MOLE CONCEPT

Mole Concept Mole Concept TG: @Chalnaayaaar SOLUTIONS Exercise-I (Conceptual Questions) 5. Ans. (1) 1. Ans. (4) Mass Moles Number of atoms 16 g oxygen (O2) 4g H2 4 = 2 2 × NA × 2 = 4NA 2 nO2 16 1 = 32 = 2 16g O2 16 1 1 32 2 2 =  NA  2 = NA Number of O-atoms 28g N2 1 × NA × 2 = 2NA = nO2 × NA × atomicity 28 =1 28 = 1  6.022  1023  2 18g H2O 18 =1 1 × NA × 3 = 3NA 2 18 = 6.022 × 1023 6. Ans. (3) 2. Ans. (4) Number of neutrons nNH3 = 4.25 = 1 = n × NA × (number of neutrons) 17 4 = 1.7  NA  7 Number of atoms 17 = nNH3 × NA × atomicity = NA  7 10 1 = 4  NA  4 7. Ans. (2) = NA = 6 × 1023 Number of atoms of oxygen 3. Ans. (4) = n × NA × number of atoms Mass Mole Number of atoms = 5.6  6.022  1023  2 22.4 [n × NA × atomicity] = 3.01 × 1023 atoms. 1gO 1/16 1  NA  1 = NA 8. Ans. (2) 16 16 Number of atoms in [O3] 1 g O2 1/32 1  NA  2 = NA = n × NA × number of atoms 32 16 = 8  NA  3 1 g O3 1/48 48 1 N 3 NA 48  A  = 16 NA 6.0221023 2 2 = = All have the same number of atoms. 9. Ans. (3) 4. Ans. (3) Sum of protons, electrons and neutrons in 12g Number of atom in 0.5g atom of nitrogen of 162C = n × NA × [sum of e, p & n] = 0.5 × NA × 2 = NA = 12 × NA × 18 Number of atom in 8g oxygen 12 = 8 × NA × 2 = NA = 18 × 6.022 × 1023 16 = 1.08 × 1025 TG: @Chalnaayaaar www.allendigital.in [1]  Digital

NEET : Chemistry 10. Ans. (2) 9 TG: @Chalnaayaaar 0.05 1 “U” atom = 238 amu 1 mol of B2A3 = =180g 1 amu = 1.66 × 10–24g 2b + 3a = 180 …(1) 1 “U” atom = 238 × (1.66 × 10–24g) molar mass of B2A = (2b + a) ≃ 3.94 × 10–22g 0.1 mol of B2A → 10g 11. Ans. (2) 1 mol of B2A → 10 = 100 0.1 1 H2O molecule = 18 amu Actual mass of 1 H2O molecule  2b + a = 100 …(2) = 18 × (1.66 × 10–24g) On solving a = 40, b = 30 ≃ 2.99 × 10–23 17. Ans. (3) 12. Ans. (3) 5.6 1 1 molecule CH4 = 16 amu nO2 = 22.4 = 4 1 amu = (1.66 ×10–24g) 18. Ans. (4) mass of 1 molecule of CH4 = 16×(1.66 × 10–24g) = 2.66 × 10–23g ngas = 2.24 = 0.1 22.4 13. Ans. (1) Molar mass = wgas 1g atom of C = 1 mole of C n gas = 12g 14. Ans. (4) = 4.4 = 44 0.1 Mass Moles Number of atoms 4.4g CO2 4.4 = 0.1 0.1 NA CO2 & N2O 44 19. Ans. (1) 3.4g NH3 3.4 = 0.2 0.2 NA Mass Moles Number of atoms 17 1g CO2 1 1 NA 44 44 1.6g CH4 1.6 = 0.1 0.1 NA 16 1g N2 1 1 NA 28 28 3.2g SO2 3.2 = 0.05 0.05 NA 64 1g O2 1 1 NA 32 32 15. Ans. (1) nNH3 = 4.25 = 1 1g H2 1 1 NA 17 4 2 2 Number of NH3 molecules 20. Ans. (1) = 1 × 6.022 ×1023 mass of gas = w(g) 4 Vol. of gas = V mL (at STP) = 1.505 ×1023 moles of gas = mass = w mol.wt. mol.wt. 16. Ans. (3) Let atomic wt. of A = a moles of gas = VSTP(ml) 22400 Atomic wt. of B = b Molar mass of w VSTP(ml) mol.wt. 22400 B2A3 = (2 × b) + (3 × a) = = mass of 1 mole substance Mol. wt. = w  22400 0.05 mol B2A3 → 9g V(ml) TG: @Chalnaayaaar www.allendigital.in  Digital [2]

