Chapter 1
Some Basic Concepts of Chemistry
Solutions
SECTION - A Objective Type Questions (One option is correct) 1. The total number of ions present in 1 ml of 0.1 M barium nitrate solution is (1) 6.02 × 108 (2) 6.02 × 1010 (3) 3.0 × 6.02 × 1019 (4) 3.0 × 6.02 × 108 Sol. Answer (3)
Number of molecules of Ba(NO3)2 1 0.1 N = 1000 A –4 = 10 NA ∵ 1 molecule Ba(NO3)2 gives 3 ions –4 –4 10 NA molecules will give = 10 NA × 3 = 3 × 6.02 × 1019 = 1.806 × 1020 2. Which of the following have lowest weight? (1) 6.023 × 1022 molecules of glucose (2) 18 ml of water at 4°C
(3) 11200 ml of CH4 at STP (4) 5.6 litre of CO2 at STP Sol. Answer (3)
11200 ml of CH4 at STP has lowest weight i.e., 8 g. 3. The largest number of molecules are in
(1) 28 g of CO (2) 46 g of C2H5OH (3) 36 g of H2O (4) 54 g of N2O5 Sol. Answer (3) Sample Number of molecules 28 NA 28 g CO 28 = NA 46 NA 46 g C2H5OH 46 = NA 36 NA 36 g of H2O 18 = 2NA
54 NA NA 54 g N2O5 108 2
36 g H2O contains maximum number of molecules 2 Some Basic Concepts of Chemistry Solution of Assignment
4. The volume of 0.1 M Ca(OH)2 needed for the neutralization of 40 ml of 0.05 M oxalic acid is (1) 10 ml (2) 20 ml (3) 30 ml (4) 40 ml Sol. Answer (2)
HCO224 Ca(OH)2 CaCO24 2HO 2
Number of milli equivalents of H2SO4 = Number of milliequivalents of Ca(OH)2
N1 × V1 = N2 × V2
M1 × n-factor × V1 = M2 × n-factor × V2
0.05 × 2 × 40 = 0.1 × 2 × V2
V2 = 20 ml 5. Number of oxalic acid molecules in 100 ml of 0.02 N oxalic acid are (1) 6.023 × 1020 (2) 6.023 × 1021 (3) 6.023 × 1022 (4) 6.023 × 1023 Sol. Answer (1) N = M × n factor
Oxalic acid is H2C2O4 Hence n factor = Basicity = 2 0.02 = M × 2 M = 0.01 Number of moles = molarity × Volume (liter) 100 = 0.01 = 0.001 1000 23 20 Number of molecules = 0.001 × NA = 0.001 × 6.023 × 10 = 6.023 × 10 6. A compound contains 3.2% of oxygen. The minimum molecular weight of the compound is (1) 300 (2) 440 (3) 350 (4) 500 Sol. Answer (4) For molecular mass to be minimum one gram atom of O (16 g oxygen) should be present 3.2 g oxygen 100 gm compound 16 100 16 g oxygen 3.2 = 500 g Minimum molecular mass of compound = 500 7. The weight of KOH in its 50 milliequivalent is (1) 1.6 g (2) 2.2 g (3) 2.8 g (4) 4.8 g Sol. Answer (3) ⎛⎞mass Number of meq ⎜⎟1000 ⎝⎠Equivalent mass ⎛⎞ ⎜⎟mass 50 ⎜⎟1000 39 16 1 ⎜⎟ ⎝⎠1
50 56 mass 1000 = 2.8 g Solution of Assignment Some Basic Concepts of Chemistry 3
8. The mole fraction of glucose in aqueous solution is 0.2, then molality of solution will be (1) 13.8 (2) 55.56 (3) 2 (4) 12 Sol. Answer (1)
0.2 200 Molality = 13.8 m 0.8 18 14.4 1000
9. A metal oxide contains 60% metal. The equivalent weight of metal is (1) 12 (2) 60 (3) 40 (4) 24 Sol. Answer (1)
60 812 Equivalent wt. of metal = 40
10. The number of neutrons in a drop of water (20 drops = 1 mL) at 4ºC (1) 6.023 × 1022 (2) 1.338 × 1022 (3) 6.023 × 1020 (4) 7.338 × 1022 Sol. Answer (2) 20 drops = 1 ml
1 or 1 drop = 20 ml
mass of one drop = Volume × density
1g1 = ml 1 = g 20 ml 20
1 20 1 Number of moles = 18 360
1 23 6.023 10 21 Number of molecules = 360 = 1.67 × 10
∵ 1 molecule contain 8 neutrons 1.67 × 1021 molecule will contain = 1.67 × 1021 × 8 = 1.338 × 1022 neutrons
11. What volume of CO2 at STP will evolve when 1 g of CaCO3 reacts with excess of dil HCl? (1) 224 ml (2) 112 ml (3) 56 ml (4) 448 ml Sol. Answer (1)
CaCO32 CO Applying POAC on C 1n 1n CaCO3 CO2
1V 1 1 × 100 22400
V = 224 ml 4 Some Basic Concepts of Chemistry Solution of Assignment
12. 1.82 g of a metal requires 32.5 ml of 1 N HCl to dissolve it. What is the equivalent weight of metal? (1) 46 (2) 65 (3) 56 (4) 42 Sol. Answer (3)
Metal HCl Pr oduct Number of eq. of metal = Number of eq. of HCl
1.82 1 32.5 Emetal 1000
1.82 1000 E = = 56 metal 32.5
13. 1 g Ca was burnt in excess of O2 and the oxide was dissolved in water to make up one litre of solution. The normality of solution is (1) 0.04 (2) 0.4 (3) 0.05 (4) 0.5 Sol. Answer (3) Ca O CaO 2 1gm Applying P.O.A.C. on Ca
1nCa 1n CaO 1 11n 40 CaO 1 n CaO 40
CaO H O Ca OH 2 2
nn CaO Ca(OH)2
11 [Ca(OH) ] 2 40 40 1
NM2 Ca(OH)22 Ca(OH) 11 20.05 = 40 20
Normality of Ca and Ca(OH)2 will be same.
