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13., , In known elements, the maximum number is of, [CPMT 1985], , Significant figures, Units for measurement,, Matter and Separation of mixture, 1., , 2., , 3., , 4., , One fermi is, , 14., , (a) Metals, (b) Non-metals, (c) Metalloids, (d) None of these, Which one of the following is not an element, (a) Diamond, (b) Graphite, (c) Silica, (d) Ozone, , 15., , A mixture of ZnCl 2 and PbCl 2 can be separated by, [AFMC 1989], , [Haryana CEET 1994; DPMT 2004], , (a) 10 13 cm, , (b) 10 15 cm, , (c) 10 10 cm, A picometre is written as, , (d) 10 12 cm, , (a) 10 9 m, , (b) 10 10 m, , (c) 10 11 m, One atmosphere is equal to, (a) 101.325 K pa, , (d) 10 12 m, , 16., , (b) 1013.25 K pa, , 17., , (c) 10 5 Nm, (d) None of these, Dimensions of pressure are same as that of, [CBSE PMT 1995], , (a) Energy, (c) Energy per unit volume, 5., , 6., , (b) Force, (d) Force per unit volume, , The prefix 1018 is, [Kerala MEE 2002], (a) Giga, (b) Nano, (c) Mega, (d) Exa, Given the numbers : 161cm, 0.161cm, 0.0161 cm. The number, of significant figures for the three numbers are, , 18., , 19., , [CBSE PMT 1998], , 7., , 8., , 9., , 10., , 11., , (a) 3, 4 and 5 respectively, (b) 3, 3 and 3 respectively, (c) 3, 3 and 4 respectively, (d) 3, 4 and 4 respectively, Significant figures in 0.00051 are, (a) 5, (b) 3, (c) 2, (d) 4, Which of the following halogen can be purified by sublimation, (a), , F2, , (b) Cl 2, , (c), , Br2, , (d) I2, , Difference in density is the basis of, [Kerala MEE 2002], (a) Ultrafiltration, (b) Molecular sieving, (c) Gravity Separation, (d) Molecular attraction, Which of the following elements of matter would best convey, that there is life on earth, (a) Oxygen, (b) Hydrogen, (c) Carbon, (d) Iron, The compound which is added to table salt for maintaining, proper health is, (a) KCl, (b) KBr, (c), , 12., , NaI, , (d) MgBr2, , Which of the following contains only one element, (a) Marble, , (b) Diamond, , (c) Glass, , (d) Sand, , (a) Distillation, (b) Crystallization, (c) Sublimation, (d) Adding aceitic acid, A mixture of methyl alcohol and acetone can be separated by, (a) Distillation, (b) Fractional distillation, (c) Steam distillation, (d) Distillation under reduced pressure, In, the, final, answer, of, the, expression, , (29.2  20.2) (1 .79  10 5 ), . The number of significant figures, 1 .37, is, [CBSE PMT 1994], (a) 1, (b) 2, (c) 3, (d) 4, 81.4 g sample of ethyl alcohol contains 0.002 g of water. The, amount of pure ethyl alcohol to the proper number of, significant figures is, (a) 81.398 g, (b) 71.40 g, (c) 91.4 g, (d) 81 g, The unit J Pa 1 is equivalent to, (a) m 3, , 20., , (b) cm 3, , (c) dm 3, (d) None of these, From the following masses, the one which is expressed nearest, to the milligram is, (a) 16 g, (b) 16.4 g, (c) 16.428 g, (d) 16.4284 g, [Manipal PMT 2001], , 21., , 22., , The number of significant figures in 6.02  10 23 is, (a) 23, (b) 3, (c) 4, (d) 26, The prefix zepto stands for, [DPMT 2004], (a) 10 9, , 23., , 24., , (b) 10 12, , (c) 10 15, (d) 10 21, The significant figures in 3400 are, (a) 2, (b) 5, (c) 6, (d) 4, The number of significant figures in 6.0023 are, , [BHU 2004], , [Pb.CET 2001], , (a) 5, (c) 3, 25., , (b) 4, (d) 1, , Given P  0.0030m , Q  2.40m , R  3000m , Significant, figures in P, Q and R are respectively, (a) 2, 2, 1, (c) 4, 2, 1, , (b) 2, 3, 4, (d) 4, 2, 3, , [Pb. CET 2002]

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26., , The number of significant figures in 60.0001 is, , 9., [Pb. CET 2000], , 27., , (a) 5, (b) 6, (c) 3, (d) 2, A sample was weighted using two different balances. The, result’s were (i) 3.929 g (ii) 4.0 g. How would the weight of the, sample be reported, (a) 3.929 g, (b) 3 g, (c) 3.9 g, (d) 3.93 g, , 10., , Laws of chemical combination, 1., , Which of the following pairs of substances illustrate the law of, multiple proportions, [CPMT 1972, 78], (a) CO and CO2, (c), , 2., , 3., , 4., , 5., , 6., , NaCl and NaBr, , (b) H 2 O and D2 O, (d) MgO and Mg(OH ) 2, , 1.0 g of an oxide of A contained 0.5 g of A. 4.0 g of another, oxide of A contained 1.6 g of A. The data indicate the law of, (a) Reciprocal proportions, , (b) Constant proportions, , (c) Conservation of energy, , (d) Multiple proportions, , Among the following pairs of compounds, the one that, illustrates the law of multiple proportions is, (a), , NH 3 and NCl 3, , (b) H 2 S and SO 2, , (c), , CuO and Cu 2 O, , (d) CS 2 and FeSO 4, , 12., , The percentage of copper and oxygen in samples of CuO, obtained by different methods were found to be the same. This, illustrates the law of, [AMU 1982, 92], (a) Constant proportions, , (b) Conservation of mass, , (c) Multiple proportions, , (d) Reciprocal proportions, , 8., , 13., , 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, [AMU 1983], (a) Law of reciprocal proportions, (b) Law of constant proportions, (c) Law of multiple proportions, (d) Law of equivalent proportions, Chemical equation is balanced according to the law of, [AMU 1984], , 7., , 11., , (a) Multiple proportion, (b) Reciprocal proportion, (c) Conservation of mass, (d) Definite proportions, Avogadro number is, (a) Number of atoms in one gram of element, (b) Number of millilitres which one mole of a gaseous, substances occupies at NTP, (c) Number of molecules present in one gram molecular mass, of a substance, (d) All of these, Different propartions of oxygen in the various oxides of, nitrogen prove the, [MP PMT 1985], (a) Equivalent proportion, (b) Multiple proportion, (c) Constant proportion, (d) Conservation of matter, , Two elements X and Y have atomic weights of 14 and 16. They, form a series of compounds A, B, C, D and E in which the same, amount of element X, Y is present in the ratio 1 : 2 : 3 : 4 : 5. If, the compound A has 28 parts by weight of X and 16 parts by, weight of Y, then the compound of C will have 28 parts weight, of X and, [NCERT 1971], (a) 32 parts by weight of Y, (b) 48 parts by weight of Y, (c) 64 parts by weight of Y, (d) 80 parts by weight of Y, Carbon and oxygen combine to form two oxides, carbon, monoxide and carbon dioxide in which the ratio of the weights, of carbon and oxygen is respectively 12 : 16 and 12 : 32. These, figures illustrate the, (a) Law of multiple proportions, (b) Law of reciprocal proportions, (c) Law of conservation of mass, (d) Law of constant proportions, A sample of calcium carbonate (CaCO 3 ) has the following, percentage composition : Ca = 40%; C = 12%; O = 48%, If the law of constant proportions is true, then the weight of, calcium in 4 g of a sample of calcium carbonate obtained from, another source will be, (a) 0.016 g, (b) 0.16 g, (c) 1.6 g, (d) 16 g, n g of substance X reacts with m g of substance Y to form p g of, substance R and q g of substance S. This reaction can be, represented as, X  Y  R  S . The relation which can be, established in the amounts of the reactants and the products will be, (a) n  m  p  q, , (b) n  m  p  q, , (c) n  m, , (d), , pq, , Which of the following is the best example of law of, conservation of mass, [NCERT 1975], (a) 12 g of carbon combines with 32 g of oxygen to form 44 g, of CO 2, (b) When 12 g of carbon is heated in a vacuum there is no, change in mass, (c) A sample of air increases in volume when heated at, constant pressure but its mass remains unaltered, (d) The weight of a piece of platinum is the same before and, after heating in air, , 14., , 15., , The law of multiple proportions is illustrated by the two, compounds, [NCERT 1972], (a) Sodium chloride and sodium bromide, (b) Ordinary water and heavy water, (c) Caustic soda and caustic potash, (d) Sulphur dioxide and sulphur trioxide, In compound A, 1.00 g nitrogen unites with 0.57 g oxygen. In, compound B, 2.00 g nitrogen combines with 2.24 g oxygen. In, compound C, 3.00 g nitrogen combines with 5.11 g oxygen., These results obey the following law, [CPMT 1971], (a) Law of constant proportion, (b) Law of multiple proportion, (c) Law of reciprocal proportion, (d) Dalton's law of partial pressure

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16., , Hydrogen combines with oxygen to form H 2 O in which 16 g, , 4., , of oxygen combine with 2 g of hydrogen. Hydrogen also, combines with carbon to form CH 4 in which 2 g of hydrogen, , 17., , 18., , 19., , 20., , combine with 6 g of carbon. If carbon and oxygen combine, together then they will do show in the ratio of, (a) 6 : 16 or 12 : 32, (b) 6 : 18, (c) 1 : 2, (d) 12 : 24, 2 g of hydrogen combine with 16 g of oxygen to form water, and with 6 g of carbon to form methane. In carbon dioxide 12 g, of carbon are combined with 32 g of oxygen. These figures, illustrate the law of, (a) Multiple proportions, (b) Constant proportions, (c) Reciprocal proportions, (d) Conservation of mass, An element forms two oxides containing respectively 53.33 and, 36.36 percent of oxygen. These figures illustrate the law of, (a) Conservation of mass, (b) Constant proportions, (c) Reciprocal proportions, (d) Multiple proportions, After a chemical reaction, the total mass of reactants and, products, [MP PMT 1989], (a) Is always increased, (b) Is always decreased, (c) Is not changed, (d) Is always less or more, A sample of pure carbon dioxide, irrespective of its source, contains 27.27% carbon and 72.73% oxygen. The data support, , 1 amu is equal to, (a), , 1, of C  12, 12, , (c) 1g of H 2, 5., , (b), , 1, of O - 16, 14, , (d) 1.66  10 23 kg, , Sulphur forms the chlorides S 2 Cl 2 and SCl 2 . The equivalent, mass of sulphur in SCl 2 is, [EAMCET 1985; Pb. CET 2001], , 6., , (a) 8 g/mole, (b) 16 g/mole, (c) 64.8 g/mole, (d) 32 g/mole, The sulphate of a metal M contains 9.87% of M. This sulphate, is isomorphous with ZnSO 4 .7 H 2 O . The atomic weight of M, is, (a) 40.3, (c) 24.3, , 7., , [IIT 1991], , (b) 36.3, (d) 11.3, , When 100 ml of 1 M NaOH solution and 10 ml of, 10 N H 2 SO 4 solution are mixed together, the resulting solution, , 8., , will be, [DPMT 1982], (a) Alkaline, (b) Acidic, (c) Strongly acidic, (d) Neutral, In chemical scale, the relative mass of the isotopic mixture of, oxygen atoms (O 16 , O 17 , O 18 ) is assumed to be equal to, [Bihar MADT 1981], , [AIIMS 1992], , 21., , 22., , (a) Law of constant composition, (b) Law of conservation of mass, (c) Law of reciprocal proportions, (d) Law of multiple proportions, The law of definite proportions is not applicable to nitrogen, oxide because, [EAMCET 1981], (a) Nitrogen atomic weight is not constant, (b) Nitrogen molecular weight is variable, (c) Nitrogen equivalent weight is variable, (d) Oxygen atomic weight is variable, Which one of the following pairs of compounds illustrates the, law of multiple proportion, [EAMCET 1989], (a), , H 2O, Na 2O, , (b) MgO, Na 2O, , (c), , Na 2O, BaO, , (d) SnCl 2 , SnCl 4, , 9., , 10., , (a) 6.02  10 23 atoms of H, (b) 4 g atom of Hydrogen, (c) 1.81  10 23 molecules of CH 4, 11., , Atomic, Molecular and Equivalent masses, 1., , [MP PMT 1986], , 2., , 3., , [MP PMT 2002], 12, , (a), , C, , (c), , H1, , (b) O, , 16, , (d) C 13, , (d) 3.0 g of carbon, In the reaction, 2 Na2 S 2O3  I2  Na2 S 4 O6  2 NaI , the, equivalent weight of Na 2 S 2 O 3 (mol. wt. = M) is equal to, (a) M, , Which property of an element is always a whole number, (a) Atomic weight, (b) Equivalent weight, (c) Atomic number, (d) Atomic volume, Which one of the following properties of an element is not, variable, [Bihar MADT 1981], (a) Valency, (b) Atomic weight, (c) Equivalent weight, (d) All of these, The modern atomic weight scale is based on, , (a) 16.002, (b) 16.00, (c) 17.00, (d) 11.00, For preparing 0.1 N solution of a compound from its impure, sample of which the percentage purity is known, the weight of, the substance required will be, [MP PET 1996], (a) More than the theoretical weight, (b) Less than the theoretical weight, (c) Same as the theoretical weight, (d) None of these, 1 mol of CH 4 contains, , 12., , 13., , (b) M / 2, , (c) M / 3, (d) M / 4, When potassium permanganate is titrated against ferrous, ammonium sulphate, the equivalent weight of potassium, permanganate is, [CPMT 1988], (a) Molecular weight /10, (b) Molecular weight /5, (c) Molecular weight /2, (d) Molecular weight, Boron has two stable isotopes, 10 B (19%) and 11 B (81%). The, atomic mass that should appear for boron in the periodic table, is, [CBSE PMT 1990], (a) 10.8, (b) 10.2, (c) 11.2, (d) 10.0