Mole Concept 21. Ans. (2) 26. Ans. (1) TG: @Chalnaayaaar Initial mass of H2SO4 = 98mg = 98 × 10–3g Vol. of 1 mole N2 [at N.T.P.] = 22.4 L Initial moles of H2SO4 = 9810−3 = 10−3mol mass of 1 mole N2 = 28g 98 Density of N2 = mass 28g = 1.25g L−1 Number of molecules removed = 3.01 × 1020 volume = 22.4L Number of moles removed 27. Ans. (2) = 3.011020 = 0.510−3mol number of carbon atoms 6.021023 = nc × NA × atomicity Moles of H2SO4 left = 10–3 – [0.5 × 10–3] 1.210−3 = 0.5 × 10–3 = 12 × 6.022 × 1023 ×1 22. Ans. (3) = 6.022 × 1019 Formula of gas = [CO]x 28. Ans. (3) Mol. wt. of gas = [12 + 16]x XY = 28x mass w w 28x = 2 × V.D. moles 28x = 2 × 70 w w 30 20 28x = 140 x = 140 =5 simple 1  60 w  60 28 ratio × 30 20 23. Ans. (4) 60 Mol. wt. = 2 × 11.2 = 22.4g 23 moles of gas = 2.4 Molecular formula = X2Y3 22.4 29. Ans. (1) Vol. of gas (at STP) = moles × 22.42 mass moles simples = 2.4  22.4 = 2.42 ratio 22.4 S 50 50 1.5 1.5 1 24. Ans. (3) 32 = 1.5 = ngas = VSTP(ml) O 50 50 = 3.125 3.125 2 22400 16 1.5 = 1.1210−7 = 1 10−11 Empirical formula = SO2 22400 2 30. Ans. (3) molecules of gas = moles × NA Mass% atomic simples = 1  10−11  6.02  1023 ratio ratio 2 C 80 80 6.66 = 3.01 × 1012 12 = 6.66 6.66 = 1 25. Ans. (3) H 20 20 20 1 6.66 Number of electrons in = 20 3 NO3− = nNO3− NA (number of electrons) = 3.110−3 6.0221023 32 C1 H3 62 1 :3 C2 H6 = 9.6 × 1020 TG: @Chalnaayaaar  Digital www.allendigital.in [3]

31. Ans. (4) 36. Ans. (4) NEET : Chemistry TG: @Chalnaayaaar Molecular formula of glucose = C6H12O6 Mass% atomic ratio simple CH O ratio 6 : 12 : 6 P 1.24g 1.24 = 0.04 4 1: 2 :1 31 E. F. of glucose = CH2O S (2.2–1.24) 3 32. Ans. (3) = 0.96 0.96 = 0.03 32 Mass% atomic simplest E.F. = P4S3 M 60 O 40 ratio ratio 37. Ans. (4) E. F = MO 60 = 2.5 2.5 = 1 Mass% Atomic ratio simple ratio 33. Ans. (1) 24 2.5 I 254 254 = 2 2 127 40 = 2.5 2.5 = 1 16 2.5 O 80 80 5 10 = 5 Formula = I2O5 Mass% atomic simples 38. Ans. (2) ratio ratio E. f.  obtain from simples ratio of atom C 38.8 38.8 = 3.23 3.23 = 1 Gives Cr = 4.8 × 1010 , 0 = 9.6 ×1010 12 3.23 Simplest ratio 1 : 2 H 16 16 16 Empirical formula = CrO2 1 3.23 = 16 5 39. Ans. (1) N 45.2 45.2 3.23 3.23 = 1 % of S  3.4 = 132 100 14 3.23 Min.mol.wt. E.F. = CH5N or CH3NH3 Minimum molecular wt. = 32100 941.176 3.4 34. Ans. (2) 40. Ans. (3) Mass% atomic simplest % of N = mass of N  no.of atoms 100 ratio ratio molecolar wt. of Insulin X 50 50 = 5 5 = 2 28.9 = 14  n 100  n = 28.9194 4 10 2.5 194 14100 Y 50 50 2.5 2.5 1 41. Ans. (3) 20 2.5 = = Mass atomic ratio simples Simplest formula = X2Y ratio 35. Ans. (2) C 38.71 38.71 = 3.22 3.22 = 1 12 3.22 E. F. of glucose = C6H12O6 H 9.67 9.67 9.67  6 1 = 9.67 3.21 = 3 E. F. = CH2O O 51.62 51.62 3.22 16 3.22 CH3COOH  C2H4O2 = 3.22 = 1  2 E.f.