Chemistry NEET Chapter 1 Notes PDF

60 E3 Chapter 1 Chemical Arithmetic ID Chemistry is basically an experimental science. In it we study physical and chemical properties of substance and measure it upto possibility. The results of measurement can we reported in two steps, (a) Arithmetic number, (b) Unit of measurement. Every experime...

60 E3 Chapter 1 Chemical Arithmetic ID Chemistry is basically an experimental science. In it we study physical and chemical properties of substance and measure it upto possibility. The results of measurement can we reported in two steps, (a) Arithmetic number, (b) Unit of measurement. Every experimental measurement vary slightly from one another and involves some error or uncertainty depending upon the skill of person making the measurements and measuring instrument. The closeness of the set of values obtained from identical measurement called precision and a related term, refers to the closeness of a single measurement to its true value called accuracy. Rule 5. All zeros on the right of the non zero digit are not significant. Example : x  1000 has only one significant figure. Again x  378000 has three significant figures. D YG U Rule 6. All zeros on the right of the last non zero digit become significant, when they come from a measurement. Significant figures U In the measured value of a physical quantity, the digits about the correctness of which we are surplus the last digit which is doubtful, are called the significant figures. Number of significant figures in a physical quantity depends upon the least count of the instrument used for its measurement. (1) Common rules for counting significant figures Following are some of the common rules for counting significant figures in a given expression Rule 1. All non zero digits are significant. Example : Suppose distance between two stations is measured to be 3050 m. It has four significant figures. The same distance can be expressed as 3.050 km or 3.050  10 5 cm. In all these expressions, number of significant figures continues to be four. Thus we conclude that change in the units of measurement of a quantity does not change the number of significant figures. By changing the position of the decimal point, the number of significant digits in the results does not change. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true. (2) Rounding off : While rounding off measurements, we use the following rules by convention Rule 1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged. Example : x  7.82 is rounded off to 7.8, again x  3.94 is rounded off to 3.9. Example : x  1007 has four significant figures. Again x  1.0809 has five significant figures. Rule 3. In a number less than one, all zeros to the right of decimal point and to the left of a non zero digit are not significant. Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8. Example : x  0.0084 has only two significant digits. Again, x  1.0084 has five significant figures. This is on account of rule 2. Rule 4. All zeros on the right of the last non zero digit in the decimal part are significant. Example : x = 16.351 is rounded off to 16.4, again x  6.758 is rounded off to 6.8. Rule 4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even. Example : x = 3.250 becomes 3.2 on rounding off, again x  12.650 becomes 12.6 on rounding off. ST Example : x  1234 has four significant figures. Again x  189 has only three significant figures. Rule 2. All zeros occurring between two non zero digits are significant. Example : x  0.00800 has three significant figures 8, 0, 0. The zeros before 8 are not significant again 1.00 has three significant figures. Rule 2. If the digit to be dropped is more than 5, then the preceding digit is raised by one. Rule 3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one. (1) C.G.S. System : Length (centimetre), Mass (gram), Time (second) (2) M.K.S. System : Length (metre), Mass (kilogram), Time (second) (3) F.P.S. System : Length (foot), Mass (pound), Time (second) (4) S.I. System : The 11th general conference of weights and measures (October 1960) adopted International system of units, popularly known