14. The normality of solution obtained by mixing 100 ml of 0.2 M H2SO4 with 100 ml of 0.2 M NaOH is (1) 0.1 (2) 0.2 (3) 0.5 (4) 0.3 Sol. Answer (1)
Number of eq. of H2SO4 = N1V1 100 = 0.2 ×2 × 1000 = 0.04 Solution of Assignment Some Basic Concepts of Chemistry 5
Number of eq. of NaOH = N2V2 100 = 0.2 ×1 × 1000 = 0.02
Resulting equivalent = N1V1– N2V2 = 0.04 – 0.02 = 0.02 0.02 Resulting eq. 100 100 N0.1Nmix V(lit) 1000 15. The vapour density of a volatile chloride of divalent metal is 59.5 and equivalent mass of the metal is (1) 96 (2) 48 (3) 24 (4) 12 Sol. Answer (3) Volatile chlorides are chlorides of alkaline earth metal
++ – M 2Cl MCl2 Molecular weight = 2 × vapour density = 2 × 59.5 =119 weight of metal = 119 – 2 × 35.5 = 119 – 71 = 48 Atomic weight of metal 48 E24metal Valency 2
N 16. 0.45 g of a dibasic acid is completely neutralised with 100 ml 10 NaOH. The molecular weight of acid is
(1) 45 (2) 90 (3) 180 (4) 22.5 Sol. Answer (2) milliequivalents of acid = milliequivalents of NaOH 0.45 1 1000 100 E10 E = 45 Molecular weight = E × Basicity = 45 × 2 = 90 17. A sample of pure calcium weighing 1.35 g was quantitatively converted to 1.88 g of pure CaO. Atomic mass of calcium is (1) 20 (2) 40 (3) 60 (4) 30 Sol. Answer (2)
Ca CaO Applying POAC on Ca
1 × nCa = 1× nCaO 6 Some Basic Concepts of Chemistry Solution of Assignment
1.35 1.88 11 AA16 1.35 A + 1.35 × 16 = 1.88 A 1.88A – 1.35 A = 1.35 × 16 0.53A = 21.6 A = 40.75 A 40 18. For the reaction,
2+ + 3Zn + 2K4[Fe(CN)6] K2Zn3[Fe(CN)6]2 + 6K , what will be the equivalent weight of K4[Fe(CN)6], if the molecular weight of K4[Fe(CN)6] is M? M M M (1) M (2) 2 (3) 3 (4) 4 Sol. Answer (3)
+2 + 3 Zn + 2 K46 [Fe(CN) ] K 2362 Zn [Fe(CN) ] + 6K ∵ + 2 mole K4Fe(CN)6 loses 6 mole K + 1 mole K4Fe(CN)6 loses 3 mole K Number of moles of cations (monovalent) lost by one molecule is n factor. So n factor = 3 M E 3
19. Equivalent weight of H3PO2 is (M molecular weight) M M M M (1) 1 (2) 2 (3) 3 (4) 4 Sol. Answer (1)
H3PO2 has only one replaceable H-atom.
20. 0.92 g of Ag2CO3 is heated strongly beyond its melting point. After heating the amount of residue is (1) 0.36 g (2) 0.39 g (3) 0.77 g (4) 0.72 g Sol. Answer (4)
Ag23 CO Ag CO 2 O 2 Applying POAC on Ag n 2 × AgCO23 = 1 × nAg
0.92 W 2 × 1 2 108 60 108
20.92 W 216 60 108
108 2 0.92 W 276
W = 0.72 g Solution of Assignment Some Basic Concepts of Chemistry 7
21. Molality of pure water is (1) 1 (2) 55.56 (3) 20 (4) Cannot be calculated Sol. Answer (2)
1000 m18 55.56 m 1
22. When 400 g of a 20% solution by weight was cooled, 50 g of solute precipitated. The percentage concentration of remaining solution is (1) 8.57% (2) 15% (3) 12.25% (4) 9.5% Sol. Answer (1)
20 400 Weight of solute = 100 = 80 g
Amount of solute remaining = 80 – 50 = 30 g Mass of solution remaining = 400 – 50 = 350
w solute 100 30 % concentration = = 100 wsolution 350
= 8.57% 23. Which of the following solution has normality equal to molarity?
(1) H2SO4 aqueous solution (2) H3PO4 aqueous solution
(3) HNO3 aqueous solution (4) Mg(OH)2 aqueous solution Sol. Answer (3)
Basicity of HNO3 is 1. Hence, Molarity = Normality 24. For the reaction
Ba (OH)2 + 2HClO3Ba (ClO3)2 + 2H2O, calculate the number of moles of H2O formed when 0.1 mole of Ba (OH)2 is treated with 0.0250 moles of HCIO3. (1) 0.1 (2) 0.125 (3) 0.25 (4) 0.025 Sol. Answer (4)
0.1mole 0.0250 mole Ba(OH)23 2HClO Ba(ClO 322 ) 2H O
⎡ 1mole 2moleH2 O Ba(OH)2 ⎢ ⎣0.1mole 0.2 mole H2 O
⎡2mole 2moleH2 O HClO3 ⎢ ⎣0.025 mole 0.025 mole H2 O
HClO3 is limiting reagent, hence 0.025 mole H2O will be formed. 8 Some Basic Concepts of Chemistry Solution of Assignment
3 25. Ammonia gas is passed into water, yielding a solution of density 0.93 g /cm and containing 18.6% NH3 by
weight. The mass of NH3 per cc of the solution is (1) 0.17 g/cm3 (2) 0.34 g/cm3 (3) 0.51 g/cm3 (4) 0.68 g/cm3 Sol. Answer (1) Let 100 g solution is taken M = V × d
M 100 g Vsolution 3 = 107.5 cm3 d 0.93 g / cm
18.6 3 3 mass of NH3 per c.c. = 107.5 = 0.17298 g/cm = 0.17 g/cm
26. The empirical formula of a commercial ion exchange resin is C8H7SO3Na. The resin can be used to soften 2+ + water according to the reaction Ca + 2C8H7SO3Na (C8H7SO3)2Ca + 2Na . What would be the maximum uptake of Ca2+ by the resin expressed in mole /g resin ? (1) 0.00246 (2) 0.0246 (3) 0.246 (4) 0.0048 Sol. Answer (1)
2 Ca 2C87 H SO 3 Na (C 87 H SO 32 ) Ca 2Na Resin M206
∵ 2 × 206 g resin takes 1 mole Ca+2
1 1 g resin takes = = 0.00246 mole/g 2206
27. A mixture of 1.65 × 1021 molecules of X and 1.85 × 1021 molecules of Y weighs 0.688 g. The mol. wt. of X is (Assume mol. wt. of Y is 187). (1) 52 (2) 84 (3) 126 (4) 41.47 Sol. Answer (4) For Y
6.023 × 1023 molecules 187 g 1.85 × 1021 molecules 0.574 g weight of x + weight of y = 0.688 weight of x + 0.574 = 0.688
weight of x = 0.1136 0.114 ∵ 1.65 × 1021 molecules of x 0.114 6.023 × 1023 molecules of x 41.47 g Molecular weight of x = 41.47 Solution of Assignment Some Basic Concepts of Chemistry 9
28. Average atomic weight of boron is 10.10 and boron exists in two isotopic forms B10 and B11. The percentage abundance of B10 is (1) 10% (2) 90% (3) 50% (4) 20% Sol. Answer (2) Assume percentage abundance of B10 = x So percentage abundance of B11 = 100 – x x 10 (100 x)11 Apply formula, 10.10 100 10 1100 11x 1010 x = 90%
29. Vapour density of mixture of NO2 and N2O4 is 34.5, then percentage abundance of NO2 in mixture is (1) 50% (2) 25% (3) 40% (4) 60% Sol. Answer (1) Molecular weight = 2 × V.D. = 69
Assume % of NO2 = x
So % of N2O4 = 100 – x x46(100x)92 Apply formula, 69 100 46x + 9200 – 92x = 6900 x = 50%
30. A mixture of FeO and Fe3O4 when heated in air to constant weight gains 5% in its weight. The % of FeO in the sample is (1) 79.75 (2) 20.25 (3) 30.25 (4) 10.25 Sol. Answer (2)