= CH3O E. F. CH2O TG: @Chalnaayaaar www.allendigital.in  Digital [4]

Mole Concept 42. Ans. (3) 46. Ans. (4) TG: @Chalnaayaaar C3H8 + 5O2 → 3CO2 + 4H2O 2 Al(s) + 3 O2(g) → Al2O3(s) 2 n= 286 = 6.5 44 2 mol of Al reacts with 3 mol of O2 to produce 2 1 mol C3H8 → 4 mol of H2O 1 mol of Al2O3 6.5 mol C3H8 → 4 × 6.5 mol of H2O 47. Ans. (3) nH2O = 26 C(s) + CO2(g) → 2CO(g) wH2O = 26 × 18 = 468g t=0 –1 – 43. Ans. (3) t = t – 1 – x 2x aA + bB → cC + dD Vol. after = 1 – x + 2x = 1.4 mass ratio is not same as coffi. Ratio so reaction x = 0.4 statement is wrong statement for this VCO2 left = 1 – 0.4 = 0.6 L reaction. VCO prepared = 2 × 0.4 = 0.8 L 48. Ans. (4) 44. Ans. (3) C+ CO2 → 2CO 25 26 CC C8H18 + 2 O2 → 8CO2 + 9H2O Mass of C3H18 = 0.8 × 1.425 1 Vol. of CO2 → 2Vol. of CO = 1140g 26 CC of CO2 → 2 × 26 CC of CO 1140 VCO = 52CC 114 nC3H8 = = 10 49. Ans. (2) 25 4Al + 3O2 → 2Al2O3 2 3 mol O2 1 mol C8H18 → mol O2 → 4 mol of Al 1 mol O2 10 mol C8H18 → 25 × mol O2 → 4 mol of Al 2 3 250 1 mol O2 → 4  1 mol of Al 2 2 3 2 nO2 = = 125 4 45. Ans. (2) n Al = 6 3 2 9 2 2Al + O2 → Al2O3 w Al = 4  27 = 18g 6 9 1 3 n= 27 = 3 50. Ans. (4) 2 mol Al → 3 mol O2 3 BaCl2 + 2 Na3 PO4 → Ba3(PO4)2 + 6 NaCl 2 0.5 0.2 1 mol Al → 3 mol O2 L.R. 0.5 0.16 0.2 = 0.1 4 3 2 = L.R. 1 mol Al → 3  1 mol O2 2 mol Na3PO4 → 1 mol Ba3(PO4)2 3 4 3 1 mol Na3PO4 → 1 mol Ba3(PO4)2 1 2 nO2 = 4 0.2 mol Na3PO4 1 × 0.2 mol Ba3(PO4)2 → 2 wt. of O2 = 1  32 = 8g nBa3(PO4 )2 = 0.1 4 TG: @Chalnaayaaar  Digital www.allendigital.in [5]

51. Ans. (3) 56. Ans. (4) NEET : Chemistry TG: @Chalnaayaaar 2H2 + O2 → 2H2O Cx Hy +  x + y  O2 → xCO2 + y H2O 2vol. H2 react with 1 vol. of O2  4  2 4 ml H2 will react with 2 ml of O2 Total vol. of O2 = VO2(reacted) + VO2(unreacted) 10ml - 40ml 50ml VO2(total) = 2 ml + 8 ml = 10 ml From stoichiometry, x = 4, y = 5 y= 10 52. Ans. (3) 2 2H2 + O2 → 2H2O Which means C4H10 57. Ans. (2) moles 4 = 2 4 = 1 Molecular weight = v. f. × Eq. wt. 2 32 8 Eq. wt. = Mol. wt . 1 1 v.f. 1 = 8 L. R. 2 1 8 w 2 = L.R. E = 3 nH2O = 1 2 = 1 58. Ans. (1) 8 4 Eq. wt. = Mol. wt . 1 v.f. w H2O = 4 18 = 4.5 53. Ans. (2) E = A n A + 2B → C For an element A = En moles 5 8 59. Ans. (3) L.R. 5 8 = 4 Eq. wt. = Atomic wt. 1 2 valency nC = 4 × 1 = 4 For SCl2 Eq. wt. = 32 = 16 54. Ans. (3) V.f .