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1. BASIC CONCEPTS OF CHEMISTRY 1. Explain the role of redo Ans. Redox titrations in ti Titrimetric anal concentration which ¡s _Fequired to react Quantitatively with a measured volume of solution of the substance to be determined. 5) > Xx reactions in titrimetric processes and galvanic cells? trimetric quantitative analysis:- ysis involves determivation of volume of a sodium of accurately known > In titrimetric analysis, substance of known concentration is called the titrant and the substance being titrated is known as titrant. Solution of known Concentration is said to be a standard solution. which is generally filled in the burette. + The process of adding the standard solution until the reaction is just complete is called titration. v The point, where the reaction just complete is called equivalence point or stoichiometric end point. > Inredox reactions end Point is detected by 1) observing a physical change, For Ex: light pink colour of KMnO4 changes. il) by using a reagent known as indicator which change. formation of turbidity ete, gives a clear visual change like a colour (D2CH% ay +41 up > CttL sy + Laca Tt gives a deep blue colour with Starch solution (DE +25O5ag) > 217 +5 O, 12 gives blue colour with starch, which becomes colourless, when all 12 is converted to 1 Hypo. Redox reactions in Galvanic cells :- +1f Zinc rod is kept in CuSO4 solution, a redox reaction takes place, as following by, ordalion 2ng + Cu e) 20 0 +04 teduchon In a galvanic cell, Zn rod is dipped in ZnSO4 solution, Cu rod in CuSOA solution. > Both the beakers contain a redox couple Le Za? / Zn 1/Cu*? 1Cu > These 2 solutions are connected by a salt bridge which provides electric contact between two solutions NARAYANA STUDY: MATERIAL Page No: 1 (O Scanned with OKEN Scanner ¿1arion reaction takes place. >In galvanic cells, reduction $: oxidation reaction ta >Reaction atanode :- Zi > Zn" +2e [oxidation] y Reaction at cathode : - Cu* +2e > Cu[reduction] y Overall Reaction Reduction ZO Qnza? [lo] da 2. State the law of definite proportions, suggest one problem to understand the law, by working out that problem. A) > Law of definite proportions (or) Law of constant or fixed proportions:- * It states that “ A given chemical substance always contains the same elements combined ina fixed Proportion by weight”. Example: Sample of copper carbonate are obtained one from natural . sources, and the other prepared in the laboratory compositions are given as following:- Elements natural Composition Synthetic Cu 51.35 51.35 oc 9.74 9.74 Cc 38.91 38.91: * Irrespective of the place, method or Person, a chemical compound i É a A ” de cont: ame i ina Fixed ration by weight. si , Ona» same element combined 3, Define and explain molar mass, Y Ex:- (1) Molar mass of 11,50, = 98g - NARAYANA STUDY o 4 E E ¡ Page No: 2 (O Scanned with OKEN Scanner 2, State and explain law of conservation of mass y can neither be created nor destróyed Ans: Law of conservation of mass : - “It states that the malle : á . dúring'a chemical change “. ; “oducts formed during a chemical * This law can also be stated as the total mass of the pro change is exactly equal to the total mass of the reuctants. * For Ext AgNO og) + Kl ayy + Ago + ENO xp Na temcighcd. * Filtrate is completely evaporated de the mass of residuc de > de and wicighed. “* Agl is yellow precipitate. It is seperated from solution by fi e KNO3] y * Total mass of [AgNO3 +K1] is equal to the total mass of[Ag a 3. State and explain law of definite Proportions. ondii Ans: *Law of definite proportions (or) Law of constant or fized propor! 5% -lements combined in a "It states that “A given chemical substance always contains the same eler : fixed proportions by weight” : Example : 2 Samples of core carton are obtained one from natural sources, and the other prepared in the laboratory. % compositions.are given as following : Elements Natural Composition Synthetic Cu 51,35 51,35 o 9.74 9.74 ¿€ qe 38.91 38.91 * Trrespective of the place, method or person, a chemical compound contains same elements combined in a Fixed ratio by weight. 