When FeO and Fe3O4 are heated both change to Fe2O3
Let weight of FeO and Fe3O4 be x g and y g Total weight of Reactant = (x + y) g
Since weight increases 5% on heating when FeO and Fe3O4 changes completely to Fe2O3 . 105 Weight of Fe O (x y) = 1.05 (x + y) 23100
FeO Fe34 O Fe 23 O x gm y gm 1.05 (x y) Applying POAC on Fe x3y1.05(xy) 2 72 232 160 x81 y319
81 % of FeO 100 = 20.25 % 81 319
% of Fe3O4 = 79.98 % 10 Some Basic Concepts of Chemistry Solution of Assignment
31. 1 ml of gaseous aliphatic compound CnH3nOm is completely burnt in an excess of O2 and cooled to room temperature. The contraction in volume is ⎛⎞13 ⎛⎞11 ⎛⎞31 ⎛⎞31 ⎜⎟1n m ⎜⎟1nm ⎜⎟1nm ⎜⎟1nm (1) ⎝⎠24 (2) ⎝⎠44 (3) ⎝⎠44 (4) ⎝⎠42 Sol. Answer (4) ⎛⎞3n 2n m ⎜⎟2 3n CH O⎝⎠ O nCO HO(l) n3nm22 2 2 2 ⎛⎞3n ⎜⎟2n m ⎝⎠2 3n m Contraction = 1 + n1 242
32. 1 mole of aliphatic compound CnH3nOm is completed burnt is an excess of O2. The reacted oxygen number of moles of is 3n m 3n m 3n m 3n m n n n n (1) 42 (2) 22 (3) 42 (4) 44 Sol. Answer (1) As solution 31 3n m Required O = n 2 42
33. 1 mol of gaseous compound CnH3nOm is completely burnt in excess of O2. The number of moles of CO2 formed is (1) n (2) n/2 (3) 2n (4) 3n Sol. Answer (1) As solution 31
Moles of CO2 = n
34. 1 mole of gaseous aliphatic compound CxH3nOm is completely burnt in excess of O2 . The no. of moles of H2O is formed (1) n (2) 3n/2 (3) 2n (4) 3n Sol. Answer (2) As solution 31 3n Moles of H O = 2 2
35. 1 litre of gaseous aliphatic compound CxHyOz is completely burnt in excess of O2 and cooled to room temperature. The contraction in volume is yz yz yz yz (1) x (2) 1 (3) x (4) 1 42 42 22 42 Sol. Answer (2) y 2x z y CHO2 O xCO HO xy z22 2 2 2 ⎛⎞y ⎜⎟2x z ⎝⎠2 yz Contraction = 1x1 242 Solution of Assignment Some Basic Concepts of Chemistry 11
36. 1 litre of aliphatic compound CxHyOz is completely burnt in an excess of O2 and cooled to room temperature. The reacted volume of oxygen will be
yz yz yz yz x x (1) 42 (2) 22 (3) 42 (4) 84
Sol. Answer (1) 37. Given that the abundances of isotopes 54Fe, 56Fe and 57 Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is [IIT-JEE 2009] (1) 55.85 (2) 55.95 (3) 55.75 (4) 56.05 Sol. Answer (2) We will have to take weighted average
5459056557 55.95 100
38. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is [IIT-JEE 2011] (1) 1.78 M (2) 2.00 M (3) 2.05 M (4) 2.22 M Sol. Answer (3)
wt. of solute 1000 M mol. wt. of solute volume of solution
120 1000 M2.05 60 1120 1.15
SECTION - B
Objective Type Questions (More than one options are correct)
23 1. 1 mole of CO2 contains (1) 6.023 × 10 atoms of O 23 23 (2) 6.023 × 10 atoms of C (3) 6.023 × 10 molecules of CO2 (4) All of these
Sol. Answer (2, 3)
1 mole CO2 contains NA no. of molecules ∵ 1 molecule of CO2 contain 1 C atom
NA molecule of CO2 contain NA C atom ∵ 1 molecule of CO2 contains 2 O atom
NA molecule of CO2 contains = 2 × NA O atom
NA = Avogadro number = 6.023 × 1023 12 Some Basic Concepts of Chemistry Solution of Assignment
2. 1 gram atom of Na is equivalent to (1) 1 gram equivalent of Na (2) 13 g (3) 23 g (4) 1 g Sol. Answer (1, 3) 1 g atom of Na = 1 mole Na = 23 gm for Na, n factor = 1 Number of eq. = Number of moles × nf = 1 × 1 = 1 3. Which of the following contains the greatest number of atoms?
(1) 1g of O (2) 1g of O2 (3) 1g O3 (4) 1g F2 Sol. Answer (1, 2, 3) 1 N N1 A Number of atoms of O in 1 gm O = 16 A = 16
1 NA Number of atoms of O in 1 gm O = A 2N = 2 32 16
1 NA Number of atoms of O in 1 gm O = N3A = 3 48 16
1 N N2 A Number of atoms of F = 38 A = 19 (1), (2), (3) have same number of atoms and greatest number of atoms also. 4. Atomic masses of many elements are non-integral because (1) They have isotopes (2) The isotopes have non-integral atomic masses (3) Their isotopes have different masses (4) The constituents neutrons, protons and electrons combine to give fractional masses Sol. Answer (1, 3) Facts 5. Which of the following is/are independent of temperature? (1) Molarity (2) Molality (3) Mole fraction (4) Normality Sol. Answer (2, 3) Molarity and normality contain volume while molality and mole fraction contain weight hence molality and mole fraction is independent of temperature. Solution of Assignment Some Basic Concepts of Chemistry 13
6. For the reaction, A + 2B C + 2D. The correct statement is (1) Equivalents of A = 2 × equivalents of B (2) Moles of A reacted = Moles of D formed (3) Equivalents of B = Equivalents of C (4) Moles of B reacted = 2 × Moles of C formed Sol. Answer (3, 4)
A2B C2D Number of eq. of A = number of eq. of B = number of eq. of C = number of eq. of D Moles of A reacted = 2 × moles of D formed Moles of B reacted = 2 × moles of C formed 7. For the reaction, A + 2B 2C, 5 moles of A and 8 moles of B are reacted, then (1) Whole A is consumed (2) Whole B is consumed (3) 8 moles of C are formed (4) 8 equivalents of C are formed Sol. Answer (2, 3)
A2B 2C ⎡1mole⇒ 2moleC A ⎢ ⎣5 mole⇒ 10 mole C ⎡2mole⇒ 2moleC B ⎢ ⎣8mole⇒ 8moleC B forms least number of moles of product hence B is limiting reagent. Moles of C formed = 8
8. 10 g carbon reacts with 100 g Cl2 to form CCl4. The correct statement is
(1) Carbon is the limiting reagent (2) Cl2 is the limiting reagent
(3) 107.8 g CCl4 is formed (4) 0.833 moles of CCl4 are formed Sol. Answer (2, 3)
C 2Cl24 CCl 10 moles of C = 12 = 0.833 100 moles of Cl = = 1.40 2 71
⎡1moleC⇒ 1moleCCl4 C ⎢ ⎣0.833 mole⇒ 0.833 mole CCl4
⎡2moleCl24⇒ 1moleCCl O ⎢ ⎣1.40 mole Cl24⇒ 0.70 mole CCl
Cl2 is limiting reagent because it forms least number of moles of product. W CCl4 = 0.70 × 154 = 107.8 g 14 Some Basic Concepts of Chemistry Solution of Assignment
