SUBJECTIVE. Potassium selenate is isomorphous to K2SO4 and thus its molecular formula is

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1 Solved Problems SP-ST-CH-1 SUBJECTIVE Problem 1: P and Q are two elements which forms P Q and PQ. If 0.15 mole of P Q weighs 15.9g and 0.15 mole of PQ weighs 9.g, what are atomic weights of P and Q? Solution: Problem : Solution: Problem : Solution: Let atomic weight of P and Q are a and b respectively Molecular weight of PQ = a + b and Molecular weight of PQ = a + b Now given that 0.15 mole of PQ weigh 15.9g 15.9 wt. ( a b) mole 0.15 mol. wt. 9. Similarly, ( a b) 0.15 Solving these two equations b = 18 a = 6 Potassium selenate is isomorphous with potassium sulphate and contains 45.5% selenium by weight. Calculate the atomic weight of selenium. Also report the equivalent weight of potassium selenate. Potassium selenate is isomorphous to KSO4 and thus its molecular formula is KSeO4. Now molecular weight of KSeO4 = (9 + a ) = (14 + a) where a is atomic weight of Se (14 + a)g KSeO4 has Se = ag a g KSeO4 has Se = 14 a % of Se = 45.5 a 100 = a a = 118. Mol. wt Also equivalent of KSeO4 = = 10.1 A sample of H SO 4 (density 1.787g ml 1 ) is labeled as 86% by weight. What is molarity of acid? What volume of acid has to be used to make 1 litre of 0.m H SO 4? HSO4 is 86% by weight Weight of HSO4 = 86g Weight of solution = 100g

2 SP-ST-CH- Problem 4: Solution: Volume of solution = ml litre M H SO Let V ml of this HSO4 areused to prepare 1 litre of 0. M HSO4 mm of conc. HSO4 = mm of dilute HSO4 V = V = 1.75 ml The molecular mass of an organic acid was determined by the study of its barium salt 4.90g of salt was quantitatively converted to free acid by the reaction with 1.64 ml of M H SO 4. The barium salt was found to have two mole of water of hydration per Ba + ion and the acid is mono basic. What is molecular weight of anhydrous acid? Meq. of barium salt = Meq. of acid = M / Molecular weight of salt = Molecular weight of anion = Molecular weight of acid = = 1.1 Problem 5: 5 ml of a solution of Na CO having a specific gravity of 1.5g ml 1 required.9 ml of a solution of HCl containing 109.5gm of the acid per litre for complete neutralization. Calculate the volume of 0.84 N H SO 4 that will be completely neutralized by 15g of Na CO solution Solution: N HCl Since NaCO is completely neutralized by HCl Meq. of NaCO = Meq. of HCl N 5 =.9 N Na. 948 CO Now NaCO fresh solution reacts with HSO4 Volume of NaCO solution = ml 1.5 Meq. of HSO4 = Meq. of NaCO 0.84 V = Volume of HSO4= 470 ml Problem 6: Borax in water gives: O 7H O 4H BO B4 7 OH How many gram of borax (Na B 4 O 7.10H O) are required to? a) Prepare 50 ml of 0. M solution b) neutralize 5 ml of M of HCl and H SO 4 separately

3 SP-ST-CH- Normality Solution: Molarity - No. of replaceable OH N = M Thus Meq. of borax in solution = = 0 w M / w w =.8g 8 / For neutralization of HCl Meq. of HCl = Meq. of borax w / Weight of borax = 0.095g For neutralization of HSO4 Meq. of borax = Meq. of HSO4 w / Weight of borax = 1.847g Problem 7: A mixture containing As O and As O 5 required 0.10 ml of 0.05N iodine for titration. The resulting solution is then acidified and excess of KI was added. The liberated iodine required 1.111g hypo (Na S O 5H O) for complete reaction. Calculate mass mixture. The reactions are: As O + I + H O As O 5 + 4H + + 4I As O 5 + 4H + + 4I As O + I + H O Solution: Meq. of I used = = Let Meq. of AsO and Meq. of AsO5 in mixture be a and b respectively. On addition of I to mixture, As is converted to As. Meq. of AsO5 = Meq. of I to mixture used = Meq. of As 5+ formed. or a = (1) After the reaction with I, mixture contains all the arsenic in +5 oxidation state which is then titrated using KI + hypo. Thus, Meq. of AsO as As +5 + Meq. of AsO5 as As +5 = Meq. of liberated I = Meq. of hypo used 1.11 or a b or a + b = By equations (1) and (), b = =.476 Wt. of AsO = Meq. Eq.Wt g and Wt. of AsO5 = g Wt. of mixture = = 0.496g 5

4 SP-ST-CH-4 Problem 8: Chile salt peter, a source of NaNO also contains NaIO. The NaIO can be used as source of iodine, produced in the following reactions. IO + HSO I + H + +SO 4 (1) 5I + IO + 6H + I (g) + H O () One litre of chile salt peter solution containing 5.80g NaIO is treated with stoichiometric quantity of NaHSO. Now an additional amount of same solution is added to reaction mixture to bring about the second reaction. How many grams of NaHSO are required in step I and what additional volume of chile salt peter must be added in step II to bring in complete conversion of I to I? 5.8 Solution: Meq. of NaHSO = Meq. of NaIO = N V = / 6 [Et. wt. of NaI = M/6 because I e I ] Meq. of NaHSO = w NaHSO = 9.14g 000 Also Meq. of I formed in I step using valence factor 6 = In II step valence factor of I is 1 and valence factor IO is 5 Thus, Meq. of I formed using valence factor 1 = Also Meq. of NaIO used in step II = N V = or V V NaIO 198/ 5 6 = 00 ml Problem 9: Solution: For estimating ozone in the air, a certain volume of air is passed through an acidified or neutral KI solution when oxygen is evolved and iodide is oxidized to give iodine. When such a solution is acidified, free iodine is evolved which can be titrated with standard Na S O solution. In an experiment, 10 litre of air at 1 atm and 7 C were passed through an alkaline KI solution, at the end, the iodine entrapped in a solution on titration as above required 1.5 ml of 0.01 N Na S O solution. Calculate volume % of O in sample. The reactions are HO + KI + O KOH + I + O Also e + I I And S + S4 +5/ + e Meq. of I = Meq. of NaSO = = or mm of I = =