as the SI units. The SI has seven basic units from which all other units are derived called derived units. The standard prefixes which helps to reduce the basic units are now widely used. Rule 5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd. Example : x = 3.750 is rounded off to 3.8, again x  16.150 is rounded off to 16.2. (3) Significant figure in calculation (i) Addition and subtraction : In addition and subtraction the following points should be remembered (a) Every quantity should be changed into same unit. (b) If a quantity is expressed in the power of 10, then all the quantities should be changed into power of 10. (c) The result obtained after addition or subtraction, the number of figure should be equal to that of least, after decimal point. (ii) Multiplication and division (a) The number of significant figures will be same if any number is multiplied by a constant. (b) The product or division of two significant figures, will contain the significant figures equal to that of least. 60 Dimensional analysis : The seven basic quantities lead to a number of derived quantities such as pressure, volume, force, density, speed etc. The units for such quantities can be obtained by defining the derived quantity in terms of the base quantities using the base units. For example, speed (velocity) is expressed in distance/time. So the unit is m / s or ms 1. The E3 unit of force (mass  acceleration) is kg ms 2 and the unit for acceleration is ms 2. Units for measurement The chosen standard of measurement of a quantity which has essentially the same nature as that of the quantity is called the unit of the quantity. Following are the important types of system for unit, ID Table 1.1 Seven basic S.I. units Mass Time Temperature Electric Current Luminous Intensity Amount of substance metre (m) Kilogram (kg) Second (s) Kelvin (K) Ampere (A) Candela (Cd) Mole (mol) U Length Table 1.2 Derived Units Unit Symbol D YG Physical quantity square metre m Volume cubic metre m Velocity metre per second ms Acceleration metre per second square ms Density kilogram per cubic metre kg m Molar mass kilogram per mole kg mol Molar volume cubic metre per mole m mol Molar concentration mole per cubic metre mol m Force newton (N) kg m s Pressure pascal (Pa) Nm Energy work joule (J) kg m s , Nm ST U Area 2 3 –1 –2 –3 –1 3 –3 –2 –2 2 Prefix Symbol Submultiple 10 yotta Y 10 zetta Z 10 exa E 10 peta P 10 tera T 21 18 15 12 –2 Table 1.3 Standard prefixes use to reduce the basic units Multiple 24 –1 Prefix Symbol 10 deci d 10 centi c 10 milli m 10 micro  10 nano n –1 –2 –3 –6 –9 10 giga G 10 pico p 10 mega M 10 femto f 10 kilo k 10 atto a 10 hecto h 10 zeto z 10 deca da 10 yocto y 9 6 3 2 1 –12 –15 –18 –21 –24 Table 1.4 Conversion factors 1 e.s.u. = 3.3356  10 C 1 mole of a gas = 22.4 L at STP 1 dyne = 10 N 1 mole a substance = N molecules 1 m = 39.37 inch 1 cal = 4.184 J 1 inch = 2.54 cm 1 eV = 1.602  10 J 1 litre = 1000 mL 1 eV/atom =96.5 kJ mol 1 gallon (US) = 3.79 L 1 amu = 931.5016 MeV 1 bar = 1  10 N m 1 lb = 453.59237 g 1 kilo watt hour = 3600 kJ 1 litre atm = 101.3 J 1 horse power = 746 watt 1 year = 3.1536  10 s 1Å = 10 m 1 joule = 10 erg 1 debye (D) = 1  10 esu cm 1nm = 10 m –10 –2 1 J = 1 Nm =1 kg m s 2 –2 7 0 1 atm = 101325 Pa 5 1 g atom = N atoms 0 t ( F) = 9/5 t ( C) + 32 o –2 o 1 g cm = 1000 kg m –3 –3 E3 1 newton =1 kg m s –1 60 –5 –19 7 –18 –9 (iii) Atoms of different elements possess different properties (including different masses). ID Laws of chemical combination –10 Various chemical reactions take place according to the certain laws, known as the Laws of chemical combination. (1) Law of conservation of mass : It was proposed by Lavoisier and verified by Landolt. According to this law, Matter is neither created nor (v) Atoms of elements take part to form molecules i.e., compounds are formed when atoms of more than one element combine. (vi) In a given compound, the relative number and kinds of atoms are constant. (2) Modern atomic hypothesis : The main modifications made in Dalton’s hypothesis as a result of new discoveries about atoms are, (i) Atom is no longer considered to be