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CS– 1 STOICHIOMETRY C1 C2 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 MOLE CONCEPT : 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. Number of moles of a substance(n) weight of substance atomic or molecular weight Also, Number of moles of a substance(n) Given number of molecules Avogadro number In gas phase reaction number of moles of a gas (n) = 1. 2. C3 PV , RT At STP/NTP one mole of any gas contains 22.4 L i.e. at 273 K and 1 atm pressure. In aq. solution n = MV [M - molarity, V - volume of solution] Practice Problems : Chlorine can be produced by reacting H2SO4 acid with a mixture of MnO2 and NaCl. The reactions follows the equation : 2NaCl + MnO2 + 3H2SO4 2NaHSO4 + MnSO4 + Cl2 + H2O what volume of chlorine at STP can be produced from 100 g of NaCl ? (At. wt. Na = 23, Cl = 35.5) (a) 19.15 lt (b) 30 lt (c) 29 lt (d) 5 lt A solution contains 5 g of KOH was poured into a solution containing 6.8 g of AlCl3, find the mass of precipitate formed [At. wt. : H-1, Al-27, Cl-35.5, K-39] (a) 2.3 g (b) 23 g (c) 32 g (d) 0.32 g [Answers : (1) a (2) a] DIFFERENT WAYS OF EXPRESSING THE CONCENTRATION TERMS : Important Definitions : mass percent mass of solute 100 mass of solution Molarity(M) No. of moles of solute , unit of molarity are mol/lit., M or molar.. Vol. of solution in L Normality ( N ) Molality (m ) No. of gramequiva lents of solute , unit of normality are g-eq./lit., N or normal. Vol. of solution in L No. of moles of solute , unit of molality are mol/kg, m or molal. wt . of solvent Mole fraction( x A ) ppm nA . n A nB mass of solute 10 6 mass of solution Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS – 2 No. of gram equivalents of solute(neq ) Equivalent weight Weight of solute Equivalent weight Molecular weight (or ) Atomic weight (or ) Ionic weight n factor The relation between different concentration terms : 1. neq = nmol × n-factor 2. neq = Normality × Volume (L) 3. Number of moles(nmol) = Molarity × Volume (L) 4. Normality = Molarity × n-factor 5. M 10xd M 6. m 7. m x B 1000 (1 x B )M A M 1000 1000d MM (d density of solution in g/ml, M molar mass of solute, xB and xA are mole fraction of solute and solvent respectively, MA molar mass of solvent) Calculation of ‘n’ Factor for Different Compounds : 1. 2. Acids : n = basicity H3PO4 n = 3 H3PO3 n = 2 H3PO2 n = 1 H3BO3 n = 1 Bases : n = acidity of base e.g. Ammonia and all amines are monoacidic bases, NaOH(n = 1), Na2CO3(aq) n = 2, NaHCO3(n = 1) 3. Salt : (Which does not undergo redox reactions) n factor = Total cationic or anionic charge, e.g. Na3PO4 n = 3, Ba3(PO4)2 n = 6 4. C4 Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Or number of electron lost or gained from one mole of the compound. EQUIVALENT CONCEPT It is based on law of equivalence which is explained as follows : Law of chemical equivalents : 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. N1V1 = N2V2 (when normalities and volumes are given). If the number of equivalence of both the reactants are different then reactant with the lesser number of equivalence will be the limiting reagent. Application of equivalent concept : It is used in acid base titration, back titration and double titration, similarly in redox titration. Equivalent concepts can be used on all reaction whether they are balanced or not balanced but mole concept is used in solving the problems when the reactions are balanced. Basic principles of tirations : 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 (neq)acid = (neq)base (ii) In case of redox titration (neq)oxidant = (neq)reductant Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS– 3 (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, (neq) acid1 + (neq) acid2 + ... = (neq) base (v) Similarly if a mixture contains more than one oxidising agents then at equivalence point, (neq) O.A1 + (neq) O.A2 +... = (neq) reducing agent. (vi) If it is a difficute to solve the problem through equivalence concept then use the mole concept. Back titration : 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. Double titration : This is a titration of specific compound using different indicators. