Mole Concept Hints & Solutions

HINTS & SOLUTIONS wt. of carbon + hydrogen = 2.4 + 0.6 = 3 = Initial wt. of hydrocarbon  It illustrate law of conservation of mass. EXERCISE - 1 Single Choice 13. 2. mole = number of molecules / NA Hence : n = N/NA 4. Gram mol. wt. of C60H22 = 742 gm i.e. wt. of 6.023 × 1023 molecules = 742 So wt. of 1 molecules = 5. Compound 742 Elements % = 1.24 × 10 6.023  1023 mole of mole of atom molecule C4H10 1 58 –21 gm. No. of atom Ca 20 20/40 = 0.5 1 1 Br 80 80/80 = 1 2 2 Hence : Empirical formula = CaBr2 200 =1 200 Hence : Molecular formula = CaBr2 n= 1 58 14 N 58 A 1 28 1 N 14 A 15 8% sulphur by mass means – 8 g sulphur is present in 100 g solid. 1 N 108 A  32 g sulphur (1 mole atom) will be present in = 1 28 2× Ag 1 108 1 108 H2O 1 18 1 3× 18 NA 6 No. of mole No. of molecule ratio mass of C 16. % of C = molar mass  100 CO2(g) 2w N2O(g) 5w 2w 44 5w 44 2w × NA 44 2:5 5w × NA 44 7 Ratio of total no. = of molecules So ratio of total = no. of atoms 21 12  100 M M = 360.1. 69.98 = H2 : 1 : He : 1 : O2 : 1 : O3 1 2 1 2 3 : : : 10×19×81×11 190 + 891 1081 = = 100 100 100 = 10.81. 10 Atomic mass = 11 weight of CO2 = 8.8 gm mole of CO2 = 8 .8 = 0.2 , 44 mole of C = 0.2 wt. of C = 0.2  12 = 2.4 g , mole of H2O = So mole of ‘H2’ = 0.3 wt. of hydrogen = 0.3  2 = 0.6 100 × 32 8 = 400 g [ compound must be having at least one atom of sulphur]  min. mol. mass = 400 g. Hence C4H10 has maximum number of atoms. Weight Simplest whole no. 14 × N2 6. % Simple Atomic mass ratio 5 .4 = 0.3 18 17 3M + N2  M3N2 Let Atomic wt. of metal = a So (3a + 28) g nitride contains metal = 3a gram  14.8 g nitride contains metal = So a = 40. 3a × 14.8 = 12 3a  28 21. By applying POAC for C atoms moles of ethylene × 2 = mole of polythene × n × 2 100 g wt. of polethene ×2= ×n×2 28 28  n wt. of polyethene = 100 g 22. KClO3  KCl + 3 O 2 2 3 mole or 33.6 litre O2 from 1 mole KClO3 2 1 11.2 litre of O2 formed by mole KClO3 3 MOLE CONCEPT 25 C + CO2  2CO  6   44  =1 Given moles   = 0.5   12   44  So C is limiting reagent  CO formed = 1 moles Now moles of Ni need to react with 1 moles of CO are 1 × 58.7 = 14.675 gm. 4 26 H2SO4 + 2NaOH  Na2SO4 + 2H2O 1 for 1 mole NaOH mole H2SO4 required 2 70 g H2SO4 in 100 g solution 100 98 g H2SO4 in =  49 = 70 g solution. 70 2 27. TiCl4 + 2Mg  358 96 Initial mole = 1.88 =4 190 24 final mole 0 4 – 2  1.88 358 wt of Ti obtained =  48 190 32 ×100 % yield = 358 × 48 = 35.38 % 190 29. Weight Mole L.R.  No. one is L.R. So weight of P4O6 = 2  1.88  X2Y3 2X + 3Y W gram W gram 0 w w 36 24 w w 1 × 36 24  3 2 1 1 72 72 2X + 3Y  X2Y3 w w 2 [ 36] 24 w Weight of X2Y3 = [72 × 2] = 2 w 2  36 So weight of X2Y3 = 2 [weight of X Taken ] O2  P4O6 + X2  X2  Na2CO3  + + + mole of NX = w 36 31. P4 + 33 M MX2 M3X8 P4O10 31 gram 32 gram According to question weight of P is conserved so Let Mole of P4O6 = a Mole of P4O10 = b Initial weight of P = Final weight of P. 1 × 220 = 27.5 8 284 = 35.5. 8 P4O10 = Ti + 2MgCl2 1.88 31 = [a × 4] × 31 + [b × 4] × 31 4 a + 4b = 1] (1) × 3 Initial weight of oxygen = Final weight of oxygen 32 = [ a × 6] × 16 + [a × 10] × 16 3a + 5b = 1] (2) × 4 12 a + 20 b = 4 1 12 a + 12 b = 3 So b = 8 8b=1 1 Similarly a = 8 MX2 M3X8 NX + CO2 + M3O4 206 =2 103 POAC for X Atom : No. of X atom in M3X8 8 [No. of mole of M3X8] = = No. of X Atom in NX 1 [No. of mole of NX] 2 1 No. of mole of M3X8 =   = mole 8  4 Now POAC for M Atom 3 [No. of mole of M3X8] = 1 × [No. of Mole of M]  3× 1 = No. of mole of M 4 weight of M atom = 3 × 56 = 42 gram 4 Ether 34. Li AlH4 + t-butyl alcohol  LiAlH C12H27O3 0.05 mole (M.W. = 254) 12.7 gram = 12.7 254 = 0.05 mole Li atom remain conserved so No. of mole of LiAlH4 = No. of mole of LiAlHC12H27O3 So No. of mole of LiAlHC12H27O3 = 0.05 % yield = 0.05 × 100 = 100% 0.05 1.12 = 0.02 mole CaO 56 35. CaCl2  CaCO3  CaO  Moles of CaCl2 = 0.02 Mole Mass of CaCl2 = 0.02 × 111 = 2.22 g % of CaCl2 = 2.22 × 100 = 50 % 4.44 36. SO32–  1(x) + 3(–2) = – 2 x = + 4 S2O42–  2 (x)+ 4 (–2) = – 2 x = + 3 S2O62–  2 (x)+ 6 (–2) = – 2 x = + 5 37 52. Mole fraction of A i.e. XA = So XH O = 2 Total moles and molality = nA  1000 X A  1000 0.2  1000  = 0.8  18 nH2O  18 XH2O  18 = 13.9 Ans. 40. Fe3O4 can be written as FeO.Fe2O3. In FeO , Fe has oxidation state + 2 , in Fe2O3 has oxidation state + 3. 1 2  2  3 8 = . 3 3 –10 e   2Nx+ N2H4   43. nH2O XA n  A XH2O nH2O Now 2(+2) + 2x + 7 (–2) = 0  x = +5 resultant oxidation number = nA Total moles (– 2)  2x – 2 (–2) = 10.  2x = 6  x = + 3. 