=2 2 2H2S + SO2 → 3S + 2H2O 32 1 Moles 1.6 = 0.04; 1.51022 = 0.025 For S2Cl2 Eq. wt. = = 32 34 61023 V.f .=1 0.04 0.02 0.025 0.025 60. Ans. (1) 2 1 L. R. = = If Eq. wt. of S in SO2 is 8 then L.R. E.R. Eq. wt. of S in SO3 = 32 = 82 6 3 SO2 will remain in excess 55. Ans. (3) 61. Ans. (2) H2 + Cl2 → 2HCl Atomic weight is not variable for an element. Vol. 12 L 11.2 L 62. Ans. (1) 12 11.2 1g Eq. of a substance is present in 8g O. L.R. 1 1 So in 0.25 mol of O2 E.R. L.R. Mass of O2 = 1  32 = 8g. 11.2 L of Cl2 will react with 11.2 L H2 to 4 product 22.4 L HCl. Now we can say 0.25 mol of O2 contains 1g Eq. Remaining H2 = 12 – 11.2 = 0.8 L of a substance. TG: @Chalnaayaaar www.allendigital.in  Digital [6]

Mole Concept 63. Ans. (2) 70. Ans. (3) TG: @Chalnaayaaar Equivalents of acid = equivalents of base wM = 0.80g x  0.45 = 0.5 20 wH2 = 0.04g 90 1000 0.8  1 x=2 E = 0.04 = 20 64. Ans. (3) 71. Ans. (1) Equivalents of base = equivalents of acid. Eq. wt. of element = mass of element  8 0.5 = 100  0.2 mass of oxygen w 1000 80  8 = 32g w =25 = 20 65. Ans. (4) 72. Ans. (2) Eq. of acid = Eq. of base WCu = WFe ECu EFe 0.126 = 200.1 w 1000 3.2 2.8 w= 0.126  1000 = 63g/Eq. ECu = 28 20  0.1 ECu = 32 66. Ans. (2) Equivalents of acid = equivalents of base. 73. Ans. (3) 2 = 3 m1 = m2 40 w E2 E2 w= 403 = 60g/Eq. E2 = m1  E1 2 m2 67. Ans. (2) 74. Ans. (2) M + 2HA ⎯→ MA2 + H2 moles of H2 = 1.12 1 Molecular wt. of metal = 24 × 2 = 48 22.4 = 20 nmetal = 12 = 1 = 0.25 mass of H2 = 1 2 = 1 = 0.1 48 4 20 10 1 mole of metal = 1 mole of H2 2.4 0.1 w 1 nH2 = 0.25 = VH2 = 0.2522.4 =5.6L w = 2.4 = 24 0.1 68. Ans. (3) Number of Equivalents of metal carbonate 75. Ans. (2) reacts = Eq. of H2SO4 ml Eq. of Acid = ml Eq. of caustic potash 0.84 = 40  0.5 45  1000 = 200 5 w 1000 90/ n w = 42 n 1000 = 1000 69. Ans. (2) 2 Number of Eq. of metal = Number of Eq. of Br n=2 1 = 8.89 76. Ans. (3) w 80 The wt. of two elements which combine with 80 w = 8.89 = 8.99 9 one another are in ratio of their Eq. wt. TG: @Chalnaayaaar  Digital www.allendigital.in [7]

77. Ans. (3) 82. Ans. (3) NEET : Chemistry TG: @Chalnaayaaar mass of metal = mass of oxygen Atomic wt. of element = 6.4 Eq.wt. of metal Eq.wt. of oxygen specific heat 68 = 32 = 6.4 = 29.9 30 w 8 0.214 w = 68 = 17g 83. Ans. (3) 4 w1 = w2 78. Ans. (2) E1 E2 Since wt. of hydrogen produced is equal in 0.638 2.16 E1 = 108 both. So w Ca = w Zin E1 = 32.69 ECa EZn Hence atomic wt. will be ECa = w Ca  EZn = 32.69 × 2 w Zn = 65.38 1.6 ECa = 2.6 32.6 = 20g 84. Ans. (3) 79. Ans. (3) Molecular mass of carbonate = (m + 12 + 48) At. wt. of