5 mm of O = mm of I = ( Mole ratio of O : I :: 1: 1) PO = atm 10 Volume % of O = = % SP-ST-CH-5 Problem 10: Solution: Problem 11: 1.5g of brass containing Cu and Zn reacts with M HNO solution, the following reactions take place. Cu + HNO Cu + + NO (g) + H O Zn + H + + NO NH Zn + + H O The liberated NO (g) was found to be 1.04 litre at 5 C and one atm. a) Calculate the percentage composition of brass. b) How many ml of M HNO will be required for completely reacting 1g of brass? (a) Cu 0 Cu + + e N 5+ + e N 4+ Eq. of Cu = Eq. of NO w 6.6 / = 10.4 mole of NO wcu = 1.5g wzn = = 0.15g Eq. of NO % of Cu = = 90% and % of Zn = = 10% (b) Thus, 1 g brass contains 0.9g Cu and 0.1g Zn Meq. of HNO = Meq. of Zn and Meq. of HNO = Meq. of Cu or 8V or V / 66 / V1 = 0.18mL V = 9.4 ml 5 ( N 8e N ) Total volume of HNO used = = ml PV RT 1.49 g of a sample of pure BaCO and impure CaCO containing some CaO was treated with dil.hcl and it evolved 168 ml of CO at NTP. From this solution, BaCrO 4 was precipitated, filtered and washed. The precipitate was dissolved in dilute sulphuric acid and diluted to 100 ml. 10 ml of this solution, when treated with KI solution, liberated iodine which required exactly 0 ml of 0.05N Na S O. Calculate the percentage of CaO in the sample. 168 Solution: n n = n CaCO BaCO CO = 400 = (1) BaCO BaCrO4 H BaCrO7 KI I + NaSO

6 SP-ST-CH eq. of NaSO = eq. of I = eq of BaCrO7 = 10 = Moles of BaCrO7 = 10 6 Moles of BaCrO4 = (1 10 ) 6 1 Moles of BaCO = 10 = () Weight of BaCO = gm From equation (1) and () we get n = CaCO weight of CaCO = = g weight of CaO = = % of CaO = 100 = 14.09% 1.49 Problem 1: In a quality control analysis for sulphur impurity 5.6g steel sample was burnt in a stream of oxygen and sulphur was converted into SO gas. The SO was then oxidized to sulphate by using H O solution to which had been added 0 ml of 0.04M NaOH. The equation for reaction is: SO (g) + H O (aq) + OH (aq) SO 4 (aq) + H O (l).48 ml of 0.04M HCl was required to neutralize the base remaining after oxidation reaction. Calculate % of sulphur in given sample. Solution: Meq. of alkali added = = 1. Meq. of alkali left = = 0.54 Meq. of alkali for SO and HO = = Weight of alkali used = = g NaOH reacts with 64g SO g NaOH reacts = = 0.01g SO 80 Now 64g SO required = g S g SO required = = g 64 % of S = = % Problem 1: Solution: A granulated sample of aircraft alloy (Al, Mg, Cu) weighing 8.7g was first treated with alkali and then with very dilute HCl, leaving a residue. The residue after alkali boiling weighed.10g and the acid insoluble residue weighed 0.69g. What is the composition of the alloy? Let Al, Mg and Cu be a, b and c g respectively. HO Al + NaOH NaAlO + H

7 SP-ST-CH-7 Mg + HCl MgCl + H Cu + HCl No reaction i.e., only Al reacts with NaOH and then only Mg reacts with HCl a + b + c = 8.7 b + c =.10 (Residue left after alkali treatment) c = 0.69 (Residue left after acid treatment) b = 6.6g % of Al = = 75.9 % of Mg = = 16. % of Cu = = 7.9 Problem 14: Solution: 5 ml from a stock solution containing NaHCO and Na CO was diluted to 50 ml with CO free distilled water. 5 ml of the diluted solution when titrated with 0.1 M HCl required 8 ml when phenolphthalein was used as an indicator. When 0 ml of diluted solution was titrated with same acid required 18 ml when methyl orange was used as an indicator. Calculate concentration of NaHCO in the stock solution in gm/litre and in mole/litre. To 100 ml of the stock solution how much NaOH should be added so that all bicarbonate will be converted into carbonate? When phenolphthalein used, NaHCO remains unaffected and NaCO will be converted into NaHCO NaCO + HCl NaHCO + NaCl 1 eq of NaCO = eq. of HCl 1 w = w = gm in 5 ml = 40.7 gm / litre = 0.8 mole / litre When methyl orange is used NaHCO and NaCO both will be converted into CO Eq. (NaCO) + eq(nahco) = eq (HCl) W = W + = For NaHCO W = gm in 50 ml = 6. gm /litre = 0.1 M eq. of bicarbonate = eq of NaOH W = 5 40 W = 1.48 gm Problem 15: One litre of a mixture of O and O at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 ml of

8 SP-ST-CH-8 M/10 sodium thiosulphate solution for titration. What is the weight per cent of ozone in the mixture? Ultraviolet radiation of wavelength 00 nm can decompose ozone. Assuming that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture? Solution: O KI H O KOH I O I Na SO Na S4O6 NaI 1 Milli mole of O = milli mole of I = mm of Na S O (mm M V in ml ) 1 1 = 40 mm 0.00mole 10 Total milli mole of O and O in mixture are calculated from PV = nrt 1 1 n n = mole Mole of O = = 0.04 Now weight of O = 0.04 g 1.44g Weight of O = g 0.096g Problem 16: Solution: Problem 17: % of O = = 6.7% No. of photon or molecules of ozone = = g of KClO are dissolved in conc. HCl and the solution was boiled. Chlorine gas evolved in the reaction was then passed through a solution of KI and liberated iodine was titrated with 100 ml of hypo. 1. ml of same hypo solution required 4.6 ml of 0.5 N iodine for complete neutralization. Calculate % purity of KClO sample. KClO + 1HCl KCl + 6HO + 6Cl Cl + KI KCl + I Also Meq. of I = Meq. of Hypo = NHypo ; NHypo Also mm of Cl mm of I 50 mm of Cl Also mm of KClO 6 6 w 50 Wt milli-mole (mm)= 1.5 mol.wt. w KClO % of KClO %.48 P and Q are two elements which forms PQ and PQ. If 0.15 mole of PQ weights 15.9g and 0.15 mole of PQ weights 9.g, what are atomic weights of P and Q?