9. Which of the following pairs have same number of molecules?
(1) 2 g of O2, 4 g of SO2
(2) 2 g CO2, 2 g of N2O
(3) 224 ml O2 at STP, 448 ml of He at 0.5 atm and 273 K (4) 2 g oxygen and 2 g ozone Sol. Answer (1, 2, 3) ⎛⎞2 N NA molecules (1) 2 g O = ⎜⎟A 2 ⎝⎠32 16 ⎛⎞4 N N A 4 g SO = ⎜⎟A molecules 2 ⎝⎠64 16 ⎛⎞2 N NA molecules (2) 2 g CO = ⎜⎟A 2 ⎝⎠44 22 ⎛⎞2 N NA molecules 2 g N O = ⎜⎟A 2 ⎝⎠44 22 ⎛⎞224 N NA molecules (3) 224 ml O at STP = ⎜⎟A 2 ⎝⎠22400 100
–2 = 10 NA 448 ml of He at 0.5 atm and 273 K PV = nRT 448 PV 0.5 n 1000 32 RT = 10 10 1 10 moles 0.0821 273 –2 molecules = 1 × 10 NA
⎛⎞2 NA ⎜⎟NA (4) 2 g oxygen = ⎝⎠32 16
⎛⎞2 NA ⎜⎟NA 2 g ozone = ⎝⎠48 24 10. 100 ml of 0.02 N oxalic acid is equivalent to (1) 100 ml of 0.01M oxalic acid (2) 6.023 × 1020 molecules of oxalic acid (3) 50 ml of 0.02 M oxalic acid (4) 100 ml of 0.02 M oxalic acid Sol. Answer (1, 2, 3) Number of eq. of oxalic acid 100 0.02 1000 = 0.002
(1) Number of eq. of H2C2O4 ⎛⎞100 ⎜⎟0.01 2 ⎝⎠1000 = 0.002 (2) Number of eq. of oxalic acid
⎛⎞6.023 1020 ⎜⎟ 2 –3 ⎜⎟23 = 2 × 10 = 0.002 ⎝⎠6.023 10 Solution of Assignment Some Basic Concepts of Chemistry 15
(3) Number of eq. of oxalic acid ⎛⎞50 ⎜⎟0.02 2 = 0.002 ⎝⎠1000 (4) Number of eq. of oxalic acid ⎛⎞100 ⎜⎟0.02 2 = 0.004 ⎝⎠1000 (1), (2), (3) are correct 11. 12 g of Mg will react completely with an acid to give 1 (1) 1 mole of O (2) mole of H (3) 1 equivalent of H (4) 2 equivalents of H 2 2 2 2 2 Sol. Answer (2, 3)
Number of eq. of Mg = Number of eq. of H2
12 WH 2 2 24 E H2
WH 1 2 1 W1 H2
WH 1 nH 2 2 M2 mole H2 Number of eq. of Mg = 1
Hence number of eq. of H2 = 1
Magnesium does not liberate O2 on reaction with an acid.
12. 0.5 g of fuming H2SO4 oleum is diluted with 100 mL of water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. The correct statement is/are
(1) Mass of SO3 is 0.104 g (2) % of free SO3 = 20.7
(3) Normality of H2SO4 for neutralization is 0.2 N (4) Weight of H2SO4 is 0.104 g Sol. Answer (1, 2)
Let 0.5 gm fuming H2SO4
Contains x gm SO3
SO32 H O H 24 SO
1 mole SO3 gives 1 mole H2SO4 x x 80 mole SO3 give, 80 mole H2SO4 Total number of moles of x0.5x H2SO4 = 80 98
Number of eq. of H2SO4 = Number of eq. of NaOH ⎛⎞x0.5x 26.7 ⎜⎟2 = 0.4 ⎝⎠80 98 1000
On solving x = 0.104 gm
Percentage of free SO3 = 20.7% 16 Some Basic Concepts of Chemistry Solution of Assignment
13. A given solution of H2SO4 is labelled as 49% (w/w), then correct statement regarding the solution is (d = 1.3 g/ml) 500 1000 m N (1) 51 (2) 51 (3) % w/v = (49 × 1.3)% (4) M = 6.5 Sol. Answer (1, 3, 4) 49 1000 500 m 98 51 51
49 1000 N d 10d (where d = density) = 13 49 100 14. Out of following, which molecules has same weight under identical volume at STP?
14 13 18CO (1) CO2 (2) NO2 (3) CO2 (4) 2 Sol. Answer (1, 2) If molecular weight for different gas is same then under identical volume, weight will be same. 15. 4 gm of NaOH can be neutralised by N (1) 100 ml of 1N HCl (2) 200 ml of H SO 2 2 4 N M (3) 1000 ml of 10 KOH (4) 2000 ml of 20 H3PO4 Sol. Answer (1, 2) 41 Equivalent of NaOH 40 10 100 1 1 Eq. of HCl = 1000 10 200 1 1 Eq. of H2SO4 = 1000 2 10 2000 1 3 3 Eq. of H3PO4 = 1000 20 10 16. Choose the correct statement regarding equivalent weight (1) Equivalent weight of a substance always remain same (2) Equivalent weight of substance depends on reaction (3) Equivalent weight may be greater than atomic weight (4) Equivalent weight may be less than atomic weight Sol. Answer (2, 3, 4) Fact.