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14., , What is the concentration of nitrate ions if equal volumes of 0.1, M AgNO 3 and 0.1 M NaCl are mixed together, , 27., , [CPMT 1983; NCERT 1985], , 15., , 16., , 17., , (a) 0.1 M, (b) 0.2 M, (c) 0.05 M, (d) 0.25 M, Total number of atoms represented by the compound, CuSO4.5H2O is, [BHU 2005], (a) 27, (b) 21, (c) 5, (d) 8, 74.5 g of a metallic chloride contain 35.5 g of chlorine. The, equivalent weight of the metal is, [CPMT 1986], (a) 19.5, (b) 35.5, (c) 39.0, (d) 78.0, 7.5 grams of a gas occupy 5.8 litres of volume at STP the gas is, (a) NO, (b) N 2 O, (c), , 18., , [CBSE PMT 1999; MH CET 2003], , (b) 2  10 23, , [AMU 1992], , 20., , 21., , 28., , C2 H 2, , (b) CO, , (c), , O2, , (d) CH 4, , 29., , 23., , 24., , 25., , (a) 100, (b) 150, (c) 120, (d) 200, The oxide of a metal has 32% oxygen. Its equivalent weight, would be, [MP PMT 1985], (a) 34, (b) 32, (c) 17, (d) 8, The mass of a molecule of water is, [Bihar CEE 1995], (a), , 3  10, , 26, , kg, , (b) 3  10, , 26, , 26., , 25, , (c) 1.5  10 kg, (d) 2.5  10, 1.24 gm P is present in 2.2 gm, (a), , P4 S 3, , (b) P2 S 2, , (c), , PS 2, , (d) P2 S 4, , kg, , 26, , kg, , (d) 2y, , [CBSE PMT 2000], , (b) 2.24 L, (d) 1.12 L, , If N A is Avogadro’s number then number of valence electrons, in 4.2 g of nitride ions ( N 3 ), , 30., , (a) 2.4 N A, , (b) 4.2 N A, , (c) 1.6 N A, , (d) 3.2 N A, , The weight of 1  10 22 molecules of CuSO 4 .5 H 2 O is, [IIT 1991], , (a) 41.59 g, (c) 4.159 g, 31., , (b) 415.9 g, (d) None of these, , Rearrange the following (I to IV) in the order of increasing, masses and choose the correct answer from (a), (b), (c) and (d), (Atomic mass: N=14, O=16, Cu=63)., I., , 1 molecule of oxygen, , II. 1 atom of nitrogen, III. 1  10 10 g molecular weight of oxygen, IV. 1  10 10 g atomic weight of copper, , [MP PMT 1995], , 22., , y, 4, , Assuming fully decomposed, the volume of CO 2 released at, will be, (a) 0.84 L, (c) 4.06 L, , The vapour density of a gas is 11.2. The volume occupied by, 11.2 g of the gas at ATP will be, [Bihar CET 1995], (a) 11.2 L, (b) 22.4 L, (c) 1 L, (d) 44.8 L, Equivalent weight of crystalline oxalic acid is, (a) 30, (b) 63, (c) 53, (d) 45, The equivalent weight of an element is 4. Its chloride has a V.D, 59.25. Then the valency of the element is, [BHU 1997], (a) 4, (b) 3, (c) 2, (d) 1, 1.25 g of a solid dibasic acid is completely neutralised by 25 ml, of 0.25 molar Ba(OH ) 2 solution. Molecular mass of the acid is, , (b), , STP on heating 9.85g of BaCO 3 (Atomic mass of Ba=137), , (c) 4  10 23, (d) 6  10 23, One litre of a gas at STP weight 1.16 g it can possible be, (a), , y, 2, , (c) y, , The number of atoms in 4.25 g of NH 3 is approximately, (a) 1  10 23, , 19., , (a), , (d) CO 2, , CO, , The atomic weights of two elements A and B are 40 and 80, respectively. If x g of A contains y atoms, how many atoms are, present in 2x g of B, , 32., , (a) II<I<III<IV, (b) IV<III<II<I, (c) II<III<I<IV, (d) III<IV<I<II, 1.520 g of the hydroxide of a metal on ignition gave, gm of oxide. The equivalent weight of metal is, , 0.995, , [DPMT 1984], , (a) 1.520, (c) 19.00, 33., , 34., , (b) 0.995, (d) 9.00, , How much coulomb charge is present on 1g ion of N 3 , (a), , 5.2  10 6 Couloumb, , (b) 2.894  10 5 Couloumb, , (c), , 6.6  10 6 Couloumb, , (d) 8.2  10 6 Couloumb, , Ratio of C p and C v of a gas X is 1.4, the number of atom of, the gas ‘X’ present in 11.2 litres of it at NTP will be, [CBSE 1999], , (a), 35., , 6.02  10 23, , (b) 1.2  10 23, , (c) 3.01  10 23, (d) 2.01  10 23, If we consider that 1/6, in place of 1/12, mass of carbon atom is, taken to be the relative atomic mass unit, the mass of one mole, of a substance will, [AIEEE 2005], (a) Decrease twice, (b) Increase two fold, (c) Remain unchanged, (d) Be a function of the molecular mass of the substance

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36., , 37., , What should be the equivalent weight of phosphorous acid, if, P=31; O=16; H=1, (a) 82, (b) 41, (c) 20.5, (d) None of these, The number of molecule at NTP in 1 ml of an ideal gas will be, (a), , 38., , 39., , 6  10 23, , (b) 2.69  10 19, , (c) 2.69  10 23, (d) None of these, The specific heat of a metal is 0.16 its approximate atomic, weight would be, (a) 32, (b) 16, (c) 40, (d) 64, The weight of a molecule of the compound C 60 H 122 is, [AIIMS 2000], , (a) 1.4  10 21 g, 40., , 41., , 42., , 46., , 48., , Mn 2 O 3, , (b) MnO 2, , (c), , MnO4, , (d) MnO42 , , 100 mL of PH3 on decomposition produced phosphorus and, , hydrogen. The change in volume is [MNR 1986], , 52., , 6.4 kg, 96 kg, , (a), , 50 mL increase, , (b) 500 mL decrease, , (c), , 900 mL decrease, , (d) Nil., , 12 g of Mg (at. mass 24) on reacting completely with acid, gives hydrogen gas, the volume of which at STP would be, [CPMT 1978], , (a), , 22.4 L, , (b) 11.2 L, , (c), , 44.8 L, , (d) 6 .1 L, , 53., , 6.02  10 23 cm 3, , (b) 22400 cm 3, , 2 g atom of nitrogen, , (c) 1 mole of S, 54., , (d) 3.0  10 23 cm 3, , Caffeine has a molecular weight of 194. If it contains 28.9% by, mass of nitrogen, number of atoms of nitrogen in one molecule, of caffeine is, (a) 4, (b) 6, (c) 2, (d) 3, A 400 mg iron capsule contains 100 mg of ferrous fumarate,, (CHCOO ) 2 Fe . The percentage of iron pasent in it is, , Which of the following has least mass, (a), , If the density of water is 1 g cm 3 then the volume occupied, by one molecule of water is approximately, [Pb. PMT 2004], , 55., , 56., , (b) 3  10, , [Pb. PET 1985], 23, , atoms of C, , (d) 7 .0 g of Ag, , How many mole of helium gas occupy 22.4 L at 0 o C at 1, atm. pressure, (a) 0.11, (c) 1.0, , [Kurukshetra CEE 1992; CET 1992], , (b) 0.90, (d) 1.11, , Volume of a gas at STP is 1.12  10 7 cc. Calculate the number, of molecules in it, [BHU 1997], (a), , 3.01  10 20, , (b) 3.01  1012, , (c), , 3.01  10 23, , (d) 3.01  10 24, , 4 .4 g of an unknown gas occupies 2.24 L of volume at, , standard temperature and pressure. The gas may be, , (b) 25%, (d) 8%, , [MP PMT 1995], 26, , The element whose a atom has mass of 10.86  10 kg is, (a) Boron, (b) Calcium, (c) Silver, (d) Zinc, The number of gram atoms of oxygen present in 0.3 gram mole, of (COOH ) 2 .2 H 2 O is, (a) 0.6, (c) 1.2, , 47., , 51., , [IIT 1988; CPMT 1994], , (a), , 100ml of 2.5 molal (2.5m) ammonium hydroxide solution, (a) 0.056 litres, (b) 0.56 litres, (c) 5.6 litres, (d) 11.2 litres, , approximately, (a) 33%, (c) 14%, 45., , when it is converted to, , 16.023  10 23 g, required for the complete, [CBSE PMT 1989], , The equivalent weight of MnSO 4 is half its molecular weight, , What volume of NH 3 gas at STP would be needed to prepare, , (c), , 44., , 50., , (b) 1.09  10 21 g, , (c) 5.025  10 23 g, (d), What is the weight of oxygen, combustion of 2.8 kg of ethylene, (a) 2.8 kg, (b), (c) 9.6 kg, (d), , (a) 18 cm 3, , 43., , 49., , (a) 21, (b) 54, (c) 27.06, (d) 2.086, One gram of hydrogen is found to combine with 80g of, bromine one gram of calcium valency=2 combines with 4g of, bromine the equivalent weight of calcium is, (a) 10, (b) 20, (c) 40, (d) 80, , 57., , proportion. The weight of 2.24 litres of this mixture at NTP is, (a) 4.6 g, (b) 1.6 g, (c) 2.3 g, (d) 23 g, Vapour density of a metal chloride is 66. Its oxide contains, 53% metal. The atomic weight of the metal is, [Bihar MADT 1982], , (b) Carbon monoxide, , (c) Oxygen, , (d) Sulphur dioxide, , The number of moles of oxygen in 1 L of air containing 21%, oxygen by volume, in standard conditions, is, [CBSE PMT 1995; Pb. PMT 2004], , (b) 1.8, (d) 3.6, , A gaseous mixture contains CH 4 and C 2 H 6 in equimolecular, , (a) Carbon dioxide, , 58., , (a) 0.186 mol, , (b) 0.21 mol, , (c) 2.10 mol, , (d) 0.0093 mol, , The number of molecules in 8.96 L of a gas at 0 o C and 1, atmosphere pressure is approximately, (a), , 6.02  10, , 23, , (c) 18.06  10, , 23, , (b) 12.04  10, , [BHU 1993], 23, , (d) 24.08  10 22