9 SP-ST-CH-9 Solution: Problem 18: Solution: Let atomic weight of P and Q are a and b respectively Molecular weight of PQ = a + b and Molecular weight of PQ = a + b Now given that 0.15 mole of PQ weigh 15.9g 15.9 wt. ( a b) mole 0.15 mol. wt. 9. Similarly, ( a b) 0.15 Solving these two equations b = 18 a = 6 5 ml of a solution containing Fe + and Fe + sulphate acidified with HSO4 is reduced by g of metallic zinc. The solution required 4.5 ml of N/10 solution of KCrO7 for oxidation. Before reduction with zinc, 5 ml of the same solution required.45 ml of same KCrO7 solution. Calculate the strength of FeSO4 and Fe(SO4) in g/litre of solution. Redox changes are Zn dust+hso4 Case I: Fe Fe (i.e. no change) Zn dust + H SO 4 e Fe Fe Zn dust is used as reducing agent and thus, Zn Zn e Let a meq. of Fe + and b Meq. of Fe + be present in 5 ml solution. In case I, after reduction with Zn. Meq. of Fe + + Meq. of Fe + from Fe + = a + b Now these are oxidized by KCrO7 Total meq. of Fe + = Meq. of KCrO7 1 a b a + b =.45 (1) Case II: If reduction is not made, the solution contains Fe + and Fe + of which only Fe + are oxidized by KCrO7. Meq. of Fe + = Meq. of KCrO7 a = a =.45 By equation (1) b = = 1.18 Meq. of FeSO4 = a =.45 Meq. of Fe(SO4 =) = b = 1.18 Meq. of FeSO4 a.45 (in 5 ml) (in 5 ml) w w M/ M/1 M. wt. of Fe (SO 4) 400 M. wt. of FeSO 15 Wt. of Fe (SO ) in 5 ml=0.6g 4 4

10 SP-ST-CH-10 Wt. of FeSO 4 in 5 ml = 0.41g Strength of FeSO 1.64g / litre 4 Strength of Fe (SO 4) 9.45g / litre Problem 19: Solution: 50 ml solution of HO was treated with excess KI (s) and the solution was acidified with acetic acid. The liberated iodine required 40 ml 0.5 M NaSO solution for the end point using starch as indicator. Find the molarity and volume strength of the HO solution. Milli equivalent of HO = milli equivalent of I = milli equivalent of NaSO Milli equivalent of HO = milli equivalent of NaSO 50N N N M n factor 0.4 M (molarity) 0. Volume strength = 5.6 N = =.4 Problem 0: Solution:.0g sample of KMnO4 (MW = 158) containing some inert materials was dissolved in water acidified with HSO4 resulting solution was treated with 6 ml 0.5 M oxalic acid solution. The excess of oxalic acid was back titrated with 0 ml 0.1 M KCrO7. Calculate percent purity of KMnO4 sample KMnO MnSO, H C O CO n-factor = 7 = 5 n-factor = 6 4 = K Cr O Cr (SO ) n-factor = (6 ) = 6 So, 0.5 M oxalic aid = 0.5 i.e. 1 N oxalic acid 0.1 M KCrO7 = i.e. 0.6N KCrO7 No. of milli equiv. of KMnO4 in the sample = No. of milli equivalents of oxalic acid reacted with it = (no. of milli equivalents of oxalic acid taken) (no. of milli equivalents of oxalic acid remained unreacted) = (no. of milli equivalents of oxalic aid taken) (no. of milli equivalents of KCrO7 consumed. = = 50 No. of milli equivalents KMnO4 = Weight of KMnO4 = = 1.58g Hence i.e. 79%

11 Assignments (New Pattern) SP-ST-CH-11 SECTION I Single Choice Questions 1. The number of moles of CaCl needed to react with excess of AgNO to produce 4.1 gram of AgCl. (a) 0.00 (b) (c) (d) How many moles of electron is needed for the reduction of each mole of Cr in the reaction,. CrO5 H SO 4 Cr (SO 4 ) H O O (a) 4 (b) (c) 5 (d) g of a metal were dissolved in HNO to prepare its nitrate. The nitrate on strong heating gives g oxide. The equivalent weight of metal is (a) 16 (b) (c) 48 (d) 1 4. A metal oxide has 40% oxygen. The equivalent weight of the metal is (a) 1 (b) 16 (c) 4 (d) The density of 1 M solution of NaCl is g/ml. The molality of the solution is: (a) (b) 1.00 (c) 0.10 (d) Which of the following gives the molarity of a 17% solution of NaOAc in HO? Given the density is 1.09 g/ml (a) M (b) 0.07M (c).07m (d).6m 7. A 15 volume sample of an HO solution is equivalent to (a) 5.0 N (b) 1.77N (c).68n (d) 7.50 N ml of N/5 NaOH will neutralize (a) g of HBO (c) 1.68g of HBO (b) g of HBO (d) 0.009g of HBO 9. The relation between molarity (M) and molality (m) is given by ( = density of solution, M 1 = molecular weight of solute) 1000M 1000M (a) m (b) m 1000 M MM MM1 1000M (c) m (d) m 1000 MM MM 1

