molecules = 32 amu × 6.02 × 1023 –24 = 32 × 1.66 × 10 gm × 6.02 × 10 23 Note : Measuring the volume is equivalent to counting the number of molecules of the gas. = 32 gm 3 Some Basic Concept of Chemistry Ex.-4 Calculate the volume in litres of 20 g hydrogen gas at STP. Sol. No. of moles of hydrogen gas Ex.-5 H2 (g) + 1 O2 (g)  2 H2 (g) + Sol. H2O ( ) 1 O2 (g)  H2O ( ) 2 Mass = Molecular mass 20 gm = 2 gm = 10 mol volume of hydrogen gas at STP = 10 × 22.4 lt. Before 1 mole 2 1 mole reaction initially After 0 the reaction 0 0 1 mole mass before reaction = mass of 1 mole H 2 (g) + 1 mole O2 (g) 2 Y-MAP = 2 + 16 = 18 gm Interconversion of mole - volume, mass after reaction = mass of 1 mole water mass and number of particles : = 18 gm The laws of chemical combination : Atoine Lavoisier, John Dalton and other scientists formulate certain law concerning the composition of matter and chemical reactions. These laws are known as the law of chemical combination. (i) The law of conservation of mass : In a chemical change total mass remains conserved. i.e. mass before reaction is always equal to mass after reaction. (ii) Law of constant or Definite proportion : All chemical compounds are found to have constant composition irrespective of their method of preparation or sources. Example : In water (H2O), Hydrogen and Oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is tap water, river water or sea water or produced by any chemical reaction. Ex.-6 1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion. Sol. In the first sample of the oxide, Wt. of metal = 1.80 g, Wt. of oxygen = (3.0 – 1.80) g = 1.2 g wt. of metal 1.80g  wt.of oxygen  1.2g  1.5 In the second sample of the oxide, Wt. of metal = 1.50 g, Wt. of oxygen = (2.50 – 1.50) g = 1 g. 4 wt. of metal Some Basic Concept of Chemistry  1 part by mass of oxygen will combine with carbon 1.50 g  wt.of oxygen  1g  1.5 42 .9 Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen a fixed. Hence, the results follow the law of constant proportion. = 57 .1 = 0.751. n the second oxide. 72.7 parts by mass of oxygen combine with carbon = 27.3 parts.  1 part by mass of oxygen will combine with carbon (iii) The law of multiple proportion : When one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other, bear a simple ratio to one another. Note : Simple ratio here means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3, later on this simple ratio becomes the valency and then oxidation state of the element. See oxidation number of carbon also have same ratio 1 : 2 in both the oxide. = 27.3 = 0.376 72.7 Step-3. To compare the masses of carbon which combine with the same mass of oxygen in both the oxides. The ratio of the masses of carbon that combine with the same mass of oxygen (1 part) is . 0.751 : 0.376 or 2 : 1 Since this is simple whole number ratio, so the above data illustrate the law of multiple proportions. PERCENTAGE COMPOSITION Ex.-7 Carbon is found to form two oxides, which contain 42.9% and 27.3% of carbon respectively. Show that these figures illustrate the law of multiple proportions. Sol. Step-1 Here we are going to find out the percentage of each element in the compound by knowing the molecular formula of compound. We know that according to law of definite proportions any sample of a pure compound always possess constant ratio with their combining elements. To calculate the percentage composition of carbon and oxygen in each of the two oxides. First oxide Second oxide Carbon 42.9 % 27.3 % Oxygen 57.1% 72.7 % (Given) (by difference) Step-2 Ex.-8 Every molecule of ammonia always has formula NH3 irrespective of method of preparation or sources. i.e. 1 mole of ammonia always contains 1 mol of N and 3 mole of H. In other words 17 gm of NH3 always contains 14 gm of N and 3 gm of H. Now find out % of each element in the compound. To calculate the masses of carbon which combine Sol. Mass % of N in NH3 with a fixed mass i.e., one part by mass of oxygen in Mass of N in 1 mol NH3  100 = Mass of 1 mol of NH3 each of the two oxides. n the first oxide, 57.1 parts by mass of oxygen combine with carbon = 42.9 parts. = 14 gm × 100 = 82.35 % 17 5 Some Basic Concept of Chemistry Mass % of H in NH3 Mass of H is 1 mol NH3 = Mass of 1 mol e of NH  100 3 = 3 × 100 = 17.65 % 17 Step-2 To calculate the value of ‘n’ Molecular mass 26 n = Empirical formula mass = =2 13 Step-3 To calculate the molecular formula of the compound. EMPIRICAL AND MOLECULAR Molecular formula FORMULA = n × (Empirical formula of the compound) i.e.molecular formula = empirical formula × n where molecular formula mass n = empirical formula mass = 2 × CH = C2 H2 Thus the molecular formula is C2 H2 Similarly for benzene To calculate the value of ‘n’ Ex.-9 Acetylene and benzene both have the empirical formula CH. The molecular masses of acetylene and benzene are 26 and 78 respectively. Deduce their molecular formulae. Sol.  