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities. When the solution containing NaOH and Na2CO3 is titrated using phenolphalein indicator following reaction takes place at the phenolphthalein end point – NaOH + HCl NaCl + H2O Na2CO3 + HCl NaHCO3 + H2O Here, eq. of NaOH 1 eq. of Na 2 CO 3 eq. of HCl 2 (n 2) When methyl orange is used, Na2CO3 is converted into NaCl + CO2 + H2O Hence, eq. of NaOH + eq. of Na 2 CO 3 = eq. of HCl (n 2) TITRATION OF MIXTURE OF BASES WITH TWO INDICATORS Every indicator has a working range Indicator pH range Behaving as Phenolphthalein 8 — 10 weak organic acid Methyl orange 3 — 4.4 weak organic base Thus methyl orange with lower pH range can indicate complete neutralisation of all types of bases. Extent of reaction of different bases with acid (HCl) using these two indicators summarised below Phenolphthalein Methyl Orange NaOH 100% reaction is indicated 100 % reaction is indicated NaOH + HCl NaCl + H2O NaOH + HCl NaCl + H2O Na2CO3 50% reaction upto NaHCO3 100% reaction is indicated stage is indicated Na2CO3 + 2HCl 2NaCl + H2O Na2CO3 + HCl NaHCO3 + NaCl + CO2 NaHCO3 No reaction is indicated NaHCO3 + HCl NaCl + H2O + CO2 100% reaction is indicated Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS – 4 Electron exchanged Eq. wt. or change in O.N. Species Changed to Reactions 1. MnO4— (O.A.) Mn2+ in acidic medium MnO4— + 8H+ + 5e— Mn2+ + 4H2O 5 E 2. MnO4— (O.A.) MnO2 in basic medium MnO4— + 3e— + 2H2O MnO2 + 4OH— 3 E 3. MnO4— (O.A.) MnO42— in MnO4— + e— + 2H2O MnO22— neutral medium 1 E 4. Cr2O72—(O.A.) Cr3+ in acidic medium Cr2O72— + 14H+ + 6e— 2Cr3+ + 7H2O 6 E 5. MnO2(O.A.) Mn2+ in acidic medium MnO2 + 4H+ + 2e— Mn2+ + 2H2O 2 E 6. Cl2(O.A.) (in bleaching powder) CuSO4 (O.A.) (in iodometric titration) S2O32— (R.A.) Cl— Cl2 + 2e— 2Cl— 2 E Cu+ Cu2+ + e– Cu+ 1 E S4O62— 2S2O32— S4O62— + 2e— 7. 8. + 5 — H2O2(O.A.) H2 O H2O2 + 2H + 2e 2H2O 2 10. H2O2(R.A.) O2 2 11. Fe2+ (R.A.) Fe3+ H2O2 O2 + 2H+ + 2e— (O.N. of oxygen in H2O2 is (–1) per atom) Fe2+ Fe3+ + e— M 2 M E 2 E E 1 Estimation of Reaction Relation between O.A. and R.A. I2 I2 + 2Na2S2O3 2NaI + Na2S4O62— I2 + 2S2O32— 2I— + S4O62— I2 2I— 2Na2S2O3 Eq. wt. (Na2S2O3) = E 2. 3. 4. 5. CuSO4 CaOCl2 MnO2 IO3— 2CuSO4 I2 2Na2S2O3 CaOCl2 + H2O Ca(OH)2 + Cl2 Cl2 + 2KI 2KCl + I2 Cl2 + 2I— 2Cl— + I2 CaOCl2 Cl2 I2 2I— 2Na2S2O3 IO3— + 5I— + 6H+ 3I2 + 3H2O Eq. wt. of CuSO4 = M 1 M 2 Eq. wt. of CaOCl2 = MnO2 Cl2 I2 2I— 2Na2S2O3 Eq. wt. of MnO2 = M 2 IO3— 3I2 6I 6Na2S2O3 Eq. wt. IO3— = Einstein Classes, M 1 M 1 2CuSO4 + 4KI Cu2I2 + 2K2SO4 + I2 or 2Cu2+ + 4I— Cu2I2 + I2 white ppt. MnO2 + 4HCl (conc.) MnCl2 + Cl2 + 2H2O Cl2 + 2KI 2KCl + I2 or MnO2 + 4H+ + 2Cl— Mn2+ + 2H2O + Cl2 Cl2 + 2I— I2 + 2Cl— M 1 2 (for two molecules) E M M 9. 1. M 5 M 3 M 1 M 6 M 2 M 2 M 6 Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS– 5 6. — + — H2O2 + 2I + 2H I2 + 2H2O H2 O 2 H2O2 I2 2I 2Na2S2O3 Eq. wt. H2O2 = 7. Cl2 + 2I— 2Cl— + I2 Cl2 M 2 Cl2 I2 2I— 2Na2S2O3 M 2 Eq. wt. of Cl2 = 8. O3 + 6I— + 6H+ 3I2 + 3H2O + O2 O3 O3 3I2 Eq. wt. of O3 = ClO— 9. ClO— + 2I— + 2H+ H2O + Cl— + I2 M 2 ClO— I2 2I 2Na2S2O3 Eq. wt. of ClO— = Cr2O72— 10. Cr2O72— + 14H+ + 6I— 3I2 + 2Cr3+ + 7H2O M 2 Cr2O72— 3I2 6I— Eq. wt. of Cr2O72— = M 6 Practice Problems : 1. [Na+] in a solution prepared by mixing 30.00 mL of 0.12 M NaCl with 70 mL of 0.15 M Na2SO4 is (a) 2. (c) 0.210 M (d) 0.246 M (b) MnO2 (c) MnO4– (d) MnO42– 6.20 g (b) 7.75 g (c) 10.5 g (d) 21.0 g – When BrO ion reacts with Br ion in acid solution Br2 is liberated. The equivalent weight of KBrO3 in this reaction is M/8 (b) M/3 (c) M/5 (d) M/6 The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous oxalate in acidic solution is (a) 6. Mn2O3 – 3 (a) 5. 0.141 M The anion nitrate can be converted into ammonium ion. The equivalent mass of NO3– ion in this reaction would be (a) 4. (b) The equivalent mass of MnSO4 is half of its molar mass when it is converted to (a) 3. 0.135 M 3/5 (b) 2/5 (c) 4/5 (d) 1 5 ml of N-HCl, 20 ml of N/2-H2SO4 and 30 ml of N/3 – HNO3 are mixed together and the volume made to 1 litre. (i) The normality of the resulting solution is (a) N/5 (b) N/10 (c) N/20 (d) N/40 (d) 2.5 g (d) 7. (ii) The wt. of pure NaOH required to neutralize the above solution is (a) 10 g (b) 2g (c) 1g 0.7 g of a sample of Na2CO3.xH2O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x is (a) 7 Einstein Classes, (b) 3 (c) 2 (d) 5 Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS – 6 8. – 4 100 mL of 1 M KMnO4 oxidised 100 mL of H2O2 in acidic medium (when MnO is reduced to Mn2+); volume of same KMnO4 required to oxidise 100 mL of H2O2 in basic medium (when MnO4–. is reduced to MnO2) will be (a) 9. 