indivisible. (ii) Atoms of the same element may have different atomic weights. U destroyed in the course of chemical reaction though it may change from one form to other. The total mass of materials after a chemical reaction is same (iv) Atoms are indestructible i.e., atoms are neither created nor destroyed in chemical reactions. D YG as the total mass before reaction. (2) Law of constant or definite proportion : It was proposed by Proust. According to this law, A pure chemical compound always contains the same elements combined together in the fixed ratio of their weights whatever its methods of preparation may be. (3) Law of multiple proportion : It was proposed by Dalton and verified by Berzelius. According to this law, When two elements A and B combine to form more than one chemical compounds then different weights of A, which combine with a fixed weight of B, are in proportion of simple whole numbers. (4) Law of equivalent proportion or law of reciprocal proportion : It was proposed by Ritcher. According to this law, The weights of the two or ST U more elements which separately react with same weight of a third element are also the weights of these elements which react with each other or in simple multiple of them. (5) Gay-Lussac’s law : It was proposed by Gay–Lussac and is applicable only for gases. According to this law, When gases combine, they do so in volumes, which bear a simple ratio to each other and also to the product formed provided all gases are measured under similar conditions. The Gay-Lussac’s law, was based on experimental observation. Important hypothesis (1) Atomic hypothesis : Keeping in view various law of chemical combinations, a theoretical proof for the validity of different laws was given by John Dalton in the form of hypothesis called Dalton's atomic hypothesis. Postulates of Dalton's hypothesis is as followes, (i) Each element is composed of extremely small particles called atoms which can take part in chemical combination. (ii) All atoms of a given element are identical i.e., atoms of a particular element are all alike but differ from atoms of other element. e.g., isotopes of oxygen O16 , O17 , and O18. (iii) Atoms of different element may have same atomic weights. e.g., isobars Ca 40 and Ar 40. (iv) Atom is no longer indestructible. In many nuclear reactions, a certain mass of the nucleus is converted into energy in the form of ,  and  rays. (v) Atoms may not always combine in simple whole number ratios. e.g., in sucrose (C12 H 22 O11 ) , the elements carbon, hydrogen and oxygen are present in the ratio of 12 : 22 : 11 and the ratio is not a simple whole number ratio. (3) Berzelius hypothesis : “Equal volumes of all gases contain equal number of atoms under same conditions of temperature and pressure”. When applied to law of combining volumes, this hypothesis predicts that atoms are divisible and hence it is contrary to Dalton's hypothesis. (4) Avogadro’s hypothesis : “Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.” Avogadro hypothesis has been found to explain as follows, (i) Provides a method to determine the atomic weight of gaseous elements. (ii) Provides a relationship between vapour density (V.D.) and molecular masses of substances. Molecular mass  2  vapour density (iii) It helps in the determination of mass of fixed volume of a particular gas. Mass of 1 ml gas = V.D.  0.0000897 gm.  V.D.  0.0000897 gm. gas has volume = 1 ml 2  V.D.(i.e., molecular 1  2  V.D. volume  ml  22400 ml V.D.  0.0000897 Valency of the element = gm. mass) = (v) It helps in determination of molecular formulae of gases and is very useful in gas analysis. By knowing the molecular volumes of reactants and products of reaction, molecular composition can be determined easily. Atomic, Molecular and Equivalent masses 1/12  Mass of an atom of C 12 (ii) Mass of an element in gm. = No. of gm. atom (iii) Number of atoms in 1 GAM = 6.02  GAM  10 23  Number of atoms in a given substance U Mass = GAM (iv) Number of atoms in 1gm of element =  6.02  10 23 6.02  10 Atomic mass 23 ST GAM (v) Mass of one atom of the element (in gm.)  6.02  10 23 Methods of determination of atomic mass (i) Dulong and Pettit's method : According to Dulong and Pettit's law Atomic mass  Specific heat = 6.4 (approx.) Atomic mass (approx.) = constant volume). Their ratio is known as  whose value is constant (1.66 for monoatomic, 1.40 for diatomic and 1.33 for triatomic gases). 