56. Weight of KOH = 2.8 gram Volume of solution = 100 ml M= 2.8  1000 28 = = 0.5 M 56  100 56 57. Mfinal = M1V1  M2 V2 0.6  250  0.2  750 ; 0.25 = ; V1  V2  Vwater 250  750  Vwater So Vwater = 200 ml. 60. Moles of Cl– in 100 ml of solution 45. MnO4– + C2O42– + H+  Mn2+ + CO2 + H2O V.f. = 5 V.f. = 2  Balanced equation : 2MnO4– + 5C2O42– + 16H+  2Mn2+ + 10CO2 + 8H2O = 2 4 6  ×2+ = 0.2184 58 .5 111 53 .5 Molarity of Cl– = 0.2184 × 1000 = 2.184. 100 400  300  200 ; 400 61. Conc. of cation = 46. conc. of anion = 200  300  400 400  ratio of the conc. = 1 (V.f.)P4 : (V.f.) HNO3 = 20 : 1  x = 1; y = 20 48. Equating number of O-atoms on both sides : 3b = a + b + 2a + c/2  4b=6a+c Also, from charge balance : c = a + b  a : b : c = 3 : 7 :10 20 = 0.5 40 Mole of maginesium = Mole of carbon = 24 =1 24 12 =1 12 Mole oxygen atom in 8 g = 50. Let, nH2O = nNaCl = n m= 64. Mole of calcium = n 1 Mole of solute = × 1000 = ×1000 18 wt. of solvent (kg) n  18 = 55.55 m. 8 = 0.5 16 Mole of oxygen atom in 16 g = Hence option (C) is correct 16 =1 16 MOLE CONCEPT 66. C : O : S = 3 : 2 : 4 Hydrogen is = 7.7%  100 – 7.7 = 92.3 % contains C,O & S 3    92.3 %C=  324 4 %S=  92.3 9 ; Elements % % / Atomic mass H 7.7 7.7 %O= Given x+y+z=1 Put the value of z in eq. (1) 28x + 46y + 92 + 0.4 = 55.4 28x + 46y = 18.6 By equation (3) & (4) y = 0.1  x = 0.5 , y = 0.1 , z = 0.4 2  92.3 ; 9 Simple Simplest ratio whole no. 6 6 C 30.76 30.76/12 = 2.56 2 2 O 20.51 20.15/16 = 1.28 1 1 S 41.02 41.02/32 = 1.28 1 1 73. (A) Explanation : 2 Ag + S  Ag2 S 2 × 108 g of Ag reacts with 32 g of sulphur 32 320 × 10 = >1g 216 216 It means ‘S’ is limiting reagent 32 g of S reacts to form 216 + 32 = 248 g of Ag2S 1 g of S reacts to form = ... (2) From equation (1) mole of CaC2 = mole of C2H4 64  10 3 = mole of C2H4 64 From equation (2) mole of C 2H4 mole of polymer = n 1 10 1 = 0.0925 neq of S = = 0.0625 16 108 (neq = number of equivalents) Since neq of S is less than neq of Ag  0.0625 eq of Ag will react with 0.0625 eq of S to form 0.0625 eq of Ag2S Hence , amount of Ag2S = neq× Eq. wt. of Ag2S = 0.0626 × 124 = 7.75 g neq of Ag = 76. Let of mol of Fe undergoing formation of FeO = x Let mol of Fe undergoing formation of Fe2O3 = 1 – x then, mol of N2 = x, mol of NO2 = y, therefore If 28 x  46 y  92 z = 55.4 1 mol of N2 O 4 = z 2Fe + 3 O 2 2  Fe2O3 1–x 3 1 x (1 – x) 4 2 1 x = 0.3 = moles of of Fe2O3 2  ...........(2) 4 Mole of FeO = 3 Mole of Fe 2 O 3 By deviding equation (1) by equation (2) 1+z= 55 .4 = 1.4 39 .6 z = 0.4 mol x x 3  (1 – x) = 0.65 = Total moles of oxygen 24 4 x = 0.4 = moles of FeO 28 x  ( y  2z )46 = 39.6 xyzz  x/2 As given, ...........(1) N2O4  2NO2 28 x  46 y  92 z = 39.6 1 z  FeO x 1 O 2 2 Fe + wt of polymer 10 3 = n (28 ) n wt of polymer = 28  103 g = 28 Kg 70. Let 248 = 7.75 g 32 Alternately 68. CaC2 + H2O  Ca(OH)2 + C2H2  C2H4 ... (1)  ...........(4) 10 g of Ag reacts with  empirical formula C2H6OS minimum molar mass = 24 + 6 + 16 + 32 = 78 nC2H4 ...........(3) 78. C Initial mole x 12 + 1 O 2 2 y 32  CO 0 .... (1) final mole y  x 1 –   32  12  2 0 CO + 6. (B) Volume = 100 + 100 = 200 ml 1 O  CO2 2 2 .... (2) For no solid residue C should be zero in eq. (1) y x 1 For that –  >0 32 12 2 y x > 32 24 200 ×1 × 98 = 19.6 gm. 1000 (A) , (B) and (D). Explanation : (A) 1.0 mol of O2 = 32 g (B) 6.02 × 1023 molecules of SO2 = 64 g , 3.01 × 1023 molecules of SO2 = 32 g (C) 0.5 mole of CO2 = 0.5 × 44 = 22 g is not correct answer. (D) 1 g atom of sulphur = 32 g 2 .8 70 × = 0.1 × 0.7 = 0.07 28 100 (in moles) No. of moles of N atoms formed = 2 × EXERCISE - 2 Part # I : Multiple Choice (A) , (C) and (D) Explanation : 4 Ag + 8 KCN + O2 + 2 H2O  4 K[Ag(CN)2] + 4 KOH  4 × 108 g of Ag reacts with 8 × 65 g of KCN 100 g of Ag reacts with 8  65 × 100 = 120 4  108 1 g of Ag require 7 .4 Hence, volume of O2 required = × 22.4 = 5.20 litre 32 Hence, (A) , (C) , (D) are correct while (B) is incorrect. initial mole final mole initial mole A3B2 + 2C  1 1 final mole 1– 1 2 0 28 .96 = 14.48 2 So (A) and (B) are correct answer. 11 H2 + S + 2O2  nH2 = Hence 100 g of Ag require = 7.4 g Hence, choice (C) is correct. 2B  3 3–2 10 (A) and (B) Explanation : M. Wt. = 0.001293 × 22400 = 28.96 M.Wt. = d × volume of 1 mole of gas at STP 5. 6 1 = 22 .4 4 H2SO4 ns = 8 1 = 32 4 nO2 = 1 2 As all reactants are in stoichiometric ratios, none will be left behind. 32 = 0.0740 g 4  108 3A + 3 0 30 × 0.1 = 0.06 100 (A) Total no . of moles = 0.07 + 0.06 = 0.13 (B) Total number of molecules = 0.07 ×6.023 × 10 23 = 4.2 × 1022 molecule = 0.421 × 1023  We have to calculate molecule of nitrogen not atoms. V. D = Hence ,, to dissolve 100 g of Ag , the amount of KCN required = 120 g Hence , statement (A) is correct.  4 × 108 g of Ag require 32 g of O2 4. 8 Amount of N2 left = y > 1.33 x  (D) Mass of H2SO4 = 10× d × x =1M M 9 (A) and (B) Explanation : 30% of molecule dissociated N2  2N y 32 > x 24 2. (A) Molarity of second solution is = A3B2 0 1 A3B2C2 0 1 2 1 4 mole of H2SO4 is formed. 