12 SP-ST-CH mol of HCl and 0.1 mol of barium chloride were dissolved in water to produce a 500 ml solution. The molarity of the Cl is (a) 0.06M (b) 0.09M (c) 0.1M (d) 0.80M 11. To prepare a 0.5 M KCl solution from 100 ml of 0.40M KCl we need to add (a) 0.75g of KCl (b) 0 ml of water (b) 0.1 mol of KCl (d) 0. mol of KCl 1. What would be the normality of a 0.1 M KCrO7 solution used as a precipitating agent for Pb +? (a) 0.1 N (b) 0.6N (c) 0.4N (d) 0.N g of a colourless oxide of nitrogen occupies 4 ml at STP. The compound is: (a) NO (b) NO (c) NO5 (d) NO 14. HPO4 is a tribasic acid and one of its salt is NaHPO4. What volume of 1 M NaOH would be added to 1 g NaH PO4 (molecular weight of 10) to exactly convert it into NaPO4. (a) 100 ml (b) 00 ml (c) 00 ml (d) 80 ml g of a metallic chloride contains 5.5g of chlorine. The equivalent weight of the metal is (a) 19.5 (b) 5.5 (c) 8.9 (d) It takes equivalent of KOH to neutralize 0.154g HXO4. The number of neutrons in is (a) 16 (b) 8 (c) 7 (d) 17. How many grams of sodium bicarbonate are required to neutralize 10.0 ml of 0.90 M vinegar? (a) 8.4g (b) 1.5g (c) 0.75g (d) 1.07g 18. A sample of hard water contains 44 ppm of HCO ions. What is the minimum mass of CaO required to remove HCO ions completely from 1 kg of such water sample (a) 56 mg (b) 11 mg (c) 168 mg (d) 44 mg ml of each of 0.5 N NaOH, N/5 HCl and N/10 HSO4 are mixed together. The resulting solution will be (a) Acidic (b) Neutral (c) Alkaline (d) None

13 SP-ST-CH-1 0. The chloride of a metal (M) contains 65.5% of chlorine. 100 ml of the vapour of the chloride of the metal at STP weight 0.7g. the molecular formula of the metal chloride is (a) MCl (b) MCl (c) MCl (d) MCl4 1. If 0.5 mole of BaCl is mixed with 0. mole of NaPO4 the maximum number of moles of Ba(PO4) that can be formed is (a) 0.7 (b) 0.5 (c) 0. (d) 0.1. The percent loss in weight after heating a pure sample of potassium chlorate (Molecular weight = 1.5) will be (a) 1.5 (b) 4.50 (c) 9.18 (d) 49. For the reaction MnO4 CO4 H Mn CO HO the correct coefficients of the reactants for the balanced reaction are MnO 4 C O 4 H (a) 5 16 (b) 16 5 (c) 5 16 (d) In an experiment, 50 ml of 0.1 M solution of a salt reacted with 5 ml of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion: SO(aq) HO SO4(aq) H(aq) If the oxidation number of the metal in the salt was, what would be the new oxidation number of the metal? (a) 0 (b) 1 (c) (d) 4 5. The chloride of a metal contains 71% chlorine by weight and the vapour density of it is 50. The atomic weight of the metal will be (a) 9 (b) 58 (c) 5.5 (d) 71 SECTION II May be more than one choice 1. The number of moles of Cr O 7 needed to oxidize 0.16 equivalent of reaction. NH 5 CrO 7 N Cr HO is (a) 0.16 (b) 0.7 (c) (d) 0.07 N H 5 by the. The hydrated salt NaSO4nHO undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be (a) 5 (b) (c) 7 (d) 10

14 SP-ST-CH litres of HS and.4 litres of SO both at STP are mixed together. The amount of sulphur precipitated as a result of chemical reaction is: (a) 16g (b) g (c) 48g (d) 96g 4. What volume of 0.1 M HSO4 will be required to produce 17.0g of HS by the reaction 5HSO4 + 8NaI 4NaSO4 + 4I + HS + 4HO? (a) 70.0 L (b) 50.0L (c) 5.0 L (d) 5.0 L ml of 10% NaOH (w/v) is added to 15 ml of 10% HCl (w/v). The resultant solution becomes (a) alkaline (b) strongly alkaline (c) acidic (d) neutral 6. How many grams of copper will be replaced in L of a 1.50 M CuSO4 solution if the later is made to react with 7.0g of aluminium? (Cu = 6.5, Al = 7.0) (a) g (b) 95.5g (c) 1.75g (d) 10.65g 7. A sample of HSO4 (density 1.8g ml 1 ) is 90% by weight. What is the volume of the acid that has to be used to make 1 L of 0. M HSO4? (a) 16 ml (b) 10 ml (c) 1 ml (d) 18 ml (d) 100 ml 8. An aqueous solution of urea containing 18g urea in 1500cm of solution has a density of 1.50 g/cm. If the molecular weight of urea is 60, then the molality of solution is (a) 0. (b) 0.19 (c) (d) For preparing M/10 solution of HSO4 in one litre we need HSO4 (a) 9.8g (b) 49.0g (c) 4.8g (d) 0.09g 10. The normality of 1% (weight/volume) HPO4 is nearly (a) 0.0 (b) 0. (c) 0.1 (d) Number of molecules of oxalic acid in 100 ml of 0.0 N oxalic acid solution are (a) (b) (c) (d) molal NaOH solution has a density of g/ml. The molarity of the solution is (a).97 (b).05 (c).64 (d) A mixture of magnesium chloride and magnesium sulphate is known to contain 0.6 moles of chloride ions and 0. moles of sulphate ions. The number of moles of magnesium ions present is (a) 0.4 (b) 0.5 (c) 0.8 (d) 1.0