6.4 Specific heat (in cals.) This law is applicable to solid elements only except Be, B, C and Si because their specific heat is variable with temperature. Atomic mass = Equivalent mass  Valency Molecular mass Atomicity (iv) Isomorphism method : It is based on law of isomorphism which states that compounds having identical crystal structure have similar constitution and chemical formulae. Example : K 2 SO 4 , K 2 CrO 4 and K 2 SeO 4 ID Mass of an element GAM  10 23 P (valency of S, Cr, Se = 6), ZnSO 4. 7 H 2 O, MgSO4.7 H 2 O, FeSO 4. 7 H 2 O (valency of Zn, Mg, Fe = 2). U D YG Gram atomic mass (GAM) : Atomic mass of an element expressed in grams is called Gram atomic mass or gram atom or mole atom. = No. of GAM  6.02 (iii) Specific heat method : It is suitable only for gases. The two types of specific heats of gases are C (at constant pressure) and C v (at g. One atomic mass unit (amu) is 1 th of the mass of an atom of carbon-12 isotope. 12 (i) Number of gram atoms =  Valency E3 Average mass of an atom Average atomic mass : If an elements exists in two isotopes having atomic masses ‘a’ and ‘b’ in the ratio m : n, then average atomic mass (m  a)  (n  b). Since the atomic mass is a ratio, it has no units and is = m n equal to Atomic mass = Equivalent mass of metal Atomic mass of a gaseous element  (1) Atomic mass : It is the average relative mass of atom of element as compared with an atom of carbon –12 isotope taken as 12. expressed in amu, 1 amu = 1.66  10 2  Vapour density of chloride Equivalent mass of metal  35.5 60 S.T.P. 24 Molecular mass of chloride Equivalent mass of chloride has  Molar mass of a gas or its 1 mole occupies 22.4 L volume at Atomic mass  Approximat e atomic mass Equivalent mass (ii) Vapour density method : It is suitable for elements whose chlorides are volatile. (iv) It also helps in the determination of molar volume at N.T.P.  Valency  (2) Molecular mass : Molecular mass of a molecule, of an element or a compound may be defined as a number which indicates how many times 1 heavier is a molecule of that element or compound as compared with of 12 the mass of an atom of carbon–12. Molecular mass is a ratio and hence has no units. It is expressed in a.m.u. Molecular mass  Mass of one molecule of the substance 1 / 12  Mass of one atom of C - 12 Actual mass of one molecule = Mol. mass 1.66 10 24 gm. Molecular mass of a substances is the additive property and can be calculated by adding the atomic masses present in one molecule. Gram molecular mass (GMM) and Gram molar volume : Molecular mass of an element or compound when expressed in gm. is called its gram molecular mass, gram molecule or mole molecule. Number of gm molecules = Mass of substances GMM Mass of substances in gm = No. of gm. molecules  GMM Volume occupied by one mole of any gas at STP is called Gram molar volume. The value of gram molar volume is 22.4 litres. Volume of 1 mole of any gas at STP = 22.4 litres Expression for mass and density Mass of 11.2 L of any gas at STP = V.D. of that gas in gm. Density of a gas at NTP = Mol.mass in gm. 22400 ml Important generalisations Number of atoms in a substance  6.02  10  Atomicity 23 (iv) EM of a salt = Number of electrons in given substance = Number of GMM  6.02  10  Number of electrons 23 (v) EM of an oxidising agent Methods of determination of molecular mass  Following methods are used to determine molecular mass, (i) Diffusion method (For gases) : The ratio of rates of diffusion of two gases is inversely proportional to the square root of their molecular masses. (ii) Vapour density method (For gases only) : Mass of a fixed volume of the vapour is compared with the mass of the same volume of hydrogen under same conditions. The ratio of these masses is called Vapour density or Relative density. Molecular mass  2  Vapour desity (iii) Victor Meyer method (For volatile liquids or solids) (iv) Colligative property method (For non-volatile solids) W Mass of metal  1.008 g  1.008  M Mass of H 2 displaced  Mass of metal W 11200 11200 = V Vol.