17. 11.2 L of CH4 and 22.4 L of C2H6 at STP are mixed. Then choose correct statement/statements (1) Vapour density of the mixture is 12.67 (2) Average molecular wt. will be less than 16 (3) Average molecular wt. will be greater than 16 and less than 30 (4) Average molecular weight will be greater then 30 Sol. Answer (1, 3) 12 16 30 Average molecular wt. = 33 = 5.33 + 20 = 25.33 Solution of Assignment Some Basic Concepts of Chemistry 17
18. Which of the following pairs follow law of multiple proportion?
12 14 (1) CO2, CO (2) Fe2O3, Fe0.90O (3) CO2, CO (4) N2O4, N2O3 Sol. Answer (1, 4) For isotope law of multiple proportions is not valid. 19. Choose the correct match/matches
(1) 18 ml of H2O at 4°C NA molecule of H2O
NA (2) 11.2 L of CO at 273°C molecule of CO 2 2 2 and 1°C
57 (3) 56 g of Fe NA atom
(4) 5.6 L of CH4 at STP 4 g weight Sol. Answer (1, 4) 1 mole of Fe57 = 57 g 20. Largest number of bond pair will be in
(1) 30 g C2H6 (2) 24 g C2H2 (3) 14 g H2 (4) 88 g CO2 Sol. Answer (1, 3)
Mole of H2 = 7 No. of bond pair = 7 × 1 = 7
SECTION - C Linked Comprehension Type Questions Comprehension-I Atoms of same element having same atomic number and different atomic mass are known as isotopes. If atomic masses of two isotopes of an element are A1 and A2 and they exist in the ratio P1 : P2, then average atomic mass PAPA A 2211 avg PP 21 1. Which isotope can be used to decide the scale of atomic mass?
12 15 24 14 (1) 6C (2) 7N (3) 11Na (4) 6C Sol. Answer (1)
12 6C is not radioactive hence can be used to decide the scale of atomic mass. 12 14 2. If % abundance of two isotopes of carbon 6C and 6C are 90% and 10% respectively then number of C-12 atoms in 12 g of sample will be approximately
(1) 0.44 NA (2) 0.88 NA (3) 0.22 NA (4) 0.11 NA
Sol. Answer (2) Average atomic mass AP AP 12 90 14 10 1220 11 2 2 12.2 100 = 100 100 12 0.98 Moles of carbon present in 12 g sample = 12.2 18 Some Basic Concepts of Chemistry Solution of Assignment
Total number of C-atoms present in 12 g = 0.98 NA Since C-12 atoms are 90% of total atoms, hence number of C-12 atoms present in the sample
90 0.98N = A 100
= 0.88 NA 35 37 3. If average atomic mass of Cl is 35.5. Chlorine exist in nature in the form of two isotopes 17Cl and 17Cl , then ratio in which they exist in nature will be
(1) 1 : 1 (2) 3 : 1 (3) 2 : 1 (4) 3 : 2 Sol. Answer (2)
35 37 Let % of 17Cl = x and 17Cl = 100 – x AP AP 11 2 2 35.5 100 35x 37(100 x) 35.5 100
35x + 3700 – 37x = 3550
37x – 35x = 3700 – 3550 2x = 150
x = 75 100 – x = 25
35 37 Ratio of 17Cl : 17Cl = 75 : 25 = 3 : 1 Comprehension-II Avogadro’s law states that under similar condition of T and P, equal volumes of gases contain equal number of particles. Experiments show that at one atmosphere pressure and at a temperature 273 K (i.e. at STP) one mole of any gas occupies a volume approximately 22.4 litres. Therefore number of moles of any sample of gas can be found by comparing its volume at STP with 22.4.
23 1 mole of any species contains 6.023 × 10 particles which is denoted by symbol NA. Number of atoms present
in 1 gm-atom of an element or number of molecules present in 1 gm-molecule of any substance is equal to NA. Hence it is number of particles present in one mole of the substance.
1. If NAV is Avogadro’s number, then 10 amu will be equal to ______gram
NAV 10 (1) 10 NAV (2) (3) (4) NAV 10 NAV Sol. Answer (3) 1 mass of one C 12 atom 1 amu = 12 1121 = 12 NAVAV N 10 10 amu = NAV Solution of Assignment Some Basic Concepts of Chemistry 19
2. At STP 11.2 L of CO2 contains (1) 1 mol (2) 2 mol (3) 0.5 mol (4) 3 mol Sol. Answer (3) Volume Moles of CO 2 Molar volume
11.2 0.5 mole = 22.4
3. The number of gm atoms of oxygen present in 0.2 mole of H2S2O8 is (1) 0.2 (2) 8 (3) 1.6 (4) 0.8 Sol. Answer (3) ∵ 1 mole H2S2O8 contains 8 g atoms of O
0.2 mole H2S2O8 will contain = 0.2 × 8 = 1.6 g atoms of O Comprehension-III All chemical reactions take place under certain laws out of which three laws are given here. (i) Law of conservation of mass: According to this law, total mass of reactants is equal to total mass of products. (ii) Law of constant composition: According to this law, a chemical compound is always found to be made up of same elements combined together in fixed proportion by weight. (iii) Law of multiple proportion: According to this law, when two elements are combined to form two or more chemical compounds, the weight of one of the elements which combine with a fixed weight of another bear a simple whole number ratio to one another. 1. Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates (1) Law of reciprocal proportions (2) Law of constant proportions (3) Law of multiple proportions (4) Law of conservation of mass Sol. Answer (3) Fact
2. 3 g of a hydrocarbon on combustion with 11.2 g of oxygen produces 8.8 g CO2 and 5.4 g H2O. The data illustrates (1) Law of conservation of mass (2) Law of multiple proportions (3) Law of definite proportions (4) Law of reciprocal proportions Sol. Answer (1)
mLHS = mRHS
3. The percentage of carbon and oxygen in samples of CO2 obtained by different methods were found to be the same. This illustrates (1) Law of conservation of mass (2) Law of constant proportions (3) Law of multiple proportions (4) Law of reciprocal proportions Sol. Answer (2) Fact. 20 Some Basic Concepts of Chemistry Solution of Assignment
Comprehension-IV On being heated in oxygen 5.72 g of red metallic oxide A was converted to 6.36 gram black metallic oxide B. When 4.77
g of B was heated in a stream of H2 gas, 3.81 g of metal M was formed. (Given, atomic weight of metal is 63.50) 1. The formula of red metallic oxide A is
(1) MO (2) MO2 (3) M2O (4) M2O3 Sol. Answer (3) In B, wt. of oxygen = 4.77 – 3.81 = 0.96 g 0.96 0.12 Equivalent of oxygen = 8 So, equivalent of metal = 0.12 3.81 381 31.8 Eq. wt. of metal = 0.12 12 By considering eq. wt. of metal in B 5.72 6.36 , where x is the equivalent weight of metal in A x 839.8 x = 63.5 2. The formula of black metallic oxide is
(1) MO (2) M2O5 (3) M2O3 (4) M2O Sol. Answer (1) In B, wt. of oxygen = 4.77 – 3.81 = 0.96 g 0.96 0.12 Equivalent of oxygen = 8 So, equivalent of metal = 0.12 3.81 381 31.8 Eq. wt. of metal = 0.12 12 By considering eq. wt. of metal in B 5.72 6.36 , where x is the equivalent weight of metal in A x 839.8 x = 63.5 3. The equivalent weight of metal M in B is (1) 118.9 (2) 65.4 (3) 63.5 (4) 31.8 Sol. Answer (4) In B, wt. of oxygen = 4.77 – 3.81 = 0.96 g 0.96 0.12 Equivalent of oxygen = 8 So, equivalent of metal = 0.12 3.81 381 31.8 Eq. wt. of metal = 0.12 12 By considering eq. wt. of metal in B 5.72 6.36 , where x is the equivalent weight of metal in A x 839.8 x = 63.5 Solution of Assignment Some Basic Concepts of Chemistry 21
Comprehension-V 36 ml of a gaseous mixture consisting of a gaseous organic compound A and just sufficient amount of oxygen required
for complete combustion gives 16 ml of CO2, 24 ml water vapour and 8 ml of N2. The volumes are measured at same temperature and pressure.