metal = molecular wt. – At. wt. of Cl % of C in compound = 6.091% = 74 – 35.5 6.091 (m + 12 + 48) = 12 100 = 39 M = 12100 − 60 wM w Cl 6.091 EM = ECl M = 137 EM = 39 35.5 = 39 85. Ans. (3) 35.5 WCl2 = 71g 80. Ans. (2) Wmetal = 111 – 71 = 40 g Molecular mass = 2 × V.D. 4035.5 = 2 × 59.25 Em = 71 = 20 = 118.50 g Atomic wt. = Em × V.F. = 20 × 2 = 40 amu let x atoms of Cl in molecule of chloride mass of Cl = x ×35.5 86. Ans. (4) 4 g of element combine with 1 mole Cl At. wt. Eq. wt. mass of element = x × 4 V.F. = Total mass of 1 mole chloride 27 9 = (x × 4) + (x × 35.5) = 118.50 g = =3 39.50x = 118.50 x=3 MCl3 81. Ans. (2) 87. Ans. (1) Atomic weight = 6.4 = 64 Molecular wt. = 2 × V.D. 0.1 = 2 × 50 = 100 g/mol nf = A 64 =2 71 % by wt. Cl Ew = 31.8 = x At.wt. of Cl 71 mass of metal chloride = 100 Exact atomic wt. = nf × Ew = 2 × 31.8 = 63.6 TG: @Chalnaayaaar [8] www.allendigital.in  Digital

Mole Concept Mass of metal chloride 92. Ans. (1) TG: @Chalnaayaaar = 35.5n +x Molecular wt. because it is wt. of 1 mole 35.5n + x = 100 substance also 22.4 L is wt. of 1 mol n 35.5 = 71 n = 2 substance. 100 100  93. Ans. (3) 35.5 × 2 + x = 100 moles of vapour = 112 = 0.005 22400 x = 29 moles of vapour = moles of liquid. 88. Ans. (3) wt. Atomic wt. = 6.4 Molecular wt. of liq. = moles 0.25 0.39 = 25.6 = 0.005 = 78 g V.F. = At. wt. 94. Ans. (3) Eq. wt. wt. = 25.6 2.13 2 Molecular wt. = moles 12 = Exact atomic wt. = 2 × 12 = 0.2  22400 = 24 56 89. Ans. (2) = 80 g/mole. 95. Ans. (4) V.D. = density of gas mole = 67.2 density of the H2 gas 22400 = 0.001293 = 14.3 molecular mass = 51010−3 22400 0.000089 67.2 90. Ans. (3) = 170 Relative density = density of substance 96. Ans. (3) density of CH4 moles of gas = 5 22.4 mass of substance = mass of CH4 molecular mass = 6.25  22.4 5 4 = x = 28 g 16 97. Ans. (1) x = 64 moles = 224 = 10−2 22400 91. Ans. (2) CP =R = 1.4 molecular mass = 0.44 = 44 CV 10−2 Now the gas is diatomic = 2 × V. D. So, gas is N2O = 2 × 16 98. Ans. (1) = 32 moles of gas = 1 22.4 molecular wt. Atomic wt. = atomicity molecular mass = 1.16  22.4 1 = 32 = 16 = 25.98 ≃ 26 2 C2H2 TG: @Chalnaayaaar  Digital www.allendigital.in [9]

99. Ans. (3) 109. Ans. (3) NEET : Chemistry TG: @Chalnaayaaar Atomic wt. of metal = Eq. wt. × valency H2 : He : O2 : O3 = 32.7 × 2 = 65.4 moles 1 : 1 : 1 : 1 22.4 22.4 22.4 22.4 For bivalent metal MCl2 65.4 + 71 = 136.4 moleculs 1 NA : 1 NA : 1 NA : 1 NA 22.4 22.4 22.4 22.4 100. Ans. (3) 2 NA NA 2 NA 3 NA Let M is the molar mass of element. atoms 22.4 : 22.4 : 22.4 : 22.4 2m g element combine with 48g oxygen 110. Ans. (2) 2m  8 = m