SOME BASIC CONCEPTS IN CHEMISTRY

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1 CS 1 Syllabus : SOME BASIC COCEPTS I CHEMISTRY Matter and its nature, Dalton s atomic theory; Concept of atom, molecule, element and compound; Physical quantities and their measurement in Chemistry, precision and accuracy, significant figures, S.I. Units, dimensional analysis; Laws of chemical combination; Atomic and molecular masses, mole concept, molar mass, percentage composition, empirical and molecular formulae; chemical equations and stoichiometry.

2 CS 2 COCEPTS C1 In this chapter we will discuss the calculations based on chemical equations. It has been classified into two parts : 1. Mole Concept 2. Equivalent Concept C2 MOLE COCEPT : In mole concept we deal with different types of relations like weight-weight, weight-volume, or volume-volume relationship between reactants or products of the reaction. Mole concept is based on balanced chemical chemical reaction. Some basic definitions used in mole concept are as follows : Limiting Reagent : A reagent which is consumed completely during the chemical reaction. umber of moles of weight of substance a substance(n) atomic or molecular weight Also, umber of moles of a substance(n) Given number of molecules Avogadro number PV In gas phase reaction number of moles of a gas (n) =, RT At STP/TP one mole of any gas contains 22.4 L i.e. at 27 K and 1 atm pressure. In aq. solution n = MV [M - molarity, V - volume of solution] Practice Problems : 1. Chlorine can be produced by reacting H 2 acid with a mixture of MnO 2 and acl. The reactions follows the equation : 2aCl + MnO 2 + H 2 2aH + Mn + Cl 2 + H 2 O what volume of chlorine at STP can be produced from 100 g of acl? (At. wt. a = 2, Cl =.) 19.1 lt 0 lt 29 lt lt 2. A solution contains g of KOH was poured into a solution containing 6.8 g of AlCl, find the mass of precipitate formed [At. wt. : H-1, Al-27, Cl-., K-9] 2. g 2 g 2 g 0.2 g [Answers : (1) a (2) a] C EQUIVALET COCEPT : Volumetric Analysis is based on acid base titration and redox titration mainly. Important Definitions : o. of moles of solute Molarity(M) Vol. of solution in L o. of gramequivalents of solute ormality() Vol. of solution in L o. of gram equivalents of solute(neq ) Equivalent weight Weight of solute Equivalent weight Molecular weight (or) Atomic weight (or) Ionic weight nfactor n eq = n mol n-factor n eq = ormality Volume (L) umber of moles(n mol ) = Molarity Volume (L) ormality = Molarity n-factor Calculation of n Factor for Different Class of Compounds

3 CS 1. Acids : n = basicity H PO 4 n = H PO n = 2 H PO 2 n = 1 2. Bases : n = acidity e.g. Ammonia and all amines are monoacidic bases. Salt : (Which does not undergo redox reactions) n factor = Total cationic or anionic charge 4. Oxidizing Agents or Reducing Agents : n factor = change in oxidation number Or number of electron lost or gained from one mole of the compound. ote : In a balance equation n factor ratio of two compounds is reverse of their molar ratio. Practice Problems : 1. [a + ] in a solution prepared by mixing 0.00 ml of 0.12 M acl with 70 ml of 0.1 M a 2 is 0.1 M M M M 2. The equivalent mass of Mn is half of its molar mass when it is converted to Mn 2 O MnO 2 MnO 4 MnO 4 2. The anion nitrate can be converted into ammonium ion. The equivalent mass of O ion in this reaction would be 6.20 g 7.7 g 10. g 21.0 g 4. When BrO ion reacts with Br ion in acid solution Br 2 is liberated. The equivalent weight of KBrO in this reaction is M/8 M/ M/ M/6. The number of moles of KMnO 4 that will be needed to react completely with one mole of ferrous oxalate in acidic solution is / 2/ 4/ 1 [Answers : (1) d (2) b () b (4) c () a] C4 LAW OF CHEMICAL EQUIVALETS : In a chemical reaction the equivalents of all the species (reactants or products) are equal to each other provided none of these compounds is in excess. 1 V 1 = 2 V 2 (when normalities and volumes are given) Relation between percentage weight by weight (x), density and strength in gm / litre (s) S = 10 xd C BACK TITRATIO This is a method in which a substance is taken in excess and some part of its has to react with another substance and the remaining part has to be titrated against standard reagent. C6 BASIC PRICIPLE OF TITRATIOS : In voltmetric analysis, a given amount (weight or volume) of an unknown substance is allowed to react with a known volume of a standard solution slowly. A chemical reaction takes place between the solute of an unknown substance and the solute of the standard solution. The completion of the reaction is indicated by the end point of the reaction, which is observed by the colour change either due to the indicator or due to the solute itself. Whether the reactions during the analysis are either between an acid and or base or between O.A. and R.A., the law of equivalence is used at end point. Following are the different important points regarding this process : (i) In case of acid base titration at the equivalence point (n eq )acid = (n eq )base (ii) In case of redox titration (n eq )oxidant = (n eq )reductant (iii) If a given volume of solution is diluted then number of moles or number of equivalence of solute remains same but molarity or normality of the solution decreases. (iv) If a mixture contains more than one acids and is allowed to react completely with the base then at the equivalence point, (n eq ) acid 1 + (neq) acid = (n eq ) base

