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>< 1. STATES OF MATTER Solid & ned S Gases | 8 CLA SSIELCATION Of MATTER MATTER” ~- ELEMENT.” Se PROPERTIES. OF MATTER” ® Physical properties e Chemical properties 4... MEASUREMENT a SECA PROPERTIES - ~e Englim system ~ e Metric system SI system has seven base units. | Symbol “COMPOUND. ‘HOMOGENOUS.- HETERO Base physical .| Symbol...) ST unit units pos si nee Length | | Metre m MaS$ ee HM Kilogram Kg i a8 j id micre fd — {Xt lampere_ : [Kelvin Mole Sapa | Candela Cd | awit ——__—~PREFIXEs USED IN SI SYST EM retin Symbol __ Yocto ¥ ane. 1977? Zepto Zz eee jg7'8 Atto {j@ Pe a _ 190 Pico Pe io ° nano n 19° Centi Li io? Milli m ~1€ d | 101 deci 10 Deca ga 107 Hecto ee = G7 Kilo gyi ; j Th TCANT TGURE agely = fl ect and prec? a dig! o represent a measurement cor iy and algo {a : Jeyain ans ich include all digits which are certal h is uncertain, & used, pits WE gitS aning ful ag : Significant figures are mee J red or unc knov estimate?” , cnown with certainty plus one whic his certar ber of So significant figure ! mint igits plus one doubtful digit. | | nthe Value of n the rece of significant figures | e greater number greater precision i sured quantity indicates ¢ value. ules for determing the nu mber of significant figures . All non zero digits are significant. 453 Three significant figures Ove 5 5 significant figures Zero preceeding to first non zero digit are not signivica (2) 0.025-2Two significant gay coe i irte i (3) Zero between non-zero digits will be significant we significant figures (4) Zero at the end or right of a number are significant rovided they are on theip side of the decimal point. example 0.0100 -) three Significant ) En Case of exponential notation, 6-023 x10** the cant figure will be four . ) Numbers that are obtained by counting ob jects: are known S €xact number . or example:- 20 apples. yu's number can be-represent by ing infinite number of zero after decimal point. rms, The results should be reported to the same-no, of laces. as that of the term with the least number of (L.) while carrying out addition or subtraction of anumber of 1 bp F, inch= 27 cm. ach = 2. 54 cin Given, t= 2.54/1 inch ch= 15 « 2.54/1= 3.81 cm. 15 in quired= Inf formation So Information re Given * conversion 3 of eness in value jects for — ne on samme ob the same ee, veiental value RACY: It denotes the Jement 15 a numbe! % s taken as in comparision to an atom of another element . standard, wie a \ * ; To measure the atomic mass Hydrogen, Carbon, oxygen are taken as standard seperately .The-scale are known Hydrogen scale , oxygen scale,carbon scale . TOMIC ; : The atomic Weight with respect to 160 isotopes of | relative Weight 16.000 taken. as standard reference is called the physical atomic weight. | The atomic weight with respect to the weightof natural oxygen as the weight 16.0005 Known as Chemical atomic Weight. ne ey The physical atomic mass of natural oxygen is slightly larger than chemical atomic scale . Physical atamic weight/ chemical atomic Weight = 16.004/16.000 = 1.000275 ATOMIC MASS UNTT (amu): Tt is'defined-as the mass of. one atom which is exactly equal to one- atom which is exactly equal to oné -twenth of the mass of one carbon- 12 atom. | 6.023x10*? atoms of c-12 has mass 12 gm 12 6.923%10" the atomic mass Every element occur as a voix ulate the atomic mass of the element,’ h of the isotopes must be cons sidered and for thi she atomic mass of the elements are generally in f ractiot t calc 1. The elements chlorine contains two isotopes cl whic has the mass of 34.98 U and Cl which has a mass of 36.98 uy. The atomic Weight of chlorine is 35.46 u. Caiculate the percentage of each of the isotopes in chiorine . 2. Using the data in the table calculate the atomic fi of naturally occurring Argon. [39.96 |99.600% Tsotopic mass Abundance. ] {35.96 0.337% , | \37.96 | 0.063% : 3. Atomic weight of ordinary hydrogen is 1.008 ordinary hydrogen contains two isotopes H and D Calculate the percentage of D isotopes of the element. MOLECULAR MASS: It is the sum of atomic mass of the elements present ina molecule It is obtained by multiplying the atomic mass of each element by the number of its atom and adding them together . For example, Mwt of Co 12+2x16 = 12+32 = 44, Mole:- One mole is the amount of asubstance that contains as many particles or entities as there are atoms in exactly 12 gm or 01.2 kg of the 12¢ isotopes . The number of particles in one mole of a substance is equal $0 6.023x10* ? or 6.02213 10° For example, one mole of atom must contain 6.023x10* * atom. One mole of molecules must contain 6.023% 10? 