Empirical Formula is CH Step-1 The empirical formula of the compound is CH  Empirical formula mass = (1 × 12) + 1 = 13. Molecular mass = 26 Molecular mass 78 n = Empirical formula mass = =6 13 thus the molecular formula is 6 × CH = C6H6 Ex.-10 An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition. C = 40.684% ; H = 5.085% and O = 54.228% The molecular weight of the compound is 118. Calculate the molecular formula of the compound. Sol. Step-1 To calculate the empirical formula of the compound.  Empirical Formula is C2 H3 O2 Step-2 To calculate the empirical formula mass. The empirical formula of the compound is C2 H3 O2  Empirical formula mass = (2 × 12) + (3 × 1) + (2 × 16) = 59. 6 Some Basic Concept of Chemistry Step-3 To calculate the value of ‘n’ Molecular mass 118 n = Empirical formula mass = =2 59 Step-4 To calculate the molecular formula of the salt. Molecular formula = n × (Empirical formula) = 2 × C2 H3 O2 = C4 H6 O4 Thus the molecular formula is C4 H6 O4. Ex.-11 What is the V.D. of SO2 with respect to CH4 M.W. SO 2 Sol. V.D. = M.W. CH 4 Ex.-12 7.5 litre of the particular gas at S.T.P. weighs 16 gram. What is the V.D. of gas Sol. 7.5 litre = 16 gram moles = DENSITY It is of two type. (i) Absolute density, (ii) Relative density. For Liquid and Solids mass volume  Absolute density =  Relative density or specific gravity V.D = CHEMICAL EQUATIONS For Gases : Molar mass Absolute density (mass/volume) = Molar volume Relative density or Vapour density : Once we get a balanced chemical equation then we can interpret a chemical equation by following ways  Mass - mass analysis  Mass - volume analysis  Mole - mole analysis  Vol - Vol analysis (separately discussed as eudiometry or gas analysis) Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature and pressure. d gas Vapour density = d H2 Mgas 48 = 24 2 INTERPRETATION OF BALANCED density of the subs tance V.D. = M = H2 7. 5 16  22.4 M M = 48 gram = density of water at 4C We know that density of water at 4ºC = 1 g/ml.  64 V.D = 16 = 4 M gas 2 Mgas = 2 V.D. Relative density can be calculated w.r.t. to other gases also. Now you can understand the above analysis by following example  Mass-mass analysis : Consider the reaction 2KClO3  2KCl + 3O2 According to stoichiometry of the reaction mass-mass ratio : 2 × 122.5 : 2 × 74.5 : 3 × 32 or Mass of KClO 3 2  122 .5 = Mass of KCl 2  74 .5 Mass of KClO 3 2  122 .5 Mass of O 2 = 3  32 7 Some Basic Concept of Chemistry Ex.-14 367.5 gram KClO3 (M = 122.5) when heated. How many gram KCl and oxygen is produced. Ex.-15 367.5 gm KClO3 (M = 122.5) when heated, how many litre of oxygen gas is produced at STP. Sol. Balance chemical equation for heating of KClO3 is Sol. You can use here equation (1) 2KClO3  2KCl + 3O2 mass of KClO 3 2  122 .5 = volume of O 2 at STP 3  22. 4 lt mass-mass ratio : 2 × 122.5 gm : 2 × 74.5 gm : 3 × 32 gm  mass of KClO 3 2  122 .5 mass of KCl = 2  74.5  367.5 W V = 3 × 3 × 11.2   Mole-mole analysis : 2KClO3  2KCl + 3O2 Moles of KClO3 Moles of KCl = 2 2 2  122 .5 367.5 = 3  32 W = 144 gm W =  V = 100.8 lt 122 .5 = 74 .5 W = 3 × 74.5 = 223.5 gm Mass of KClO3 2  122 .5 Mass of O 2 = 3  32  2  122 .5 367.5 = 3  22.4 lt V Moles of O 2 3 Mass - volume analysis : Now for any general balance chemical equation like Now again consider decomposition of KClO3 a A + b B  c C + d D 2KClO3  2KCl + 3O2 you can write. mass volume ratio : Moles of A reacted a 2 × 122.5 gm : 2 × 74.5 gm : 3 × 22.4 lt. at STP we can use two relation for volume of oxygen Mass of KClO 3 volume of O 2 at STP 2  122 .5 = 3  22.4 lt ...(i) and Mass of KCl 2  74 .5 = volume of O 2 at STP 3  22.4 lt = = moles of B reacted b moles of C produced c = moles of D produced d Note : In fact mass-mass and mass-vol analysis are also interpreted in terms of mole-mole analysis you can use following chart also. ...(ii) 8 LIMITING REAGENT The reactant which is consumed first and limits the amount of product formed in the reaction, and is therefore, called limiting reagent. Limiting reagent is present in least stoichiometric amount and therefore, controls amount of product. The remaining or left out reactant is called the excess reagent. When you are dealing with balance chemical equation then if number of moles of reactants are not in the ratio of stoichiometric coefficient of balanced chemical equation, then there should be one reactant which is limiting reactant. Some Basic Concept of Chemistry Sol. From the reaction : Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O given mole of reactant 6 : 4 given molar ratio 3 : 2 Stoichiometric coefficient ratio 1 : 2 See here given number of moles of reactants are not in stoichiometric coefficient ratio. Therefore there should be one reactant which consumed first and becomes limiting reagent. But the question is how to find which reactant is limiting, it is not very difficult you can easily find it. According to the following method. How to find limiting reagent : Step : I Ex.