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59., , The equivalent weight of a metal is 9 and vapour density of its, chloride is 59.25. The atomic weight of metal is, , (d) 12  10 23, 6  10 23, The volume occupied by 4.4 g of CO 2 at STP is, (c), , 6., , [Pb. CET 2002], , 60., , (a) 23.9, , (b) 27.3, , (c) 36.3, , (d) 48.3, , The molecular weight of a gas is 45. Its density at STP is, , [AFMC 1997, 2004; Pb. CET 1997, 2002], , 7., , [Pb. PMT 2004], , 61., , 62., , (a) 22.4, (b) 11.2, (c) 5.7, (d) 2.0, Equivalent weight of a bivalent metal is 37.2. The molecular, weight of its chloride is, [MH CET 2003], (a) 412.2, (b) 216, (c) 145.4, (d) 108.2, On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2, g of metal. If the vapour density of metal is 32, the simplest, formula of the oxide would be, , (a), 8., , 9., , [DPMT 2004], , 63., , (a), , MO, , (b) M 2 O 3, , (c), , M 2O, , (d) M 2 O 5, , 10., , 0.5  10, , (c), , 3.5  10 23, , (b) 1.5  10, , 2., , (d) 1.8  10 32, , 11., , Which one of the following pairs of gases contains the same, number of molecules, [EAMCET 1987], (a) 16 g of O 2 and 14 g of N 2, , 12., , (b) 2.0478  10 24, , (c) 3.0115  10 24, (d) 4.0956  10 24, The number of molecules in 16 g of methane is, (b) 6.02  10 23, 3.0  10 23, 16, 16,  10 23,  10 23, (c), (d), 6 .02, 3 .0, Number of molecules in 100 ml of each, [Bihar MADT 1985], O 2 , NH 3 and CO 2 at STP are, (a) In the order CO 2  O2  NH 3, , (c) 28 g of N 2 and 22 g of CO 2, , (b) In the order NH 3  O 2  CO 2, , (d) 32 g of O 2 and 32 g of N 2, , (c) The same, (d) NH 3  CO 2  O 2, , Number of gm of oxygen in 32.2 g Na 2 SO 4 .10 H 2 O is, (a) 20.8, (b) 22.4, (c) 2.24, (d) 2.08, 250 ml of a sodium carbonate solution contains 2.65 grams of, Na 2 CO 3 . If 10 ml of this solution is diluted to one litre, what, , (a) 0.1 M, , (b) 0.001 M, , (c) 0.01 M, , (d) 10, , 4, , M, , 13., , 14., , g mol 1, , (d) mol g 1, , The number of water molecules in 1 litre of water is, [EAMCET 1990], , (b) 18  1000, (d) 55.55 N A, , (a) 18, (c) N A, 15., , The number of electrons in a mole of hydrogen molecule is, [CPMT 1987], , A molar solution is one that contains one mole of a solute in, (a) 1000 g of the solvent, (c) One litre of the solution, , (b) One litre of the solvent, (d) 22.4 litres of the solution, , The number of oxygen atoms in 4.4 g of CO 2 is approx., [CBSE PMT 1990], , (a) 1.2  10 23, , (b) 6  10 22, , of, , The molecular weight of hydrogen peroxide is 34. What is the, unit of molecular weight, [MP PMT 1986], (a) g, (b) mol, (c), , [IIT 1986], , 5., , (c) 6.02  10 24, (d) 6.02  10 25, The total number of protons in 10 g of calcium carbonate is, , (b) 8 g of O 2 and 22 g of CO 2, , is the concentration of the resultant solution (mol. wt. of, [EAMCET 2001], Na 2 CO 3 =106), , 4., , (b) 6.02  10 23, , (a), , [Haryana PMT 2000], , 3., , (c) 4.84  10 17, (d) 6.023  10 23, One mole of calcium phosphide on reaction with excess of, water gives, [IIT 1999], (a) One mole of phosphine, (b) Two moles of phosphoric acid, (c) Two moles of phosphine, (d) One mole of phosphorus pentoxide, 19.7 kg of gold was recovered from a smuggler. How many, atoms of gold were recovered (Au =197), [Pb. CET 1985], , (a) 1.5057  10 24, , 23, , The mole concept, 1., , (b) 1.084  10 18, , ( N 0  6.023  10 23 ), , [Pb. CET 2000], , (a), , 6.023  10 19, , (a) 100, , The number of molecules in 4.25 g of ammonia are, , 23, , (a) 22.4 L, (b) 2.24 L, (c) 0.224 L, (d) 0.1 L, The number of water molecules present in a drop of water, (volume 0.0018 ml) at room temperature is, [DCE 2000], , (a), , 6.02  10, , (b) 12.046  10, , 23, , 23, , (c) 3.0115  10, (d) Indefinite, The numbers of moles of BaCO 3 which contain 1.5 moles of, 23, , 16., , oxygen atoms is, (a) 0.5, , (b) 1, , (c) 3, , (d) 6.02  10 23, , [EAMCET 1991]

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17., , Which of the following is Loschmidt number, (a), , 6  10, , (b) 2.69  10, , 23, , 19, , (c) 3  10, (d) None of these, How many molecules are present in one gram of hydrogen, 23, , 18., , 2., , [AIIMS 1982], , (a), 19., , 20., , 3., , (b) 3.01  10 23, , (c) 2.5  10 23, (d) 1.5  10 23, The total number of gm-molecules of SO 2 Cl 2 in 13.5 g of, sulphuryl chloride is, [CPMT 1992], (a) 0.1, (b) 0.2, (c) 0.3, (d) 0.4, The largest number of molecules is in, [BHU 1997], (a) 34 g of water, (b) 28 g of CO 2, (c), , 21., , 6.02  10 23, , 46 g of CH 3 OH, , 4., , The number of moles of sodium oxide in 620 g of it is, (a) 1 mol, (c) 18 moles, , 22., , [MP PMT 1986], , (d) 54 g of N 2 O5, [BHU 1992], , 5., , (b) 10 moles, (d) 100 moles, [BHU 1992], , 0 .5 g of hydrogen, , (c) 7 g of nitrogen, 23., , (b) 4 g of sulphur, , (a) Different percentage composition, (b) Different molecular weights, (c) Same viscosity, (d) Same vapour density, A compound (80 g) on analysis gave C = 24 g, H = 4 g, O =, 32 g. Its empirical formula is, [CPMT 1981], (a), , C2 H 2 O2, , (b) C 2 H 2 O, , (c), , CH 2 O 2, , (d) CH 2 O, , The empirical formula of a compound is CH 2 O. 0.0835 moles, of the compound contains 1.0 g of hydrogen. Molecular, formula of the compound is, (a) C 2 H 12 O 6, (b) C 5 H 10 O 5, , 2 g of oxygen contains number of atoms equal to that in, , (a), , (a) 40, (b) 60, (c) 8, (d) 10, The percentage of nitrogen in urea is about, [KCET 2001], (a) 46, (b) 85, (c) 18, (d) 28, If two compounds have the same empirical formula but, different molecular formula, they must have, , (c), 6., , (d) 2 .3 g of sodium, , The empirical formula of an acid is CH 2O2 , the probable, (a), , CH 2 O, , (b) CH 2O2, , (c), , C 2 H 4 O2, , (d) C3 H 6 O4, , [CBSE PMT 2001], , 24., , (a), , 7., , (c) Fructose and sucrose, , [BCECE 2005], , (a), , 6.0  10, , (c) 12  10 23, , (b) 3  10, , In which of the following pairs of compounds the ratio of C, H, and O is same, , (b) Glucose and acetic acid, , (c) 5  6.02  10 23 atoms/mole, (d) None of these, The number of molecules of CO2 present in 44g of CO2 is, 23, , [AFMC 2000], , (a) Acetic acid and methyl alcohol, , 45  6.02  10 23 atoms/mole, , (b) 5  6.62  10 23 atoms/mole, , 25., , (d) C 3 H 6 O 3, , molecular formula of acid may be, , Molarity of liquid HCl with density equal to 1.17 g / cc is, (a) 36.5, (b) 18.25, (c) 32.05, (d) 4.65, How many atoms are contained in one mole of sucrose, [Pb. PMT 2002], (C12 H 22 O11 ), , C 4 H 8 O8, , 23, , (d) All of these, , Chemical stoichiometry, 1., , (d) 3  10 10, , How much of NaOH is required to neutralise 1500 cm 3 of 0.1, [KCET 2001], N HCl (Na = 23), (a) 40 g, , 26., , A sample of phosphorus trichloride (PCl 3 ) contains 1.4 moles, , 27., , (e) 2.409  10 24, The number of sodium atoms in 2 moles of sodium, ferrocyanide is, [BHU 2004], , (b) 4 g, , (c) 6 g, (d) 60 g, of the substance. How many atoms are there in the sample[Kerala PMT 2004], 2., How much water should be added to 200 c.c of semi normal, (a) 4, (b) 5.6, solution of NaOH to make it exactly deci normal, (c) 8.431  10 23, (d) 3.372  10 24, , [AFMC 1983], , (a) 12  10 23, , (b) 26  10 23, , 34  10 23, , (d) 48  10 23, , (c), , 3., , Percentage composition & Molecular formula, 1., , The percentage of oxygen in NaOH is, , [CPMT 1979], , (a) 200 cc, , (b) 400 cc, , (c) 800 cc, , (d) 600 cc, , 2.76 g of silver carbonate on being strongly heated yield a, residue weighing, [Pb. CET 2003], (a) 2.16 g, , (b) 2.48 g, , (c) 2.64 g, , (d) 2.32 g