(ml) of H 2 displaced at STP This method is useful for metals which can displace hydrogen from acids or can combine with hydrogen (Mg, Zn, Na, Ca etc.) U D YG Where  Mi Xi  X total  (ii) Oxide formation method : The mass of the element which combines with 8 grams of oxygen is the equivalent mass of the element. Average atomic mass and molecular mass  Ai X i  X total (a) Equivalent mass of metal ID 22400 ml of vapours of a substance = Molecular mass of that substance Discussed in colligative properties of solutions. (i) Hydrogen displacement method : The mass of metal which displaces 11200 ml of hydrogen at NTP from an acid, alkali or alcohol is the equivalent mass of the metal. (b) Equivalent mass of metal It is based on Dalton's law of partial pressure and Avogadro's hypothesis (gram molar volume). M (Average molecular mass) = Equivalent mass of common oxidising agent changes with the medium of the reaction. Methods of determination of equivalent mass M2 M1 A (Average atomic mass) = Formula mass Number of electrons gained per molecule or Total change in O.N. E3 r1  r2 Formula mass Total positive or negative charge 60 = Number of GMM Mass of metal 8 Mass of oxygen (a) Equivalent mass of metal = (b) Equivalent mass of metal = A1 , A2 , A3..... are atomic mass of species 1, 2, 3,…. etc. with % ratio as X1 , X 2 , X 3..... etc. Similar terms are for molecular masses. (3) Equivalent mass : The number of parts by mass of a substance that combines with or displaces 1.008 parts by mass of hydrogen or 8.0 parts of oxygen or 35.5 parts of chlorine is called its equivalent mass (EM). On the other hand quantity of a substance in grams numerically equal to its U equivalent mass is called its gram equivalent mass (GEM) or gram equivalent. (iii) Chloride formation method : The mass of an element which reacts with 35.5 gm. of chlorine is the equivalent mass of that element. (a) Equivalent mass of metal  ST (i) EM of an element  (ii) EM of an acid  Atomic mass Valency Molecular mass Basicity (Basicity of acid is the number of replaceable hydrogen atoms in one molecule of the acid). Molecular mass (iii) EM of a base  Acidity (Acidity of a base is the number of replaceable– OH groups in one molecule of the base). Mass of metal  35.5 Mass of chlorine (b) Equivalent mass of metal  Mass of the substance in grams Number of GEM  GEM of the substance Expressions for equivalent mass (EM) Mass of metal  5600 Vol. of O 2 at S.T.P. in ml Mass of metal  11200 Vol. of Cl 2 (in ml.) at STP (iv) Neutralisation method : (For acids and bases). Equivalent mass of acid (or base)  W VN Where , W = Mass of acid or base in gm., V = Vol. of base or acid in litre required for neutralisation N is Normality of base or acid (v) Metal displacement method : It is based on the fact that one gm. equivalent of a more electropositive metal displaces one gm equivalent of a less electropositive metal from its salt solution. Mass of metal added Eq. mass of metal added  Mass of metal displaced Eq. mass of metal displaced Element W1 E  1 W2 E2 % Relative no. of atoms = %/at. wt. Simplest Ratio Empirical Formula (vi) Electrolytic method : The quantity of substance that reacts at electrode when 1 faraday of electricity is passed is equal to its gram equivalent mass. Gram equivalent mass = Electrochemical equivalent  96500 The ratio of masses of two metals deposited by the same quantity of electricity will be in the ratio of their equivalent masses. Relative no. of atoms : Divide the percentage of each element present in compound by its at. weight. This gives the relative no. of atoms of element in molecule. W1 E  1 W2 E2 It the simplest ratio obtained are not complete integers, multiply them by a common factor to get integer values of simplest ratio. (vii) Double decomposition method AB  CD  AD   CB Empirical formula : Write all constituent atoms with their respective no. of atoms derived in simplest ratio. This gives empirical formula of compound. Mass of compound AB Eq. mass of A  Eq. mass of B  Mass of compound AD Eq. mass of A  Eq. mass of D Molecular formula : Molecular formula  n  empirical formula where ' n' is the whole no. obtained by 60 Mass of salt taken (W1 ) Eq. mass of salt (E1 )  Mass of ppt. obtained (W2 ) Eq. mass of salt in ppt. (E 2 ) Mass of compound I (W1 ) E  Eq. mass of radical I  Mass of compound II (W2 ) E  Eq. mass of radical II (E = Eq. mass of the metal) (ix) Volatile chloride method E  2  V.D. of Chloride 2  V.D.  