15 SP-ST-CH The reaction between aluminium metal and dilute hydrochloric acid produces H(g) and Al + ions. The molar ratio of aluminium used to hydrogen produced is (a) 1: (b) :1 (c) : (d) : 15. Element X reacts with oxygen to produce a pure sample of XO. In an experiment it is found that 1.00g of X produces 1.16g of XO. Calculate the atomic weight of X. Given: atomic weight of oxygen, 16.0 g mol 1. (a) 67 (b) 100. (c) 15 (d) In an aqueous solution of barium nitrate, the [NO ] is 0.80M. This solution is labelled as (a) N Ba(NO) (b) M Ba(NO) (b) M Ba(NO) (d) MNO mole of HPO and 0.01 moles of HPO are titrated with V ml of 0.1 M NaOH solution in presence of phenolphthalein indicator, V will be (a) 00 ml (b) 00 ml (c) 400 ml (d) 500 ml ml of a solution of HOOCCOONa.HO is completely neutralized by 10 ml. of decinormal solution of caustic soda. In a separate titration, 10ml of 0.M kmno4 solution in acidic medium be required for complete oxidation of x ml of above HOOCCOONa.HO solution. What is the value of x? (a) 100 ml (b) 50 ml (c) 10 ml (d) 0 ml ml 0.5 M Br solution upon being made alkaline undergoes complete disproportionation into Br and BrO. The resulting solution requires 45 ml As(III) solution to reduce BrO to Br. Given that As(III) is oxidised to As(V), what is the molarity of As(III) solution? (a) 0. (b) 0.1 (c) 0.4 (d) A sample of hard water contains 19 ppm of SO 4 ions and 05 ppm of HCO ions with Ca + with as the only cations. The concentration in ppm of Ca + ions (Atomic mass = 40) in the hard water is: (a) 100 (b) 180 (c) 140 (d) Hydroxyl amine reduces iron (III) according to following equation NHOH Fe (SO 4) N (g) HO FeSO4 HSO4 which statement is correct (a) n-factor for Hydroxyl amines is 1 (b) equivalent weight of Fe(SO4) is M/ (c) 6 meq of Fe(SO4) is contained in millimoles of ferric sulphate (d) all of these.

16 SP-ST-CH g of KMnO4 is dissolved in water and the solution is made upto 1 litre. An unknown salt containing 6.88g Fe + ion was dissolved in water and solution was made upto 100 ml. It was found that 0 ml of salt solution decolourised 7.5 ml of the above permanganate solution. Which of the following statements are correct? (a) Normality of ferrous ion = 0.15 N (b) Actual strength of ferrous ion 68.8 g liter 1 (c) Percentage of ferrous ion is (d) Normality of ferrous ion is 0.1 N. 1 mol of HSO4 will exactly neutralize (a) mol of ammonia (b) 1 mol of Ca(OH) (c) 0.5 mol of Ba(OH) (d) mol of NaOH 4. During the titration of a mixture of NaOH, NaCO and inert substances against HCl. (a) Phenolpthalein is used to detect the end point when half equivalent of NaCO and full equivalent NaOH is consumed (b) Phenolpthalein is used to detect the second end point (c) Methyl orange is used to detect the final end point (d) Methyl orange is used to detect the first end point g KI reacts with excess of KIO to produce I, which converts NaSO into SO 4. If the hypo solution was decimolar, then the volume required to reach equivalent point will be (a) litre (b) 15 ml (c) 0.0 litre (d) 0 ml SECTION III Comprehension Type Questions Write-up I Double Titration 1. Methyl orange as an indicator The volume of 0.1 NHCl used in the titration will correspond the neutralization as directed. NaHCO + HCl NaCl + CO + HO (colour change) NaCO + HCl NaCl + CO + HO (colour change) NaOH + HCl NaCl + HO (colour change) The volume of 0.1M NaOH in the titration will neutralize as directed HA (mono basic acid) + NaOH NaA (colour change) HA (di basic acid) + NaOH NaHA (colour change) HA (tribasic acid) + NaOH NaHA (colour change). Phenolphthelien as an indicator The volume of 0.1 NHCl used in the titration will correspond the neutralization as directed. NaHCO + HCl (no colour change) NaCO + HCl NaHCO (colour change) NaOH + HCl NaCl (colour change) The volume of 0.1M NaOH in the titration will neutralize as directed

17 SP-ST-CH-17 HA (mono basic acid) + NaOH NaA (colour change) HA (di basic acid) + NaOH NaA (colour change) HA (tribasic acid) + NaOH NaHA (colour change) mole of HPO4 and 0.01 mole HPO reacts with V ml of standard NaOH in presence of phenolpthelien indicator volume of NaOH used is (a) 500 ml (b) 400 ml (c) 00 ml (d) 00 ml. 5 ml of NaCO solution requires 100 ml of 0.1 N HCl to reach end point with Phenolphthalein as indicator. Molarity of resulting solution with respect to HCO ion (a) M (b) M (c) 0.16 M (d) 0.08 M. 0 ml of x M HCl neutralizes completely 10 ml of 0.1 M NaHCO solution and a further 5 ml of 0. M Na CO solution to methyl orange end point. The value of x is: (a) M (b) 0.1M (c) 0.15 M (d) 0. M 4. Certain moles of CO is dissolved in excess of NaOH. The resulting solution is divided into two equal parts. One part needs 0 ml of.5 N HCl and other part needs 40 ml of.5 N HCl using phenolphthalein and methyl orange indicator respectively. The mass of CO dissolved is: (a).5 10 (b) 5 10 (c) (d) Write-up II Iodine Titration All such titration which involves the direct titration of Iodine with a reducing agent are grouped under Iodimetry. Iodimetry is employed to determine the strength of reducing agent such as sodium those sulphate I + NaSO I + SO 4 6 If iodine is liberated as a result of chemical reaction involving oxidation of an idodide ion by a strong oxidizing agent in neutral or acidic medium the liberated iodine is then titrated with reducing agent. Iodometry is used to estimate the strength of oxidizing agent. For example the estimation of Cu ++ with thiosulphate. Cu ++ + I CuI + I SO SO I I Starch used as indicator near the end point which form blue colour complex with I. The blue colour disappears when there is no more of free I. 5. In Iodine titration Iodine remains in solution in the form of (a) I (b) I (c) I (d) I