500 mL 3 (c) 300 mL 3 (d) 100 mL 0.2 g (b) 0.4 g (c) 0.6 g (d) 1.0 g 3 mol of a mixture of FeSO4 and Fe2(SO4)3 required 100 mL of 2 M KMnO4 solution is acidic medium. Hence mol fraction of FeSO4 in the mixture is (a) 11. (b) 100 mL of a mixture of NaOH and Na2SO4 is neutralised by 100 mL of 0.5 M H2SO4. Hence amount of NaOH in 100 mL mixture is (a) 10. 100 mL 3 1 3 (b) 2 3 (c) 2 5 (d) 3 5 5.3 g of M2CO3 is dissolved in 150 mL of 1 N HCl. Unused acid required 100 mL of 0.5 N NaOH. Hence equivalent weight of M is (a) 23 (b) 12 (c) 24 (d) 13 [Answers : (1) d (2) b (3) b (4) c (5) a (6) (i) d (ii) c (7) c (8) b (9) b (10) a (11) a] C5 VOLUME STRENGTH OF H2O2 x volume of H2O2 means x litre of O2 is liberated by 1 volume of H2O2 on decomposition 2H 2 O 2 2H 2 O O 2 68 gm 22.4 lit at STP Volume strength of H2O2 solution = N × 5.6...... (where N is the normality of the H2O2 solution Practice Problems : 1. (a) Calculate the strength of ‘20 V’ of H2O2 in terms of : (i) (b) 2. normality (ii) grams per litre (iii) molarity and (iv) percentage Calculate the volume strength of 2.0 N H2O2 solution. In a 50 ml solution of H2O2 an excess of KI and dilute H2SO4 were added. The I2 so liberated required 20 ml of 0.1 N Na2S2O3 for complete reaction. Calculate the strength of H2O2 in grams per litre. [Answers : (a) (i) 3.58 N (ii) 60.86 g/lit. (iii) 1.79 M (iv) 6.086% (W/V) (b) 11.2 V (2) 0.68 g/litre] Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS– 7 SINGLE CORRECT CHOICE TYPE 1. 2. 1 g of the carbonate of a metal was dissolved in 25 ml of N-HCl. The resulting liquid required 5 ml of N-NaOH for neutralization. The eq. wt. of the metal carbonate is 8. 6. Molality of 18 M H2SO4 (d = 1.8 gmL–1) is 30 (a) 36 mol kg–1 (b) 200 mol kg–1 (c) 20 (d) None (c) 500 mol kg–1 (d) 18 mol kg–1 If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na3PO4, the maximum amount of Ba 3(PO4)2 that can be formed is (a) 0.70 mol (b) 0.50 mol (c) 0.20 mol (d) 0.10 mol 9. When one gram mol of KMnO4 reacts with HCl, the volume of chlorine liberated at NTP will be (a) 11.2 litres (b) 22.4 litres (c) 44.8 litres (d) 56.0 litres 34 g of hydrogen peroxide is present in 1120 ml of solution. This solution is called (a) 10 vol solution (b) 20 vol solution (c) 30 vol solution (d) 32 vol solution To prepare a solution that is 0.50 M KCl starting with 100 mL of 0.40 M KCl (a) add 0.75 g KCl (b) add 20 mL of water (c) add 0.10 mol KCl (d) evaporate 10 mL water In hot alkaline solution, Br2 disproportionates to Br– and BrO3– 13. 3Br2 + 6OH– 5Br– + BrO3– + 3H2O hence equivalent weight of Br 2 is (molecular weight = M) (a) (c) 7. 0.08 M KCl and 0.01 M HCl (b) 12. 5. (d) 50 11. 4. 0.08 M KCl and 0.01 M KOH (a) 10. 3. (c) M 6 (b) 3M 5 (d) M 5 14. 5M 5 When 80 mL of 0.20 M HCl is mixed with 120 mL of 0.15 M KOH, the resultant solution is the same as a solution of 15. 1 g equiv. of a substance is the weight of that amount of a substance which is equivalent to (a) 0.25 mol of O2 (b) 0.50 mol of O2 (c) 1 mol of O2 (d) 8 mol of O2 The molality of a H2SO4 solution is 9. The weight of the solute in 1 kg H2SO4 solution is (a) 900.0 g (b) 468.65 g (c) 882.0 g (d) 9.0 g The density of 1 M solution of NaCl is 1.0585 g/mL. The molality of the solution is (a) 1.0585 (b) 1.00 (c) 0.10 (d) 0.0585 Which is false about H3PO2 (a) it is tribasic acid (b) one mole is neutralised by 0.5 mol Ca(OH)2 (c) NaH2PO2 is normal salt (d) it disproportionates to H3PO3 and PH3 on heating. When KMnO 4 acts as an oxidising agent and ultimately forms [MnO4], MnO2, Mn2O3, Mn2+ then the number of electrons transferred in each case respectively is (a) 3, 5, 7, 1 (b) 1, 5, 3, 7 (c) 4, 3, 1, 5 (d) 1, 3, 4, 5 With increase of temperature, which of these changes ? (a) mole fraction (b) Fraction of solute present in water (c) molality (d) weight fraction of solute In a compound C, H and N atoms are present in 9:1:3. 5 by weight of compound is 108. Molecular formula of compound is (a) 0.16 M KCl and 0.02 M HCl (a) C2H 6N2 (b) C 3H 4N (b) 0.08 M KCl (c) C9H12N3 (d) C6H 8N2 Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS – 8 16. 17. 8 g of sulphur are burnt to form SO2 which is oxidised by Cl2 water. The solution is treated with BaCl2 solution. The amount of BaSO4 precipitated is (a) 1 mol (b) 0.5 mol (c) 0.24 mol (d) 0.25 mol (c) 19. 20. 21. 22. 24. One mole of a mixture of CO and CO2 requires exactly 20 gram of NaOH in solution for complete conversion of all the CO2 into Na2CO3. How many grams of NaOH would it require for conversion into Na2CO3 if the mixture (one mole) is completely oxidised to CO2 (a) 18. 