12 m = 0.2 mole / kg weight of solvent = 1000 gram weight of solute = 0.2 × 98 = 19.6 gram Total weight of solution = 1000 + 19.6 = 1019.6 ml. 13 Let W gas of SO2 and O2 are taken moles of SO2 = W 64 ; moles of O2 = W ; 32 WN A WN A ; molecules of SO2 = 32 64 hence molecules of O2 > molecules of SO2 since moles of O2 > moles of SO2 , hence volume of O2 at STP > volume of SO2 at STP. molecules of O2 = MOLE CONCEPT 2P + Q  14. initial mole 12 8 final mole 0 8–6  moles of R formed = 6 % of Q left behind = 24. Mole fraction of H2O = 1 – 0.25 = 0.75 R x C2H5OH 0 6 X C2H5OH  XH2O 2 × 100 = 25% 8 wt. % = 25. Mole of Al = As NH3 is very less hence we can say 100 g water has 10–3 g NH3  1000 gm water has = = 10–2 g NH3 = 26. Atomic mass given weight 10 2 mole NH3 = 5.88  10–4 mole NH3. 17 No. of mole 17. Let wg water in added to 16 g CH3OH No. of Atom 16  1000 500 molality = = W  32 W W = 27 gm. 26.88 mass = = 29.09 0.924 density  30  29.09    100  3% . % decrease in volume =  30   28. % of Na = M= (H2O2  O2 + 2H+ + 2e–) × 3 ___________________________________________ Cr2O72– + 8H+ + 3H2O2  2Cr3+ + 7H2O + 3O2 The reaction practically occurs with this stoichiometry. . Let volume of solution is 1000 ml moles of H2SO4 = 18 mass of H2SO4 = 18  98 = 1764 g mass of solution = 1000  1.8 = 1800 g mass of solvent = 1800 – 1764 = 36 g 18 = 500.  36   1000    A 40 x gram x 40 x × NA 40 B 80 2x gram 2x 80 x × NA 40 x × NA = y 40 27. (B) Explanation : 1 mole of electrons weighs = 9.1 × 10–31 kg × 6.023 ×1023 = 54 × 10–8 kg = 54 × 10– 8 × 1000 = 54 × 10– 5 g = 54 × 10– 5 g × 103 mg = 54 × 10– 2 mg = 0.54 mg (A) is not correct because charge on 1 e is 1.6 × 10–19 C and not on 1 mole of electrons.(C) and (D) ruled out as explained above. 20. Cr2O72– + 14H+ + 6e–  2Cr3+ + 7H2O molality = 54 =2 27 But according to question = 19. Mass of ethyl alcohol = 1.5 × 0.792 g Mass of water = 15  1 Total mass of solution = 15 + 0.792  15 = 26.88 Volume of solution = or n C2H5OH  nH2O  mass of Mg = 2 × 24 = 48 gm 10 3  1000 g 100 x A  1000 500 0.25  1000 = = (1  x A )mB 0.75  18 W nC 2H5OH 0.25  46 × 100 = 46%. 0.25  46  0.75  18 . 10–3 g NH3 in 100 g solution one litre water has mass = 1000  1 g = 23 mass of sodium  100  7 =  100 M molecular mass 23  100 = 328.6 7 29. mass moles C 24 H 8 O 32 24 12 8 1 32 16 ratio 2 8 Simple integer ratio 1 4 Hence empirical formula is CH4O 2 1 30. KI is limiting reagent  3 mole of KI will give 33 mole of NO2 according to stoichiometry. 31. C + O2  CO2 + CO POAC on ‘C’ atom 1 (mole of C) = 1 (mole of CO2) + 1 (mole of CO) 38. (A) Explanation : 2 Ag + S  Ag2 S 2 × 108 g of Ag reacts with 32 g of sulphur 10 g of Ag reacts with 280 240 = mole of CO2 + 28 12 Mole of CO2 = 20 – 10 = 10 It means ‘S’ is limiting reagent 32 g of S reacts to form 216 + 32 = 248 g of Ag2S 10 Mole % of CO2 = ×100 = 50%. 20 1 g of S reacts to form = 32. Ag2CO3 (s)  2Ag (s) + CO2 (g) + 1 2 O2 (g) Alternately neq of Ag = C2H2 + 5 2 O2  2CO2 + H2O By Stoichiometry of reaction Moles of CO2 formed = 11.2 1  22.4 2 Moles of O2 required = 5 5 1 × = 8 4 2  Moles of Ag2CO3 required = 2 × neq of S = 5 5 = 8 4 5 Mass of Ag2CO3 required = × 276 = 345 g 4 33. (1) It is a fact. (2) 2H2 + O2  2H2O Initial mole 2 3 0 final mole 0 3-1 =2 2 (3) C 1 = 0.0625 16 35. 1 gm molecule is 1 mole Mole of SO42– 4 × 1 = 4 gm ion. 36. Mass of NaCl = 10 × 0.96 = 9.6 gm 9 .6 moles of NaCl = 58 .5 9 .6 no. of molecules = × 6.023 × 1023 ~ 1023 58 .5 32 37. = 0.2 2x + 3y ; Hence: x = 56 & y = 16. 92.8 = 0.4 3x + 4y 10 = 0.0925 ; 108 (neq = number of equivalents) 39. CaCl2 + Na2CO3  CaCO3 + 2 NaCl  CaCO3  CaO + CO2  1.62   Mole of CaCl2 = mole of CaCO3 = mole of CaO =   56   1.62   Molar mass of CaCl Mass of CaCl2 =  2  56  + O2  CO2 4A + 2B + 3C  A4B2C3 Initial mole 2 1.2 1.44 0 final mole 0 0.48 C is limiting reagent.  moles of A4B2C3 is 0.48. 248 = 7.75 g 32 Since neq of S is less than neq of Ag  0.0625 eq of Ag will react with 0.0625 eq of S to form 0.0625 eq of Ag2S Hence , amount of Ag2S = neq× Eq. wt. of Ag2S = 0.0626 × 124 = 7.75 g  1.62   × 111 gm. =  56  w w 32 12 Here C is limiting reagent. 34. 32 320 × 10 = >1g 216 216 % of CaCl2 = 3.21 × 100 = 32.1 %. 10 40. (i) 2e + 6H+ + BiO3–  Bi3+ + 3H2O (ii) 4H2O + Mn2+  MnO4– + 8H+ + 5e ––––––––––––––––––––––––––––––––––––––––– (i) × 5 + (ii) × 2, we get 14 H+ + 5 BiO3– + 5Mn2+  5Bi3+ + 2MnO4– + 7 H2O Hence, (B) is the correct balanced reaction. MOLE CONCEPT 41. Explanation : M. wt. of NaNO3 = 85 70 mg of Na+ are present in 1 mL 50 ml of solution contains 50 × 70 = 3500 mg = 3.5 g Na+ ion 23 g of Na+ are present in 85 g of NaNO3 85 3.5 g of Na+ are present in × 3.5 = 12.934 g of NaNO3 23 49. 