4 CS 4 (v) Similarly if a mixture contains more than one oxidising agents then at equivalence point, (n eq ) O.A 1 + (n eq ) O.A = (n eq ) reducing agent. (vi) If it is a difficute to solve the problem through equivalence concept then use the mole concept. Practice Problems : 1. ml of -HCl, 20 ml of /2-H 2 and 0 ml of / HO are mixed together and the volume made to 1 litre. (i) The normality of the resulting solution is / /10 /20 /40 (ii) The wt. of pure aoh required to neutralize the above solution is 10 g 2 g 1 g 2. g g of a sample of a 2 CO.xH 2 O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of /10 HCl for complete neutralization. The value of x is ml of 1 M KMnO 4 oxidised 100 ml of H 2 O 2 in acidic medium (when MnO 4 is reduced to Mn 2+ ); volume of same KMnO 4 required to oxidise 100 ml of H 2 O 2 in basic medium (when MnO 4. is reduced to MnO 2 ) will be 100 ml 00 ml 00 ml 100 ml ml of a mixture of aoh and a 2 is neutralised by 100 ml of 0. M H 2. Hence amount of aoh in 100 ml mixture is 0.2 g 0.4 g 0.6 g 1.0 g. mol of a mixture of Fe and Fe 2 ( ) required 100 ml of 2 M KMnO 4 solution is acidic medium. Hence mol fraction of Fe in the mixture is g of M 2 CO is dissolved in 10 ml of 1 HCl. Unused acid required 100 ml of 0. aoh. Hence equivalent weight of M is [Answers : (1) c (2) c () b (4) b () a (6) a]

5 CS IITIAL STEP EXERCISE 1. 1 g of the carbonate of a metal was dissolved in 2 ml of -HCl. The resulting liquid required ml of -aoh for neutralization. The eq. wt. of the metal carbonate is one 2. If 0. mol of BaCl 2 is mixed with 0.20 mol of a PO 4, the maximum amount of Ba (PO 4 ) 2 that can be formed is 0.70 mol 0.0 mol 0.20 mol 0.10 mol. When one gram mol of KMnO 4 reacts with HCl, the volume of chlorine liberated at TP will be 11.2 litres 22.4 litres 44.8 litres 6.0 litres 4. 4 g of hydrogen peroxide is present in 1120 ml of solution. This solution is called 10 vol solution 20 vol solution 0 vol solution 2 vol solution. To prepare a solution that is 0.0 M KCl starting with 100 ml of 0.40 M KCl add 0.7 g KCl add 20 ml of water add 0.10 mol KCl evaporate 10 ml water 6. In hot alkaline solution, Br 2 disproportionates to Br and BrO 0.08 M KCl and 0.01 M KOH 0.08 M KCl and 0.01 M HCl 8. Molality of 18 M H 2 (d = 1.8 gml 1 ) is 6 mol kg mol kg 1 00 mol kg 1 18 mol kg g equiv. of a substance is the weight of that amount of a substance which is equivalent to 0.2 mol of O mol of O 2 1 mol of O 2 8 mol of O The molality of a H 2 solution is 9. The weight of the solute in 1 kg H 2 solution is g g g 9.0 g 11. The density of 1 M solution of acl is 1.08 g/ml. The molality of the solution is Which is false about H PO 2 it is tribasic acid one mole is neutralised by 0. mol Ca(OH) 2 ah 2 PO 2 is normal salt it disproportionates to H PO and PH on heating. Br 2 + 6OH Br + BrO + H 2 O hence equivalent weight of Br 2 is (molecular weight = M) M 6 M M M 7. When 80 ml of 0.20 M HCl is mixed with 120 ml of 0.1 M KOH, the resultant solution is the same as a solution of 0.16 M KCl and 0.02 M HCl 0.08 M KCl

6 CS 6 FIAL STEP EXERCISE 1. 8 g of sulphur are burnt to form SO 2 which is oxidised by Cl 2 water. The solution is treated with BaCl 2 solution. The amount of Ba precipitated is 1 mol 0. mol 0.24 mol 0.2 mol 2. One mole of a mixture of CO and CO 2 requires exactly 20 gram of aoh in solution for complete conversion of all the CO 2 into a 2 CO. How many grams of aoh would it require for conversion into a 2 CO if the mixture (one mole) is completely oxidised to CO 2 60 grams 80 grams 40 grams 20 grams. One gram of a mixture of a 2 CO and ahco consumes y gram equivalents of HCl for complete neutralisation. One gram of the mixture is strongly heated, the cooled and the residue treated with HCl. How many gram equivalents of HCl would be required for complete neutralization? 2 y gram equivalent y gram equivalents y/4 gram equivalents y/2 gram equivalents 4. If equal volumes of 1 M KMnO 4 and 1 M K 2 Cr 2 O 7 solutions are allowed to oxidise Fe (II) to Fe(III), then Fe (II) oxidised will be more by KMnO 4 more by K 2 Cr 2 O 7 equal in both cases none of these. 0. g of fuming H 2 S 2 O 7 (Oleum) is diluted with water. This solution is completely neutralized by 26.7 ml of 0.4 aoh. The percentage of free SO in sample is 0.6% 40.6% 20.6% 0% 6. One mole of 2 H 4 loses 10 mol of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound. What is the oxidation state of nitrogen in Y ASWERS (IITIAL STEP EXERCISE) 1. a 2. d. d 4. a. a 6. c 7. c 8. c 9. a 10. b 11. b 12. a ASWERS (FIAL STEP EXERCISE) 1. d 2. a. b 4. b. c 6. c