23. 60 grams (b) 40 grams (d) 80 grams 20 grams One gram of a mixture of Na2CO3 and NaHCO3 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 ? (a) 2 y gram equivalent (b) y gram equivalents (c) 3y/4 gram equivalents (d) 3y/2 gram equivalents 25. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorus acid (H3PO3), the volume of 0.1 M aqueous KOH solution required is (a) 10 mL (b) 20 mL (c) 40 mL (d) 60 mL Excess of KI reacts with CuSO4 solution and then Na 2S 2O 3 solution is added to it. Which of the statements is incorrect for this reaction ? (a) Cu2I2 formed (b) CuI2 is formed (c) Na2S2O3 is oxidised (d) evolved I2 is reduced Two solutions of a substance (non electrolyte) are mixed in the following manner 480 ml of 1.5 M first solution and 520 mL of 1.2 M second solution. What is the molarity of the final mixture ? (a) 1.20 M (b) 1.50 M (c) 1.344 M (d) 2.70 M If equal volumes of 1 M KMnO4 and 1 M K2Cr2O7 solutions are allowed to oxidise Fe (II) to Fe(III), then Fe (II) oxidised will be (a) more by KMnO4 (b) more by K2Cr2O7 (c) equal in both cases (d) none of these 0.5 g of fuming H2S2O7(Oleum) is diluted with water. This solution is completely neutralized by 26.7 ml of 0.4 N NaOH. The percentage of free SO3 in sample is (a) 30.6% (b) 40.6% (c) 20.6% (d) 50% One mole of N2H4 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. (a) –1 (b) –3 (c) +3 (d) +5 25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 ml. The molarity of barium hydroxide solution was (a) 0.35 (b) 0.07 (c) 0.14 (d) 0.28 Einstein Classes, ANSWERS (SINGLE CORRECT CHOICE TYPE) 1. a 11. b 2. d 12. a 3. d 13. d 4. a 14. b 5. a 15. d 6. c 16. d 7. c 17. a 8. c 18. b 9. a 19. b 10. b 20. c 21. 22. 23. 24. 25. Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 c b c b c CS– 9 EXCERCISE BASED ON NEW PATTERN 1. 2. COMPREHENSION TYPE Comprehension-1 A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO2 ceases. The volume of CO2 at 750 mm Hg pressure and at 298 K is measured to be 123.9 ml. A 1.5 g of the same sample requires 150 ml of M/10 HCl for complete neutralisation. The percentage composition of sodium carbonate in the mixture is (a) 42% (b) 18% (c) 53% (d) none The percentage composition of sodium bicarbonate in the mixture is (a) 42% (b) 18% (c) 53% (d) none 7. 8. (A) (B) (C) 3. 4. 5. 6. The amount of HCl used for complete neutralization of 2g of sample is (a) 0.73 (b) 0.42 (c) 0.53 (d) 0.18 Comprehension-2 Reducing sugars are sometimes characterized by a number RCu, which is defined as the number of mg of copper reduced by 1 g of the sugar, in which the half-reaction for the copper is Cu2+ + OH— Cu2O + H2O (unbalanced) It is sometimes more convenient to determine the reducing power of a carbohydrate by an indirect method. In this method 43.2 mg of the carbohydrate was oxidized by an excess of K3Fe(CN)6. The Fe(CN)64— formed in this reaction required 5.29 cm3 of 0.0345 N Ce (SO 4) 2 for reoxidation to Fe(CN)63— [the normality of the cerium(IV) sulfate solution is given with respect to the reduction of Ce4+ to Ce3+]. The number of milimole of Ce(SO 4) 2 used for re-oxidation to Fe(CN)63– is (a) 0.183 (b) 0.0915 (c) 1.83 (d) 9.14 43.2 mg of sugar was reduced by (a) 11.01 mg Cu2+ (b) 11.6 mg Cu2+ 2+ (c) 1.101 mg Cu (d) 1.16 mg Cu2+ The RCu value for the sample is (a) 269 (b) 2.69 (c) 26.9 (d) 0.269 Comprehension-3 An acid solution of a KReO4 sample containing 26.83 mg of combined rhenium was reduced by passage through a column of granulated zinc. The effluent, including was washing from the column, was then titrated with 0.1000 N KMnO4; 11.45 mL of the standard permanganate was required for the reoxidation of all the rhenium to the perrhenate ion, Einstein Classes, (D) (A) (B) (C) (D) 1. 2. ReO 4 —. Assuming that rhenium was the only element reduced. Number of equivalence of KMnO4 used (a) 1.145 (b) 1.145 × 10–3 –3 (c) 11.45 × 10 (d) 3.25 × 10–3 What is the oxidation state to which rhenium was reduced by the zinc column ? (a) 0 (b) –1 (c) –2 (d) –3 MATRIX-MATCH TYPE Matching-1 Column - A 0.1M of MnO4– in acidic medium 0.6 mol of KMnO4 in acidic medium Molarity of pure water (density of water = 1g/ml) 0.083 molar Cr2O72– in acidic medium Column - B (P) oxidised 0.25M C2O42– (Q) oxidised 0.5M Fe2+ (R) oxidised 0.166M FeC2O4 (S) oxidises 1 mol of Fe(C2O4)2 (T) 5.55 × 10 Matching-2 Column-A Column-B Total no. of electrons in (P) 3.01 × 1021 1.6 g methane are No. of sulphate ions (Q) 6.02 × 1023 present