50. 42. 3I2 + OH–  IO3– + 5I– (balance reaction) So, ratio is 1 : 5. 43. MClx + x AgNO3  xAgCl + M (NO3)x Mole of MCl x Mole of AgNO 3  1 x 1 0.1 = (0.5 × 0.8) x 0 .4 x= =4 0 .1 M×1000 m= (1000×d - M× M.Wt.) 44. (A) Explanation : where ‘m’ is molality, M is molarity. -2 = = 10 ×1000 (1000×1.1-10-2 ×106) 10 10 = = 9.00 × 10– 3 1100  1.6 1099.4 [Take 1099.4 = 1100] 45. At 4ºC i.e. 277 K density of water = 1 gm/ml  1 kg water  1000 ml water = 1 lit.  Molality & molarity remains same. 46. Molarity = (% w / w)×density ×10 Molar mass of solute = 98 ×1.84 ×10 = 18.4 M 98 47. Mole of NaCl = Molarity = 5.85 = 0.1 58 .5 0 .1 = 0.1 M 1 Moles in 1 ml of solution = MV = 0.1 × 10–3 = 10–4 mole. Number of ions in 1 ml = 2 × 10–4 × 6.023 × 1023 = 1.204 × 1020. 48. Molarity = M Let volume of be 1 ltr.  mass of solvent = 1000 d – M × M2 M Molality = m = × 1000 1000 d  MM2 Molarity = 10  1.8  98 = 18 M 98 For reaction with 2 moles NaOH 1M H2SO4 1 lit. volume required ; 1M HCl 2 lit. volume required  cheapest will be 1 M H2SO4 1 lit. Part # II : Assertion & Reason 5 Statement-1 and Statement-2 both are two formula idependent to each other. 7 Due to temperature change volume get changed. Hence concentration units dependent on volume get changed. 10 Molality & mole fraction are mass dependent terms while molarity is volume dependent term. EXERCISE - 3 Part # I : Matrix Match Type 1. Zn(s) + 2HCl (aq)  ZnCl2(s) + H2 (g) (A) Initial mole 2 2 final mole (2-1=1) 0 0 1 0 1 2 1 × 100 = 50% 2 Volume of H2 = 22.4 lit. Solid product obtained = 1 mole Limiting reagent is HCl. Excess reagent left = HCl  AgCl(s) + HNO3(g) (B) AgNO3(aq) + Initial mole 170 =1 170 1- 18.25 1  36.5 2 1 1 = 2 2 0 1 2 1 2 0 1 Excess reagent = 0 1 2 1 Volume of gas = 11.2 lit. × 100 = 50% 1 mole 2 Limiting reagent is HCl. Solid product = (C) CaCO3(s)  CaO(s) 100 1 0 100 0 1 Excess reagent not present Volume of gas = 22.4 lit. at STP Solid product is 1 mole Initial mole + CO2(g) 0 1  2KCl (D) 2KClO3(s) Initial mole 0 0 0 2 3 2 (A) Molarity of cation = = 3O2(g) 2 3 No excess reagent left Volume of gas = 44.8 lit. 2 Solid product is mole. 3 3. + Comprehension # 4 : 2. Mass of solute = 60 × 0.4 + 100 × 0.15 = 24 + 15 = 39 gm Mass of solvent = 160 – 39 = 121 gm 3. Volume of solution = M1V1  M2 V2 V1  V2 3(0.2)100  0.1 400 0 .6  0. 4 = 500 5 = 0.2 (B) Molarity of cation = 50  0.4  0 = 0.2 100 0.4  50  0 = 0.2 100 2(0.2)30  0 (C) Molarity of cation = = 0.12 100 Molarity of Cl— = 30  0.2 = 0.06 100 (D) 24.5 gm H2SO4 in 100 ml solution Molarity of SO42– = Molarity = 25.4 = 2.5 98 0.1  Concentration of cation = 2 × 2.5 M Concentration of SO42– = 2.5 M. Comprehension # 5 : 1. amt mole fraction C 0.2732 0.0227 1 6 H 0.0382 0.0382 1.68 10 Ca 0.152 0.0038 0.167 1 O 0.3540 0.0227 1 6 Simplest formula C6H10CaO6 CaO6C6H10 2. Formula weight 3. The molecular mass of lactate pentahydrate = 308 218 gm anhydrous salt recovered = 308 g lactate pentahydrate 1 gm anhydrous salt recovered =  % w/w O2 = 3. w O2  100 w O 2  w N2 D= m 22.8 = = 1.2857 gm/L V 22.4 50 × 74.5 3.73 mol–1 1000 The price of K2SO4 = 2. = 0.2  32  100 = = 22.2% 0.2  32  0.8  28 Density of air at NTP 1 mole of air = 0.8 mole N2 + 0.2 mole O2 = 0.8 × 28 + 0.2 × 32 = 28.8 gm = 22.4 Ltr volume. 308 = 1.41gm 218 Comprehension # 6 : 1. The cost of 1000 gm KCl is 50 kg The cost of 74.5 g KCl is Part # II : Comprehension Comprehension # 1 : 2. 1 mole of air  0.8 mole of N2 = 0.8 × 28 g N2  0.2 mole of O2 = 0.2 ×32 g O2 160 = 100 ml 1 .6  39     58.5   Molarity = = 6.67 M 100  10 3 0 .6 0.2  100  0.1 400 = = 0.12 5 500 Molarity of Cl— =  39     58 .5  Molality = = 5.509 = 5.5 m. 121  10 3 Mass of solute = 39 gm 3. 50 × 74.5 × 2 174 mole of K in KCl =  Rs. 42.82 kg–1 1000 74.5  13.42 mole of K2O form 13.42 mole of K = mass of K2O = 6.71 × 94 = 630.8 gm = 0.631 kg 13.42 = 6.71 2 MOLE CONCEPT Comprehension # 7 : 1. Ba(OH)2 + 2HNO3  Ba(NO3)2 + 2H2O 0.4 mole 0.4mole In resultant sol. Ba(OH2) is remaning, therefore nature of sol. basic. 2. 2 74 2 31 M.F. = (CH3O)2 = C2 H 6 O 2 n= EXERCISE - 4 Subjective Type Vol. of Ba(OH)2 = 342  600 mL 0 .57 1. mole of OH– = 0.2 × 2 = 0.4 molarity of OH– = 0.4  0.5 0.8 Comprehension # 8 : 1. 8 mole NaBr obtain from = 3 mole Fe 2.06 × 10 3 ×3 103 × 8 2.06 × 10 3 3 × 56 × = 420 kg mass of Fe = 103 8 mole of Fe = mole NaBr = 2. mole of Fe3Br8 = 100 × 2.06 ×10 6 103 × 70 × 8 mole of Fe = mole FeBr2 = mass of Fe = 2.06 ×10 3 ×100 ×100 ×3 103 × 70 × 60 × 8 2.06 × 10 3 ×100 × 100 3 × 56 × 103 × 70 × 60 8 mass of Fe = 103 kg mole of NaBr 2 2.06 × 103 = 10 = 103 × 2 Comprehension # 9 : 1. CO2 = 22 g = 0.5 mol 13.5 mol. H2O = 13.5 g = 18 3. 