-16 Three mole of Na2 CO3 is reacted with 6 moles of HCl solution. Find the volume of CO2 gas produced at STP. The reaction is Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O Sol. From the reaction : Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O given moles 3 mol 6 mol given mole ratio 1 : 2 Stoichiometric coefficient ratio 1 : 2 See here given moles of reactant are in stoichiometric coefficient ratio therefore none reactant left over. Divide the given moles of reactant by the respective stoichiometric coefficient of that reactant. Step : II See for which reactant this division come out to be minimum. The reactant having minimum value is limiting reagent for you. Step : III Now once you find limiting reagent then your focus should be on limiting reagent From Step  &  6 =6 1 Now use Mole-mole analysis to calculate volume of CO2 produced at STP Moles of Na 2CO 3 Mole of CO 2 Pr oduced = 1 1 = 67.2 L Ex.-17 6 moles of Na2 CO3 is reacted with 4 moles of HCl solution. Find the volume of CO2 gas produced at STP. The reaction is HCl 4 =2 2 (division is minimum)  HCl is limiting reagent From Step  Moles of CO2 produced = 3 volume of CO2 produced at STP = 3 × 22.4 L Na2 CO3 From Mole of HCl Moles of CO 2 produced = 2 1  mole of CO2 produced = 2 moles  volume of CO2 produced at S.T.P. = 2 × 22.4 = 44.8 lt. Na2 CO3 + 2HCl  2 NaCl + CO2 + H2O 9 Some Basic Concept of Chemistry PRINCIPLE OF ATOM CONSERVATION or 3 × (POAC) POAC is conservation of mass. Atoms are conserved, moles of atoms shall also be conserved in a chemical reaction (but not in nuclear reactions.) wt. of KClO 3 mol. wt. of KClO 3 =2× vol. of O 2 at NTP s tan dard molar vol. ( 22 .4 lt.) This principle is fruitful for the students when they  The above equations thus gives the mass-volume relationship of reactants and products. don’t get the idea of balanced chemical equation in the problem. The strategy here will be around a particular atom. Ex.-18 27.6 g K CO was treated by a series of 2 3 We focus on a atom and conserve it in that reaction. reagents so as to convert all of its carbon to K2 Zn3 This principle can be understand by the following [Fe(CN)6]2. Calculate the weight of the product. example. [mol. wt. of K2CO3 = 138 and mol. wt. of K2Zn3 Consider the decomposition of KClO3 (s)  [Fe(CN)6]2 = 698] KCl (s) + O2 (g) (unbalanced chemical reaction) Sol. Here we have not knowledge about series of chemical reactions Apply the principle of atom conservation (POAC) for K atoms. but we know about initial reactant and final product accordingly Moles of K atoms in reactant = moles of K atoms in products Several K2CO3    K2Zn3 [Fe(CN)6]2 Steps or moles of K atoms in KClO3 Since C atoms are conserved, applying POAC for C atoms, moles of C in K2CO3 = moles of K atoms in KCl. Now, since 1 molecule of KClO3 contains 1 atom of K or 1 mole of KClO3 contains 1 mole of K, similarly,1 mole of KCl contains 1 mole of K. = moles of C in K2Zn3 [Fe(CN)6]2 Thus, moles of K atoms in KClO3 = 12 × moles of K2Zn3 [Fe(CN)6]2 = 1 × moles of KClO3 ( 1 mole of K2CO3 contains 1 moles of C) and moles of K atoms in KCl = 1 × moles of KCl. wt. of K 2CO 3 wt. of the product mol. wt. of K 2CO 3 = 12 × mol. wt. of product 1 × moles of K2CO3  moles of KClO3 = moles of KCl or wt. of KCl in g wt. of KClO 3 in g = mol. wt. of KCl mol. wt. of KClO 3  The above equation gives the mass-mass relationship between KClO3 and KCl which is important in stoichiometric calculations. Again, applying the principle of atom conservation for O atoms, moles of O in KClO3 = 3 × moles of KClO3 moles of O in O2 = 2 × moles of O2  3 × moles of KClO3 = 2 × moles of O2 27 .6 698 wt. of K2Zn3 [Fe(CN)6]2 = 138 × = 11.6 g 12 Concentration terms : The following concentration terms are used to expressed the concentration of a solution. These are  Molarity (M)  Molality (m)  Mole fraction (x)  % calculation 10 Some Basic Concept of Chemistry  Normality (N)  ppm Ex.-19 149 gm of potassium chloride (KCl) is dissolved  Remember that all of these concentration terms in 10 Lt of an aqueous solution. Determine the are related to one another. By knowing one molarity of the solution concentration term you can also find the other (K = 39, Cl = 35.5) concentration terms. Let us discuss all of them one Sol. Molecular mass of KCl = 39 + 35.5 = 74.5 gm by one. 149 gm  Moles of KCl = 74 .5 gm = 2 MOLARITY (M) : The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity of the solution.  Molarity of the solution = 2 10 = 0.2 M MOLALITY (m) : The number of moles of solute dissolved in1000 gm (1 kg) of a solvent is known as the molality of the solution. i.e., Molarity of solution number of moles of solute = volume of solution in litre w  1000 number of moles of solute Molarity (M) = (Mol. wt of solute )  V ml i.e., molality = mass of solvent in gram  1000 Some other relations may also useful. Let Y gm of a solute is dissolved in X gm of a solvent. The molecular mass of the solute is M0. Then Y/M0 mole of the solute are dissolved in X gm of the solvent. Hence  Molarity of solution may also given as : Number of millimole of solute Total volume of solution in ml  Molarity is a unit that depends upon temperature. It varies inversely with temperature . 