in 50mL of 0.1M H2SO4 solution are 500cm3 of 0.2 molar NaCl (R) 6.02 × 1022 3 is added to 100cm of 0.5 molar AgNO3. Thus no. of ions of AgCl formed are The no. of of Fe2+ ions (S) 3.01 × 1022 formed when excess of iron is treated with 5 ml of 0.04 N HCl (T) 6.02 × 1020 MULTIPLE CORRECT CHOICE TYPE Assuming complete dissociation of H 2 SO 4 as (H2SO4 2H+ + SO42–), the number of sulphate ions present in 50 ml of 0.1 M H2SO4 solution are (a) 5 × 10–3 mol (b) 3.01 × 1021 23 (c) 5 × 10 (d) 3.01 × 10–3 mol In the following redox reaction 2MnO4– + 10Cl– + 16H+ 2Mn2+ 5Cl2 + 8H2O. Pick up the correct statements. (a) MnO4– is reduced (b) Cl– is oxidising agent (c) MnO4– is an oxidising agent (d) Cl– is reduced Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS – 10 3. 4. 5. Which of the following represents redox reaction (s) ? (a) Cu + Cu2+ 2Cu+ (b) MnO4– + Mn2+ + OH– MnO2 + H2O (c) Cr2O72– + 2OH– 2CrO42– + H2O (d) 2CrO42– + 2H+ Cr2O72– + H2O Which of the following are redox reactions ? (a) Zn + 2HCl ZnCl2 + H2 (b) Al(OH)3 + 3HCl AlCl3 + 3H2O (c) Disproportionation of Cu+ ions in a given solution (d) Ag+(aq.) + I–(aq.) Ag I (s) Which of the following is disproportionation ? (a) 2Cu+ Cu + Cu2+ (b) 3Cl2 + 6OH– ClO3– + 5Cl– + 3H2O (c) 2H2S + 8O2 2H2O + 3S 1 Cl NaCl 2 2 Oxidation number of C is zero in (a) CHCl3 (b) CH2Cl2 (c) C6H12O6 (d) CO Among the species given below which can act as oxidising as well as reducing agent ? (a) SO2 (b) SO3 (c) H2 O 2 (d) H 2S Which of the following changes involve oxidation ? (a) change of Zn to ZnSO4 by reaction with H2SO4 (b) change of Cl2 to chloride ion (c) change of H2S to S (d) change of sodium sulphite to sodium sulphate. (d) 6. 7. 8. Na + 1. 2. 3. 4. 5. Assertion-Reason Type Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True STATEMENT-1 : 22.4 L of ethane at N.T.P. contains one mole of hydrogen molecules. STATEMENT-2 : One mole of hydrogen molecules at N.T.P. occupies 22.4 L of volume. STATEMENT-1 : The masses of oxygen which combine with fixed mass of nitrogen in N2O, NO, N2O3 bears a simple multiple ratio. STATEMENT-2 : The combination according to law of multiple proportions. STATEMENT-1 : 8.075 × 10–2 kg of Glauber’s salt is dissolved in water to obtain 1 dm3 of solution of density 1077.2 kg m–3 . The molarity of the resultant solution is 0.25 M . STATEMENT-2 : The volume in mL of 0.5 M H2SO4 needed to dissolve 0.5 g of copper (II) carbonate is 24.3 mL STATEMENT-1 : Basicity of an acid can change in different reactions. STATEMENT-2 : As the equivalent weight always remains same. STATEMENT-1 : If equal volume of C M KMnO4 and C M K2Cr2O7 solutions are allowed to oxidised Fe2+ to Fe3+ in acidic medium, then Fe2+ oxidised by KMnO 4 is more then K 2 Cr 2 O 7 of same concentrations. STATEMENT-2 : Number of moles of electrons gained by 1 mole of K2Cr2O7 is more than the 1 mole of KMnO4 (Answers) EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE 1. c 2. a 7. b 8. b 3. a 4. b 5. b 6. a [A-P, Q, R ; B-S ; C-T ; D-P] 2. [A-Q ; B-P ; C-S ; D-R] 3. a, b 4. a, c 5. a, b 6. b, c 3. B 4. C 5. D MATRIX-MATCH TYPE 1. MULTIPLE CORRECT CHOICE TYPE 1. a, b 2. a, c 7. a, c 8. a, c, d ASSERTION-REASON TYPE 1. D 2. Einstein Classes, A Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS– 11 INITIAL STEP EXERCISE (SUBJECTIVE) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 2.68 × 10–3 mol of a solution containing an ion An+ require 1.61 × 10–3 mol of MnO4– for the oxidation of An+ to AO3– in acid medium. What is the value of n? How much 1.00 M HCl should be mixed with what volume of 0.250 M HCl in order to prepare 1.00 L of 0.500 M HCl ? Calculate the final concentration of HNO 3 if 0.20 mol HNO3 is added to a beaker containing 2.0 L of 1.1 M HNO3 and enough pure water is added to give a final volume of 3.0 L. If 40.00 mL of 1.600 M HCl and 60.00 mL of 2.000 M NaOH are mixed, what are the molar concentrations of Na + , Cl – , and OH – in the resulting solution ? Assume a total volume of 100.00 mL. Calculate the molarity of the original H 3PO 4 solution if 20.0 mL of H3PO4 solution is required to completely neutralize 40.0 mL of 0.0500 M Ba(OH)2 solution. What volume of 96.0% H2SO4 solution (density 1.83 g/mL) is required to prepare 2.00 L of 3.00 M H2SO4 solution ? How many mL of 0.5000 M KMnO4 solution will react completely with 20.00 g of K 2C 2O 4.H 2O according to the following equation ? 