element combine with 8g of Mgas 0.44 48 3 MN2 0.28 = oxygen m = 9 Mgas = 0.44  28 3 0.28 m = 27 = 44 molecular mass = 2(27) + 3(16) = 102g/ 111. Ans. (2) mole. Entire mass of ethylene is converted to 101. Ans. (2) polyethylene so according to LOMA 100 g Dalton 112. Ans. (4) 102. Ans. (4) Conservation of mass SnCl2 113. Ans. (3)  show law of multiple proportions. AB SnCl4 NH3 SO2 103. Ans. (4) Moles x x 22.4 22.4 Gaseous volumes Same moles. 104. Ans. (2) Multiple proportions. Exercise-II (Previous Year Questions) 105. Ans. (4) 1. Ans. (4) None of these. [law of conservation of mass] 2H2 + O2 ⎯→ 2H2O 106. Ans. (2) Nuclear reaction moles 10 = 5 64 = 2 2 32 107. Ans. (3) L.R. x1 = 5 x2 = 2 = 2 Number of molecule = n × NA 2 1 n is same for all O2, NH3 & CO2. E. R. L. R. 108. Ans. (1) nH2O = 2 × 2 = 4 mole volume  moles 2. Ans. (3) number of atoms= n × NA × atomicity n= molecular mass = 28 =2 0.1 × NA × 3 Empirical mass 14 0.1 × 6.022 × 1023 × 3 1.806 × 1023 M. F. = 2 × CH2 = C2H4 TG: @Chalnaayaaar www.allendigital.in  Digital [ 10 ]

Mole Concept 3. Ans. (4) 1 TG: @Chalnaayaaar 2 number of molecules = n × NA nHCl =  2 = 1 (1) 64  NA = NA (2) 44  NA = NA Mg left in excess = 0.0416 – (0.0175) × 2 64 44 wt. = 0.0066 × 24 = 0.16g (3) 48  NA = NA (4) 8  NA = 4NA 8. Ans. (1) 48 2 H2 + O2 4. Ans. (2) moles 100 = 50 400 = 12.5 2 32 [Cr(H2O)4(Cl2)]Cl ⎯E⎯xces⎯s Ag⎯NO⎯3→ molar ratio = 50 : 12.5 [Cr(H2O)4(Cl)2]+ + AgCl 4: 1 1000.01 1000 = 0.001 9. Ans. (2) 1 mol [Cr(H2O)4(Cl2)]Cl → 1 mol AgCl number of molecules = n × NA 0.001 [Cr(H2O)4(Cl2)]Cl → 0.001 mol AgCl nAgCl = 0.001 (1) 18  NA = NA 5. Ans. (3) 18 H2 O2 CH4 (2) 18NA (3) 18 x x x 32 16 (4) 1.8  NA = NA 2 1.8 10 moles 10. Ans. (4) Vol. x 22.4 x  22.4 x  22.4 If NA change from 6.022 × 1023 to 6.027 × 1020 2 32 16 then the mass of 1 mole of carbon will also 11 1 change. 2 32 16 11. Ans. (2) 16 : 1 : 2 MgCO3 ⎯⎯→ MgO + CO2 6. Ans. (1) moles 20 8 = 1 84 40 5 H2 + Cl2 → 2HCl moles 22.4 = 1 11.2 1 1 mol MgCO3 ⎯→ 1 mol MgO 22.4 22.4 = 2 nMgO = nMgCO 3 = 1 1 5 L.R. x1 = 1 x2 = 21 Mass of MgCO3 = 1  84 = 84 = 16.8 1 L.R. 5 5 nHCl = 1 2 = 1 % Purity = 16.8 100 = 84% 2 20 7. Ans. (1) 12. Ans. (3) X + 2Y → XY2 2Mg + O2 → 2MgO moles 1 0.56 0.1→ 10g mass of 1 mol XY2 = 100g 24 32 3X + 2Y → X3Y2 = 0.041 = 0.0175 0.05 → 9g mass of 1 mol X3Y2 = 180g 0.041 0.0175 X + 2Y = 100 2 3X + 2Y= 180 1 L.R. x1 = x2 = L.R. – 2X = – 80  X = 40 40 + 2Y= 100 = 0.02 2Y= 60  Y = 30 TG: @Chalnaayaaar www.allendigital.in [ 11 ]  Digital