18 SP-ST-CH In the reaction, CuSO4 + 4KI CuI + KSO4 + I the ratio of equivalent weight of CuSO4 to its molecular weight is: (a) 1/8 (b) ¼ (c) ½ (d) 1 7. When g of CuSO 4 in a solution is reacted with KI, then the liberated iodine required 100 ml 1 M Na SO for complete reaction, then what is the percentage purity of CuSO 4 sample used in making the solution. (a) 10% (b) 0% (c) 5% (d) None of these ml of 0.1 N hypo decolourised iodine by the addition of x g of crystalline blue vitriol to excess of KI. The value of x is (a) 5g (b).5g (c) 10g (d) 1.5g Write-up III Like acid base titration, in redox titration also, the equivalence point is reached when the reducing agent is completely oxidized by the oxidizing agent. But contrary to the acidbase titrations, oxidizing agents can themselves be used as internal indicator in redox titration e.g. CrO7 (orange yellow), Cr + (green), MnO4 (purple), Mn + (light pink), where strength of the solution may be expressed as molarity i.e. number of moles of solute per litre of solution. 9. In a titration experiment, a student finds that.48 ml of a NaOH solution are needed to neutralize g of KHP (molecular formula KHC8H4O4). What is the concentration in molarity of NaOH solution? (a) M (b) 0.8M (c) 0.057M (d) 0.08 M 10. A 16.4 ml volume of 0.17M KMnO4 solution is needed to oxidize 5.00 ml of a FeSO4 solution in an acidic medium. What is the number of moles of FeSO4 being oxidized for the reaction 5Fe MnO4 8H Mn 5Fe 4HO (a) ml (b) ml (c) ml (d) ml 11. A purple coloured solution is added from a burette to FeSO4 solution kept in the flask. After sometime, the purple colour changes to light pink. The ion formed from that solution is (a) MnO 4 (b) Fe + (c) Fe + (d) Mn + 1. Concentrated aqueous sulphuric acid is 98% HSO4 by mass and has a density of 1.84 g/ml. What volume of the concentrated acid is required to make 5 litre of 0.50M HSO4 solution. (a) 71.7 ml (b) 1.5 ml (c) ml (d) 7.1 ml

19 SP-ST-CH-19 Write-up IV KIO is an oxidant which is reduced to I or ICl depending on the medium whether it is dil. HCl or 6 M HCl. KI on the other hand is a reductant, which is oxidized to I or ICl depending on the condition as mentioned above. In IIT-JEE 199, a question based on this concept was asked and the same in slightly modified form but basic spirit remaining the same, reads as follows: 1.0g sample of AgNO containing some background inert impurities was dissolved in water and solution was treated with 0 ml KI solution. The precipitated AgI was filtered off. The filtrate containing unreacted KI was back titrated with 7.5 ml 0.1 M KIO in 6 M HCl. In another titration with 10 ml fresh stock solution of KI exactly required 1.0 ml 0.1 M KIO solution for complete oxidation in 6 M HCl solution. Find percent purity of AgNO in the sample. A JEE aspirant solved the problem in the following manner. KIO vs. KI titration: 5 1 KIO I Cl n-factor = M KIO is 0.4N Strength of stock solution of KI V1N1 = VN 10 N1 = N1 = 0.4 N No. of m. equiv. of AgNO in the sample = No. of milli equiv. of KI added No. of milli equiv. of KIO conserved in the back titration = = 5 n-factor of AgNO in its reaction with KI (reaction being double displacement) is 1 (= charge carried by cation or anion). So, No. of m. mole of AgNO in the sample = % Mass of AgNO in the sample = Correct statement among the following is? (a) The solution is correct (b) The solution is wrong as KIO under the conditions given in the figure must be reduced to I (c) The solution is wrong as KI must have oxidized to I under the condition as laid down in the question. (d) The solution is wrong as there is fallacy in the calculation of no. of m. equiv. of AgNO 14. The correct final answer according to your calculation should be (a) the same as calculated above (b) 4.5% (c) 50% (d) 7.5% 15. The volume (ml) of 0.5 M KIO needed to oxidize completely 40.0 ml 0.1 M KI in dil. HCl medium is: (a) 8.0 (b) 10.4 (c). (d) 4.8

20 SP-ST-CH An aqueous solution containing 0.10g KIO (Formula weight = 14) and an excess of KI was acidified with HCl. The liberated I consumed 45.0 ml of thiosulphate. The molarity of sodium thiosulphate solution is IO I H I HO (a) 0.06M (b) 0.01M (c) 0.16M (d) 0.5M Write-up V Titration is a method of finding out the strength of an unknown solution by reacting with a known solution using a suitable indicator. The choice of the indicator depends on the nature of the reaction. The method is carried out by taking a definite volume of solution of unknown strength in the beaker, indicator is added and then from the burette the known solution is added till the indicator changes the colour which indicates the reaction is complete. At this point the number of equivalence of the solution in the beaker is equal to the number of equivalence of the solution added from the burette cc M phosphorous acid is titrated with 0.5M NaOH. The volume of NaOH needed to reach equivalence point to completely react with phosphorus acid is (a) 10cc (b) 80cc (c) 40cc (d) 0cc 18. If in the above titration NaOH is replaced by 0.M Fe(CrO7) then the volume of Fe(CrO7) required to completely react with phosphorous acid is (a) 8cc (b) 11.1cc (c) 10cc (d) 10.5cc 19. A solution containing x m mole of HSO4, y millimole of HCO4 requires A ml 1M Ca(OH) for complete neutralisation. B ml of 0.1M Mg(MnO4) is required for complete oxidation of the same solution. x and y can be expressed as (a) B A, A (c) A B, A (b) A B, B (d) A B, B 0. The volume of M H4PO7 needed to neutralise cc M Ca(OH) is (a) cc (b) 1cc (c) 4cc (d) 1.5cc SECTION IV Subjective Questions LEVEL I 1. What is the normality and nature of a mixture obtained by mixing 0.6g of NaCOHO to 100 ml of 0.1N HSO4?. How much AgCl will be formed by adding 00 ml of 5N HCl to a solution containing 1.7g AgNO?