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4., , In the reaction, 4 NH 3 (g)  5O 2 (g)  4 NO(g)  6 H 2 O(g) ,, When 1 mole of ammonia and 1 mole of O 2 are made to react, , (c) 80 grams, 13., , (d) 320 grams, , What is the normality of a 1 M solution of H 3 PO4, , to completion, , [AIIMS 1982], , (a) 1.0 mole of H 2 O is produced, (b) 1.0 mole of NO will be produced, , 14., , (c) All the oxygen will be consumed, , (a) 0.5 N, , (b) 1.0 N, , (c) 2.0 N, , (d) 3.0 N, , Normality of 2M sulphuric acid is, (a) 2N, , (d) All the ammonia will be consumed, 5., , 6., , 7., , 8., , Haemoglobin contains 0.33% of iron by weight. The molecular, weight of haemoglobin is approximately 67200. The number of, iron atoms (At. wt. of Fe = 56) present in one molecule of, haemoglobin is, [CBSE PMT 1998], , 15., , How many g of a dibasic acid (Mol. wt. = 200) should be, present in 100 ml of its aqueous solution to give decinormal, strength, [AIIMS 1992], (a) 1 g, , (b) 2 g, , (d) 2, , (c) 10 g, , (d) 20 g, , What quantity of ammonium sulphate is necessary for the, production of NH 3 gas sufficient to neutralize a solution, , 16., , The solution of sulphuric acid contains 80% by weight, H 2 SO 4 . Specific gravity of this solution is 1.71. Its normality, , containing 292 g of HCl ? [HCl=36.5; (NH 4 )2 SO 4 =132;, , is about, , NH 3 =17], , (a) 18.0, , (b) 27.9, , (c) 1.0, , (d) 10.0, , [CPMT 1992], , (a) 272 g, , (b) 403 g, , (c) 528 g, , (d) 1056 g, , 17., , [CBSE 1991], , Mohr's salt is dissolved in dil. H 2 SO 4 instead of distilled, , The percentage of P2 O 5 in diammonium hydrogen phosphate, , water to, , (NH 4 ) 2 HPO4 is, , (a) Enhance the rate of dissolution, , [CPMT 1992], , (a) 23.48, , (b) 46.96, , (c) 53.78, , (d) 71.00, , (b) Prevent cationic hydrolysis, (c) Increase the rate of ionisation, , 1, moles of oxygen combine with Al to form Al2 O 3 the, 2, weight of Al used in the reaction is (Al=27), [EAMCET 1980], If 1, , (d) Increase its reducing strength, 18., , (b) 54 g, 19., , (d) 31 g, , The percentage of Se in peroxidase anhydrous enzyme is 0.5%, by weight (atomic weight=78.4). Then minimum molecular, weight of peroxidase anhydrous enzyme is, (a) 1.568  10 4, , (b) 1.568  10 3, , (c) 15.68, , (d) 3.136  10 4, , 20., , Acidified potassium permanganate solution is decolourised by, (a) Bleaching powder, , (b) White vitriol, , (c) Mohr's salt, , (d) Microcosmic salt, , Approximate atomic weight of an element is 26.89. If its, equivalent weight is 8.9, the exact atomic weight of element, would be, [DPMT 1984], (a) 26.89, , (b) 8.9, , (c) 17.8, , (d) 26.7, , Vapour density of a gas is 22. What is its molecular mass, [AFMC 2000], , H 2 evolved at STP on complete reaction of 27 g of, , Aluminium with excess of aqueous NaOH would be, , 21., , (a) 33, , (b) 22, , (c) 44, , (d) 11, , Equivalent weight of KMnO4 acting as an oxidant in acidic, medium is, , [CPMT 1991], , 12., , N, 4, , (c) 4, , [CBSE PMT 2001], , 11., , (d), , (b) 1, , (c) 49.5 g, , 10., , N, 2, , (a) 6, , (a) 27 g, 9., , (c), , [AIIMS 1992], , (b) 4N, , [CPMT 1990; MP PET 1999], , (a) 22.4, , (b) 44.8, , (a) The same as its molecular weight, , (c) 67.2, , (d) 33.6 litres, , (b) Half of its molecular weight, (c) One-third of its molecular weight, , What is the % of H 2 O in Fe(CNS )3 .3 H 2O, (a) 45, , (b) 30, , (c) 19, , (d) 25, , What weight of SO 2 can be made by burning sulphur in 5.0, moles of oxygen, (a) 640 grams, , (b) 160 grams, , (d) One-fifth of its molecular weight, 22., , 0.16 g of dibasic acid required 25 ml of decinormal NaOH, solution for complete neutralisation. The molecular weight of, the acid will be, [CPMT 1989], (a) 32, , (b) 64, , (c) 128, , (d) 256

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23., , 24., , [CPMT 1992], , To neutralise 20 ml of M / 10 sodium hydroxide, the volume, of M / 20 hydrochloric acid required is, , (a) 0.6 g, , (b) 1.0 g, , [Andhra MBBS 1980], , (c) 1.5 g, , (d) 2.0 g, , (a) 10 ml, , (b) 15 ml, , (c) 20 ml, , (d) 40 ml, , 32., , In the preceeding question, the amount of Na 2CO 3 present in, the solution is, , Hydrochloric acid solutions A and B have concentration of 0.5, N and 0.1 N respectively. The volume of solutions A and B, required to make 2 litres of 0.2 N hydrochloric are, [KCET 1993], , 33., , (b) 1.060 g, , (c) 0.530 g, , (d) 0.265 g, , How many ml of 1 (M) H 2 SO 4 is required to neutralise 10 ml, of 1 (M) NaOH solution, , (a) 0.5 l of A + 1.5 l of B, , [MP PET 1998; MNR 1982; MP PMT 1987], , (b) 1.5 l of A + 0.5 l of B, (c) 1.0 l of A + 1.0 l of B, (d) 0.75 l of A + 1.25 l of B, 25., , 5 ml of N HCl, 20 ml of N / 2 H 2 SO 4 and 30 ml of, , 34., , The normality of the resulting solution is, , 27., , (a), , N /5, , (b), , N / 10, , (c), , N / 20, , (d), , N / 40, , [MNR 1991], , 35., , 29., , (b) 5.0, , (c) 10.0, , (d) 20.0, , Which of the following cannot give iodometric titrations, 3, , (a), , Fe, , (c), , Pb2 , , (b) Cu, (d), , 2, , Ag , , KMnO4 reacts with ferrous ammonium sulphate according to, , Under similar conditions of pressure and temperature, 40 ml of, slightly moist hydrogen chloride gas is mixed with 20 ml of, ammonia gas, the final volume of gas at the same temperature, and pressure will be, [CBSE PMT 1993], , the equation, , (a) 100 ml, , (b) 20 ml, , (a) 20 ml of 0.1 M FeSO 4, , (c) 40 ml, , (d) 60 ml, , MnO4  5 Fe2   8 H   Mn2   5 Fe3   4 H 2O , here 10, ml of 0.1 M KMnO4 is equivalent to, , 2 MnO4  5 C2O42   16 H   2 Mn2   10CO 2  8 H 2O , here, , (c) 40 ml of 0.1 M FeSO 4, , 20 ml of 0.1 M KMnO4 is equivalent to, , (d) 50 ml of 0.1 M FeSO 4, , [CBSE PMT 1996], , 36., , Ca(OH ) 2  H 3 PO4  CaHPO4  2 H 2 O, , (c) 50 ml of 0.5 M H 2C2O4 (d) 20 ml of 0.1 M H 2C2O4, , weight of H 3 PO4 in the above reaction is, , In order to prepare one litre normal solution of KMnO4 , how, , (a) 21, , (b) 27, , many grams of KMnO4 are required if the solution is used in, , (c) 38, , (d) 49, , acidic medium for oxidation, , [MP PET 2002], , (a) 158 g, , (b) 31.6 g, , (c) 790 g, , (d) 62 g, , [CPMT 1999], , (b) 30 ml of 0.1 M FeSO 4, , KMnO4 reacts with oxalic acid according to the equation,, , (a) 20 ml of 0.5 M H 2C2O4 (b) 50 ml of 0.1 M H 2C2O4, , 28., , (a) 2.5, , [AIIMS 1997], , N / 3 HNO3 are mixed together and volume made to one litre., , 26., , [CPMT 1992], , (a) 2.650 g, , 37., , the, , equivalent, , [Pb. PMT 2004], , The mass of BaCO 3 produced when excess CO 2 is bubbled, through a solution of 0.205 mol Ba(OH ) 2 is, [UPSEAT 2004], , What is the concentration of nitrate ions if equal volumes of 0.1, M AgNO3 and 0.1 M NaCl are mixed together, , (a) 81 g, , (b) 40.5 g, , (c) 20.25 g, , (d) 162 g, , [NCERT 1981; CPMT 1983], , 30., , 31., , (a) 0.1 N, , (b) 0.2 M, , (c) 0.05 M, , (d) 0.25 M, , 38., , solution of NaOH to give a concentration of 10 mg per ml is, , 30 ml of acid solution is neutralized by 15 ml of a 0.2 N base., The strength of acid solution is [CPMT 1986], (a) 0.1 N, , (b) 0.15 N, , (c) 0.3 N, , (d) 0.4 N, , 39., , (a) 100, , (b) 200, , (c) 250, , (d) 500, , Number of moles of KMnO 4 required to oxidize one mole of, Fe(C 2 O 4 ) in acidic medium is, , A solution containing Na 2CO 3 and NaOH requires 300 ml of, 0.1 N HCl using phenolpthalein as an indicator. Methyl orange, is then added to the above titrated solution when a further 25 ml, of 0.2 N HCl is required. The amount of NaOH present in, solution is (NaOH  40, Na 2CO3  106), , The amount of water that should be added to 500 ml of 0.5 N, , 40., , [Haryana CEE 1996], , (a) 0.6, , (b) 0.167, , (c) 0.2, , (d) 0.4, , A hydrocarbon contains 86% carbon, 488ml of the hydrocarbon, weight 1.68 g at STP. Then the hydrocarbon is an

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41., , (a) Alkane, , (b) Alkene, , (c) Alkyne, , (d) Arene, , (c) 18 g, 50., , The ratio of amounts of H 2 S needed to precipitate all the metal, CuSO 4 will be, , 42., , 43., , 44., , 45., , 46., , 47., , 48., , 49., , (b) 1:2, , (c) 2:1, , (d) None of these, , in, , M, FeSO 4 was titrated with, 10, acidic medium. The amount of, , KMnO4 used will be, , [CPMT 1984], , A, , solution, , of, , KMnO4 solution, , ions from 100 ml of 1 M AgNO 3 and 100 ml of 1 M, , (a) 1:1, , (d) 19 g, , 51., , 10, , ml, , (a) 5 ml of 0.1 M, , (b) 10 ml of 1.1 M, , (c) 10 ml of 0.5 M, , (d) 10 ml of 0.02 M, , c.c. of O 2 and 50 c.c. of H 2 . The volume of the gases formed, , 1.12 ml of a gas is produced at STP by the action of 4.12 mg of, alcohol, with methyl magnesium iodide. The molecular mass of, alcohol is, [Roorkee 1992; IIT 1993], , (i) at room temperature and (ii) at 1100C will be, , (a) 16.0, , (b) 41.2, , (c) 82.4, , (d) 156.0, , An electric discharge is passed through a mixture containing 50, , (a) (i) 25 c.c. (ii) 50 c.c., , (b) (i) 50 c.c. (ii) 75 c.c., , (c) (i) 25 c.c. (ii) 75 c.c., , (d) (i) 75 c.c. (ii) 75 c.c., , 52., , 100 ml of 0.1 N hypo decolourised iodine by the addition of x g, of crystalline copper sulphate to excess of KI. The value of ‘x’, is (molecular wt. of CuSO 4 .5 H 2 O is 250), (a) 5.0 g, , (b) 1.25 g, , (c) 2.5 g, , (d) 4 g, , 53., , How many grams of caustic potash required to completely, neutralise 12.6 gm HNO 3, (a) 22.4 KOH, , (b) 1.01 KOH, , (c) 6.02 KOH, , (d) 11.2 KOH, , (b) 9 kg, , (c) 27 kg, , (d) 1.8 kg, , (a), , XY, , (b), , X 2Y, , (c), , XY 3, , (d), , X 2Y3, , A compound contains atoms of three elements in A, B and C. If, the oxidation number of A is +2, B is +5 and that of C is – 2, the, possible formula of the compound is, [CBSE PMT 2000], , If isobutane and n-butane are present in a gas, then how much, oxygen should be required for complete combustion of 5 kg of, this gas, (a) 17.9 kg, , The simplest formula of a compound containing 50% of, element X (atomic mass 10) and 50% of element Y (atomic, mass 20) is, [Roorkee 1994], , 54., , (a), , A 3 (BC 4 ) 2, , (b), , A 3 (B 4 C)2, , (c), , ABC 2, , (d), , A 2 (BC 3 ) 2, , What will be the volume of CO 2 at NTP obtained on heating, 10 grams of (90% pure) limestone, , 16.8 litre gas containing H 2 and O 2 is formed at NTP on, , (a) 22.4 litre, , electrolysis of water. What should be the weight of electrolysed, water, , (b) 2.016 litre, , (a) 5 g, , (b) 9 g, , (d) 20.16 litre, , (c) 10 g, , (d) 12 g, , [Pb. CET 2001], , (c) 2.24 litre, 55., , The ratio of the molar amounts of H 2 S needed to precipitate, , On electrical decomposition of 150 ml dry and pure O 2 , 10%, , the metal ions from 20mL each of 1M Cd (NO 3 )2 and, , of O 2 gets changed to O, then the volume of gaseous mixture, , 0.5 M CuSO 4 is, , after reaction and volume of remaining gas left after passing in, turpentine oil will be, , (a) 1 : 1, , (a) 145 ml, , (b) 149 ml, , (c) 128 ml, , (d) 125 ml, , What should be the weight of 50% HCl which reacts with 100 g, of limestone, (a) 50% pure, , (b) 25% pure, , (c) 10% pure, , (d) 8% pure, , What should be the weight and moles of AgCl precipitate, obtained on adding 500ml of 0.20 M HCl in 30 g of, AgNO 3 solution? ( AgNO 3 = 170), (a) 14.35 g, , (b) 15 g, , [CPMT 1997], , (b) 2 : 1, (c) 1 : 2, (d) Indefinite, 56., , 12 g of Mg (at. mass 24) will react completely with acid to, give, [MNR 1985], , (a) One mole of H 2, (b) 1/2 mole of H 2, (c) 2/3 mole of O 2, (d) Both 1/2 mol of H 2 and 1/2 mol of O 2