3 no. of molecules. One mole of electrons must contain 6.023x10°? electron. tn case of elements, the atomic weight of the element _papresent one mole. val 13 To i 9 get the simplest molar ratio, following method is followed. LWhen % wi is divi : ; sles n% wt ig divided by atomic mass then simplest whole no. ig not obtained then consider the smallest no.and divide P . Doig ly al! of the no. determined by % wt/At.mass ratio by that smallest no. 2 I¢ it is possible To consider the simplest whole no.then consider that no.2.923 3.1 to get the whole no.there be any common multiple, then multiply all of the digits by common multiple 15:25 = 1.5x2: 2.5%2 = 3:5 Molecular Formuta:- Tt can be determined by following : formula wt. . mula)wT. = Molecular X(empirical for * © At STP 22.4 L of gas will contain its molecular wt. Of 5 es! 1. An organic compounds contain carbons, hydrogen and oxygen according to following percentage—c = 40. 48° H= 5.08% V.D =59 calculate the molecular formule, 2. _An oxide of nitrogen contains 30.43% of nitrogen .T? molecular weight of the compound is equal to 92 am calculate the molecular formula of the compound. A compound on analysis gave the following percentage composition Na= 14.31% s=9.97% H=6.22%, 0-69. 5% 3 Calculate the mole molecular formula of the compound on The assumption that all the ‘hydrogen in the compound are | present in combination with oxygen as water of § F crystallization M.wt. of the compound is 322. % / ‘4 ENTS:- i : urunes* lly used up ina chemical reaction a T which is ToT The reagents which is . and which denote the amount of of a rea as a limiting reagents. 2H2+O2—-2H20 H: and O2 react with each other in 2:1 ratio to produce water. Tf in this reaction 4 mole of and and one mole of be allowed to undergo reaction then O: will be the limiting reagent because one mole of will react with émole of and the produc} will be 2 mole of H.0. N: +32 2NH; Here one mole of Np react with 3mole of He to produce 2 mole of NHs, but if 2 mole of N: be given 6 mole Of H will be required and N: will be limiting reagents. Strength of + olution: Tt denotes the amount of the solute which will b definite amount of solvent. & present in 1 Massy = Mass of solute . oS x100 Mass of solution 1 tion should behave Mass Soy Mass of solute Yy x wees POE 100 ‘volur . ne volume of solution 3. Volume %~- = Zones x 100 v volume of solution . 4. Molarity (M) > It denote the no. of moles of solute present in 1 litre of the solution. e determined in i Molar solution: Lf one mole of solute b lution. ( Unit mole litre of solution then it is known as molar se a) . No. “No.of m snibles of solute Molarity = “vol. of “of solution in litre 5 Molality(m) : Ti denotes the no. of moles of solute present in 1 kg of the solvent. No.0. ‘moles af solute _ Molality (mm) = = weight of the solvent in kg/1000gm ne mole of solute be present in one kg 0 Molal solution: If o ol Kg *). the solvent then it is known as molal solution (7m » 7. Normality( N): It denote the no. of gm equivalent weight « @ Substance present in 1 litre of the solution. (N) _ No.of gm equivalent weight ~ vol.of the solution litre 8. Equivalent weight: It denote the amount of substance that can combine with 1.008 gm of H: or 8 gm of oxygen or 35.5 gm chlorine or the amount of substance that will be able to liberate 1.008 gm of H: or 8 gm of O; or 35.5 gm 0: Cl: from its compound. Atomic weight valency Equivalent weight of an element = Example,E wt. of Mg = 24/2 = 12 Equivalent of an acid . Molecular weight of the acid “ No.of replaceable H-atom present m the molecule of an acid. weight of acid ically of the acid Molecular weight of base Acidity of the base Equivalent of a base = - _ Molecular weight of salt No, of cations X its valency Equivalent wt. of salt = ~ - : —. weight takens No. of gm-equivalent weight = “Equivalent weight Examples:- LA solution is prepared by dissolving 2 gm of A to 18 of water. Calculate the mass percent Mass of A= =<. x100 2 = 37 X100 = 10% 2. Calculate the molarity of NaoH in the solution prepared by dissolving 4 gm of NaoH in 250 ml of the Solution. Muwt. of NaoH = 23+ 16 + 1 = 40. 4 Mole of NaoH = 49 250 _ 4 1000 40 * “250 3. The density of 3M solution of Nacl 1.25 gm mL * Caicuiate the molality of the solution . 