21 SP-ST-CH-1. The reaction, C + O CO is carried out by taking 4g of carbon and 96g O, find out: a) Which reactant is left in excess? b) How much of it is left? c) How many mole of CO are formed? d) How many g of other reactant should be taken so that nothing is left at the end of reaction? g of a metallic carbonate was heated and the CO evolved was found to measure 16 ml at 7 C and 700mm pressure. What is equivalent weight of metal? 5. A sample of chalk is contaminated with impurity of CaSO4. 1 g of the chalk is dissolved in 0 ml of N/10 HCl. The solution required 40 ml of N/10 NaOH to neutralize the excess acid. What is the percentage of CaCO in chalk? 6. A solution contains a mixture of sulphuric acid and oxalic acid, 5 ml of the solution requires 5 ml of 0.1 M NaOH for neutralisation and 4 ml of 0.0 M KMnO4 for oxidation. Calculate the molarity of solution with respect to sulphuric acid and oxalic acid g of a mixture of KCO and LiCO required 0 ml of 0.5 N HCl solution for neutralization. What is % composition of mixture? g sample of NaCO and KCO was dissolved in water to form 100 ml of a solution. 0 ml of this solution required 40 ml of 0.1 N HCl for complete neutralization. Calculate the weight of NaCO in mixture. If another 0 ml of this solution is treated with excess of BaCl, what will be the weight of precipitate 9. 5 ml of HO solution were added to excess of acidified solution of KI. The iodine so liberated required 0 ml of 0.1 N NaSO for titration. Calculate the strength of HO in terms of normality, percentage and volume. 10. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO) with aqueous hydrochloric acid according to the reaction 4HCl(aq) + MnO(s) HO(l) + MnCl(aq) + Cl(g) What mass of 95% pure MnO is required for the production of 1.0 kg of Cl? LEVEL II ml of 1 M HSO4, 5 ml of 4 M HNO and 0 ml of HCl of unknown normality were mixed and made upto 1 litre. 0 ml of this solution required 6 ml of Ba(OH) solution prepared by dissolving 4.75g of pure Ba(OH).8HO in 50 ml solution in water. Calculate the normality of HCl.. An aqueous solution containing 0.10 g KIO (formula wt = 14.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I consumed 45 ml of thiosulphate solution to decolourise the blue-starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

22 SP-ST-CH-. 40 ml of HSO4 is exactly neutralized by 0 ml of Ca(OH) solution in water. The resulting solution is evaporated to dryness and the mass of anhydrous residue left is 1.6g. Calculate the molarity HSO4 and Ca(OH) solution ml of 8 N HNO, 4.8 ml of 5N HCl and a certain volume of 17 M HSO4 are mixed together and made upto litre. 0 ml of this acid mixture exactly neutralizes 4.9 ml of NaCO solution containing 1 g NaCO.10HO in 100 ml of water. Calculate the amount of sulphate ions in g present in solution ml of a solution containing 0. g of impure sample of HO reacts with 0.16 g of KMnO4 (acidic). Calculate: a) Purity of HO. b) Volume of dry O evolved at 7 C and 750 mm of Hg pressure g mixture of KCrO7 and KMnO4 was treated with excess of KI in acidic medium. Iodine liberated required 100 cm of 0.15 N sodium thiosulphate solution for titration. Find the per cent amount of each in the mixture. 7. HO is reduced rapidly by Sn +, the products being Sn 4+ and water. HO decomposes slowly at room temperature to yield O and water. Calculate the volume of O produced at 0 o C and 1 atm. when 00 gm. of 10% by mass HO in water is treated with 100 ml. of M Sn + and then the mixture is allowed to stand until no further reaction occurs. 8. A 1.85 g. sample of a mixture of CuCl and CuBr was dissolved in water and mixed thoroughly with a 1.8 g. portion of AgCl. After the reaction the solid, a mixture of AgCl and AgBr, was filtered, washed, and dried. Its mass was found to be.05 g. What percent by mass of the original mixture was CuBr? 9. A 1. g. of a mixture containing HCO4HO and KHCO4HO and impurities of a neutral salt, consumed 18.9 ml of 0.5 N NaOH for complete neutralization. On titration with KMnO4 solution 0.4 g of the same substance needed 1.55 ml of 0.5 N KMnO4. Calculate the percentage composition of the substance. 10. A mixture of KMnO4 and KCrO7 weighing 0.4 gms on being treated with KI in acid solution, liberated sufficient iodine to react with 60 ml of 0.1 N NaSO. Find the percentage of chromium and manganese in the mixture. LEVEL III (Judge yourself at JEE level) 1. A sample of hard water contains 96 ppm. of SO 4 and 18 ppm of HC O, with Ca + as the only cation. How many moles of CaO will be required to remove HC O from 1000 kg of this water? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca + ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its ph (One ppm means one part of the substance in one million part of water, weight/weights)?. Hydrogen peroxide solution (0 ml) reacts quantitatively with a solution of KMnO4 (0 ml) acidified with dilute HSO4. The same volume of the KMnO4 solution is just decolourised by 10 ml of MnSO4 in neutral medium simultaneously forming a dark