NEET : Chemistry 13. Ans. (3) 1 NA TG: @Chalnaayaaar HCOOH(ℓ) ⎯H⎯2SO⎯4→ H2O(ℓ) + CO(g) 108 108 (2)  NA  1 = n= 2.3 1 (3) 1  NA  1 = NA 46 = 20 24 24 H2C2O4 ⎯H⎯2SO⎯4→ H2O(ℓ) + CO(g) + CO2(g) (4) 1  NA  2 = NA 32 16 n= 4.5 1 90 = 20 18. Ans. (4) 1 mole HCOOH produce 1 mole CO(g) Number of C–atom = 12 × NA ×1 20 20 12 1 mole H2C2O4 produce 1 mole CO & 1 = NA = 6.022 × 1023 20 20 20 19. Ans. (3) mole CO2. C 78 78 = 6.5 6.5 = 1 KOH consume CO2, So gas left are 12 6.5  1 + 1  CO H 22 22 = 22 22 = 3.3 3  20 20  1 6.5 nCO = 1 CH3 10 20. Ans. (1) 1 CaCO3(s) + 2HCl(aq.) → CaCl2(aq.) + CO2(g) + H2O(ℓ) 10 mass of CO = × 28 = 2.8g 500.5 = 0.025 1000 14. Ans. (1) 1 mol CaCO3 react with 2 mole HCl Number of molecules = n × NA 1 (1) 18  NA = NA 0.025 mol HCl react with 2 × 0.025 mol CaCO3 18 nCaCO3 = 0.0125 (2) 0.18  N A = NA mass of CaCO3 = 0.0125 × 100 18 100 (3) 0.00224  NA = NA = 1.25 g 22.4 10000 95 = 1.25 100 (4) 10–3 NA w 15. Ans. (3) w = 125 = 1.32g N2 + 3H2 → 2NH3 95 2 mole NH3 produce by 3 mole H2 21. Ans. (2) so 20 mole NH3 will produce by 60 mole of H2. mass = Vol. × density 2 mass = 2.5 × 2.15 nH2 = 30 = 5.375 = 5.4g 16. Ans. (4) 22. Ans. (1) PV = nRT Fe0.96 O v= 1.8 0.083647 = 5.372 Fe+x3 Fe(+02.96−x) O 18  1 3x × 2(0.96 – x) – 2 = O 3x + 1.92 – 2x –2 = O 17. Ans. (1) x = 0.08 Number of atoms = n × NA × atomicity 0.08 1 @(C1h)a7l1naaNyaA aa1r = NA fraction of Fe = 0.96 = 0.083 = 12 7 TG: [ 12 ] www.allendigital.in  Digital

Mole Concept Exercise-III (Analytical Questions) 6. Ans. (4) TG: @Chalnaayaaar 1. Ans. (3) 2H2 + O2 → 2H2O Number of atoms = n × NA × atomicity 10L 10L Number of atoms of O2 = 1  NA  2 = NA L.R. 10 = 5 10 = 1 32 16 2 1 Number of atoms of O3 = 1  NA  3 = NA VH2O 10  2 = 10L. 48 16 2 2. Same number of atoms in 1 g O2 & O3 but 7. O2 will also left as excess reagent. different number of molecules. Stoichiometric coefficients ratio is same as Statement-I is incorrect Statement-II is volume ratio. incorrect. Ist incorrect, IInd correct. Ans. (2) Ans. (1) Mass of 1 butane (C4H10) molecule is 58 amu. 2H2 + O2 → 2H2O Mass of 1 mole butane = 58 amu × NA H2 + O2 → H2O2 = NA × amu = 1 g = 58g Both reaction follows law of multiple proportion A & R both are correct & R is 1 mole has = 6.022 × 1023 molecules. correct explanation of A. 3. Ans. (1) 8. Ans. (1) 12 27 → nC = 12 = 1 & n Al = 27 = 1 → Average mass of C atom is 12.011 amu & actual mass of C is 12.011 × 1.67 × 10–24g. Moles are same so number of atoms also → Average mass can be calculate by the same number of atoms help of taking Average of Isotopes mass. n × NA × 1 12C, 13C, 14C. 1 × NA × 1 = NA atoms. A & R both are correct also R is correct → Gram atomic mass represents mass of 1 explanation of A. mole of atoms & 1 mol has 6.022 × 1023 9. Ans. (3) atoms. → Equal volume of gas. 4. Ans. (2) 3B → C n = v moles are same for all gases. → 2A + 22.4(L) n= 3 4 But number of atoms can be different for L.R. 3 = 1.5 4 = 1.33 different gases. 