Eq. mass of metal chloride E  35.5 D YG (x) Silver salt method (For organic acids) 108  Mass of silversalt  107 Mass of Ag metal Molecular mass of acid = Equivalent mass of acid  Basicity The mole concept One mole of any substance contains a fixed number (6.022  10 ) of any type of particles (atoms or molecules or ions) and has a mass equal to the atomic or molecular weight, in grams. Thus it is 23 U correct to refer to a mole of helium, a mole of electrons, or a mole of Na  , meaning respectively Avogadro’s number of atoms, electrons or ions. Weight(grams) Weight of one mole (g/mole) ST  Number of moles   Chemical stoichiometry Stoichiometry (pronounced “stoy-key om-e-tree”) is the calculation of the quantities of reactants and products involved in a chemical reaction. That means quantitative calculations of chemical composition and reaction are referred to as stoichiometry. Basically, this topic involves two types of calculations. (a) Simple calculations (gravimetric analysis) and (b) More complex calculations involving concentration and volume of solutions (volumetric analysis). 2  V.D. of Chloride  35.5 Valency Equivalent Mass of acid = molecular weight of compound empirical formula weight of compound U Valency of metal  n ID (viii) Conversion method : When one compound of a metal is converted to another compound of the same metal, then E3 or Simplest ratio : Find out lowest value of relative no. of atoms and divide each value of relative no. of atoms by this value to estimate simplest ratio of elements. Weight Atomic or molecular weight Percentage composition & Molecular formula (1) Percentage composition of a compound Percentage composition of the compound is the relative mass of each of the constituent element in 100 parts of it. If the molecular mass of a compound is M and B is the mass of an element in the molecule, then Mass of element X Percentage of element   100   100 Molecular mass M (2) Determination of empirical formula : The empirical formula of a molecule is determined using the % of elements present in it. Following method is adopted. There is no borderline, which can distinguish the set of laws applicable to gravimetric and volumetric analysis. All the laws used in one are equally applicable to the other i.e., mole as well as equivalent concept. But in actual practise, the problems on gravimetric involves simpler reactions, thus mole concept is convenient to apply while volumetric reactions being complex and unknown (unknown simple means that it is not known to you, as it’s not possible for you to remember all possible reactions), equivalent concept is easier to apply as it does not require the knowledge of balanced equation. (1) Gravimetric analysis : In gravimetric analysis we relate the weights of two substances or a weight of a substance with a volume of a gas or volumes of two or more gases. Problems Involving Mass-Mass Relationship Proceed for solving such problems according to the following instructions, (i) Write down the balanced equation to represent the chemical change. (ii) Write the number of moles below the formula of the reactants and products. Also write the relative weights of the reactants and products (calculated from the respective molecular formula), below the respective formula. (iii) Apply the unitary method to calculate the unknown factor (s). Problems Involving Mass-Volume Relationship For solving problems involving mass-volume relationship, proceed according to the following instructions, (i) Write down the relevant balanced chemical equations (s). (iii) Gases are usually expressed in terms of volumes. In case the