1 1 Molarity  temperature  volume Molality = Y  1000 M0  X  Molality is independent of temperature changes.  If a particular solution having volume V1 and molarity Ex.-20 225 gm of an aqueous solution contains 5 gm of urea. What is the concentration of the solution in = M1 is diluted upto volume V2 mL than terms of molality. (Mol. wt. of urea = 60) M1V1 = M2V2 Sol. Mass of urea = 5 gm M2 : Resultant molarity Molecular mass of urea = 60  If a solution having volume V1 and molarity M1 is 5 mixed with another solution of same solute having Number of moles of urea = 60 = 0.083 volume V2 mL & molarity M2 Mass of solvent = (255 – 5) = 250 gm then  Molality of the solution M1V1 + M2V2 = MR (V1 + V2) Number of moles of solute = Mass of solvent in gram × 1000 M1V1  M2 V2 MR = Resultant molarity = V  V 1 2 = 0.083 250 × 1000 = 0.332. 11 MOLE FRACTION (x) : The ratio of number of moles of the solute or solvent present in the solution and the total number of moles present in the solution is known as the mole fraction of substances concerned. Let number of moles of solute in solution = n Number of moles of solvent in solution = N  Mole fraction of solute (x1) = Some Basic Concept of Chemistry Ex.-22 20 cm of an alcohol is dissolved in80 cm3 of water. Calculate the percentage of alcohol in solution. 3 Sol. Volume of alcohol = 20 cm3 Volume of water = 80 cm3  Percentage of alcohol 20 = 20  80  100 = 20. n nN  Mole fraction of solvent (x2) = MISCELLANEOUS N nN  AVERAGE/ MEAN ATOMIC MASS : also x1 + x2 = 1  Mole fraction is a pure number. It will remain independent of temperature changes. The weighted average of the isotopic masses of the element’s naturally occuring isotopes. % calculation : M athematically, average atomic mass of X (A x) = a. The concentration of a solution may also expressed in terms of percentage in the following way. a1x 1  a 2 x 2  .....  an x n 100 % weight by weight (w/w) : It is given as mass of solute present in per 100 gm of solution. Where : mass of solute in gm i.e. % w/w = mass of solution in gm  100 b. % weight by volume (w/v) : It is given as mass of solute present in per 100 ml of solution. i.e., % w/v = c. mass of solute in gm  100 volume of solution in ml % volume by volume (v/v) : It is given as volume of solute present in per 100 ml solution. i.e., % v/v = volume of solute in ml volume of solution in ml × 100 Ex.-21 0.5 g of a substance is dissolved in 25 g of a solvent. Calculate the percentage amount of the substance in the solution. a1, a2, a3 ........... atomic mass of isotopes. and x1, x2, x3 ........... mole % of isotopes. Ex.-23 Naturally occuring chlorine is 75% Cl35 which has an atomic mass of 35 amu and 25% Cl37 which has a mass of 37 amu. Calculate the average atomic mass of chlorine (A) 35.5 amu (C) 71 amu (B) 36.5 amu (D) 72 amu Sol. (A) Average atomic mass = % of  isotopex its atomsmass % of I isotopex its atomicmass 100 = 75 x 35  25 x 37 100 = 35.5 amu Sol. Mass of substance = 0.5 g Mass of solvent = 25 g  percentage of the substance (w/w) = 0. 5  100 0.5  25 Note : (a) In all calculations we use this mass. (b) In periodic table we report this mass only. = 1.96 12 Some Basic Concept of Chemistry  MEAN MOLAR MASS OR MOLECULAR MASS: The average molar mass of the different substance present in the container = n1M1  n2M2  ......nnMn . n1  n 2  ....nn Where : M1, M2, M3 ........... are molar masses. CONCEPT OF EQUIVALENTS Equivalent mass of element Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of chlorine, is known as the equivalent weight of that element. e.g. n1, n2, n3 ........... moles of substances. Ex.-24 The molar composition of polluted air is as follows Gas At. wt. mole % composition Oxygen 16 16% Nitrogen 14 80% Carbon dioxide - 03% Sulphurdioxide - 01% What is the average molecular weight of the given polluted air ? (Given, atomic weights of C and S are 12 and 32 respectively. 48g 32g 12g 8g  32 g of O2 reacts with 48 g of Mg  8 g of O2 =  Equivalent weight of Mg = 12 48  8 32 n M j  Equivalent weight of Zn = Al + j j1 n = = 3 × 71 g 2 27  35.5  35.5 chlorine reacts with 111 .5 = 9.0 g of Al j = 100 j1  Mavg = 3 Cl  AlCl3 2 2  111.5 g chlorine reacts with 27 g of Al. j n Here 65.5 = 32.75 g 2 j j1 j n n = 12 g Similarly, Zn + H2SO4  ZnSO4 + H2 65.5 g 32.75 27 g j n Sol. Mavg = 2Mg + O2  2MgO 16 x 32  80 x 28  44 x 3  64 x 1 100 512  2240  132  64 100 2948 100  Equivalent weight of aluminium = 27 3 = 9.0 As we can see from the above examples that equivalent weight is the ratio of atomic weight and a factor (say n- factor or valency factor) which is in above three cases is their respective valencies. Equivalent weight (E) : In general, Eq. wt. (E) = 29.48 Ans. Atomic weight or Molecular weight valency factor(v.f ) = 13 Some Basic Concept of Chemistry  Valency factor = number of replacable OH– ions per base molecule. Number of Equivalents mass of species For a solution, Number of equivalents = N1V1, where N is the normality and V is the volume in litres   NaOH, 1 Ex.