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57., , 1 .5 mol of O 2 combine with Mg to form oxide MgO . The, , (b) 22.4 L of CO 2 at STP, , mass of Mg (at. mass 24) that has combined is, , (c) 0.44 g of CO 2, [KCET 2001], , 58., , (a) 72 g, , (b) 36 g, , (c), , (d) 24 g, , 48 g, , 8., , 100 g CaCO3 reacts with 1litre 1 N HCl. On completion of, reaction how much weight of CO 2 will be obtain, [Kerala CET 2005], , (a), , 5 .5 g, , (b) 11 g, , (c), , 22 g, , (d) 33 g, , (e), , 44 g, , 9., , (d) None of these, In a mole of water vapour at STP, the volume actually occupied, or taken by the molecules (i.e., Avogadro’s, No.  Volume, of one molecule) is, [Kerala EEE 2000], (a) Zero, (b) Less than 1% of 22.4 litres, (c) About 10% of the volume of container, (d) 1% to 2% of 22.4 litres, (e) Between 2% to 5% of 22.4 litres, If 10 21 molecules are removed from 200mg of CO 2 , then the, number of moles of CO 2 left are, (a), , 10., , 2.85  10 3, , [IIT 1983], , (b) 28.8  10 3, , (c) 0.288  10 3, (d) 1.68  10 2, The set of numerical coefficient that balances, equation K 2 CrO4  HCl  K 2 Cr2 O7  KCl  H 2 O is, , the, , [Kerala CEE 2001], , 1., , (a) 1, 1, 2, 2, 1, (c) 2, 1, 1, 2, 1, , Mixture of sand and sulphur may best be separated by, [Kerala CET 2001], , (a) Fractional crystallisation from aqueous solution, (b) Magnetic method, (c) Fractional distillation, , 11., , (d) Dissolving in CS 2 and filtering, 2., , 3., , 4., , 13., , The molar heat capacity of water at constant pressure is 75, JK 1 mol 1 . When 1.0 kJ of heat is supplied to 100 g of water, , 6., , 7., , [JIPMER 2000], , (a), , 6.023  10, , 21, , molecules of CO 2, , of, , Na 2 S 2O3, , using, , K2Cr2O7, , (a), , MW / 2, , (b) MW / 3, , (c), , MW / 6, , (d) MW / 1, , by, , [IIT 2000], , 3.92 g of ferrous ammonium sulphate crystals are dissolved in, 100 ml of water, 20 ml of this solution requires 18 ml of, KMnO4 during titration for complete oxidation. The weight of, [Tamilnadu CET 2002], , 14., , which is free to expand, the increases in temperature of water is, (a) 6.6 K, (b) 1.2 K, (c) 2.4 K, (d) 4.8 K, A compound possesses 8% sulphur by mass. The least, molecular mass is, [AIIMS 2002], (a) 200, (b) 400, (c) 155, (d) 355, Which of the following contains maximum number of atoms, , standardization, , KMnO4 present in one litre of the solution is, , (d) NO, , H2, , In, , (d) 12.15  10 3, , iodometry, the equivalent weight of K2Cr2O7 is, , 3, , contain the same number of molecules. The gas X is, (a) CO, (b) CO 2, 5., , 12., , 10 dm of N 2 gas and 10 dm of gas X at the same temperature, , (c), , One litre hard water contains 12.00 mg Mg 2  milli equivalent, of washing soda required to remove its hardness is, (a) 1, (b) 12.15, (c) 1  10 3, , Irrespective of the source, pure sample of water always yields, 88.89% mass of oxygen and 11.11% mass of hydrogen. This is, explained by the law of, [Kerala CEE 2002], (a) Conservation of mass, (b) Constant composition, (c) Multiple proportions, (d) Constant volume, Zinc sulphate contains 22.65% of zinc and 43.9% of water of, crystallization. If the law of constant proportions is true, then, the weight of zinc required to produce 20 g of the crystals will, be, (a) 45.3 g, (b) 4.53 g, (c) 0.453 g, (d) 453 g, 3, , (b) 2, 2, 1, 1, 1, (d) 2, 2, 1, 2, 1, , 15., , 16., , (a) 3.476 g, (b) 12.38 g, (c) 34.76 g, (d) 1.238 g, A 100 ml solution of 0.1 n HCl was titrated with 0.2 N NaOH, solution. The titration was discontinued after adding 30 ml of, NaOH solution. The remaining titration was completed by, adding 0.25 N KOH solution. The volume of KOH required for, completing the titration is, [DCE 1999], (a) 70[CBSE, ml PMT 2003], (b) 32 ml, (c) 35 ml, (d) 16 ml, What volume of Hydrogen gas, at 273 K and 1 atm pressure, will be consumed in obtaining 21.6 g of elemental boron, (atomic mass = 10.8) from the reduction of boron trichloride by, Hydrogen, [AIEEE 2003], (a) 22.4 L, (b) 89.6 L, (c) 67.2 L, (d) 44.8 L, The mass of 112 cm 3 of CH 4 gas at STP is, [Karnataka CET 2001], , [

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17., , (a), , 0.16 g, , (b) 0 .8 g, , (c), , 0.08 g, , (d) 1 .6 g, , Complete combustion of 0.858 g of compound X gives, , 7., , Reason, , :, , Under similar conditions of temperature and, pressure all gases contain equal number of, molecules., , Assertion, , :, , Reason, , :, , One atomic mass unit (amu) is mass of an, atom equal to exactly one-twelfth the mass of, a carbon-12 atom., Carbon-12 isotope was selected as standard., , Assertion, , :, , Molecular mass of A is, , 2.63 g of CO 2 and 1.28 g of H 2 O . The lowest molecular, , mass X can have, (a), , [Kerala MEE 2000], , (b) 86 g, , 43 g, , (c) 129 g, 18., , 8., , (d) 172 g, , In the following reaction, which choice has value twice that of, the equivalent mass of the oxidising agent, SO 2  H 2O  3 S  2 H 2O, , (a) 64, (c) 16, , [DPMT 2000], , :, :, , Reason, , :, , Assertion, , :, , Reason, , :, , Assertion, , :, , Reason, , :, , Assertion, , :, , Reason, , :, , Equivalent, weight, of, an, Atomic weight of the element, , Valency of the element, , Assertion, , :, , Reason, , :, , 14., , Assertion, , :, , 15., , Reason, Assertion, , :, :, , Reason, , :, , Assertion, , :, , Mass spectrometer is used for the, determination of isotopes., Isotopes are the atoms of same element, differing in mass numbers., Gases combine in simple ratio of their, volume but, not always., Gases deviate from ideal behaviour., Isomorphous substances form crystals of, same shape and can grow in saturated, solution of each other., They have similar constitution and chemical, formulae., Atomicity of oxygen is 2., , 10., , 1., , 2., , Assertion, , :, , Volume of a gas is inversely proportional to, the number of moles of a gas., , Reason, , :, , The ratio by volume of gaseous reactants and, products is in agreement with their mole, ratio., [AIIMS 1995], , Assertion, , :, , Molecular weight of oxygen is 16., , Reason, , :, , Atomic weight of oxygen is 16., , Assertion, , :, , Atoms can neither be created nor destroyed., , Reason, , :, , Under similar condition of temperature and, pressure, equal volume of gases does not, contain equal number of atoms., , 11., , 12., , [AIIMS 1996], , 3., , 13., , [AIIMS 1994,2002], , 4., , Assertion, , :, , One mole of SO 2 contains double the number of, molecules present in one mole of O 2 ., , Reason, , :, , Molecular weight of SO 2 is double to that of, O2 ., , 5., , 6., , Assertion, , :, , 1.231 has three significant figures., , Reason, , :, , All numbers right to the decimal point are, significant., , Assertion, , :, , 22.4 L of N 2 at NTP and 5.6 L O 2 at NTP, contain equal number of molecules., , mass of B is M., Vapour density of A four times that of B., Pure water obtained from different sources, such as, river, well, spring, sea etc. always, contains hydrogen and oxygen combined in, the ratio 1 : 8 by mass., A chemical compound always contains, elements combined together in same, proportion by mass, it was discovered by, French chemist, Joseph Proust (1799)., As mole is the basic chemical unit, the, concentration of the dissolved solute is, usually specified in terms of number of moles, of solute., The total number of molecules of reactants, involved in a balanced chemical equation is, known as molecularity of the reaction., A certain element X, forms three binary, compounds, with, chlorine, containing, 59.68%,68.95% and 74.75% chlorine, respectively. These data illustrate the law of, multiple proportions., According to law of multiple proportions, the, relative amounts of an element combining, with some fixed amount of a second element, in a series of compounds are the ratios of, small whole numbers., Equivalent weight of Cu in CuO is 63.6, and in Cu 2 O 31.8., , Reason, Assertion, , 9., , (b) 32, (d) 48, , Read the assertion and reason carefully to mark the correct option out, of the options given below :, (a), If both assertion and reason are true and the reason is the, correct explanation of the assertion., (b), If both assertion and reason are true but reason is not the, correct explanation of the assertion., (c), If assertion is true but reason is false., (d), If the assertion and reason both are false., (e), If assertion is false but reason is true., , M, if the molecular, 4, , 16., , element