2. Ist exp. CuO = 1.375 gm Cu = 1.098 gm O = 0.277 gm IInd exp. Cu = 1.179gm CuO = 1.4476 gm O = 0.2686 gm Cu Cu  3.96 38 ; 4 ; 4 O O In both the cases ratio of Cu/O is same  Y  0.471  X   0 .324  1.4 537  r1    Y  0 .5 09  X   0 .11 7  4 .35 0  r2   r2  2.992 6 ; 3 r1 So satisfy law of multiple proposition. 3. = 35.125 × 28 = 983.5 gm 4.  0.0 7  molecular =    N A  3 =2.34 × 1021  18  5. mole of CO2 = 6. 1 06 .5  1 mole 1 06 .5 NO. of atom of Na=1 × NA Cl = 1 × NA O = 1 × NA n NaClO 3  n P4 = 92.9 = 0.75 mole 4 × 31 N P4 = 0.75 × N A = 4.52 ×10 23 molecules C = 0.5 mol = 6 g N P = 18.04 ×10 23 molecules H = 1.5 mol = 1.5 g O = 8 gm = 0.5 mol E.F. = CH3O let molar mass = M 7. n Na = 8. 27 41 .75   M = –107 + 167 = 60 108 M  1  108 E.F. mass = 12 + 3 + 16 = 31 (a) 1 × 23 gm (c) 1 × 63.5 gm 9. mHg = 13.6 × 1000 gm nHg = mHg/200 = 68 mole 5.75 = 0.25 mole 23 (b) 1 × 35.5 gm 10. 3CaCO3 + 2H3PO4  Ca3(PO4)2 + 3H2O + 3CO2 50/100mole 70/98 mole = 0.5 0.7142 -- 0.7142 – Limiting reactant m CaCO3 = 17. C4H8 + 6O2  4CO2 + 4H2O x 4x 5x 13 O  4CO2 + 5H2O 2 2 y 4y 5y (4x + 5y) × 44 = 8.8 x + y = 0.05 ......(1) (4x + 5y) × 18 = 4.14 4x + 5y = 0.23 ......(II) 4x + 4y = 0.2 .......(I) y = 0.03 % by mass of C4H10 C4H10 + 2  0.5  × 0.5 = 0.3808   3  3  0.5 × M Ca 3 (PO 4 )2 = 51.66 gm 3 m H 3PO 4 = 0.3808 × M H 3PO 4 = 31.31gm 11. ClNH2 + 2NH3  N2H4 + NH4Cl 1000 mole excess 51.5 = 19.417 19.417 mole % yield = = y  y 18. CxHy +  x   O 2  xCO2 + HO 4  2 2 v excess 14.781 ×100 = 76.125% 19.417 y  x + v 4  8 2% 12. 5C + 2SO2   CS2 + 4CO 450 = 7.03 Kmole 64 excess 0.82 ×   +v –  x + 7.03 = 2.88 Kmole = 219.09 kg 2 15. M = × x ×153 × 100 = 65.65% 29 21.  x=2 C2H6 1 × 6.023 × 1023 = 7.09 × 107 gm 0.75 cm 3 / gm Mgas Mair  1.1 7  Mgas = 1.17 × 29 = 33.93 gm 22. Y3A5O12 200 × 200 × 10–3 (a) y = 44.95%, Al = 22.73%, O = 32.32% (b) 17.98 gm (27 / 98) ×1000 = 3.8 (100 / 1.2) (3n  1) O 2  nCO2 + (n + 1) H2O 2 (3n  1) / 2 7   6n  2  7n n = 2 n 4  y=6 20. M =  × (75 × 10–8 cm)2 × (5000 × 10–8 cm) 2.5 ×106 = 278.947gm 0.95 16. CnH2n+2 + y v 2 19. Molar mass = 3.2707 × 10–22 × 6.023 × 1023 = 196.99426gm x × 0.95 = 5 × 0.5 14. 106 x= xv y y  v + xv + v = 2.5v 4 2 y = 1.5 4 xv = 2v 13. BaO + CaO x × [153] + y × [56] = 28 ......(I) BaO + 2HCl  BaCl2 + H2O x 2x CaO + 2HCl  CaCl2 + H2O y 2y 2x + 2y = 6 × 0.1008 = 0.6048 ......(II) % of BaO = 0.03 × 58 × 100 = 60.8% 2.86 C2H6 28.3 × 1× 10 -4 100 = 8.8 × 10 -8 mole 23. n = [12 × 12 + 4 + 35.5 × 4 + 16 × 2] 24. 6LiH + 8BF3  6LiBF4 + B2H6 2 2 0.25 MOLE CONCEPT 25. Al + 3HCl  AlCl3 + 3 H2  2 1.5x + y = 0.04925 x mole 1.5x x + 27 + y × 24 = 1 Mg + 2HCl  MgCl2 + H2 x × 9 + y × 8 = 0.33 y mole y 30. n N 2 ........(I) m N2  0.0946 gm % N2 = ........(II) 31. (a) M = CaCl2 + NaCl Na2 CO3 32. Xethanol = CaCO3 33. CO + CaO mNaCl = 6.7889 gm % NaCl = 67.9% 27. n O 2  625 nC = 1 mole n O2 nC  0.6 2 5 28. 1 O  CO2 2 2 34. O3  2  n O 2  n CO  2n CO 2 .......(I) 2  n C  n CO  n CO 2 .......(II) 35. n CO  2 8 2 1  n CO 2  4 4 11 element mass per 100 gm mole s imples t ratio C 58.77 58.77 / 12 5 H 13.81 13.81 / 1 14 N 27.42 27.14 / 2 2 E.F. = C5H14N2 = 102 = M.F. 3 O 2 2 80 20 3  2 0  80  1 10 2 Increase in volume =110–100=10ml O2 + C  CO + CO2  46 / 46 = 0.25 46 / 46 + 54 / 18 x –x/2 x Cl+4 + 2O2  CO2 + 2H2O () y –2y y x  2 y = 6.5 .....(I) 2 x+y=7 .....(II) x + y + z = 10 .....(III) x = 5 ml y = 2 ml z = 3 ml 1.62 = 0.02892 56  0.02892  111 = 3.211 g n CaCl2  n CaO = m CaCl2 0 .0 946  10 0  66.7% 0.14 4 / 40 = 0.5 0.2 5.3 / 106 = 0.5 (b) M = 0.1 0.365 / 36.5 = 0.2 (c) M = 0.05 x  27  100  54.6% 1 Mg = 45.4% % Al = 26. (774.5 - 14.5) 82.1 × 760 1000 = 3.3786 × 10 -3 mole = 0.081 × 300 Empirical formula Al K S  10.5 15.1 24.8 = KAlS2O8     0.388  1 0.387 1 0.775 2 O Elements 49.6 Mass percentage 3.1 Mole ratio 8 Simple ratio Empirical formula weight = 258 Fromweight lossinformation: 54.4ganhydroussalt  45.6 g H2O  258 g anhydrous salt  216.26 g = 12 mol H2O  Empirical formula of hydrated salt= KAlS2O8 . 12 H2O 90 36. 1.0 mole of KClO3  3.0 mole of Zn Empirical formula = CaC2N2S2, Empirical formula weight =156 Hence, molecular formula =CaC2N2S2 5.104 3× 5.104 mole KClO3  122.5 122.5 mole of Zn = 3× 5.104 × 65 = 8.124 g Zn 122.5 37. Apply conservation of moles of silver before and after precipitate exchange reaction as : 1.8 x 2.052 - x = + 143.5 188 143.5 where, x is mass of AgBr in mixed precipitate.  x = 1.064 1 1 x Also, moles of CuBr2= moles of AgBr = × 2 2 188 1 x  Mass of CuBr2 = × × 223.5 = 0.6324 2 188 (on substituting x) Mass % of CuBr2 = 34.18 38. Moles of NaCl in sample = 0.01 = moles of AgCl from NaCl in precipitate 2 = 0.01393 143.5  Moles of AgCl from KCl= 0.00393 = moles of KCl  Mass of KCl in sample = 0.00393× 74.5= 0.2928g Mass % of KCl in the sample = 29.28 Total moles of AgCl precipitate = 39. Let the mixture contain x g CuSO4 . 