4. State and explain law of multiple proportions. “If two elements chemically combine to give two or more compounds, then the weight ofone element which , . combines with fixed weight of the other element in those compounds will be a simple multiple ratio to oneanother “ * For Ex :- N2 8: 02 combine to give Nitric oxide (NO) and nitrogen dioxide (NO2) * In NO, 14 gm of Nitrogen combines with 16g of oxygen to give nitric oxide (NO). * In NO2 14gmofNitrogen combines with 32 gm ofoxygen ; * In both the compounds the weights of nitrogen are fixed as 14 gm * The weight of oxygen that combines with'14 gm ofnitrogen are 162min NO and 32gm in NO2 poe : . Wt ofoxygen in these compounds are in the ration of 16:32 or 1:2 a simple multiple ratio 5. There is no need that a given species must always posses equivalent wei ana 4 W 2 a ? - Isitgrue? Explain. a cight always:constant Ans. Yes, jt is true. : . * The equivalent weight of some substance d ; , A Pr . bi on. pe epends upon the type of reaction, and medium of *For example :- KMnO4 acts as an oxidisi in acidi solutions also, ng agent in acidic, neutral as well as basic * In acid medium :- NARAYANA. STUDY MATERIAL : Page No: 4 (O Scanned with OKEN Scanner KMnO, +8H* +50” > Mn” 44H,0 mol.wtof KMno, no.ef electrons involved in reaction equivalent weight ofKMnO, = 1 wtof K Mi mola of MO, 158, 0% 608 Su 5 *1n strong basic medium:- * MnQ¡e” > Mno;? mol, wtof-KMnO, no.of electrons involved in reaction y. equivalent weight of KmnO, = _ mol.wtof KMno,: RECAE * In neutral medium or mild alkaline medium KMnO, +2H,0+3e: >K" 4+40H" + Mno, 158.04 Equivalent weight of KMnO,= EUROS .68 6. The equivalent weight of sulphuric acid is given as 98. How far it is true when sulphuric acid is a dibasic acid. Ans. No, Itis not True. * Equivalent weight of sulphuric acid pá upon basicity, of acid molecular weight basicity. =158.04 -Equivalent veightof H,SO, = Equiválent Weight of H,SO,=49 7. How are the end points of titrations detected in the following reactions. (i) MnO; Oxidises Fe” (ii) Cro; Oxidises Fe”? (ii) Cu? Oxidises Il” Ans: 1) MnO; Oxidises Fer? “The end point of this reaction is identified by a colour change in this reaction at the be end point, colour of solution turns to pink (ii) Cr,O7? Oxidises Fe? | soil loz A The end point of this reaction is identified by a colour change in this reaction na point. For this reaction, at end point the colour of solution changes to violet. : (iii) Cu” Oxidises IT” : The reaction is given as 2Cy” +41" —>Cu,l, +1, á In this reaction iodine is released which gives deep blue colour with starch solution at the end point. NARAYANA STUDY MATERIAL — DRSNES Page No: 5 (O Scanned with OKEN Scanner 4. L(a)+2570,(aq) >21 (aq)+S, O aq) For the above titrimetric reaction, how is the end point determined? Ans: L(aq)+255'0,(a9) >21 (aq) +50 (aq) lodine gives blue colour with Starch solution but when all 12 gets converted to I by hypo (sodium thiosulphate) the blue colour disappear at end point. PROBLEMS: 1.The percentage composition ofan organic compound is given below. Its molecular weight is 136. ; Calculate its molecular formula. C=70.59%, H= 5.88%; O=23.53% Sol: Element Composition Atomic Ratio Simple Ratió 70.59 5.88 bmp Sa ¡e 70.59 a 5.88 147 5.88 5.88 ——=5.88 ——=4 E E 1 1.47 23.53 : 1.47 == =1.47 —=1 3 29-53 16 LAT Empirical formula C,H,O Molecular formula =(£.F)n Molecular weight 136 >n= n =2 Empirical formula weight 68 Molecular formula =(C,H,0), = C¿Hj0, 2. Analysis ofan organic compound gave the following results. a) 0.100 gm. of the compound gave 0.2282 g. of CO2 and 0:076 gm. of H,O b) 0.200 gm. of the compound at 150C and 760 mm. pressure gave 21.8 mi. of N2. The molecular weight ; of compound is 107. Calculate its molecular formula. 