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17., , Reason, , :, , 1 mole of an element contains 6.023  10 23, atoms., , Assertion, , :, , 1 amu equals to 1.66  10 24 g ., , Reason, , 1.66  10, , :, , 24, , 1, th of mass of a, g equals to, 12, , C 12 atom., , 1, , a, , 2, , b, , 3, , b, , 4, , c, , 5, , a, , 6, , b, , 7, , a, , 8, , c, , 9, , d, , 10, , c, , 11, , b, , 12, , c, , 13, , c, , 14, , d, , 15, , a, , 16, , a, , 17, , b, , 18, , b, , 19, , a, , 20, , a, , 21, , b, , 22, , b, , 23, , c, , 24, , a, , 25, , a, , 26, , c, , 27, , d, , Percentage composition & Molecular formula, 1, , a, , 2, , a, , 6, , b, , 7, , b, , Significant figures, Units for measurement,, Matter and Separation of mixture, , 3, , b, , 4, , d, , 5, , a, , Chemical stoichiometry, , 1, , a, , 2, , d, , 3, , a, , 4, , c, , 5, , d, , 1, , c, , 2, , c, , 3, , a, , 4, , c, , 5, , c, , 6, , b, , 7, , c, , 8, , d, , 9, , c, , 10, , c, , 6, , c, , 7, , c, , 8, , b, , 9, , a, , 10, , d, , 11, , c, , 12, , b, , 13, , a, , 14, , c, , 15, , b, , 11, , c, , 12, , d, , 13, , d, , 14, , b, , 15, , a, , 16, , b, , 17, , b, , 18, , a, , 19, , a, , 20, , c, , 16, , b, , 17, , b, , 18, , c, , 19, , d, , 20, , c, , 21, , b, , 22, , d, , 23, , a, , 24, , a, , 25, , b, , 21, , d, , 22, , c, , 23, , d, , 24, , a, , 25, , d, , 26, , b, , 27, , d, , 26, , b, , 27, , b, , 28, , b, , 29, , c, , 30, , a, , 31, , b, , 32, , c, , 33, , b, , 34, , c, , 35, , d, , 36, , d, , 37, , b, , 38, , d, , 39, , a, , 40, , b, , 41, , b, , 42, , c, , 43, , c, , 44, , d, , 45, , a, , 46, , b, , 47, , a, , 48, , a, , 49, , a, , 50, , d, , 51, , c, , 52, , b, , 53, , a, , 54, , b, , 55, , b, , 56, , b, , 57, , a, , 58, , c, , Laws of chemical combination, 1, , a, , 2, , d, , 3, , c, , 4, , a, , 5, , c, , 6, , c, , 7, , c, , 8, , b, , 9, , b, , 10, , a, , 11, , c, , 12, , b, , 13, , a, , 14, , d, , 15, , b, , 16, , a, , 17, , c, , 18, , d, , 19, , c, , 20, , a, , 21, , c, , 22, , d, , Critical Thinking Questions, , Atomic, Molecular and Equivalent masses, , 1, , d, , 2, , b, , 3, , b, , 4, , a, , 5, , c, , 1, , c, , 2, , b, , 3, , a, , 4, , a, , 5, , b, , 6, , b, , 7, , b, , 8, , b, , 9, , a, , 10, , d, , 6, , c, , 7, , d, , 8, , b, , 9, , a, , 10, , b, , 11, , a, , 12, , c, , 13, , a, , 14, , d, , 15, , c, , 11, , a, , 12, , b, , 13, , a, , 14, , c, , 15, , b, , 16, , c, , 17, , a, , 18, , b, , 16, , c, , 17, , a, , 18, , d, , 19, , a, , 20, , a, , 21, , b, , 22, , b, , 23, , d, , 24, , c, , 25, , a, , 26, , a, , 27, , c, , 28, , d, , 29, , a, , 30, , c, , 1, , e, , 2, , e, , 3, , c, , 4, , e, , 5, , d, , 31, , a, , 32, , d, , 33, , b, , 34, , a, , 35, , c, , 6, , d, , 7, , a, , 8, , c, , 9, , a, , 10, , b, , 36, , b, , 37, , b, , 38, , c, , 39, , a, , 40, , b, , 11, , a, , 12, , e, , 13, , e, , 14, , a, , 15, , a, , 41, , c, , 42, , d, , 43, , a, , 44, , d, , 45, , d, , 16, , b, , 17, , a, , 46, , b, , 47, , c, , 48, , c, , 49, , b, , 50, , b, , 51, , a, , 52, , b, , 53, , b, , 54, , c, , 55, , b, , 56, , a, , 57, , d, , 58, , d, , 59, , a, , 60, , d, , 61, , c, , 62, , d, , 63, , b, , The mole concept, , Assertion & Reason

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Significant figures, Units of measurement,, Matter and Separation of mixture, , 6., , (c) As the given sulphate is isomorphous with ZnSO 4 .7 H 2 O, its formula would be MSO 4 .7 H 2 O .m is the atomic, weight of M, molecular weight of, , 4., , Force [MLT 2 ], (c) Pressure , ,  [ML1 T  2 ], Area, [L2 ], Energy per unit volume , , [ML2 T 2 ],  [ML1 T  2 ], [L3 ], , (29.2  20.2) (1 .79  10 5 ) 9 .0  1 .79  10 5, , 1 .37, 1 .37, Least precise terms i.e., 9.0 has only two significant figures., Hence, final answer will have two significant figures., , 17., , (b), , 18., 19., , (a) Pure ethyl alcohol  81.4  0.002  81.398 ., (a) JPa–1; Unit of work is Joule and unit of pressure is Pascal., Dimension of Joule i.e. work  F  L  MLT 2  L, , , ,  ML2 T 2, , , , , , 1, 1, 1 1 A, , , ,  MLT 1, Pa Pressure F, F, A, , , ,  , , , ,   , , So, JPa–1  ML2 T 2  L2  L  L3 ., 22., , (d) 1 zepto  10 21, , 23., , (a) As we know that all non zero unit are significant number., Therefore significant figure is 2., , 24., , (a) Number of significant figures in 6.0023 are 5 because all, the zeroes stand between two non zero digit are counted, towards significant figures., , 25., , (b) Given P  0.0030m , Q  2.40m & R  3000m In, P(0.0030) initial zeros after the decimal point are not, , significant. Therefore, significant figures in P(0.0030) are 2., Similarly in Q(2.40) significant figures are 3 as in this case, final zero is significant. In R  (3000) all the zeroes are, significant hence, in R significant figures are 4., 26., , (b) All the zeroes between two non zero digit are significatn., Hence in 60.0001 significant figures is 6., , 27., , (d) Round off the digit at 2nd position of decimal 3.929, = 3.93., , Laws of chemical combination, 12., , (b), , X Y ⇌ R S, , ng, , mg, , pg, , qg, , n  m  p  q by low of conservation of mass., , Atomic, Molecular and Equivalent masses, 5., , (b) The atomic weight of sulphur =32, In SCl 2 valency of sulphur =2, So equivalent mass of sulphur , , 32,  16 ., 2, , MSO 4 .7 H 2 O, ,  m  32  64  126  m  222, m, Hence % of M ,  100  9.87 (given) or, m  222, 100m  9.87m  222  9.87 or 90.13m  222  9.87, 222  9 .87, or m ,  24.3 ., 90.13

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7., ,  22.4gm of gas occupies 22.4L at S.T.P., , (d) For NaOH, M  N, , N1 V1  100ml  1 N  100ml(N ), For H 2 SO 4 , N 2 V2  10ml  10 N  100ml(N ), , 21., , (b) Equivalent weight , , Hence, N 1 V1  N 2 V2 ., 10 ., , Molecular weight of, , (b) 1 mole of CH 4 contains 4 mole of hydrogen atom i.e. 4g atom, of hydrogen., 2, , 11., , (a), , Na 2 SO 3  I 2  Na 2 S 4 O6  NaI, , Molecular weight, Valency, COOH, |, , C OOH, 22., , 2.5, , (b) Valency of the element , , , n  2  0. 5  1, E, , M, M, , M, n  factor, 1, , E, , M, 5, , 23., , (b), , 13., , (a) Atomic mass , , 10  19  81  11 190  891, 1081, , , 100, 100, 100, ,  10.81, 14., , (c) 0.1M, , (d) Molarity , , 0 .25 , , 12., , AgNO3 will, , react, , 0.1 M, , with, , 24., , 16., , , , NaCl to form, , (c) wt. of metallic chloride  74.5, , Equivalent weight of metal , , Molecular weight , , weight of metal  100  32  68 gm, , ", , ", , wt. of metal, 8, wt. of oxygen, , =, , 26., , 39,  35.5  39, 35.5, , 6  10 23 molecules has mass  18 gm, , , , 27., , So molecular weight = 29, So, molecular formula of compound is NO, (d)  17gm NH 3 contains 6  10 23 molecules of NH 3, , 6  10 23,  4 .25, 17, , 28., ,  22.4 L of gas at S.T.P. weight  22.4  1.16, , 1.24gm P is present in =, , 220,  1 .24  2 .2 gm, 124, , x, 40, x,  Avogadro no.  y (say), Number of atoms of A , 40, 40y, Or x , Avogadro no., , (c) Number of moles of A , , 2x, 80, , Number of atoms of B, 2x, 2, 40 y, ,  Av.no. , ,  Av.no.  y, 80, 80 Av.no., (d) BaCO3  BaO  CO 2 , Molecular weight of BaCO3  137  12  3  16 =197,  197gm produces 22.4L at S.T.P., 22.4,  9 .85  1 .12 L at S.T.P.,  9.85 gm produces, 197, ,  25.984  26, This molecular weight indicates that given compound is, C2 H 2 ., (a) Molecular weight  2  V.D  2  11.2  22.4, , 31  4, gm P is present in 220 gm P4 S 3, (124 ), , Number of moles of B , , 6  10 23  4 .25,  4  6  10 23 .,  No. of atoms , 17, (a)  1L of gas at S.T.P. weight 1.16g, , 18,  3  10 23 gm, 6  10 23, , (a) Choice (a) is P4 S 3, , , , 7.5,  22.4  28.96, 5.8, ,  4.25 gm NH 3 contains =, , (a), , 68,  8  17 ., 32, ,  3  10 26 kg ., , (a)  5.8L of gas has mass  7.5 gm, ,  22.4L ", , 20 ., , 1 .25  1000, 25  molecular weight, , 1 molecules has mass , , weight of metal,  35.5, weight of chlorine, , , 19., , W (gm)  1000, V (ml )  molecular weight, , , ,  wt. of metal  74.5  35.5  39, , 18., , 118.50, =3., 39.5, , Equivalent weight of oxide , , 25 ., , 126,  63 ., 2, , 2  V .D, 2  59.25, , E  35.5, 4  35.5, , 0.1,  0 .05 M ., 2, , wt. of chlorine = 35.5, , 17., ,  2H 2O , , 1 .25  1000,  200 ., 0 .25  25, (c) Let weight of metal oxide = 100 gm, Weight of oxygen, = 32gm, , , , 0.1M NaNO 3 . But as the volume doubled, conc. of, NO 3 , , 22.4,  11.2  11.2 L ., 22.4, ,  11.2gm of gas occupies, , 29., , (a) 14 gm N 3  ions have  8 N A valence electrons