-30 Sol. v .f.  = eq. wt. of that species  For Bases : Eq. wt.  M 1 KOH 1 M 1  Bases may be defined as the substances in which OH group is/are directly attached with group I Equivalent mass is a pure number which, when elements (Li,Na, K,Rb,Cs), group II elements (Be, expressed in gram, is called gram equivalent mass. Mg,Ca,Ba ) or group III elements (Al, Ga,In,Tl), The equivalent mass of substance may have different transition metals, non-metallic cations like PH 4+ , NH4+ etc. values under different conditions. There in no hard and fast rule that equivalent weight  will be always less than the molecular mass. Valency factor calculation  For Elements : Valency factor = valency of the element.  For Acids : Valency factor = number of replaceable H+ ions per acid molecule Ex.-31 2NaOH + H2 SO4  Na2 SO4 + 2H2O Base Acid Sol. Valency factor of base = 1 Here, two molecule of NaOH replaced 2H+ ion from the H2 SO4. Therefore, each molecule of NaOH replaced only one H+ ion of acid, so v.f. = 1. Ex.-29 HCl , H2SO4 H3PO4 H3PO 3 + {see there are only two replaceable H ions} Sol. Valency factor  1 2 3 2 (assume 100% dissicoiation) Eq. wt. (E) M 1 M 2 M 3 Acid - base reaction : In case of acid base reaction, the valence factor is the actual number of H+ or OH– replaced in the reaction. The acid or base may contain more number of replaceble H+ or OH– than actually replaced in reaction.  v. f. for base is the number of H+ ion from the acid replaced by each molecule of the base M 2  Replaceable hydrogen atoms are those hydrogen atoms which are attached with the atoms of group VI and group VII i.e. O,S,Se,Te, & F, Cl ,Br ,I.  v. f. for acid is the number of OH– replaced from the base by each molecule of acid Ex.-32 NaOH + H2SO4  NaHSO4 + H2O Base Acid Sol. Valency factor of acid = 1 Here, one of molecule of H2SO4 replaced one OH– from NaOH. Therefore, valency factor for H2SO4 is one  Eq. wt. of H2SO4 = Mol.wt 1 14 Some Basic Concept of Chemistry  Salts : NORMALITY (a)In non-reacting condition Normality of a solution is defined as the number of equivalents of solute present in one litre (1000 mL) solution.  Valency factor = Total number of positive charge or negative charge present in the compound. Let V mL of a solution is prepared by dissolving W g of solute of equivalent weight E in water. Ex.-33 Na2 CO3 , Sol. V.f. Eq.wt. Fe2(SO4)3 FeSO4.7H2O 2 2×3 = 6 2 M 2 M 6 M 2  Number of equivalents of solute = VmL of solution contain  1000 W  1000 EV mL W E W E equivalents of solute solution will contain equivalents of solute. Note : In case of hydrated salt, positive/ negative charge of water molecule is not counted.  Normality (N) = (b)In reacting condition  Normality (N) = Molarity x Valency factor W  1000 EV N × V (in mL) = M × V (in mL) × n Ex.-34 Na2 CO3 + HCl  NaHCO3 + NaCl Base Acid or  milliequivalents = millimoles × n Sol. It is an acid base reaction, therefore valency factor for Na2CO3 is one while in non-reacting condition, it will be two. Ex.-36 Calculate the normality of a solution containing 15.8 g of KMnO4 in 50 mL acidic solution. (c) Equivalent weight of oxidising / reducing agents in a redox reaction In case of redox change , v.f. = Total change in oxidation number per molecule . Sol. Normality (N) = Here E= Ex.-35 KMnO4 + H2O2  Mn2+ + O2 W  1000 E V W = 15.8 g , V = 50 mL molar mass of KMnO 4 = 158/5 = 31.6 Valency factor So, normality = 10 N Ex.-37 Calculate the normality of a solution containing Sol. Mn in KMnO4 is going from +7 to +2 , so change in 50 mL of 5 M solution of K2Cr2O7 in acidic oxidation number per molecule of KMnO4 is 5. So medium. the valency factor of KMnO4 is 5 and equivalent Sol. Normality (N) = Molarity × valency factor weight is M 5 . = 5 x 6 = 30 N 15 Some Basic Concept of Chemistry DRAWBACKS OF EQUIVALENT LAW OF EQUIVALENCE The law states that one equivalent of an element CONCEPT combine with one equivalent of the other. In a  Since equivalent weight of a substance (for example oxidising or reducing agent) may be variable hence chemical reaction, equivalents and milli equivalents it is better to use mole concept. of reactants react in equal amount to give same e.g. 5e– + 8H+ + MnO4–  Mn2+ + 2H2O number of equivalents or milli equivalents of products – Mol. wt. of MnO 4 separately.  Eq.wt of MnO4– = 5 Accordingly – –  e.g. 3e + 2H2O + MnO4  MnO2 + 4OH– (i) aA + bB  mM + nN  Mol. wt. of MnO 4 – Eq.wt of MnO =  4 meq of A = meq of B = meq of M = m.eq. of N 3 (ii) In a compound MxNy meq of MxNy = meq of M = meq of N  Ex.