5H2O. x 5 x × 159 + × 120 = 3  x = 3.72 249 246  Mass percentage of CuSO4.5H2O = 74.4  41. Working in backward direction In the last step moles of(AgBr+AgI) = moles of AgI 0.4881 - x x 0.5868  + =  x = 0.0933 g 188 235 235 Mass % of NaI 0.0933 100 × 150 × = 29.77 235 0.2 Now subtracting mass of AgI from 1st and 2nd precipitate gives Mass of (AgCl + AgBr) = 0.3187 g and mass of AgBr = 0.3948 g = 0.3187 - y 0.3948 y + = y= 0.245g 188 188 143.5 Mass % of NaCl Again  0.245 100 = 50 × 58.5 × 143.5 0.2 Mass % of NaBr = 20.23 = 42. Weight loss is due to conversion of NaHCO 3 into Na2CO3 : 31 g weight is lost per mole of NaHCO3.  0.3 g wt. loss from 0 .3 0 .3 mol of NaHCO3 producing 31 62 moles of Na2CO3. Total moles of carbonate =15× 10–3 3 = 0.01 620 Mass of Na2CO3 in original sample = 1.06 42.4 % Na2CO3  Moles of carbonate in original sample = 0.015 – 43. If M is molar mass of (CH3)x AlCly 40. Mass % of Ca 0 .643 x × 16 = 0.222 M 0 .6 4 3 y and m(AgCl) = × 143.5 = 0.996 M 0.16 100 × 40 × = 25.6 100 0.25 Mass % of S 0.344 32 × 100 × = 41 = 233 0.115 Mass % of N m(CH4) = 0.155 14 ×100 × = 17.9 = 17 0.712 AlSo M = 15x + 27 + 35.5 y = 15x + 27 + =  Mass % of C = 15.48 Now : Elements Ca Mass % 25.6 Mol ratio 0.64 Simple ratio 1 S 41 1.28 2 dividing :  N 17.9 1.28 2 C 15.48 1.29 2 x  2, y 3 5 .5 x = 32.75x + 27 2 0 .6 4 3 x  1 6 = 0.222  x = 1.98  2 y=1 3 2 .75 x  2 7 44. Mass of AgCl = 0.09 × 143.5 = 12.915 g which is 95.77 % of total ppt.  Total mass of precipitate = 13.485g and mass of impurity = 0.57 g MOLE CONCEPT  Mass of NaCl + KCl = 5.9 g 5.9 - x x + = 0.09 58.5 74.5  x = 2.94 g NaCl, 2.96 g KCl m (Na2O) = 1.558 g  m% (Na2O) = 31.16 m (K2O) = 1.867 g  m% (K2O) = 37.34  45. In order to obtain maximum yield from a reaction, the reactants must be supplied in stoichiometric amount So that no reactant should be left unreacted. The balanced chemical reaction is Pb(NO3)2 + 2KI PbI2 + 2KNO3 Let x g of KI is taken x   moles of Pb(NO 3) 2 present 166 x = 2  166  moles of KI =  x 5-x x = x = 2.5 g mass of PbI2 = × 2 × 166 330 332 460 = 3.464 g 46. Mass of uranium in the sample = 1.48  23 8 = 0.894 g 394 Mass % of uranium in the sample = 89.4 UO2(NO3)2+Na2C2O4+xH2OUO2(C2O4) xH2O+ 2NaNO3 m mol 3.756 2.985 Here Na2C2O4 is the limiting reagent, therefore, m mol of UO2(C2O4).xH2O formed is 2.985.  M(UO2(C2O4 )).xH2O = 1.23 × 1000 = 412 2.985 = 238 + 32 + 88 + 18 x x = 54 3 18 47. Volume of smallest cell = r2l = (60 × 10–8 cm)2 (6000 × 10–8 cm) = 6.785 × 10–17 cm3 mass of one smallest cell = 7.6 × 10–17 g  Molar mass of mother cell = 7.6 × 10–17 × 24 × 60 × 6.023 × 1023 = 6.6 × 1010 amu 48. Let the sample contain x g Mohr's salt [FeSO4 (NH4)2 SO4 . 6H2O]  x 0.5  x 0.75 2   392 132 233 Solving x = 0.23 g  Mohr's salt = (NH4)2SO4 = 54% 0.23 × 100 = 46 %, 0.50 Also moles of Fe in 0.2g sample = x 0 .2  3 92 0.5 = 2.347 × 10–4  mass of Fe 2O 3 obtained on ignition of 0.2 sample 2 .3 4 7  1 0 4 = × 160 = 18.77 mg 2 49. Smallest volume of AgNO3 would be required when the entire mass is due to highest molecular weight constituent. Hence, for smallest volume, the whole mass should be of BaCl2 . 2H2O 0 .3 m mol of BaCl2.2H2O= ×1000= 1.229 m mol 244 m mol of AgNO3 required = 2 × 1.229 = 2.458 2 .4 5 8 Volume of AgNO3 required = = 16.38 mL (smallest) 0 .15 Largest volume of AgNO3 would be required when entire mass is due to lowest molecular weight constituent, i.e., NaCl. 0 .3 m mol of NaCl = × 1000 = 5.128 = m mol of AgNO3 5 8 .5 required 5 .1 2 8  Volume of AgNO3 required = = 31.18 mL (largest) 0 .1 5 50. Mixture(N2,NO2,N2O4)has mean molar mass=55.4. x y z Given : N2O4  2NO2 z 2z  55.4 = 28x + 46(y + 2z) x+y+z wt.× mole    mean molar mass = Total mole    Given : x + y + z = 1 (mole) So 55.4 = 28x + 46 (y + 2z) ...(1) 28 x  46 (y  2 z)  39.3 = x  y  2z  39.6 (x + y + 2z) = 28 x + 46 (y + 2z) From eq (1) & x + y + z = 1 or 39.6 (1 + z) = 59.4 5 9 .4 or 1 + z = 3 9 .6 or z = 0.4 from eq. (1) 55.4 = 28x + 46(y + 2z) z  0.4 put 55.4 = 28x + 46y + 36.8 28x+ 46y = 18.6 52. (CH2)n + 4nCoF3 (CF2)n + 2nHF + 4nCoF2 ...(1) 2CoF2 + F2  2CoF3 wt.  F = 19, C = 12, Co = 59, Mwt. (CF2)n = 50n from eq. (1) (CF2)n = 4nCoF2 ...(2)  x+y+z=1 x + y + 0.4 = 1 ( z = 0.4) x + y = 0.6 ...(3) eq. (2) × 1 ..... eq. (3) × 28 28 x + 46 y = 18.6 28 x + 28 y = 16.8 – – – w E 1000 50 n  x+y+z=1 x  0 .5 = 80× 97  y z y  CxHyOz +  x    O2 xCO2 + H2O  4 2 2 10mL + 100mL 0 After reaction – + 0 20  1 .52 40 1.52 = 0.76 kg 1  53.(a) ...