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4.2gm of N 3  ions have , 30 ., , Molecular weight of H 3 PO3  3  31  48  82, , 8 N A  4 .2,  2.4 N A, 14, ,  Equivalent weight , , (c) [  Molecular weight of CuSO 4 .5 H 2O, ,  63.5  32  64  90  249.5 ], 6  10, , 23, , 37., , 1  10 22 molecules has weight , , 249.5  1  10, 6  10 23, ,  1 ml at NTP has =, ,  41.58  10 1, ,  4.158, 31., , 38., , (a) (I) 1 molecule of oxygen, , , , (b)  22400 ml at NTP has 6.023  10 23 molecule, , molecules has weight  249.5 gm, 22, ,  1 molecule of O 2 has mass , , 32, 6  10 23, , 39., , (a) Molecular weight of C60 H122  12  60  122  1, , , , 6  10, , , , 1 molecule C60 H122 has mass, , (II) 1 atom of nitrogen, 2  6  10 23 atoms of N 2 has mass = 28gm, ,  1 atom of N 2 has mass, , 28, , 2  6  10 23, , 40 ., , (III) 1  10 10 g molecular weight of oxygen, , g atomic weight  2  1  10 10  2  10 10 g, , (d), , 41., , 1 .520 x  17, , 0 .995, x 8,  1.520 x  1.520  8  0.995 x  0.995  17, , 42., , 43., , (d) d , , M, M, ; 1, or M = V; 18gm = 18ml, V, V, , 18, 6  10 23, , (a) 100gm caffeine has 28.9gm nitrogen, , , 44., , 28.9,  194  56.06 gm, 100, , No. of atoms in caffeine , , 56.06, 4., 14, , (d) Molecular weight of (CHCOO )2 Fe  170, , Fe present in 100 mg of (CHCOO )2 Fe, , CP,  1 .4 so, given gas is diatomic, CV, , 56,  100mg  32.9 mg, 170, This is present in 400 mg of capsule, 32.9,  100  8 .2 ., % of Fe in capsule , 400, , ,  No. of atoms  3.01  10 23  2  6.023  10 23 atoms, (b) The acid is dibasic., , 64,  2 .8  10 3 gm, 28, , (c) 2.5 molal NH 4 OH means 2.5 moles of NH 3 in 1000 g, , 194gm caffeine has =, , 11.2L  3.01  10 23 molecules, , 36., , 2.8  10 3 gm C 2 H 4 requires =, ,  3  10 23 cm 3 ., ,  2.89  10 5 coulomb, , (a), , , , 1 molecule of water has volume , , Then 1 gm ion N 3  (1 mole) carries, , 34., , 28gm C 2 H 4 requires 64gm oxygen, , 6  10 23 molecule of water has volume =18 cc, , (b) One ion carries 3  1.6  10 19 coulomb, ,  3  1.6  10 19  6.02  10 23, , , ,  0.25  22.4 L  5.6 L ., , 1.520 x  12.160  0.995 x  16.915, , 33., , (b) C2 H 4  2O2  2CO 2  2 H 2 O, , Hence, 100 cc solution of NH 3 requires = 0.25 mole, , or 0.525 x  4.755, 4 .755, x, 9., 0 .525, , 842, 6  10 23, , H 2 O (1000 cc of solution), , wt. of metal hydroxide, EM  EOH , , wt. of metal oxide, EM  EO , , , molecule C60 H122 has mass = 842gm, ,  6.4  10 3 gm = 6.4kg., , So, order of increasing masses II  I  III  IV ., 32., , 23, ,  140.333  10 23 gm  1.4  10 21 gm ., ,  2.3  10 23 gm, , (IV) 1  10 10 g atomic weight of copper, , 6 .4,  40 ., 0 .16, ,  720  122  842, ,  5.3  10 23 gm, , , , 6 .023  10 23, 22400, , = 0.0002688  10 23  2.69  1019 ., (c) Sp. heat × atomic wt.= 6.4, 0.16 × atomic wt.= 6.4, Atomic wt. , , 6  10 23 molecule has mass  32 gm, , Molecular weight 82, = 41., , Basicity, 2, , 45 ., , 26, (d) 1 atom has mass  10.86  10 kg

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 10.86  10 23 gm, , 3  10 23 atoms of C has mass , , 6.023  10 23 atoms has mass, , 46., , = 10.86  10 23  6.023  10 23 = 65.40 gm, This is the atomic weight of Zn., (b)  1mole (COOH )2 . 2 H 2 O has 96gm oxygen, ,  0.3 mole (COOH )2 . 2 H 2 O has 96  0.3  28.8 gm, , 47., , 28.8,  1 .8 .,  No. of gram atoms of oxygen , 16, (c) Equimolecular proportion means both gases occupied equal, 2 .24, volume ,  1 .12 L, 2, For CH 4 :, , 5 4., , (c) 1mole of S has mass = 32gm, (d) 7.0 gm of Ag, So, lowest mass = 6gm of C., (c) 1mole of any gas at STP occupies 22.4 L., , 55., , (b) , , , , 5 6. (a) , , 16,  1 .12  0 .8 gm ., 22.4, , For C 2 H 6, , 5 7., , 3 .0, 30, gm  1 .5 gm,  1 .12 , 2, 22.4, Total mass  1.5 gm  0.8 gm  2.3 gm ., (c) Let wt. of metal oxide = 100 gm, wt. of metal = 53gm, wt. of oxygen = 47gm, wt. of metal, Equivalent weight of oxygen , 8, wt. of oxygen, 53,  8  9 .02, 47, 2  V .D, 2  66, 132, , ,  2.96  3, Valency , E  35.5 9  35.5 44.5,  Atomic weight  Equivalentweight  Valency, , , (b), , 5 2., , (a), , (b), , 4.4,  22.4  44, 2 .24, , (d) 1L of air =210 cc O 2, 1,  210  0.0093 ., 22400, 22.4L of a gas at STP has no. of molecules, , (d) , ,  6.023  10 23,  8.96L of a gas at STP has no. of molecules, , 6 .02  10 23  8 .96,  2.408  10 23  24.08  10 22 ., 22.4, (a) Given equivalent weight of metal = 9, Vapour density of metal chloride = 59.25,  molecular weight of metal chloride, , , 5 9., ,  2  V.D  2  59.25  118.5,  valency of metal, , , molecular weight of metal chloride, equivalnet weight of metal  35.5, , 118.5, 118.5, ,  2.66, 9  35.5, 44.5, Therefore atomic weight of the metal, =equivalent weight  valency, Valency of metal , , 4, ,  9  2.66  23.9, , 2 PH 3  2 P  3 H 2, (solid), 2ml, 3ml, 100ml, 150ml, Increase in volume  150ml  100ml  50ml increase., , molecular wt. of metal, volume, , 60 ., , (d) The density of gas , , 61., , 45,  2 gm litre1, 22.4, (c) Equivalent weight of bivalent metal = 37.2, , ,  Atomic weight of metal  37.2  2  74.4, , Mg  2 HCl  MgCl2  H 2, ,  Formula of chloride  MCl 2, ,  24g Mg evolves 22.4L H 2 at STP, , Hence, molecular weight of chloride, , , , 5 3., , 5 8., , Mn SO 4  Mn O2, Change of valency  4  2  2, M, .,  Equivalent weight , 2, , 5 1., , 22.4L of gas has mass , , 210 cc , ,  9.02  3  27.06, (b) One gram of hydrogen combines with 80 gm of bromine., So, equivalent weight of bromine = 80 gm,  4gm of bromine combines with 1 gm of Ca, 1,  80gm of bromine combines with =  80  20 ., 4, 2, , 50 ., , 12g Mg evolves H 2 at STP, , 6  10 23  1 .12  10 7, 22400, , 22400 cc = 1 mole, , 1.12L C 2 H 6 has mass , , 49., , 1.12  10 7 of gas at STP has, , So given gas is CO 2 because CO 2 has molecular mass=44., , 22.4L C 2 H 6 has mass = 30 gm, , 48., , 22400 cc of gas at STP has 6  10 23 molecules, ,  .03  1014  3  1012 ., 2.24L of gas has mass = 4.4gm, , , , 22.4L CH 4 has mass  16 gm, 1.12L CH 4 has mass , , 12  3  10 23,  6 gm, 6  10 23, , 22.4,  12, 24, =11.2L at STP., , (b) (a) 2gm atom of nitrogen = 28gm, (b) 6  10 23 atoms of C has mass  12 gm, , (MCl2 )  74.4  2  35.5  145.4, 62., , (c) As we know that, Equivalent weight , , weight of metal, 8, weight of oxygen

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, , 32,  8  64, 0 .4, ,  1.2  10 23 atoms., , 6., , (b) 44g CO 2 occupies 22.4L at STP, , mol. wt, Vapour density , 2, , 4.4g CO 2 occupies , , Mol. wt  2  V.D  2  32  64, , mol. wt 64, As we know that n , , 1, eq. wt, 64, , 7., , Mass, g, ; 1, or g  ml, Volume, ml, 0.0018ml = 0.0018gm, , (a) D ensity , , Suppose, the formula of metal oxide be M 2 On . Hence the, , No. of moles , , formula of metal oxide  M 2 O ., 63., , 22.4,  4 .4 = 2.24L., 44, , (b) Molecular weight of NH 3 is 17, , weight, 0 .0018, ,  1  10  4, Molecular weight, 18, , No. of water molecules = 6.023  10 23  1  10 4, , , , According to the mole concept, ,  6.023  1019 ., , 17 gm NH 3 has molecules  6.02  10 23,  1 gm NH 3 has molecules , , 6 .02  10 23, 17, , Ca 3 P2  6 H 2 O  2 PH3  3Ca(OH)2, , 8., , (c), , 9., , (d) Amount of gold  19.7kg  19.7  1000 gm =19700 gm, No. of moles , ,  4.25 gm NH 3 has molecules, ,  No. of atoms  100  6.023  10 23, , 6.02  10 23  4 .25, ,  1 .5  10 23 molecule, 17, , The mole concept, ,  6.023  10 25 atoms., , 10 ., , (c) , , , 1., , (a) 16g O 2 has no. of moles , , 16 1, , 32 2, , 14 1, , 28 2, No. of moles are same, so no. of molecules are same., , (b), , 10 gm CaCO 3 =, , 6 .023  10 23,  10, 100, , 1 molecule of CaCO 3 = 50 protons, 6.023  10 22 molecule of CaCO 3  50  6.023  10 22,  3.0115  10 24, , Na 2 SO 4 . 10 H 2 O  2  23  32  4  16  10  18,  46  32  64  180  322gm, , 322gm Na 2 SO 4 .10 H 2 O contains = 224 gm oxygen, , 11., , (b) 16gm of CH 4 = 1mole  6.023  10 23 molecules., , 12., , (c), , According to avogadro's hypothesis equal volumes of all gases, under similar conditions of temperature and pressure contains, equal no. of molecules., , 14., , (d), , d, , 32.2gm Na 2 SO 4 .10 H 2 O contains, =, 3., , 100 gm CaCO 3  6.023  10 23 molecules, ,  6.023  10 22 molecule, , 14g N 2 has no. of moles , , 2., , 19700,  100, 197, , 32.2  224,  22.4 gm, 322, , (b) Molarity , , Since d = 1, , W (gm)  1000, molecular wt.  V(ml.), , So, M  V, 18gm = 18ml, 18ml = N molecules (N = avogadro's no.), , 2.65  1000,  0 .1 M, 106  250, , , , A, , 1000ml , , 10ml of this solution is diluted to 1000 ml N1 V1  N 2 V2, , 10  0.1  1000  x, x, , 0.1  10,  0 .001M ., 1000, , 4., , (c) According to definition of molar solution  A molar solution is, one that contains one mole of a solute in one litre of the, solution., , 5., , (a) 44g of CO has 2  6  10 23 atoms of oxygen, 2, , 4.4g of CO has =, 2, , 12  10 23,  4 .4, 44, , M, (d = density, M= mass, V =volume), V, , 15., , (a), , 16., , (a) , , A, , NA,  1000 = 55.555 N., 18, , This is fact., , , , 3 moles of oxygen is that in 1 mole of BaCO3, 1.5 moles of oxygen is that in mole of BaCO3, , 1, 1,  1 .5   0 .5 ., 3, 2, (b) The no. of molecules present in 1ml of gas at STP is known as, Laschmidt number., 22400 ml of gas has total no. of molecules, , , 17., , A, ,  6.023  10 23