-38 Find the number of moles of KMnO4 needed to oxidise one mole Cu2S in acidic medium. The reaction is KMnO4 + Cu2S  Mn2+ + Cu2+ + SO2 Sol. From law of equivalence, equivalents of Cu2S = equivalents of KMnO4 moles of Cu2S × v.f. = moles of kMnO4 × v.f. 1 × 8 = moles of KMnO4 × 5   moles of KMnO4 = 8/5 ( v.f. of Cu2S = 2 (2 – 1) + 1 (4 – (–2))) = 8 and v.f. of KMnO4 = 1 (7 –2) = 5) Ex.-39 The number of moles of oxalate ions oxidized by one mole of MnO4– ion in acidic medium are : 2 5 2 3 (C) 5 (A) (B) 5 5 (D) 3 Sol. Equivalents of C2O42– = equivalents of MnO4– x(mole) × 2 = 1 × 5 ( v.f. of C2O42– = 2 (4 – 3) = 2 and v.f. of MnO4– = 1 (7 – 2) = 5). x= 5 mole of C2O42– ions. 2 Thus, the number of equivalents of MnO4– will be different in the above two cases but number of moles will be same. Normality of any solution depends on reaction while molarity does not. For example : Consider 0.1mol KMnO4 dissolved in water to make 1L solution. Molarity of this solution is 0.1 M. However, its normality is NOT fixed. It will depend upon the reaction in which KMnO4 participates. e.g. if KMnO4 forms Mn2+, normality = 0.1 x 5 = 0.5 N. This same sample of KMnO4, if employed in a reaction giving MnO2 as product (Mn in +4 state), will have normality 0.1 × 3 = 0.3 N. The concept of equivalents is handy, but it should be used with care. One must never equate equivalents in a sequence which involves same element in more than two oxidation states. Consider an example, KIO3 reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo solution. The reactions are : (i) O3– + ¯  2 (ii) 2 + S2O32–  S4O62– + ¯ meq of hypo = meq of I2 = meq of IO3 = meq of I  meq of hypo = meq of IO3-. This is wrong. Note that I2 formed by equation (i) has v.f. = 5/3 & in equation (ii) has v.f. = 2.  v.f. of I2 in both the equation are different, therefore we cannot equate milli equivalents in sequence. In this type of case, students are advised to use mole concept. 16 Some Basic Concept of Chemistry TITRATIONS Titration is a procedure for determining the Ex.-40 How many millilitres of 0.02 M KMnO4 solution concentration of a solution by allowing a carefully would be required to exactly titrate 25 mL of 0.2 M measured volume to react with a standard solution Fe(NO3)2 solution in acidic medium ? of another substance, whose concentration is known. Sol. Method -1 : Mole concept method Standard solution - It is a solution whose concentration is known and is taken in burette. It is 2+ Starting with 25 mL of 0.2 M Fe , we can write : also called Titrant. Millimoles of Fe2+ = 25 x 0.2 ........(1) There are two type of titrants : and in volume V (in milliliters) of the KMnO4,  Primary titrants/standard - These reagents can Millimoles of MnO4¯ = V (0.02) ........(2) be accurately weighed and their solutions are not to The balanced reaction is : be standardised before use. Ex : Oxalic acid, K2Cr2O7, AgNO3, CuSO4, ferrous MnO4¯ + 5Fe2+ + 8H+  ammonium sulphate, hypo etc. Mn2+ + 5Fe3+ + 4H2O  Secondary titrants/standard : These reagents This requires that at the equivalent point, cannot be accurately weighed and their solutions are 2 – to be standardised before use. m.moles of Fe m.moles of MnO 4 = 5 1 Ex : NaOH, KOH, HCl, H2SO4, 2, KMnO4 etc. Titrate : Solution consisting of substance to be ( 25 )(0.2) V(0.02)  = (from (1) & (2)) estimated, generally taken in a beaker . 5 1 Equivalence point : It is the point when number of  V = 50 mL. equivalents of titrant added becomes equal to Method -2 : Equivalent Method : number of equivalents of titrate. At the equivalence point, At equivalence point : 2+ milliequivalents of MnO4¯ = milliequivalents of Fe n1V1M1 = n2V2M2 M1 × vf1 × V1 = M2 × vf2 × V2 Indicator :An auxiliary substance added for physical detection of completion of titration at equivalence 0.02 × 5 × V1 = 0.2 × 1 × 25 point. It generally show colour change on completion ( MnO4–  Mn2+ ; v.f. = 5, Fe2+ of titration. 3+  Fe ; v.f. = 1) Type of Titrations :  Acid-base titrations (to be studided in Ionic  V1 = 50 mL. equilibrium)  Redox Titrations 17 Some Basic Concept of Chemistry Some Common Redox Titrations Table of Redox Titrations : (Excluding Iodometric / Iodimetric titrations) 1. 2. Estimation of By titrating with Fe2+ MnO4¯ Fe2+ Cr2O 72– Reactions Relation*between OA and Fe2+  Fe3+ + e– 5Fe2+  MnO4¯ MnO4– + 8H+ + 5e–  Mn2+ + 4H2O Eq. wt. of Fe2+ = M/1 Fe2+  Fe3+ + e– 6Fe2+  Cr2O72– Cr2O72– + 14H+ + 6e–  Eq.wt. of Cr2O72– = M/6 2Cr3+ + 7H2O 3. C2O 42– MnO4¯ C2O42–  2CO2 + 2e– 5C2O42–  2MnO4¯ MnO4– + 8H+ + 5e–  Eq. wt. of C2O42– = M/2 Mn2+ + 4H2O 4. H2O2 MnO4¯ H2O2  2H+ + O2 + 2e– 5H2O2  2MnO4¯ nO4– + 8H+ + 5e–  Eq.wt. of H2O2 = M/2 2+ Mn + 4H2O 5. As2O3 MnO4– As2O3 + 5H2O  Eq. wt. of As2O3 = M/4 2AsO43– + 10H+ + 4e– MnO4– + 8H+ + 5e–  Mn2+ + 4H2O 6. AsO33– BrO3– AsO33– + H2O  Eq. wt. of AsO33– = M/2 AsO43– + 2H+ + 2e– BrO3– + 6H+ + 6e–  Eq.wt. of BrO3– = M/6 Br– + 3H2O Permanganate Titrations :    KMnO4 is generally used as oxidising agent in acidic Ex.