volume of the gas is measured at room temperature and pressure (or under conditions other than N.T.P.), convert it into N.T.P. by applying gas equation. (iii) Iodiometric titrations : This is a simple titration involving free iodine. This involves the titration of iodine solution with known sodium thiosulphate solution whose normality is N. Let the volume of sodium thiosulphate is V ml. I 2  2 Na 2 S 2 O 3 2 NaI  Na 2 S 4 O6 (iv) Volume of a gas at any temperature and pressure can be converted into its weight and vice-versa with the help of the relation, by n  2, n 1 g PV   RT where g is weight of gas, M is mole. wt. of gas, R is M gas constant. Equivalents of I 2  Equivalent of Na 2 S 2 O 3  Equivalents of I 2  N  V  10 3 Calculate the unknown factor by unitary method. Moles of I 2  Problems Based on Volume-Volume Relationship Such problems can be solved according to chemical equation as, E3 (iv) Iodometric titrations : This is an indirect method of estimation of iodine. An oxidising agent is made to react with excess of solid KI. The oxidising agent oxidises I  to I 2. This iodine is then made to react with Na 2 S 2 O 3 solution. ID (iii) In case volume of the gas is measured under particular (or room) temperature, convert it to volume at NTP by using ideal gas equation. N  V  10 3 2  N  V  10 3  Mass of free I 2 in the solution    254  g. 2   (i) Write down the relevant balanced chemical equation. (ii) Write down the volume of reactants and products below the formula to each reactant and product with the help of the fact that 1 gm molecule of every gaseous substance occupies 22.4 litres at N.T.P. 60 (ii) Write the weights of various solid reactants and products. Oxidising Agent Take the help of Avogadro’s hypothesis “Equal volume of different 2 Na 2 S 2 O3 ( A)  KI I 2   2 NaI  Na 2 S 4 O 6 gases under the similar conditions of temperature and pressure contain the same number of molecules”. Let the normality of Na 2 S 2 O 3 solution is N and the volume of U D YG (2) Volumetric analysis : It is a method which involves quantitative determination of the amount of any substance present in a solution through volume measurements. For the analysis a standard solution is required. (A solution which contains a known weight of the solute present in known volume of the solution is known as standard solution.) thiosulphate consumed to V ml. Equivalent of A  Equivalent of I 2  Equivalents of Na 2 S 2 O 3 Equivalents of I 2 liberated from KI  N  V  10 3 To determine the strength of unknown solution with the help of known (standard) solution is known as titration. Different types of titrations are possible which are summerised as follows, Moles of I 2 liberated from KI   N  V  10 3   254  g. Mass of I 2 liberated from KI   2   (i) Redox titrations : To determine the strength of oxidising agents or reducing agents by titration with the help of standard solution of reducing agents or oxidising agents. Examples: U K 2 Cr2 O7  4 H 2 SO 4 K 2 SO 4  Cr2 (SO 4 )3  4 H 2 O  3[O] [2 FeSO 4  H 2 SO 4  O Fe2 (SO 4 )3  H 2 O]  3 6 FeSO 4  K 2 Cr2 O7  7 H 2 SO 4 3 Fe(SO 4 )3  K 2 SO 4  Cr2 (SO 4 )3 7 H 2 O ST 2 KMnO4  3 H 2 SO 4 K 2 SO 4  2 MnSO 4  3 H 2 O  5[O] [2 FeSO 4  H 2 SO 4  O Fe2 (SO 4 )3  H 2 O]  5 10 FeSO 4  2 KMnO4  8 H 2 SO 4 5 Fe2 (SO 4 )3  K 2 SO 4  2 MnSO 4  8 H 2 O Similarly with H 2 C 2 O4 2 KMnO4  3 H 2 SO 4  5 H 2 C 2 O4 K 2 SO 4  2 MnSO 4  8 H 2 O  10CO 2 etc. (ii) Acid-base titrations : To determine the strength of acid or base with the help of standard solution of base or acid. Example: NaOH  HCl NaCl  H 2 O and NaOH  CH 3 COOH CH 3 COONa  H 2 O etc. N  V  10 3 2 (v) Precipitation titrations : To determine the anions like  CN , AsO33  , PO43  , X  etc, by precipitating with AgNO3 provides examples of precipitation titrations. NaCl  AgNO3 AgCl   NaNO 3 KSCN  AgNO3 AgSCN   KNO 3 End point and equivalence point : The point at which titration is stopped is known as end point, while the point at which the acid and base ( or oxidising and reducing agents ) have been added in equivalent quantities is known as equivalence point. Since the purpose of the indicator is to stop the