2 WAR COS Solution: a) Carbon weight percent “yg weight of hydro Carbon e 120, -12,02282 ¿09 44 0.1 NARAYANA STUDY MATERIAL E Page No: 7. (O Scanned with OKEN Scanner - Carbon welght%% = 62.24 b) Hydrogen weight percent o eajahtof WU” welg! Carbon e jeightal Me se 100. ¿E 59758 x100 a me Hydrogenweighi% =84.. ¿ cd a OR. ta itrogen Volume at STP . lo c) Nitrogen weight pércent gx weight of subs nos : We have to firid nitrogen volume by using given data BI, PY, 760x218 - 760%H - 7, 20.66ml EA 288 273 EEE : Volume of N, =20.66ml 20.66 Nitrogen weight percent = 3x0.200 =12.91 Ñ Element Composition Atomic Ratio * Simple Ratio 62.24 5.186 == =5.186 =5.64=5 Ms 6224 2 0.92 q 84 y : Ac 12.91 + N 12.91 =—=0.92 2 : j 1645 102 0 16.45 ===1:028 4 16 028 So, empirical formulae = C¿H,NO Empirical formula weight = 99 : A 99 Molecular weight =99 n= Se = So, molecular formula = C,H,NO ER * NARAYANA STUDY MATERIAL (O Scanned with OKEN Scanner sa ; Ñ Si Je Ratio" Atomic Ratio laa 443 Element e Composition 432 “e A E ; 2.23 Di 15.55 : 01555 _, H 15.55 1558. 215,55 2.23 : a 2.23 31.27 20.1 N 31007 a 2.23 empitical formula = C¿H,N empirical formula.weight = 45 molecular formula = (empirical formula), A =10molecular formula = (CHN ro = Cao Ho Nro 6. Write down molar masses of (a) One mole of mercury (b) One mole of sulphur molecules (c) One mole of potassium dichromates (K,Cr,O,) formula units. Atomic weights : Hg =200.6: S=32.6: Pb: 207: K=39: Cr=52; O=16 amu. Solution: a) One mole mercury (Hg) + Given atomic weight of Hg = 200.6 One mole mercury weight = 200.6 g b) Given atomic weight of 8=32 ,6 g One mole sulphur weight = 32.6x8 = 260.82 C) Given atomic weight of Pb=207 One mole lead weight = 207 g D) Given atomic weight of K=39, Cr=52; O=16 molar mass K,Cr,O, +2x39+2x52+7x:16=294g Te Calculate the"molar masses of Ca(NO,),, AL, 0,,C, H,,0s (atomic Se a Ca=40,N=14, Al = 27, Ha1, C=12, O=16 amu) a) Find out the number of moles of sodium bicarbonate present in 5.08 gm. of sociura bicarbonate (NaHCO,) ; b) Calculate the number of moles of helium in 6,46 gm. of helium. (at wt. of helum= 4) : €) Calculate the number of moles of zinc in 23.3 gm. of zinc, d) Calculate the number'of moles of sulphur in 16.3 gm. of sulphur.. Sol: Molar mass of Ca(NO,) =40+2[14+3x16]=164g A1,0,=2x27+3x16=102g CHO, =6x12+12x1+6x16=180g paa NARAYANA STUDY MATERIAL ————— Ñ E O A Page No::10 (O Scanned with OKEN Scanner a) 84 gm of Sodium bicarbonate = 1 mole 5.08 gms of Sodium bicarbonate = ? 5.08 Number of moles of Sodium bicarbonate in 5.08g= Ta 7 0.06 moles b) 4 gm of helium = 1 mole ; 6.46 6.46 gms of helium a =1.615 moles 0) 65.38 gms of Zinc= 1 mole ra 2 23.3 gms of Zine 65.38 =0.356 moles d) 32 gm of sulphur = 1 mole 16.3 16.3 gms of sulphur = a” 0.51 moles 8. a) Your friend requested you to weigh 0.5 moles of ammonium sulphate. How many grams do you weigh? [Hint: Formula of ammonium sulphate( NH, ),SO, Atomic weight : N= 14, H=1, S=32, 0=16] b) Calculate the weight of 0.885 Moles of Mg (NO,), [Atomic weight : Mg = 24, N=14, O=16] c) What is the weight o£ 0.059 moles of aspirin (C,¿H¿O,) [Atomic weight: O=16, C=12, H=1 Sol: .a) Formula of Ammonium Sulphate (VH,),SO, One mole of (NH,),SO, contained weight = 2[14+1 x4 ] +32+4 x4 x 16 =132g 0.5 mole of contained weight = : b) One mole of Mg contained weight = 24 +2[14+316]=148g y . ] . 148x 0.885 0.885 moles of Mg (VO,), contained weight = ———=130.98g €) One mole of asprin contained weight = 9x12+8x1+4x16=180g j . . : . 0.0590x 180 0.0590 moles of (C,H,0,) asprin contained weight = -——— =10.6g 9. Calculate the molecular weights of the following compounds. 1. Glucose C¿H,,0, 2. Calcium fluoride CaF, — 3.Magnesium oxide MgO Atomic weight of C=12, H=1, O=16, Ca=40, 1, F=19, Mg = 24.3 Ans: a) Given atomic weight of C=12, H=1, O=16 molecular weight of glucose is 6x12+12x1+6x16=180 b) Given atomic weights Ca=40,1, F=19; molecular weight of calcium fluoride is 40,1 +2 x 19=78.1 c) Given atomic weights Mg = 24,3; 0=16 molecular weight of magnesium oxide = 24.3 + 16=40.3 NARAYANA STUDY MATERIAL == z TT Page No::110: (O Scanned with OKEN Scanner 