-41 Write the balanced reaction of titration of KMnO4 Vs oxalic acid in presence of H2SO4. medium, generally provided by dilute H2SO4 . KMnO works as self indicator persistent pink color Sol. Reaction : 2KMnO 4 + 3H 2 SO 4 + 5H 2 C 2 O 4 4 is indication of end point.  Mainly used for estimation of Fe2+ , oxalic acid ,oxalates, H2O2 etc. Redox Changes : C23+  2C4+ +2e K2SO4 + 2MnSO4 + 8H2O + 10CO2 M   EH2C 2 O 4   2  5e + Mn7+  Mn2+ M   EKMnO 4   5  Indicator : KMnO4 acts as self indicator. 18 Some Basic Concept of Chemistry IODOMETRIC/IODIMETRIC Ex.-42 Write the balanced reaction of titration of KMnO4 Vs ferrous ammonium sulphate in presence TITRATIONS of H2SO4. Compound containing iodine are widely used in Sol. Reaction : 2KMnO 4 + 10[FeSO 4 (NH 4 ) 2 SO 4. titrations. 6H2O] + 8H2SO4  (i) Iodide ions can be oxidised to  by suitable oxidising 2 agent 5Fe2(SO4)3 + 10(NH4)2SO4 + K2SO4 + 2MnSO4 + 68H2O Redox Changes : 2¯ (aq)  2 (s) + 2e¯ (ii) Iodine (V) ions, O3¯ , will oxidise ¯ to 2 7+ Mn + 5e  Mn O¯ (aq) + 5¯ (aq) + 6H+ (aq)  M   EFeSO 4   1  Fe2+  Fe3+ + e 2+ M   EKMnO 4   5  32 (s) + 3H2O () (iii) Thiosulphate ions, S2O32– , can reduce iodine to iodide ions. Indicator : KMnO4 acts as self indicator 2S2O (aq)  (s)  S4 O62– + 2– colourless black colourless Iodometric Titrations (Titration Solution is of Na2S2O3 . 5H2O) –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– S.No. Estimation of Reaction Relation between O.A. and R.A. –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 1. 2. 2 CuSO4 2 + 2Na2S2O3  2Na + Na2S4O6 2  2  2Na2S2O3 or 2 + 2S2O32–  2¯ + S4O62– Eq.wt. of Na2S2O3 = M/1 2CuSO4 + 4K  2Cu + 2K2SO4 + 2 2CuSO4  2  2 = 2Na2S2O3 or 2Cu2+ + 4¯  2Cu + 2 Eq.wt.of CuSO4 = M/1 white ppt 3. 4. CaOCl2 MnO2 CaOCl2 + H2O  Ca(OH)2 + Cl2 Cl2 + 2K  2KCl + 2 CaOCl2  Cl2  2  2  2Na2S 2O 3 Cl2 + 2¯  2Cl¯ + 2 Eq.wt. of CaOCl2 = M/2  MnO2 + 4HCl(conc.)  MnCl2 + Cl2 + 2H2O MnO2  Cl2  2  2Na2S2O3 Cl2 + 2K  2KCl + 2 Eq.wt. of MnO2 = M/2 or MnO2 + 4H+ + 2Cl¯  Mn2+ + 2H2O + Cl2 Cl2 + 2¯  2 + 2Cl¯ 5. O 3 ¯ O3¯ + 5¯ + 6H+  32 + 3H2O O3¯  32  6  6Na2S2O3 Eq.wt. of O3¯ = M/6 6. H 2O 2 H2O2 + 2¯ + 2H+  2 + 2H2O H2O2  2  2  2Na2S2O3 Eq.wt. of H2O2 = M/2 7. Cl2 Cl2 + 2¯  2Cl¯ + 2 Cl2  2  2  2Na2S2O3 Eq.wt. of Cl2 = M/2 19 8. O3 O3 + 6¯ + 6H +  Some Basic Concept of Chemistry O3  32  6  6Na2S2O3 32 + 3H2O Eq.wt. of O3 = M/6 9. ClO¯ ClO¯ + 2¯ + 2H+  H2O + Cl¯ + 2 ClO¯  2  2  2Na2S2O3 Eq.wt. of OCl– = M/2 10. Cr2O 72– Cr2O72– + 14H+ + 6¯  32 + 2Cr3+ + 7H2O Cr2O72–  32  6 Eq.wt. of Cr2O72– = M/6 11. MnO4– 2MnO4– + 10¯+ 16H+  2MnO4– + 52 + 8H2O 2MnO4¯  52  10 Eq.wt. of MnO4¯ = M/5 12. BrO3– BrO3– + 6¯ + 6H+  Br– + 32 + 3H2O BrO3–  32  6 Eq.wt. of BrO3– = M/6 13. As(V) H2AsO4 + 2¯+ 3H+  H3AsO3 + H2O + 2 H3AsO4  2  2 Eq.wt. of H3AsO4 = M/2 14. HNO2 2HNO2 + 2¯  2 + 2NO + H2O 2HNO2  2  2 Eq.wt. of HNO2 = M/1 15. HClO HClO + 2¯ + H+  Cl¯ + 2 + H2O HClOI22Na2S2O3 Eq.wt. of HClO = M/2 Iodimetric Titrations –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– S.No. Estimation of Reaction Relation between O.A. and R.A. –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 1. H2S H2S + 2  S + 2¯ + 2H+ (in acidic medium) 2. SO32– Eq.wt. of H2S = M/2 SO32– + 2 + H2O  SO42– + 2¯ + 2H+ Sn2+ Sn2+ + 2  Sn4+ + 2¯ Sn2+  2  2 Eq.wt. of Sn2+ = M/2 (in acidic medium) 4. As(III) (at pH 8) SO32–  2  2 Eq.wt. of SO32– = M/2 (in acidic medium) 3. H2S  2  2 H2AsO3¯ + 2 + H2O  HAsO42– + 2¯ + 3H+ H2AsO3–  2  2 Eq.wt. of H2AsO3¯ = M/2 5. N 2H 4 + N2H4 + 22  N2 + 4H + 4¯ N2H4 = 22  4 Eq.wt. of N2H4 = M/4 20 Some Basic Concept of Chemistry (b) Basic medium : Ex.-43 The sulphur content of a steel sample is determined by converting it to H2S gas, absorbing the H2S in 10 mL of 0.005 M I2 and then back titrating the excess I2 with 0.002 M Na2S2O3 . If 10 mL Na2 S2O3 is required for the titration, how many milligrams of sulphur are contained in the sample? Reactions : 2OH– + H2O2 O2 + 2H2O + 2e– v.f = 2 Note : Valency factor of H2O2 is always equal to 2. Volume strength of H2O2 : Strength of H2O2 is represented as 10V , 20 V , 30 V etc. 20V H2O2 means one litre of this sample of H2O2 on decomposition gives 20L of O2 gas at STP. H2S + I2  S + 2I– + 2H+ Decomposition of H2O2 is given as : I2 + 2S2O32–  2I– + S4O62– Sol. Used millimoles of I2 = (m.moles of I2 taken initially) – H2O2  H2O + m. moles of hypo used 2 10 = 0.005 × 10 – 0.002 × 2 1 O 2 2 1 mole 1 × 22.4 L O2 at STP 2 = 34g = 11.2 L O2 at STP To obtain 11.2 litre O2 at STP, at least 34 g H2O2 must be decomposed. = 0.04 = millimoles of H2S  weight of sulphur = 0.04 × 10–3 × 32 × 103 mg = 1.28 mg. For 20 L O2 , we should decompose atleast 34 ×20 11.2 g H2O2 HYDROGEN PEROXIDE (H2O2) H2O2 can behave both like oxidising and reducing agent in both the mediums (acidic and basic).  