2 3 → atom is fundamental quantity which take L.R. in chemical reaction. nC = 4 1 = 4 10. Ans. (1) 3 3 [NH4]3 PO4. B is L.R. & A. is E. R. (a) number of O − atom = 4 = 1 number of Hatom 12 3 Ist correct, IInd incorrect. number of cation 3 5. Ans. (2) (b) number of anion = 1 CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) → 1mole CH4 reacts with 2mole O2 to (c) g − atom of N = 3 produce 1 mole CO2 & 2 mole H2O. g − atom of O 4 → Coefficients ratio will not be same as (d) total number of atom is 1 mol (NH4)3PO4 (5 × 3) + 1 + 4 = 20 NA mass ratio. A, B both are correct. Ist correct, IInd incorrect. TG: @Chalnaayaaar  Digital www.allendigital.in [ 13 ]

NEET : Chemistry 11. Ans. (1) 13. Ans. (1) TG: @Chalnaayaaar 3CaCO3 + 2H3PO4 → Ca3(PO4)2 + (A) nH2SO4 = 20 3H2O + 3CO2 Mass = 20 × 98 = 1960 (S) n = 50 1 49 1 (B) nH2O = 540 = 30 (R) 100 = 2 98 = 2 18 1 1 (C) Number of atoms 6 4 L.R. n1 = = 10−10 6.0221023 2 2 L.R. = 6.022 × 1013 (Q) nCa3(PO4 )2 = 1 (D) Number of amu in 20g 6 1 g = 1 1.6710−24 g 1  mass of Ca3(PO4)2 = 6 310 = 51.67 20 g = 20 1 1.6710−24 g 1  nCO2 = 6 3 = 2 = 0.5 1 = NA amu 1 nH2O = 6 3 =  = 0.5 = 20 NA (P) 2 14. Ans. (3)  mass = 1 18 = 9g Number of NA atoms = 5.3  N A  2 = NA 3 106 10 moles of unreacted H3PO4 (a) Number of NA atoms = 4  NA  1 = NA 40 10 1 2 3−2 1 = 2 − 6 = 6 = 6 (b) Number of NA atoms = 5.85  N A 1 = NA 58.5 10 mass = 1  98 = 16.33 (c) Number of NA atoms = 0.25 NA 2 = NA 6 2 A & C both are correct. (d) Number of NA atoms = 5.6  NA  3 = 0.1NA 164 12. Ans. (1) A & B are correct. (A) number of atoms = n × NA × atomicity 15. Ans. (2) Atomicity for substance can be different. (A) Number of moles of Hydrogen = 8 (B) H2 + Cl2 ⎯→ 2HCl P → 2 moles of C3H4 has 8 mol H atoms. n = 2 = 1 71 = 1 R → 4 moles of N2H2 has 8 mol H atoms. 2 71 (B) Number of moles of carbon = 6 1 mol H2 react with 1 mol Cl2 to produce P → 2 moles of C3H4 has 6 mol C atoms. 2 mol HCl nHCl = 2 Q → 3 moles of CH3COOH has 6 mol C atoms. Mass of HCl = 2 × 36.5 = 71 g (C) Number of moles of oxygen = 6 (C) Reactant which completely consume in Q → 3 moles CH3COOH has = 6 mole O atoms. reaction will act as L.R. S → 2 moles N2O3 has = 6 mole O atoms. (D) [A, B, C] (D) Number of moles of N2 moleculs = 4 R → number of molecule of N in N2H4 = 4. TG: @Chalnaayaaar www.allendigital.in  Digital [ 14 ]