1. Volume of O2 required for complete combustion (1) 8 ml (2) 28 ml (3) 74 ml (4) 22 ml Sol. Answer (2) 2. The molecular formula of compound will be
(1) CH5N (2) C2H5N (3) C2H6N2 (4) C4H10N2 Sol. Answer (3)
SECTION - D Assertion-Reason Type Questions 1. STATEMENT-1 : One mole of an ideal gas have volume of 22.4 litre at STP. and STATEMENT-2 : Under identical condition, equal weight of gases have same volume. Sol. Answer (3) Fact
2. STATEMENT-1 : H3PO4 is a tribasic acid. and
STATEMENT-2 : In H3PO4, only two H-atoms are replaceable. Sol. Answer (3)
H3PO4
H3PO4 is a tribasic acid and its 3 H atoms are replaceable. 3. STATEMENT-1 : 18 g of water vapour and 18 g of ice will not contain the same number of molecules. and STATEMENT-2 : Number of molecules are independent of temperature and pressure. Sol. Answer (4)
Since molecular mass of H2O and ice are same, hence, 18 g H2O and 18 g ice will contain same number of molecules. 4. STATEMENT-1 : Atomic mass of Mg is 24. and 1 th 12 STATEMENT-2 : An atom of magnesium is 24 times heavier than 12 of the mass of carbon atom (C ). 22 Some Basic Concepts of Chemistry Solution of Assignment
Sol. Answer (1) Atomic mass of Mg = 24 24 12Mg mass of single atom of element atomic mass 1 mass of single atom of element 12 1 12 From above relation we can say that an atom of Mg is 24 times heavier than 12 of the mass of C 5. STATEMENT-1 : Atomic weight of an atom can never be in fraction. and STATEMENT-2 : Average atomic weight of chlorine is 35.5. Sol. Answer (4) Average atomic weight may be in fraction while atomic wt. of an atom never be in fraction. 6. STATEMENT-1 : Law of conservation of mass is generally applicable to all the chemical reactions. and
STATEMENT-2 : Law of constant composition is not valid for non stoichiometric compound like Fe0.93O. Sol. Answer (2) On chemical reaction, mass of reactants is equal to mass of products. 7. STATEMENT-1 : Solvent have always same physical state as that of solution. and STATEMENT-2 : Solution contains more than one solvent. Sol. Answer (3) In a solution, solvent is always one but solute may be more than one. 8. STATEMENT-1 : Molality is equal to molarity, if density of solution is one. and STATEMENT-2 : Molality does not depend on the temperature. Sol. Answer (4) If density is one then weight of solution is equal to volume of solution. To calculate molality, weight of solvent is required. Which is independent from temperature. 9. STATEMENT-1 : On dilution, molarity of solution changes. and STATEMENT-2 : Number of moles of solute in a solution does not change on dilution. Sol. Answer (2) On dilution, molarity decreases while number of moles of solute does not change. 10. STATEMENT-1 : Equivalent weight of an acid is always less than its molecular weight. and
Molecular weight STATEMENT-2 : Equivalent weight of acid = n-factor
Sol. Answer (4) Equivalent weight of acid may be equal to molecular weight if basicity of acid is one. Solution of Assignment Some Basic Concepts of Chemistry 23
11. STATEMENT-1 : In any chemical reaction, total number of molecules are conserved. and STATEMENT-2 : Atom can neither be created nor be destroyed. Sol. Answer (4) Number of molecule may change in a reaction. 12. STATEMENT-1 : During a chemical reaction, total mass remains constant. and STATEMENT-2 : Moles may vary in a reaction. Sol. Answer (3) Moles may vary in a reaction. 13. STATEMENT-1 : Compound having same general formula may have different empirical formula. and STATEMENT-2 : Compound having same empirical formula may have different general formula. Sol. Answer (4) Fact.
14. STATEMENT-1 : 18 ml of H2O and 18 ml of CO2 at 277 K have same number of moles. and
STATEMENT-2 : Density of H2O is more than CO2. Sol. Answer (4)
H2O is liquid
15. STATEMENT-1 : In 32 g of O2, two gram atom of oxygen atom are present. and
STATEMENT-2 : Molecular weight of O2 will be 32 g. Sol. Answer (3) –1 Molecular wt. of O2 is 32 g mol
SECTION - E Matrix-Match Type Questions 1. Match the following Column-I Column-II
(A) Number of carbon atoms in 1 g molecule of CO2 (p) 0.5 N0
(B) Number of molecules in 48 g O2 (q) N0
(C) Number of molecules in 11.2 L H2 at STP (r) 3 N0
(D) Number of hydrogen atoms in 1 Mole of NH3 (s) 1.5 N0 (N0 = Avogadro’s Number) Sol. Answer A(q); B(s); C(p); D(r)
(A) No. of C-atoms in 1 g molecule of CO2 = N0 48 N0 (B) No. of molecules in 48 g of O2 = 32 = 1.5 N0
N0 (C) No. of molecules in 11.2 L H at STP = = 0.5 N 2 2 0
(D) No. of H-atoms in 1 mole of NH3 = 3 × N0 = 3N0 24 Some Basic Concepts of Chemistry Solution of Assignment
2. Match the following Column I Column II OH2OH2 (A) 2 2 2 (p) 25.5 g product g3 g66.22 (B) N3H 2NH (q) 0.25 g of a reactant is left 22 3 24.5 g 5.5 g H Cl 2 HCl (C) 22 (r) H2 is the limiting reagent 1.4 g 40g CHH2C (D) 2 4 (s) 41.1 g product g20 g375.6 Sol. Answer A(p); B(q); C(s); D(r) 3. Match the following Column I Column II (% by mass of oxygen)
(A) Na2S2O3 (p) 30%
(B) KMnO4 (q) 39%
(C) Na3PO4 (r) 57%
(D) MgCO3 (s) 40.5% Sol. Answer A(p), B(s), C(q), D(r)) 4. Match the following Column I Column II (A) 1 g molecule of chlorine gas (p) 14 L at STP
(B) 1 g equivalent of Br2 gas (q) 22.4 L at STP
(C) 32 g of CH4 (g) (r) 80 g
(D) 40 g SO2 (g) (s) 44.8 L at STP Sol. Answer A(q), B(r), C(s), D(p)
SECTION - F Integer Answer Type Questions
1. Number of hydrogen atoms in 36 ml of H2O at 277 K are x NA. x is ______. Sol. Answer (4)
36 ml H2O = 36 g H2O = 2 mole
1 mole H2O contains 2 × NA H atoms
2 mole H2O contains 4 × NA H atoms 2. 5.6 litre of the gas have 1 g weight at STP. Then atomic weight of the gas is ______. Sol. Answer (4) 5.6 1 No. of moles of gas 22.4 4
Weight No. of moles Molecular weight 11 ⇒ Mol. wt. = 4 4Mol. wt. Solution of Assignment Some Basic Concepts of Chemistry 25
3. How much amount of CaCO3 in gram having percentage purity 50 per cent produces 0.56 litre of CO2 at STP on heating? Sol. Answer (5)
CaCO32 CaO CO
5.6 1 To produce 0.56 litre or mole 22.4 10 40