(1) 4 4 4–2 4–4 2 2 0 2 2 A2B4 + A2  2 A3B4 2 2 2–2 2–1 2 1 2  A2 = 1, A2B4 = 2 Mw 2 Mw = 2 × 23 = 46 12x + y + 16z = 46 12 × 2 + y + 16z = 46 y + 16z = 22 from eq. (1) & (2) y – 2z = 4 y + 16z = 22 – – – 3 A + 2B2 A3B4 2 2 A2 + 2B2 A2B4 Initial After x 2 V.D. of compound (CxHyOz) = 23  V.D.= (CF ) moles= 1000  20 moles of (CF ) 2 n 2 n 50n n   20 20 2n HF = × 2n = 40 mol n  40 y z  + 100–10  x    10x –  4 2 y z  1 4 2 y – 2z = 4 Property of KOH has to absorbed all CO2.  10x = 20 1  38 × 80 × 97 = 1520 g = 1.52kg. 2  97 2HF  H2 + F2 y z  100 – 10  x    + 10x = 90  4 2 Mw = 46 3 A + 2B2 A3B4 2 2 (b) A2 + 2B2 A2B4 ...(2) 1 2 2 After 0 2–1 0.5 1 0.5  A2B4 = 0.5, B2  Initial – 18z = –18 (c) A2 + 2B2 A2B4 z 1 , y  6 Initial 1.25 2 After 1.25 – 0.25 – Molecular formula = C2H6O. w 4n  97 1 00 0  4n  97 5 0n w = 80 × 97 g (CoF2)  2CoF2 + F2  2CoF3 2 × 97  1 × 38 y  0.1 Given vol. = w= 18 y = 1.8 51. w E 1 1 – 1 3 A + 2B2 A3B4 2 2 MOLE CONCEPT 2 A2B4 + A2  2 A3B4 1 0.25 1–0.5 – 0.5  A2B4 = A3B4 = 0.5 54. (a) 1L KMnO4  79% (w/v) i.e. 100 mL solution contain 79 g KMnO4 wt. 79 moles of KMnO4 = = = 0.5 Mw 158 0.5  1000  5 M 100 HCl10%(w/w)i.e.100 g solution contain 10g HCl D = 1.825 g/mL Molarity (M) = V= M 100  D 1.825  1000 10  1.825  1000 =5M 36.5  100 2KMnO4+16HCl2KCl + 2MnCl2 + 8H2O + 5Cl2 M × V 1 M × V1 5×1 5×9 5 45 – 5 12.5 Molarity = 80 = 10 mol. 100 (b) 2KMnO4+16HCl  2KCl+2MnCl2 + 8H2O + 5Cl2 Cl2 = 12.5 × 1× (c)  = = 710  25 L 28.4 vol. of water treated vol. of total feed 25 25  2.5 = 1 9 vol. of KMnO 4  HCl 55. D = 1.03 g/cm3 2.8% NaCl  100 g solution contain 2.8 g NaCl. V= 100 L 1.03  1000 1 L  2.8  1.03  100 0 g 100 2.8  1 0.3  0.493 58.5 M 2V 2 = M 1V 1 0.493 × 106 = 5.45 × V1 V1 = 9 × 104 So water evaporated =106 – 9 × 104 = 9.095 × 105 L moles = 56. Let free SO3  xg SO3 in form of H2SO4  x  98 = 1.225 x 80 So total x + 1.225 x = 100 x = 449.49 4 4.94  18 = 10..11 g % oleum 80 = 100 + 10.11 = 110.11% water required = 57. 100 mL milk  4mL fat 1 L milk  40 mL fat density of fat = 875 kg/m3 = 0.875 g/mL mass of fat = 40 × 0.875 = 35g fat free milk mass = 1035 – 35 = 1000g Vol. = 1000 – 40 = 960 mL = 1000 = 1.0416 g/mL 960 H 2 SO 4  a    b 58. SO 3 SO 2  c  a+b+c=1g SO2  1.5% so, C = 0.015 g  SO2 a + b = 0.985 g H2SO4 + 2NaOH  Na2SO4 a 98 SO3 + 2NaOH  Na2SO4 + H2O b 80 SO2 + 2NaOH  Na2SO3 + H2O 0 .0 1 5 64 b 0.015   a     = 23.47 × 10–3 64   98 80 0.0102 a + 0.0125 b + 0.00234 = 0.011735 a + 1.225 b = 1.1275 a + b = 0.985 0.225 b = 0.1425 b = 0.633 g  SO3 a = 0.35 M g  H2SO4 Combined SO3 = 0 .3 5 1 4 × 80 = 0.2868g 98 59. Volume =1 × 3 × 300 × 6 × 10–10 = 5.4 × 10–7 m3 = 0.54 cm3  = 1 g/cm3 mass = 0.54 g n(CH3)2SiCl2+2nOH–2nCl–+nH2O+ [(CH3)2SiO]n w 129 EXERCISE - 5 Part # I : AIEEE/JEE-MAIN 1. w  {7 4 n} 129n 74 w = 0.54  w = 0.9413 g 129 60. CH4 a + 2O2  2a CO2 + a 3.  3n  1  CnH2n – 2 +   O  nCO2 + (n – 1) H2O  2  2  3n  1    (20–a) n(20–a)  2  For methane a + n (20 – a ) = 40 For oxygen 5. ...(1) 6.    3n  1  100  2a   2  (20  a)  40   475  4 95 147 5 +1 –1 moles of MgCl2 = 159.4 = 1.678 95 Total moles of Cl– = (5 + 1.678) × 2 = 13.356 volume of solution = 1 L Molarity of Cl– = 13.356 M + 1 +6 –2 0 +6 ...(2) 7. +4 +1 –2 8. +4 On the other hand, in this reaction, oxidation number of S decreases by two, thus this is a reduction reaction and H2SO4 behaves as oxidizing agent. 8 moles of O-atom are contained by 1 mole Mg3(PO4)2. Hence, 0.25 moles of O-atom = 1 × 0.25 = 3.125 × 10–2 8 mole Mg3(PO4)2. M molality (m) = 1000d - MM × 1000 1 M = Molarity, M1 = Molecular mass of solute, d = density = 2.05  1000 = 2.28 mol kg–1 (1000  1.02) – (2.05  60) 2Al(s) + 6HCl(aq)  2Al3+(aq) + 6Cl¯(aq) + 3H2(g) 3 mole H2 from 6 mole HCl consumed.  1 mole H2 from 2 mole HCl consumed. 1/2 mole(11.2 Lit) H2 from 1 mole HCl consumed. 10. 3.6 M solution means 3.6 mole of H2SO4 is present in 1000 ml of solution  Mass of 3.6 moles of H2SO4 is = 3.6 × 98 g = 352.8 g  Mass of H2SO4 in 1000 ml of solution = 352.8 g Given, 29g of H2 SO4 is present in 100 g of solution 9. = 159.4 g MgCl2 M1V1  M2 V2 V1  V2 1.5  480  1.2  520 = = 1.344M 480  520 In the reaction. 2 H I + H2 S O4  I2 + SO2 + 2H2 O Final molarity = Ist half reaction : 2HI  I2 –1 0 In this reaction oxidation number of I increases by one, thus this is an oxidation reaction and HI behaves as a reducing agent. IInd half reaction : H2SO4  SO2 61. CaCl2  5M = 555 g in 1 L solution or in 1050g solution wt. of (solvent + MgCl2) = 1050 – 555 = 495 g MgCl2  5 m 1000 g solvent  5 mol of MgCl2 = 5 × 95 = 475 g MgCl2 i.e., 1475 (solvent + MgCl2) 475g MgCl2 495 (solvent + MgCl2)  6.02  1120 / 6.02  10 23 = 0.01 M 100 / 1000 The mass of one mole of a substance will remain unchanged. Let the oxidation state of Cr is x. x + 4(0) + 2 (–1) = + 1 x – 2 = + 1 or, x = + 1 + 2 = + 3. 