16H+ + 2MnO4– + 5C2O42– 10CO2 + 2Mn2+ + 8H2O When 50.00 mL of a nitric acid solution was titrated with 0.334 M NaOH, it requires 42.80 mL of the base to achieve the equivalence point. What is the molarity of the nitric acid solution ? What mass of HNO3 was dissolved in 90.00 mL of solution ? The acidic substance in vinegar is acetic acid CH3COOH. When 6.00 g of a certain vinegar was titrated with 0.100 M NaOH, 40.11 mL of base had to be added to reach the equivalence point. What percent by mass of this sample of vinegar is acetic acid ? Calculate the percent of BaO in 29.0 g of a mixture of BaO and CaO which just reacts with 100.8 mL of 6.00 M HCl. BaO + 2HCl BaCl2 + H2O ; CaO + 2HCl CaCl2 + H2O Calculate the normality of each of the following solutions : (a) 7.88 g of HNO3 per L solution (b) 26.5 g of Na2CO3 per L solution (if acidified to form CO2). What volumes of 12.0 N and 3.00 N HCl must be mixed to give 1.00 L of 6.00 N HCl ? One gram of a mixture of CaCO3 and MgCO3 gives 240 ml of CO2 at N.T.P. Calculate the percentage composition of mixture (Ca = 40, Mg = 24, C = 12, O = 16). Einstein Classes, 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. (a) What volume of 5.00 N H2SO4 is required to neutralize a solution containing 2.50 g NaOH ? (b) How many g pure H2SO4 are required ? A 0.250 g sample of a solid acid was dissolved in water and exactly neutralized by 40.0 mL of 0.125 N base. What is the equivalent weight of the acid ? Exactly 50.0 mL of Na2CO3 solution is equivalent to 56.3 mL of 0.102 N HCl in an acid-base neutralization. How many g CaCO 3 would be precipitated if an excess of CaCl2 solution were added to 100 mL of this Na2CO3 solution ? How many cm3 of concentrated sulfuric acid, of density 1.84 g/cm3 and containing 98.0% H2SO4 by weight, should be taken to make 1.00 L of normal solution. Assume complete ionisation. A 40.8 mL sample of an acid is equivalent to 50.0 mL of Na2CO3 solution, 25.0 mL of which is equivalent to 23.8 mL of a 0.102 N HCl. What is the normality of the first acid ? Given the unbalanced equation KMnO4 + KI + (H)2SO4 (K)2SO4 + MnSO4 + I2 + H2 O (a) How many g KMnO4 are needed to make 500 mL 0.250 N solution ? (b) How many g KI are needed to make 25.0 mL 0.360 N solution ? A solution contains 4 g of Na2CO 3 and NaCl in 250 ml. 25 ml of this solution required 50 ml of N/10 HCl for complete neutralization. Calculate % composition of mixture. The density of a 2.0 M solution of acetic acid (MW = 60) in water is 1.02g/mL. Calculate the mole fraction of acetic acid. The density of a 2.03 M solution of acetic acid in water is 1.017 g/mL. Calculate the molality of the solution. Calculate the (a) molar concentration and (b) molality of a sulfuric acid solution of density 1.198 g/cm3, containing 27.0% H2SO4 by weight. A solution contains 57.5 mL ethyl alcohol (C2H5OH) and 600 mL benzene (C6H6). How many g alcohol are in 1000 g benzene ? What is the molality of the solution ? Density of C2H5OH is 0.800 g/mL; of C6H6, 0.900 g/mL. One gram of an alloy of aluminium and magnesium when treated with excess of dil. HCl froms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at 0 0 C has a volume of 1.20 litres at 0.92 atm. pressure. Calculate the composition of the alloy. [H = 1, Mg = 24, Al = 27]. Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS – 12 26. 27. 28. 29. 30. A gas mixture of 3.0 litres of propane and butane on complete combustion at 250C produced 10 litres of CO 2. Find out the composition of the gas mixture. In an Industrial process for producing acetic acid, oxygen gas is bubbled into acetaldehyde CH3CHO, containing manganese (II) acetate (catalyst) under pressure at 600C. 2CH3CHO(l) + O2(g) 2CH3COOH(l) In a laboratory test of this reaction, 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel. (a) How many grams of acetic acid can be produced by this reaction from these amounts of reactants ? (b) How many grams of the excess reactant remaining after the reaction is complete ? The hourly energy requirements of an astronaut can be satisfied by the energy released when 34 grams of sucrose are “burned” in his body. How many grams of oxygen would he needed to be carried in spaces capsule to meet his requirement for one day ? 1.84 g of a mixture of CaCO3 and MgCO3 are heated strongly till no further loss of weight takes place. The residue weighs 0.96 g. Find the percentage composition of the mixture. (Mg = 24, Ca = 40, C = 12, O = 16) 1.0 g of a mixture of Potassium chloride and Potassium iodide dissolved in water and precipitated with Silver nitrate, gave 1.618 g of silver halides. Calculate the percentage of each in the mixture (K = 39, Cl = 35.5, Ag = 108, I = 127). 