15 Amount of CaCO required = 100 g 3 40 2
Percentage purity is = 50% So, amount required = 5 g
4. Number of lone pairs in 18 ml H2O at 273 K is xNA. Then value of x will be Sol. Answer (2) No. of moles = 1
No. of molecules = NA
No. of lone pair = 2 NA M M 1 2 . 5. Equivalent weight of potash alum is x and equivalent weight of gypsum is y Then x – y will be Sol. Answer (6)
M Eq.wt of Potash alum = 8
M Eq.wt of Gypsum = 2
6. An Oleum is labelled as 105%. Then 100 g of this Oleum will react with y gram of water. What will be value of y? Sol. Answer (5) Oleum water H SO 24 100 g 5 g 105 g
7. The given compounds follow law of multiple proportion : N2O4 and NxO5. The value of x may be Sol. Answer (2)
N2O4 and NXO5 show law of multiple proportion therefore x can be only 2. 8. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is [IIT-JEE 2010] Sol. Answer (3)
25.2 25.25 25 Average titre value = 3
= 25.15 Significant figure is 3. 26 Some Basic Concepts of Chemistry Solution of Assignment
9. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y × 10x. The value of x is [IIT-JEE 2010] Sol. Answer (7) 4 r3 Volume of 1 Ag atom = 3 4108 rcm33 23 3 6.023 10 10.5 r = 1.6 × 10–8 cm r = 1.6 × 10–10 m Number of Ag atoms in 10–12 m2 r2 × n = 10–12 m2 10–12 n –10 2 3.14 (1.6 10 ) 10 107 n 8 n = 1.25 × 107 The value of x = 7 10. 29.2% (w/w) HCl stock solution has a density of 1.25 g mL–1. The molecular weight of HCl is 36.5 g mol–1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is [IIT-JEE 2012] Sol. Answer (8)
1.25 29.2 1000 Molarity of HCl = 10(M) 100 1 36.5 V × 10 = 200 × 0.4 V = 8 mL 11. If the value of Avogadro number is 6.023 × 1023 mol–1 and the value of Boltzmann constant is 1.380 × 10–23 JK–1, then the number of significant digits in the calculated value of the universal gas constant is [JEE(Advanced)2014] Sol. Answer (4) 6.023 × 1023 × 1.380 × 10–23 = 8.312 It has four significant figure.
–1 12. A compound H2X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g ml . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is [JEE(Advanced)2014] Sol. Answer (8)
w2 1000 m mw21 1 mL solvent having mass 0.4 g. 1000 mL solvent having mass 400 g 1000 mL solution contain 3.2 × 80 g solute = 256 g
256 1000 m8 80 400 Solution of Assignment Some Basic Concepts of Chemistry 27
SECTION - G Multiple True-False Type Questions 1. STATEMENT-1 : Solution is the example of homogeneous mixture. STATEMENT-2 : Homogeneous mixture have uniform composition. STATEMENT-3 : A solution may contain more than one solute. (1) T T T (2) F T T (3) T T F (4) F F T Sol. Answer (1) Solution is an example of homogeneous mixture. In a solution, solute may be more than one. 14 2. STATEMENT-1 : Equal weight of CO2 and NO2 have same number of molecule. 14 STATEMENT-2 : CO2 and NO2 have same molecular weight.
STATEMENT-3 : Equal volume of CO2 and NO2 have same weight. (1) T F F (2) T T F (3) F T T (4) F F T Sol. Answer (2) 14 Molecular weight of CO2 and NO2 is same. M 3. STATEMENT-1 : Equivalent weight of Mohr salt is 4 where M-molecular weight. STATEMENT-2 : Mohr salt contains only one metallic cation i.e., Fe2+. Molecular weight STATEMENT-3 : Equivalent weight of a salt = Total positive charge . (1) T T T (2) F T F (3) T F T (4) F F T Sol. Answer (1)
Chemical formula of Mohr salt is FeSO44242 .(NH ) SO .6H O (n-factor = 4) M Equivalent wt. = 4 4. STATEMENT-1 : Equivalent weight of a substance can never be greater than its molecular weight. M STATEMENT-2 : Equivalent weight of boric acid is 1 where M is the molecular weight. M STATEMENT-3 : Equivalent weight of H PO is . 3 2 3 (1) F T F (2) T F T (3) F T T (4) T F F Sol. Answer (1) 5. STATEMENT-1 : Law of conservation of mass is valid in nuclear reaction. STATEMENT-2 : On dilution, normality of the solution changes. STATEMENT-3 : Molarity of the solution does not depend on amount of solution. (1) T T T (2) F T T (3) F F T (4) T F F Sol. Answer (2) 6. STATEMENT-1 : Molarity is temperature independent. STATEMENT-2 : Formality is other name of molarity in case of ionic compounds. STATEMENT-3 : % w/w of a solution is temperature dependent. (1) T T F (2) F T F (3) T T T (4) F F T Sol. Answer (2) 28 Some Basic Concepts of Chemistry Solution of Assignment
SECTION - H Aakash Challengers Questions
1. A gaseous mixture contains 1 mole each of CH4 and C2H6 . Then the average molecular wt. of that gaseous mixture is (1) 20 (2) 23 (3) 35 (4) 15 Sol. Answer (2)
CH42 + 4Cl CCl 4 + 4HCl …(i)
C26 H + 6Cl 2 C 2 Cl 6 + 6HCl …(ii)
Mole of Cl2 consumed = 5 mole From equation (i) and (ii) Percentage composition = 50% Average molecular weight = 23
2. 4.6 g Na is dissolved in 1 litre of water. Then how much H2 gas will be evolved? (1) 2.24 L (2) 1.12 L (3) 4.48 L (4) 11.2 L Sol. Answer (1)
2Na + 2H22 O 2NaOH + H 3. A metal carbonate (0.5 kg) gives 0.28 kg of its oxide on heating. Hence, the equivalent weight of metal is (1) 20 g eq–1 (2) 40 g eq–1 (3) 25 g eq–1 (4) 30 g eq–1 Sol. Answer (1)
MCO32 MO CO
500 280 1 ; E 20 g eq E30E8
4. One mole of KClO3 is heated in presence of MnO2. The produced oxygen is used in burning of Al. Then oxide of Al that will be formed (1) 2 moles (2) 1 mole (3) 4 moles (4) 3 moles Sol. Answer (2) 3 KClO∆ KCl + O 32 2 3 2Al + O Al O 2 223 5. 24 g C is burnt in presence of air gas produced form 1M dibasic acid when passed through 1L water, then correct statement is
(1) 12 g C react to form CO2 (2) 8 g C react to form CO
(3) 16 g C react to form CO (4) 9 g C react to form CO2 Sol. Answer (1)
C+O22 CO
2C + O2 2CO Solution of Assignment Some Basic Concepts of Chemistry 29
6. When a gaseous olefinic hydrocarbon is burnt completely in excess of O2, a contraction in volume equal to double to the volume of hydrocarbon is noticed then hydrocarbon will be
(1) C2H2 (2) C2H4 (3) C2H6 (4) C3H8 Sol. Answer (2) 3x CH O xCO xHO x2x2 2 2 2 ⎛⎞3xV ⎜⎟VxV Contraction in volume ⎝⎠2 7. Choose the incorrect match regarding equivalent wt. Acid Equivalent wt.