1 L solution of H2O2 contains 34 ×20 g H2O2 11.2  1 L solution of H2O2 contains 34 20 × 11.2 17 equivalents of H2O2  Oxidising agent : (H2O2 H2O) (a) Acidic medium : Normality of H2O2 =  v.f. = 2 (b) Basic medium : 2e + H2O2  2OH  Reducing agent : (H2O2  O2) (a) Acidic medium : 20 34 20 × = 5 .6 11.2 17 Normality of H2O2 (N) =  MH2O2 = – v.f = 2 M 34  = 17) 2 2 Volume strength of H 2 O 2 5.6 2e– + 2H+ + H2O2  2H2O – ( E H2 O 2   NH2O2 v.f . = NH2O 2 2 Molarity of H2O2 (M) = Volume strength of H2 O 2 11.2 H2O2  O2 + 2H+ + 2e– Strength (in g/L) : Denoted by S v.f = 2 Strength = Molarity × Mol. wt = Molarity × 34 Strength = Normality × Eq. weight = Normality × 17 21 Ex.-44 20 mL of H2O2 after acidification with dilute H2SO4 required 30 mL of N KMnO4 for complete 12 oxidation. Final the strength of H2O2 solution. [Molar mass of H2O2 = 34] Sol. meq. of KMnO4 = meq. of H2O2 Some Basic Concept of Chemistry specified, we take ppm to be in terms of mass. Hence, a 100 ppm solution means that 100 g of solute is present in every 1000000 g of solution. ppmA = mass of A 6 10 6 = mass fraction × 10 Total mass Measurement of Hardness : Hardness is measured in terms of ppm (parts per million) of CaCO3 or equivalent to it. 1 = 20 × N 12 30 1 N = = N 12  20 8 30 × mass of CaCO 3 10 6 Hardness in ppm = Total mass of solution  strength = N × equivalent mass = 1 8 × 17 HARDNESS OF WATER (HARD Ex.-45 0.00012% MgSO4 and 0.000111% CaCl2 is present in water. What is the measured hardness of water and millimoles of washing soda required to purify water 1000 L water ? WATER DOES NOT GIVE Sol. Basis of calculation = 100 g hard water = 2.12 g/L. LATHER WITH SOAP) Temporary hardness - due to bicarbonates of Ca & Mg MgSO4 Permanent hardness - due to chlorides & sulphates of Ca & Mg. There are some method by which we can soften the water sample. CaCl2 = 0.000111g = (a) By boiling : 2HCO3–  H2O + CO2 + CO32– or Ca(HCO3)2 + Ca(OH)2  CaCO3 + 2H2O Ca + CO3 2–  CaCO 3 (b) By Washing Soda : (c) By ion exchange resins : Na2R + Ca mole 0.000111 mole 111  equivalent moles of CaCO3 =  0.00012 0.000111     111   120 mole  CaR + 2Na (d) By adding chelating agents like  0.00012 =  120   0.000111   111  × 100 = 2 × 10–4 g CaCl2 + Na2CO3  CaCO3 + 2NaCl 2+ 0.00012 120  mass of CaCO3 By Slaked lime : 2+ = 0.00012g = Hardness (in terms of ppm of CaCO3) = 2  10 4  10 6 = 2 ppm 100 + (PO3–)3 etc. PARTS PER MILLION (ppm) When the solute is present in very less amount, then this concentration term is used. It is defined as the number of parts of the solute present in every 1 million parts of the solution. ppm can both be in terms of mass or in terms of moles. If nothing has been CaCl2 + Na2CO3  CaCO3 + 2NaCl NaSO4 + Na2CO3  MgCO3 + Na2SO4  Required Na2CO3 for 100g of water  0.00012 =  120   0.000111   mole 111  = 2 × 10–6 mole 22  Required Na2CO3 for 1000 litre water = 2  10 6 2  10 6  mole 100 100 = 20 1000 Some Basic Concept of Chemistry Method of determination : CaOCl2 ( d = 1g/mL) + 2CH3COOH  Ca(CH3COO)2 + H2O + Cl2 (Sample of bleaching powder) mole = 20 m mole Cl2 + 2KI  2KCl + I2 STRENGTH OF OLEUM Oleum is SO3 dissolved in 100% H2SO4. Sometimes, oleum is reported as more than 100% by weight, say y% (where y > 100). This means that (y  100) grams of water, when added to 100 g of given oleum sample, will combine with all the free SO3 in the oleum to give 100% sulphuric acid. Hence, weight % of free SO3 in oleum  = 80(y 100)/18 I2 + 2Na2S2O3 Starch as indicator      Na2S4O6 v.f. = 2 + 2Nal v.f. = 1 End point is indicated by disappearance of blue colour. Let M = Molarity of hypo (Na2S2O3) solution millimoles of Cl2 produced = m.moles of I2 used by hypo = M V where V = vol of hypo solution 2 Ex.-46 What volume of water is required (in mL) to used in ml. prepare 1 L of 1 M solution of H2SO4 (density = M  V  10 3 1.5g/mL) by using 109% oleum and water only (Take mass of Cl2 produced = × 71 2 density of pure water = 1 g/mL). Sol. 1 mole H2SO4 in 1L solution = 98 g H2SO4 in 1500 = 35.5 × M × V × 10–3 g solution = 98 g H2SO4 in 1402 g water. 3 Also, in 109% oleum, 9 g H2O is required to form  % of available chlorine = 35.5  M  V  10 × 100 W 109 g pure H2SO4 & so, to prepare 98 g H2SO4, water needed is 9/109 × 98 = 8.09 g. where W = amount of belaching powder taken in g. Total water needed = 1402 + 8.09 = 1410.09 g 3.55  M  V or % of available Cl2 = W = 1410.09 mL CALCULATION OFAVAILABLE Ex.-47 3.55 g sample of bleaching powder suspended CHLORINE FROM A SAMPLE in H2O was treated with enough acetic acid and K OF BLEACHING POWDER solution. Iodine thus liberated required 80 mL of 0.2 The weight of available Cl2 released from the given M hypo for titration. Calculate the % of available sample of bleaching powder on reaction with dilute chlorine. acids or CO2 is called available chlorine. CaOCl 2  H2 SO 4  CaSO 4  H2O  Cl2 CaOCl 2  2HCl  CaCl 2  H2 O  Cl2 Sol. % of Cl2 = 3.55  0.2  80 = 16% 3.55 CaOCl 2  CO 2  CaCO 3  Cl2 23