titration close to the point at which the reacting substances were added in equivalent quantities, it is important that the equivalent point and the end point be as close as possible. Normal solution : A solution containing one gram equivalent weight of the solute dissolved per litre is called a normal solution ; e.g. when 40 g of NaOH are present in one litre of NaOH solution, the solution is known as normal ( N ) solution of NaOH. Similarly, a solution containing a fraction of gram equivalent weight of the solute dissolved per litre is known as subnormal solution. For example, a solution of NaOH containing 20 g (1/2 of g eq. wt.) of NaOH dissolved per litre is a sub-normal solution. It is written as N /2 or 0.5 N solution. Formula used in solving numerical problems on volumetric analysis (1) Strength of solution = Amount of substance in g litre1 (2) Strength of solution = Amount of substance in g moles litre1 (3) Strength of solution = Normality  Eq. wt. of the solute Limiting reagent or reactant In many situations, an excess of one or more substance is available for chemical reaction. Some of these excess substances will therefore be left over when the reaction is complete; the reaction stops immediately as soon as one of the reactant is totally consumed. The substance that is totally consumed in a reaction is called limiting reagent because it determines or limits the amount of product. The other reactant present in excess are called as excess reagents. 60 Let us consider a chemical reaction which is initiated by passing a spark through a reaction vessel containing 10 mole of H and 7 mole of O. = molarity  Mol. wt. of solute 2 (4) Molarity Moles of solute Volume in litre (5) Number of moles  2 H 2 (g)  O2 (g)  2 H 2O (v) Wt. in gm  M  V(inl) Mol.wt. Moles before reaction 10 7 0 0 2 10 E3 Moles after reaction Volume in litres  at NTP (only for gases) 22.4 The reaction stops only after consumption of 5 moles of O as no further amount of H is left to react with unreacted O. Thus H is a limiting reagent in this reaction. 2 (6) Number of millimoles  Wt. in gm  1000 mol. wt. (7) Number of equivalents Wt. in gm  1000  normality  Volume in ml. Eq. wt. D YG  (9) Normality  x  No. of millimoles  x  Molarity  where x  Strength in gm litre1 Eq. wt. Mol.wt. , x = valency or change in oxi. Number. Eq. wt. U (10) Normality formula, N1 V1  N 2 V2 ST Wt. of solvent  100 (11) % by weight  Wt. of solution Wt. of solvent  100 (12) % by volume  Vol. of solution Vol. of solvent  100 (13) % by strength  Vol. of solution (14) Specific gravity  (15) Formality  Wt. of solution  Wt. of 1 ml. of solution Vol. of solution Wt. of ionic solute Formula Wt. of solute Vin l (16) Mol. Wt. = V.D  2 (For gases only) 2  Law of conservation of mass does not hold good for nuclear U Wt. in gm  x  No. of moles  Normality  Volume in litre Eq. wt. (8) Number of milliequivalents (meq.) 2 2 ID  Molarity Volume in ml.  2 reactions.  Law of definite proportions do not hold good for nonstoichiometric compounds e.g., Wusitite Fe O. 0.95  Law of definite proportions, law of multiple proportions and law of reciprocal proportions do not hold good when same compounds is obtained by using different isotopes of te same element e.g. H O and D O. 2 2  The term atom was introduced by Ostwald and the term molecule was introduced by Avogadro.  The concept of element was introduced by Robert Boyle.  The number of atoms present in a molecule of a gaseous element is called Atomicity.  Both atomic mass and molecular mass are just rations and hence have no units.  1 mol of H O  22400 cc of H O (because it is a liquid). Instead, 2 2 1 mol of H O = 18cc of H O 2 2 (because density of H O = 1 g/cc) 2  1 M H SO = 2NH SO. 2 4 2 4  Minimum molecular mass of a macromolecular substance can be calculated by analysing it for one of the minar components. Minimum molecular mass is obtained when it is supposed that one molecule of the macromolecule contains only one atom or molecule of the minor component. Minimum molecular mass Atomic/molecular mass of minor component Mass of minor component per gram of macromolec ule ST U D YG U ID E3 60 