4. (20–a)  3n  1  2a +   (20  a ) = 60  2  2a + 30 n – 1.5na – 10 + 0.5 a = 60 2.5 a – 1.5na + 30 n = 70 2.5 a – 1.5n (a – 20) = 70 2.5 a + 1.5n (20 – a) = 70 from (1) & (2) a = 10 n=3 C3H4 % composition  50% Moles of solute Vlt = 2. 2H2O Molarity = MOLE CONCEPT  352.8 g of H2SO4 is present in solution 1216 Mass Now density = = = 1.216 g/mL = 1.22 g/mL 1000 Volume 11. Xethyl alcohol = 5.2 1000 = 0.086 5.2 + 18 0.01 0.01 / 60 = 60  0.3 0.3 = 5.55 × 10–4 m. 12. Molality = ; d = 1 g/ml 14. % of N = 10  7.5kg 100 1 H atoms are replaced by 2H atoms. So mass gain by person = 7.5 kg 20. Given chemical eqn. M2CO3 + 2HCl 2MCl + H2O + CO2 1 gm (0.01186 mol)  from the balanced chemical eqn. Mass due to 1H is  75  1  0.01186  M = 84.3 gm/mol M moles of solute 13. Molarity = volume of sol. (l) = Hydrogen = 10.0% and Nitrogen = 2.6% Total weight of person = 75 kg 100 × 352.8 = 1216 g of 29 21. 120 ×1.15 ×1000 = 2.05 M 60 ×1120 y  y CxHy(g) +  x   O2(g)  xCO2(g) + H2O () 4  2 no.of oxygen atom in CxHyOz = z 1.4  milliequivalents of acid consumed mass of organiccompound 1 1     Meq of acid consumed =  60   2    20  1 = 10 2   10   y  no. of oxygen atom required for CxHycombustion is  x    4  1.4 10 = 10% 1.4 15. 2C8H7SO3Na + Ca7+  (C8H7SO3)2Ca 2 mole 1 mole 2 × 206 gm take 1 mole of Ca2+  % of N = y  2 =  2x   . 2  1 mole of Ca2+. 412 16. moles of Br = 1 × moles of AgBr mass of Br =  % of Br = 3x  2x : 4x : 3x 2 2 : 4 : 3 Hence C2H4O3 Part # II : IIT-JEE ADVANCED 141  10–3  80 188 141  10 188  1. 80 250  10–3  100 = 24% 17. CH3COOH (0.06M) 50 ml m. moles = 50 × 0.06 = 3 m. moles left = 50 × 0.042 = 2.1 m. moles absorbed = 0.9 mass absorbed = y  y 2x 3x =  2x    z = x +  z = x + 2   4 4 2 x : 2x : 141  10–3 188 –3 1 2 So z =  1gm takes =1× 12x 6 y =1 2x = y for CxHyOz 54 0.9  10–3  60 × 103= = 18 mg 3 3 18. Bonus 19. Mass in the body of a healthy human adult has :Oxygen = 61.4%, Carbon = 22.9% – MnO4 ; x + 4(–2) = – 1 or x = + 7 ; CrO2Cl2 ; x + 2(–2) + 2(–1) = 0 or x = +6. 2. (i) 4.0 M, 500 ml NaCl no. of m moles of NaCl = 500 × 4 = 2000 m moles = 2 moles = 2 moles of Cl– ions as 2Cl–  Cl2 + 2e– So 1 mole of Cl2 is generated. (ii) no. of moles of Na+ = 2 moles So max. wt of Na amalgam (assuming equimolar Na & Hg) = 46 + 400 = 446 gm. (iii) Two moles of e– are required = 2 × 96500 C = 193000 C. 3. 54 Fe  5% 56 Fe  90% Fe  5% Av. atomic mass = x1A1 + x2A2 + x3A3 = 54 × 0.05 + 56 × 0.9 + 57 × 0.05 = 55.95. 57 4. 5. 6. 7. 25 .2  25.25  25.0 75.45 = 3 3 = 25.15 = 25.2 mL number of significant figures will be 3. Average titre value = Only Na & F will show one non-zero oxidation state. These are Na+ & F–. The balance chemical equation is 3Br2 + 3Na2CO3  5NaBr + NaBrO3 + 3CO2 Mole = 120 =2 60 mass of solution = 1120 g 8. V= 1120 112 = L 1.15  1000 115 M= 2  115 = 2.05 mol/litre 112 29.2% (w/w) HCl has density = 1.25 g/ml Now, mole of HCl required in 0.4 M HCl = 0.4 × 0.2 mole = 0.08 mole if v mol of orginal HCl solution is taken then volume of solution = 1.25 v mass of HCl = (1.25 v × 0.292) 1.25v  0.292 mole of HCl = = 0.08 36.5 36.5× 0.08 mol = 8 mL so, v = 0.29 ×1.25 9. Since the vessel is the thermally insulated so q=0 Pext = 0, So w = 0 So U = 0 (ideal gas) Hence T = 0  T = 0  T2 = T1  P2V2 = P1V1 The process is however adiabatic irriversible. So we cannot apply P2 V2  P1V1 Hence ans is (A), (B), (C) R  R = K.NA NA = 6.023 × 1023 × 1.380 × 10–23 J.mol–1.k–1 There are 4 significant figures in each term. Hence, these be 4 significant figure in R. 11. Given 3.2 M solution 10. K =  moles of solute = 3.2 mol Consider 1 L Solution.  volume of solvent = 1 L Psolvent = 0.4 g.mL–1  msolvent = P × V = 400 g 3.2 mol  molality = 0.4 kg = 8 molal 13. Ca(OH)2 + (NH4)2SO4  1584 g 2NH3 + CaSO4.2H2O [Ni(H2 O)6]Cl2 Complex Compound Number of Moles of (NH4)2SO4 = 1584 = 12 moles 132 Moles of NH3 released = 24 moles 952 = 4 moles 238 Number of moles of Gypsum (CaSO4.2H2O) formed = 12 moles Mass of Gypsum formed = 12 172 = 2064 Moles of moles of NiCl2.6H2O = Number of moles of complex formed [Ni(NH3)6]Cl2 = 24 6 = 4 moles Mass of complex formed = 4 232 = 928 g Total Mass = 2064 + 928 = 2992 g 14. PbS + O2  Pb + SO2 Mole 103 32 Moles of Pb formed = 103 32 Mass of Pb formed = 103  207  6468.75gm 32 = 6.46875 kg = 6.47 kg 15. 2MnCl2 + 5K2S2O8 + 8H2O  2KMnO4 + 4K2SO4 + 6H2SO4 + 4HCl .......... (1) 2KMnO4 + 5H2C2O4 + 3H2SO4  K2SO4 + 2MnSO4 + 8H2O + 10CO2 .......... (2) Mass of oxalic acid added = 225 mg Milimoles pf oxalic acid added = 225  2.5 90 From equation (2) Milimoles of KMnO4 used to react with oxalic acid = 1 and milimoles of MnCl2 required initially = 1  Mass of MnCl2 required initially = 1 × 126 = 126 mg