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Some Basic Concepts of Chemistry

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Some Basic Concepts of Chemistry 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.
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