31. 32. 33. 20% surface sites have adsorbed N2. On heating N2 gas evolved from sites and were collected at 0.001 atm and 298 K in a container of volume is 2.46 cm 3 . Density of surface sites is 6.023 × 1014/cm2 and surface area is 1000 cm2, find out the number of surface sites occupied per molecule of N2. A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm3/g. If the virus is considered to be a single particle, find its molar mass. (a) Calculate the amount of calcium oxide required when it reacts with 852 g of P4O10. (b) 34. 35. 6CaO P4O 10 2Ca 3 ( PO 4 ) 2 [At. wt. : Ca = 40, P = 31, O = 16] How many milliliters of 0.5 M H2SO4 are needed to dissolve 0.5 g of copper (II) carbonate ? CuCO 3 H 2 SO 4 CuSO 4 H 2 O CO 2 1.5 gm of an impure sample of (NH4)2SO4 was boiled with excess of Caustic soda solution in a Kjeldahl’s flask and the ammonia evolved was passed into 200 ml of semi-normal H2SO4 solution. Thepartially neutralized acid was made to 500 ml with distilled water. 25ml of this diluted acid required 40.8 ml of decinormal Caustic soda for complete neutralization. Calculate the percentage purity of Ammonium sulphate. A sample of a metal (M) carbonate was neutralized by 10 ml of 0.1 N-hydrochloric acid and the resulting chloride gave 0.0517 g of phosphate [M3(PO4)2]. Calculate the eq. wt. of M. (the formula of Phosphoric acid is H3PO4). Determine the atomic weight of M ? FINAL STEP EXERCISE (SUBJECTIVE) 1. 2. One litre of a mixture of O2 and O3 at STP was allowed to react with excess of acidified solution of KI. The iodine liberated require 40mL of M/10 sodium thiosulfate solution for titration. What is the mass percentage of ozone in mixture. 3. A mixture of FeO and Fe3O4 when heated in air to a constant weight gains 5 % in its weight. Find the composition of initial mixture. (Fe = 56, O = 16) 4. A 10.0 g sample of “gas liquor” is boiled with an excess of NaOH, and the resulting ammonia is KI passed into 60 cm3 if 0.90 N H2SO4. Exactly 10.0 2 O 3 I 2 O 2 O cm3 of 0.40 N NaOH is required to neutralize the What volume of 3.00 M HNO3 can react completely excess sulfuric acid (not neutralized by the NH3). with 15.0 g of a brass (90.0% Cu, 10.0% Zn) Determine the percent ammonia in the “gas liquor” according to the following equations ? examined. Cu + 4H+(aq) + 2NO3–(aq) 2NO2(g) + Cu2+ + 2H2O 5. A 10.0 mL portion of (NH4)2SO4 solution was treated 4Zn + 10H+(aq) + NO3–(aq) NH2+ + 4Zn2+ + 3H2O with excess NaOH. The NH 3 gas evolved was What volume of NO2 gas at 250C and 1.00 atm absorbed in 50.00 mL of 0.1000 N HCl. To pressure would be produced ? neutralize the remaining HCl, 21.50 mL of 0.0980 N NaOH was required. What is the molar concentration of the (NH4) 2SO 4 ? How many g (NH4)2 SO4 are in 1 L solution ? Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS– 13 6. 7. 8. 9. 10. 11. 12. 13. A solution contained the mixture of Na2CO3 and NaCl. 25 ml of the solution required 20.4 ml of 0.095 N-HCl to convert the carbonate into the chloride. 25 ml of the later solution (containing chloride converted from carbonate) required 38.76 ml of N/10-AgNO 3 to ppt, the chloride completely. Calculate the strength of Na2CO3 and NaCl in one litre of the original solution. 10 g of a mixture of Cu2S and CuS was treated with 200 mL of 0.75 M MnO 4 – in acidic solution producting SO2, Cu2+ and Mn2+ . SO2 was bubbled off and excess of MnO4– was titrated with 175 mL of 1.00 M Fe2+. Calculate the CuS in the mixture. Determine the volume of dilute nitric acid (density 1.11 g/mL, 19.0% HNO3 by weight) that can be prepared by diluting with water 50 mL of the concentrated acid (density 1.42 g/mL, 69.8% HNO3 by weight). Calculate the molar concentrations and molalities of the concentrated and dilute acids. A 1.00 g sample of KClO3 was heated under such conditions that a part of it decomposed according to the equation (i) 2KClO3 = 2KCl + 3O2 and the remaining underwent change according to the equation (ii) 4KClO3 = 2KClO4 + KCl If the amount of oxygen evolved was 146.8 ml at S.T.P., calculate percentage by weight of KClO4 in the residue (K = 39.1, Cl = 35.5) To a 25 ml H 2 O 2 solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 ml of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H2O2 solution. A solution contains Na2CO3 and NaHCO3. 