23 SP-ST-CH- brown precipitate of hydrated MnO. The brown precipitate is dissolved in 10 ml of 0. M sodium oxalate under boiling condition in the presence of dilute HSO4. Write the balanced equations involved in the reactions and calculate the molarity of HO.. A solution of 0. g of a compound containing Cu + and C O ions is titrated with 0.0 M KMnO4 in presence of HSO4 consumes.6 ml oxidant. The resulting solution is neutralized by NaCO, acidified with dilute CHCOOH and titrated with excess of KI. The liberated I required 11. ml of 0.05 M NaSO for complete reduction. Find out mole ratio of Cu + and C O in compound g sample of AgNO is dissolved in 50 ml of water. It is titrated with 50 ml of KI solution. The AgI precipitated is filtered off. Excess of KI in filtrate is titrated with M/10 KIO in presence of 6 M HCl till all I converted into ICl. It requires 50 ml of M/10 KIO solution. 0 ml of the same stock solution of KI requires 0 ml of M/10 KIO under similar conditions. Calculate % of AgNO in sample. The reaction is : KIO + KI + 6HCl ICl + KCl + HO g of pyrolusite ore was treated with 50 cm of 1.0 N oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 50 cm in a flask. 5 cm of this solution when titrared with 0.1 N KMnO4 required cm of the solution. Find out the percentage of pure MnO in the sample and also the percentage of available oxygen. 6. A sample of MnSO4. 4HO is strongly heated in air. The residue (MnO4) left was dissolved in 100 ml of 0.1 N FeSO4 containing dil. HSO4. This solution was completely reacted with 50 ml of KMnO4 solution. 5 ml of this KMnO4 solution was completely reduced by 0 ml of 0.1 N FeSO4 solutions. Calculate the amount of MnSO4.4HO in sample. 7. An equal volume of reducing agent is titrated separately with 1 M KMnO4 in acid, neutral and alkaline medium. The volumes of KMnO4 required are 0 ml,. ml and 100 ml in acid, neutral and alkaline medium respectively. Find out oxidation state of Mn in each reaction product. Give balance equation. Find the volume of 1 M KCrO7 consumed if same volume of reductant is titrated in acid medium. 8. A.0 g sample containing FeO4, FeO and an inert impure substance is treated with excess of KI solution in presence of dilute HSO4. The entire iron is converted to Fe + along with the liberation of iodine. The resulting solution is diluted to 100 ml. A 0 ml of dilute solution requires 11.0 ml of 0.5 M NaSO solution to reduce the iodine present. A 50 ml of the diluted solution, after complete extraction of iodine requires 1.80 ml of 0.5 M KMnO4 solution in dilute HSO4 medium for the oxidation of Fe +. Calculate the percentage of FeO and FeO4 in the original sample. 9. A steel sample is to be analysed for Cr and Mn simultaneously. By suitable treatment the Cr is oxidised to Cr O and the Mn to MnO4. A g sample of steel is used to 7 produce 50.0 ml of a solution containing CrO7 and MnO4. (a) A ml portion of this solution is added to a BaCl solution and by proper adjustment of the acidity, the chromium is completely precipitated as BaCrO4; g is obtained. (b) A second ml portion of this solution requires exactly ml of M standard Fe + solution for its titration (in acid solution). Calculate the % of Mn and % of Cr in the steel sample 4

24 SP-ST-CH A mixture of HCO4 and NaHCO4 weighing.0 g was dissolved in water and the solution made upto one litre. 10 ml of this solution required.0 ml of 0.1 N NaOH solution for complete neutralization. In another re-experiment 10 ml of same solution in hot dilute HSO4 medium required 4 ml of 0.1 N KMnO4 for complete neutralization. Calculate the amount of HCO4 and NaHCO4 in mixture. SECTION V Miscellaneous Questions True and False 1. I. n-factor of HCl in the reaction: KMnO HCl MnCl + KCl + 8HO + 5Cl is 5 8. II. 1 st I.P. of N is more than that of oxygen but the nd I.P. has the reverse order. III. Carbanion intermeidate undergoes rapid pyramidal inversion. (a) I F, II T, III T (b) I T, II F, III T (c) I T, II T, III T (d) I F, II F, III T. I. Equivalent weight of FeSO4(NH4)SO44HO is same as molecular weight in reaction with KMnO4. II. One mole of NaSO require 1/ mole of KCrO7 for oxidation in acidic medium. III. ph of an aqueous solution obtained on mixing 100 ml of 0.1M NaOH with 100 ml of 0.1M HCN will be equal to 7. (a) I T, II T, III F (b) I T, II F, III T (c) I F, II T, III F (d) I F, II F, III T The following questions (to) consists of two statements, one labelled as ASSERTION (A) and REASON (R). Use the following key to chose the correct appropriate answer. (a) If both (A) and (R) are correct, and (R) is the correct explanation of (A). (b) If both (A) and (R) are correct, but (R) is not the correct explanation of (A). (c) If (A) is correct, but (R) is incorrect. (d) If (A) is incorrect, but (R) is correct. ASSERTION (A) REASON (R). In the titration of NaCO with HCl using methyl orange indicator, the volume of the acid required at the equivalence point is twice that of the acid required using phenolphthalein indicator. Two moles of HCl are required for complete neutralization of one mole of NaCO. 4. When KMnO4 solution is added to hot oxalic acid solution, the decolourisation is slow in the beginning but becomes spontaneous after sometime. Mn + acts as autocatalyst.

25 Answers to Assignments SP-ST-CH-5 SECTION I 1. (b). (b). (b) 4. (a) 5. (b) 6. (d) 7. (c) 8. (c) 9. (d) 10. (d) 11. (a) 1. (d) 1. (a) 14. (c) 15. (c) 16. (a) 17. (c) 18. (b) 19. (c) 0. (a) 1. (d). (c). (a) 4. (c) 5. (a) SECTION II 1. (d). (d). (c) 4. (c) 5. (c) 6. (b) 7. (c) 8. (b) 9. (a) 10. (b) 11. (a) 1. (a) 1. (b) 14. (c) 15. (d) 16. (c) 17. (a) 18. (b) 19. (c) 0. (b) 1. (d). (b), (c), (d). (a), (b), (d) 4. (a), (c) 5. (a), (b) SECTION III 1. (c). (d). (c) 4. (b) 5. (a) 6. (d) 7. (a) 8. (b) 9. (a) 10. (b) 11. (d) 1. (c) 1. (d) 14. (b) 15. (c) 16. (a) 17. (b) 18. (c) 19. (b) 0. (b) SECTION IV LEVEL I g. a) O; (b) ; (c) ; (d) 7g % M, M 7. % of KCO = 96%, % of LiCO = 4% 8. Weight of NaCO = 0.6 gms; Wt. of BaCO = 0.94g N, 1.6%, volume g 1. 1.N M LEVEL II

26 SP-ST-CH-6. MHSO 0.5; M 4 Ca(OH) g 5. a) 85%, (b) ml 6. % of KCrO7 = 14.6% and % of KMnO4 = 85.4% L % 9. HCO4.HO = 14.7% KHCO4.HO = 81.67% 10. % Mn = 14.1, % Cr = 0.9 LEVEL III 1. CaO (required) = 1.5 mole, Ca ++ (ppm) = 40 ppm, ph = / 4. 85% 5. MnO = 48.88%, 9% g ml 8. FeO = 49.%, FeO4 = 4.8% 9. Cr =.8%, Mn = 1.467% 10. HCO4 = 0.90g, NaHCO4 = 1.1g SECTION V 1. (c). (a). (b) 4. (a)