(1) H3PO2 – M
M (2) H3PO4 – 3
M (3) H3BO3 – 3
M (4) H SO – 2 4 2
Sol. Answer (3)
H3BO3 is mono basic.
8. An element A reacts with compound BO3 to produce A3O4 and B2O3. The number of moles of A3O4 produced if one mole each of A and BO3 is reacted, is 1 2 (1) 3 (2) 3 (3) 3 (4) 1 Sol. Answer (2)
6A 4BO33423 2A O 2B O 9. Metal chloride contains 71% chlorine. Then calculate equivalent weight of that metal bromide (at. wt. Br = 80) (1) 14.5 (2) 85 (3) 94.5 (4) 100 Sol. Answer (3) Eq. wt. of metal = 14.5 Eq. wt. of metal Bromide = 80 + 14.5 = 94.5 10. Which of the following has highest number of neutrons? 56 (1) 44 g of CO2 (2) 28 g 26Fe (3) 80 g H2 (4) 100 g He Sol. Answer (4)
No. of neutron = 25 NA × 2 = 50 NA. 11. Which of the following has largest wt.? (1) 1 mole of e– (2) 1 mole of n (3) 1 mole p (4) 1 mole of particle Sol. Answer (4) -particle is helium nucleus 30 Some Basic Concepts of Chemistry Solution of Assignment
12. A compound contains elements X and Y in 1 : 4 mass ratio. If the atomic masses of X and Y are in ratio 1 : 2, then empirical formula will be
(1) XY (2) XY2 (3) X2Y (4) X4Y Sol. Answer (2) XY Mass ratio 1 : 4 Wt. ratio: 1 : 2
ratio = X : Y2 13. In which of the following reactions, law of mass conservation is not valid?
9 4 12 1 n (1) 4Be + 2He 6C + 0 (2) C + O2 CO2
(3) CH4 + 2O2 CO2 + 2H2O (4) NaOH + HCl NaCl + H2O Sol. Answer (1) This reaction is an example of nuclear reaction.
14. Molarity of pure D2O will be (assume density of D2O is 1 g/ml) (1) 55.56 (2) 1 (3) 50 (4) 10 Sol. Answer (3) 1000 50 Molarity = 20
15. A gaseous mixture contains 40% O2, 40% N2, 10% CO2, 10% CH4 by volume. Calculate the vapour density of the gaseous mixture. 40 32 40 28 10 44 10 16 M Sol. w 100
1280 1120 440 160 = 100 3000 30 = 100 M 30 w 15 Vapor density = 22
16. 11.2 litre of a hydrocarbon at STP produces 44.8 litre of CO2 at STP and 36 gm of H2O during its combustion. Calculate the molecular formula of hydrocarbon. m Sol. C H + O nCO + H O n m 2 2 2 2 11.2 n 11.2 m11.2 mole mole 22.4 22.4 222.4
n44.8 222.4 =n = 4
m1 2 22 m = 8
So given hydrocarbon is C4H8 Solution of Assignment Some Basic Concepts of Chemistry 31
1 17. 10 g impure NaOH is completely neutralised by 1000 ml of 10 N HCl. Calculate the percentage purity of the impure NaOH. Sol. Eq. of NaOH = eq. of HCl
W1 40 10
W = 4 gram 4 100 40% % purity = 10
18. Calculate the total number of atoms in 2 litre gaseous mixture of CH4 and SO2 at STP which have vapour density 26. Sol. Average molecular weight = 52
Percentage of SO2 = 75%
Percentage of CH4 = 25%
NA No. of atom = 1.5 3 0.5 4 22.4
N 6.5 A 6.5 N = 22.4 22.4 A
19. P and Q are two elements which form P2Q3 and PQ2. If 0.15 mole of P2Q3 weighs 15.9 g and 0.15 mole of PQ2 weighs 9.3 g, then, what are atomic weights of P and Q? Sol. Let atomic weight of P = a Atomic weight of Q = b
P2Q3 0.15 (2a +3b) = 15.9 2a + 3b = 106 …(i)
PQ2 0.15 (a + 2b) = 9.3 a + 2b = 62 … (ii) (ii) × 2 – (i)
2a 4b 124 2a 3b 106 —— b18 a + 2 × 18 = 62 a = 62 – 36 a = 26 Atomic weight of P = 26 Atomic weight of Q = 18 32 Some Basic Concepts of Chemistry Solution of Assignment
20. How many sulphur atom will be present in 2 litre of centi normal aq. solution of H2SO4? –2 Sol. Answer 10 NA 21. Calculate the equivalent weight of following assuming molecular weight M
(i) H3BO3 (ii) H3PO2
(iii) FeSO4 . (NH4)2SO4 . 6H2O (iv) K2SO4 Cr2(SO4)3 . 24H2O
(v) Al 2O3 (vi) HNO3
(vii) MgCO3 (viii) CaSO4 Sol. Answer M M (i) M (ii) M (iii) 4 (iv) 8 M M M (v) 6 (vi) M (vii) 2 (viii) 2 22. Calculate the molality of aq. glucose solution in which mole fraction of glucose is 0.05. 0.05 1000 2.92 Sol. m = 0.95 18
23. 18.4 gram mixture of MgCO3 and CaCO3 produce 4.48 litre of CO2 at STP. Then calculate the amount of MgCO3 and CaCO3 in mixture.
Sol. Weight of CaCO3 = 10 g
Weight of MgCO3 = 8.4 g
24. 31.3 gram mixture NaBr and NaCl treated with H2SO4, 28.4 gram of Na2SO4 is produced. Then calculate the amount of NaCl and NaBr in the mixture. Sol. Weight of NaCl = 13.01 g Weight of NaBr = 18.28 g