20 cm3 of this solution requires 5.0 cm3 of 0.1 M H2SO4 solution for neutralization using phenophthalein as the indicator. Methyl orange is then added and further 5.0 cm3 of 0.2 M H2SO4 was required. Calculate the masses of Na2CO3 and NaHCO3 in 1 L of this solution. A 1.2 g of a mixture containing H2C2O4.2H2O and KHC2O4.H 2O and impurities of a neutral salt, consumed 18.9 ml of 0.5 N NaOH for neutralization. On titrating with KMnO4 solution 0.4 g of the same substance needed 21.55 ml of 0.25 N KMnO4. Calculate the percentage composition of the substance. A mixture of FeO and Fe 2 O 3 is reacted with acidified KMnO4 solution having a concentration of 0.2278 M, 100 ml of which was used. The solution then was treated with Zn dust which converted the Fe3+ of the solution to Fe2+ . The Fe2+ required 1000 ml of 0.13 M K2Cr2O7 solution. Find the mass % of FeO and Fe2O3. Einstein Classes, 14. 15. 16. 17. 18. 19. 20. 10 g of a mixture of anhydrous nitrates of two metals A and B were heated to a constant weight and gave 5.531 g of a mixture of the corresponding oxides. The equivalent weights of A and B are 103.6 and 31.8 respectively. What was the percentage of A in the mixture ? (N = 14, O = 16). Equal weights of mercury and iodine are allowed to react completely to form a mixture of mercurous and mercuric iodides. Calculate the ratio of the weights of mercurous and mercuric iodides formed. (I = 127, Hg = 201). 5 ml of 8 N nitric acid, 4.8 ml of 5 N HCl acid and a certain volume of 17 M sulphuric acid are mixed together and made up to 2 litre. Thirty ml of this acid mixture exactly neutralise 42.9 ml of sodium carbonate solution containing one gram of Na2CO3. 10H2O in 100 ml of water. Calculate the mass in gram of the sulphate ions in solution. 200 ml of a solution of mixture of NaOH and Na2CO3 was first titrated with phenolphthalein and N/10 HCl. 17.5 ml of HCl was required for the end point. After this methyl orange was added and 2.5 ml of same HCl was again required for next end point. Find out amounts of NaOH and Na2CO3 in mixture. 50 ml of solution, containing 1g each of Na2CO3, NaHCO3 and NaOH, was titrated with N HCl. What will be the titre readings if (a) only phenolphthalien is used as indicator ? (b) only methyl orange is used as indicator from the very beginning ? (c) methyl orange is added after the first end point with phenolpthalein ? A mixture of calcium carbonate and sodium chloride weighing 3.20 g was added to 100 ml of 1.02 N HCl. After the reaction had ceased the liquid was filtered, the residue washed and the filtrate was made up to 200 ml. 20 ml of this dilute solution required 25 ml of N/5 NaOH for neutralization. Calculate the % of CaCO3 in the mixture. (Ca = 40). A solution contains a mixture of sulphuric acid and oxalic acid, 25 ml of the solution require 35.5 ml of N/10-NaOH for neutralization and 23.45 ml of N/10 - KMnO4 for oxidation. Calculate (a) the Normality of the solution with regard to sulphuric acid and oxalic acid (b) the number of g of each of these substances present in one litre of the solution. Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CS – 14 ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE) 1. 4. 6. 9. 11. 13. 14. 15. 18. 20. 22. 23. n=2 2. 333 mL of 1m HCl 0.56 M OH–, 0.640 Cl–, 1.2MNa+ 335 mL 7. 86.9 mL 4.01 % 10. 65.5 % BaO (a) 0.1251 N (b) 0.5 N CaCO3 = 62.5 %, MgCO3 = 37.5% (a) 12.5 mL (b) 3.07 gm 50g/eq 16. 0.575 gm 0.119 N 19. (a) 3.95 gm Na2CO3 = 66.25%, NaCl = 33.75% 2.27 m (a) 3.30 m (b) 3.78 m 3. 5. 8. 0.80 m 0.0667 M 1.62 gm, 0.286 M 12. .33 L, .667 L 17. (b) 21. 27.2 cm3 1.49 gm 0.038 24. 0.54 kg C6H6, 1.85 m 25. 26. Al% = 54.87%, Mg% = 45.13% 66.66% propane, 33.33 % butane 27. (a) (b) 28. 29. 916.2 gm CaCO3 = 54.35%, MgCO3 = 45.65 % 30. 31. 34. 2 80.96 KCl = 39.6 % KI = 60.4% (a) 1008 g (b) 8.097 mL 32. 35. 70.96 × 106 g/mol 20.03, 40.06 33. 27.24 gm 2.73 gm ANSWERS SUBJECTIVE (FINAL STEP EXERCISE) 1. 6.57 % 3. % of FeO = 20.25% % of Fe3O4 = 79.75 % 2. VHNO3 302L, VNO 2 10.4L 4. 8.5% 5. 0.145 M, 19.1 g/L 6. Na2CO3 = 4.109 gm, NaCl = 4.535 gm 7. CuS = 57.4 % 8. 15.7 M, 3.35 M, 36.8 m, 3.73 m 9. 49.83% 10. 1.344 12. KHC2O4.H2O = 80.9 %, H2C2O4.2H2O = 14.7 % 13. % FeO = 13.34 %, % Fe2O3 = 86.66 % 15. 0.513 : 1 11. Na2CO3 = 5.3 gm, NaHCO3 = 4.2 gm 16. 14. 6.528 gm 17. 32.28 WNaOH = 0.06gm WNa 2CO3 0.0265 gm 18. (a) 19. 81.25% 34.4 mL (b) 55.8 mL (c) 21.3 mL 20. H2SO4 = 0.0482N, 2.36 g/L H2C2O4 = 0.0938N, 4.225 g/L Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111