7 CS 7 AIEEE AALYSIS [2002] 1. umber of atoms in 8. gram Fe (at. wt. of Fe =.8 g mol 1 ) is half than in 8 g He twice that in 60 g carbon When KMnO 4 acts as an oxidising agent and ultimately forms [MnO 4 ], MnO 2, Mn 2 O, Mn 2+ then the number of electrons transferred in each case respectively is,, 7, 1 1,,, 7 4,, 1, 1,, 4,. With increase of temperature, which of these changes? mole fraction Fraction of solute present in water molality weight fraction of solute 4. In a compound C, H and atoms are present in 9:1:. by weight of compound is 108. Molecular formula of compound is C 2 H 6 2 C H 4 C 9 H 12 C 6 H 8 2 AIEEE AALYSIS [200]. 2 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of ml. The molarity of barium hydroxide solution was molecules of urea are present in 100 ml of its solution. The concentration of urea solution is M 0.01 M 0.02 M 0.1 M (Avogadro constant, A = mol 1 ) [2004] 7. To neutralise completely 20 ml of 0.1 M aqueous solution of phosphorus acid (H PO ), the volume of 0.1 M aqueous KOH solution required is 10 ml 20 ml 40 ml 60 ml [2004] 8. Excess of KI reacts with Cu solution and then a 2 S 2 O solution is added to it. Which of the statements is incorrect for this reaction? [2004] Cu 2 I 2 formed CuI 2 is formed a 2 S 2 O is oxidised evolved I 2 is reduced AIEEE AALYSIS [2004/200/2006] 9. Two solutions of a substance (non electrolyte) are mixed in the following manner 480 ml of 1. M first solution and 20 ml of 1.2 M second solution. What is the molarity of the final mixture? 1.20 M 1.0 M 1.44 M 2.70 M [200] 10. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will [200] decrease twice increase two fold remain unchanged be a function of the molecular mass of the substance 11. The oxidation state of chromium in the final product formed by the reaction between Kl and acidified potassium dichromate solution is [200]

8 12. The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is zero two one three [200] CS 8 1. How many moles of magnesium phosphate, Mg (PO 4 ) 2 will contain 0.2 mole of oxygen atoms? [2006] AIEEE AALYSIS [2007] 14. In the reaction, 2Al (s) + 6HCl (aq) 2Al + (aq) + 6Cl (aq) + H 2(g), 67.2 L H 2 (g) at STP is produced for every mole Al that reacts 11.2 L H 2 (g) at STP is produced for every mole HCl (aq) consumed 6 L HCl (aq) is consumed for every L H 2 (g) produced.6 L H 2 (g) is produced regardless of temperature and pressure for every mole Al that reacts 1. The density (in g ml 1 ) of a.60 M sulphuric acid solution that is 29% H 2 (Molar mass = 98 g mol 1 ) by mass will be ASWERS AIEEE AALYSIS 1. b 2. d. b 4. d. b 6. b 7. c 8. b 9. c 10. a 11. d 12. b 1. b 14. b 1. a

9 CS 9 TEST YOURSELF 1. umber of moles of a substance can be calculated by molecular weight of substance weight of substance given number of molecules Avogadro number 9. Change in number of moles when temperature increase from 27 0 C to 2 0 C, pressure increase from atm to.44 atm and volume remain the same is no change PT RV all 2. n-factor in the 2H 2 is The molarity of 8. g acl in litre solution is What weight of silver is present in 4. g Ag 2 S? [At. weight of Ag = 107.9, S = 2] The number of molecule present in 4.48 litre solution at STP is If 20 ml of aoh is titrated with 0 ml of HCl, 0 then normality of HCl is The normality of g of H 2 in 2 litre solution is Which of the following relation is wrong? n eq = n mol n-factor n eq = ormality Volume umber of moles = Molarity volume Molarity = ormality n-factor 1. b 2. a. c 4. a. a ASWERS 6. b 7. d 8. d 9. d

Syllabus : CONCEPTS. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18 CS 1

Syllabus : CONCEPTS. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18 CS 1 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, E-mail einsteinclasses00@gmail.com, CS S T I C H I E T RY Syllabus : C CNCEPTS In this chapter