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1ml of gas has total no. of molecules , , 6 .023  10 23, 22400, , 25 ., , (a) wt of CO 2  44, mol wt of CO 2  44, ,  2.69  1019 ., , 18., , (b) , , , 6 .02  10 23, ,  3 .01  10 23 molecule., 2, 19., , , , 13.5 gm of SO 2 Cl 2 , , 44,  6 .02  10 23  6.02  10 23, 44, (c) No. of atoms in one molecule, , 26., , = no. of moles  6.022  10 23, 27., ,  1.4  6.022  10 23  8.432  10 23, (d) As we know that four sodium atom are present in sodium, , ferrocyanide [ Na 4 Fe(CN )6 ], , 1,  13.5  0 .1 ., 135, , Hence, number of Na atoms = No. of moles  number of, atom  Avogadro’s number, , (a) (a) 34gm of water, , , , , 2  4  6.023  10 23  48  10 23, , 18gm H 2 O = 6.023  10 23 molecule, , 6 .023  10 23,  34, 18, , 34gm H 2 O =, ,  11.37  10 23 mole, , Percentage composition & Molecular formula, 1., , (a) , ,  44gm CO 2  6  10 23 molecules, ,  32gm CH 3 OH  6  10 23 molecules,  46gm CH 3 OH , , 3., 4., , 23, , molecules, , (a) Urea- NH 2  CO  NH 2, , 5., , 23., , , , 24., , 7., , (b) Glucose - C 6 H 12 O6, Ratio of C, H and O  1 : 2 : 1, In acetic acid CH 3  C  O  H, ||, , O, , 1170gm, Mol.wt., , 1170,  32.05mole / litre, 36.5, , 6.023  10 23 molecule of sucrose has, , 45  6.023  10 23 atoms/mole, , Ratio of C, H and O 1 : 2 : 1 ., , Chemical stoichiometry, , (Mol. wt. of HCl =36.5), , (a) 1 mole of sucrose contains 6.023  10 23 molecules,  1 molecule of sucrose has 45 atoms, , , , 1,  11.97, 0 .0835, , (Empirical formula) n  Molecular formula, n = whole no. multiple i.e. 1,2,3,4.............., If n  1 molecular formula CH 2O2 ., , (c) Molarity = mole/litre,  1cc contains 1.17gm, ,  1000 cc contains 1170 gm, , 28,  100  46.66 ., 60, , (b) Empirical formula of an acid is CH 2O2, , 2, 1,  mole, 16 8, , 4, 1,  mole., 32 8, , 100 gm of urea contains, , =12gm of hydrogen., 12 gm of H 2 is present in C 2 H12 O6, , 620gm of Na 2 O = 10 mole., , also 4g of sulphur , , , ,  1gm mole of compound contain =, , 6., , (b) 2gm of oxygen contains atom , , 60 gm of urea contains 28gm of nitrogen, , So, Empirical formula  CH 2O, (Simplest formula)., (a)  0.0835 mole of compound contains 1 gm of hydrogen, , Molecular weight = 46 +16 = 62, , 22., , , , (b) Based on facts., (d) C  24 gm , H  4 gm , O  32 gm, , 6  10 23,  54  3  10 23 molecules.,  54gm of N 2 O5 , 108, (b) Sodium oxide  Na 2 O, 62gm of Na 2 O = 1 mole, , 16,  100 =40% oxygen., 40, , So, Molecular formula  C2 H 4 O2, , 6  10 23,  46  8.625  10 23, 32, , (d)  108gm of N 2O5  6  10, , 21., , 2., , 6  10 23, ,  28  3 .8  10 23, 44, , (c) 46gm of CH 3 OH, , 40 gm NaOH contains 16gm of oxygen, ,  100 gm of NaOH contains, , (b) 28gm of CO 2, ,  28gm CO 2, , wt. of CO 2,  6 .02  10 23, mol wt of CO 2, , , (a) Molecular weight of SO 2 Cl 2, ,  32  32  2  35.5 = 135 gm,  135 gm of SO 2 Cl 2 = 1gm molecule, , 20 ., , No. of molecule , , 2gm of hydrogen  6.02  10 23 molecules, 1gm of hydrogen, , 1., , (c), , W(gm)  1000, V  Eq.wt., 1500 ml of 0.1N HCl = 150 ml (N), N 

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1, , 2., , (c), , W(gm)  1000, 150  40, , W (gm ) ,  6 gm ., 150  40, 1000, , 1, 1,  V2 ; V2  1000ml, N 1 V1  N 2 V2 ;  200 , 2, 10, , 10 ., , 27 gm, , (a), , 11., , (c) In Fe(CNS )3 . 3 H 2 O, , 12., , 3  18,  100  19% ., 284, (d) 5 S  5 O 2  5 SO 2 ; 5 O 2  5 SO 2 ; 5  64  320 gm ., , 13., , (d) H 3 PO4 is tribasic so N  3 M  3  1  3 ., , 14., , (b) H 2 SO 4 is dibasic N  2 M  2  2  4 ., , 15 ., , (a) For Dibasic acid E , , % of H 2 O , , , , 2 Ag2CO 3  4 Ag  2CO 2  O2, , 2  276 gm, , 4  108 gm, ,  2  276 gm of Ag2CO 3 gives 4  108 gm,  1 gm of Ag2CO 3 gives , , 4  108, 2  276, ,  2.76 gm of Ag2CO 3 gives, , N, , 4  108  2.76,  2.16 gm, 2  276, , 4., , 16., , (b), , Oxygen is limiting reagent, So, X , , 1,  0 .2 all oxygen consumed, 5, , 18., , 67200 gm Hb =, , 6., , (c), , (NH 4 )2 SO 4  2 NH 3 , 132, , N , , 10  1.71  80  98,  27.9, 98  49, , 2 KMnO4  10 FeSO 4  8 H 2 SO 4 , K 2 SO 4  2 MnSO 4  5 Fe2 (SO 4 )3  8 H 2 O, , 2 HCl, , 2(36 .5 )73 gm, , 10  sp. gr. of the solution wt.% of solute Mol.wt., Molecular wt. of solute Eq. wt., , 2 FeSO 4  H 2 SO 4  [O]  Fe2 (SO 4 )3  H 2 O]  5, [Mohrsalt], , 67200  0 .33, gm Fe, 100, , 672  0 .33, 4., 56, , gm atom of Fe =, , N, , K2 SO 4  2 MnSO4  3 H 2 O  [O], , (c)  100gm Hb contain = 0.33gm Fe, , , , W  1000, E  V (in ml ), , (c) 2 KMnO4  3 H 2 SO 4 , , Left NH 3  1  4  0.2  0.2 ., 5., , M 200, ,  100, 2, 2, , 1, W  1000, ,  W  1gm ., 10 100  100, , (c), , 4 NH 3(g )  5 O2(g )  4 NO( g)  6 H 2 O( g ), t0, 1, 1, 0, 0, t  t 1  4x 1  5x, 4x, 6x, , 3, H2, 2, , 3, 22 .4 33 .6 L, 2, , Volume of water added  1000  200  800ml ., 3., , (d) H 2 O  Al  NaOH  NaAlO2 , , 19., , Mohr-salt reducing agent KMnO4 / H   oxidising agent, (d) Atomic weight = Equivalent weight × Valency, , 73 g HCl  132 g(NH 4 )2 SO 4, 292 g HCl  528 g(NH 4 )2 SO 4, 7., , (c), , 2(NH 4 )2 HPO4  P2 O5, 2(36 1  31  64 ) 264, , % of P2 O 5 , , wt. of P2 O 5,  100, wt of salt, , , 8., , (b), , 2 Al , , 62  80 142, , 26.89, , ,  8.9  3  26.7  Valency ,  3 ., 8.9, , , , 20 ., 21., , MW  2  V.D.  2  22  44 ., (d) 2 KMnO4  3 H 2 SO 4  K 2 SO 4  2 MnSO 4  3 H 2 O  5[O], (c), , Change by 5, , 142,  100  53.78% ., 264, , 3, O 2  Al2 O 3, 2, , According to equation, , Eq. wt. , , 22., , 0 .16, 25, W, 1,  1000 ,  25 ;,  1000 , E, 10, E, 10, , M  2  E  2  64  128 ., HCl, , Al ., 9., , 100  78.4,  1.568  10 4, 0 .5, Minimum m.w.  molecule at least contain one selenium., , Mol.wt., 5, , (c) Dibasic acid NaOH; N 1 V1  N 2 V2, , 3, mole of O 2 combines with 2 mole, 2, , 2 mole Al = 54gm, (a) 0.5 gm Se  100 gm peroxidase anhydrous enzyme, , 2, , 7, , 23., , (d) NaOH, , N 1 V1  N 2 V2 ; 20 , , 78.4gm Se , , 24., , (a), , 1, 1, ,  V ; V = 40 ml., 10 20, , NV  N 1 V1  N 2 V2, 0.2  2  0.5 x  0.1(2  x ), , 0. 4  0. 5 x  0. 2  0. 1 x, 0.2  0.4 x

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x, , 25 ., , (d), , 35 ., , 1, L  0.5 L, 2, , M1 V1 M 2 V2 0 .1  10 M 2 V2, ;, ; M 2 V2  5 ., , , 1, 5, n1, n2, , NV  N1V1  N 2 V2  N 3 V3, N  1000  1  5 , N  0 .025 , , 26., , (d) KMnO4  Mohr salt, , (b), t0, t t, , 1, 1,  20   30  5  10  10  25, 2, 3, , (d) The equivalent weight of H 3 PO4 , , 37., , 98,  49, 2, (b) Ba(OH)2  CO 2  BaCO3  H 2O, , N, ., 40, , NH 3(g )  HCl( g )  NH 4 Cl(s), 20ml, 40ml, 0, 0, 20 ml, solid, , molecular weight, 2,  mole wt of H 3 PO4 = 3 + 31 + 64 = 98, , 36., , , , Atomic wt. of BaCO3 = 137  12  16  3 = 197, , Final volume = 20 ml ., 27., , (b), , No. of mole , , M 1 V1, M 2 V2 20  0 .1 M 2 V2, ;, ; M 2 V2  5 ., , , 2, 5, n1, n2, ,  205 mole of Ba(OH )2 will give .205 mole of BaCO3, , (b) Acidic medium E , , 29., , (c) 0.1 M AgNO3 will react with 0.1 M NaCl to form 0.1 M, , NaNO 3 . But, ,  wt. of 0.205 mole of BaCO3 will be, , M 158, ,  31.6 gm ., 5, 5, , 28., , 30 ., , wt. of substance, mol wt., 1, mole, of, Ba(OH )2 gives 1 mole of BaCO3, , , KMnO4 Oxalic acid, , .205  197  40.385 gm  40.5 gm, 38., , 10  10 3 gm,  1000 =0.25 N, 40  1, V1  500ml ,, V2  ?, , N1V1  N 2 V2 ; 0.5  500  0.25  V2, , (b) (I) Phenopthalein indicate partial neutralisation of, , Na 2 CO 3  NaHCO 3, Meq. of Na 2 CO 3 + Meq. of NaOH = Meq. of HCl, W, W,  1000   1000  NV, E, E, , 39., , V2  1000mL final volume water added = 1000 – 500, = 500 mL., (a) eq. of KMnO4 = eq. of Fe(C2 O4 ), , x 5  1 3, x  0.6, 40 ., , (b), , (Suppose Na 2 CO 3  a gm , NaOH = b gm), , Element, , At.wt., , Mole, , Ratio, , Empirical, formula, , a, b,  1000 ,  1000  300  0 .1 .....(1), 106, 40, (II) Methyl orange indicate complete neutralisation, , C =86%, H =14%, , 12, 1, , 7.1, 14, , 1, 2, , CH, , HCl HCl, , 41., , N 1 V1  N 2 V2 , 25  0.2  0.1  V2 so V2  50ml excess, , 32., , N1  0.5 N  10mg per mL, N2 , , 0 .1, NO 3 ,  0 .05 M, 2, (a) Acid, base, , N 1 V1 = N 2 V2 ; N 1  30  0.2  15 ; N1  0.1 N, 31., , (d), , as the volume is doubled, conc. of, , CuSO 4  Cu 2  S 2   CuS, (H 2 S ), [100×1=100 millimole],  1 mole  1 mole,  100 millimole  100 millimole H 2 S required, , a  Na 2 CO 3  0.53 gm ., M1V1 M 2 V2, , ( NaOH ), n1, n2, , 33., , 34., , 1  V1 1  10, , ; V1  5ml ., 1, 2, (c) Atom in highest oxidation state can oxidize iodide to liberate, I2 which is volumetrically measured by iodometric titration, using hypo., 2 I   I2, , Pb2  Lowest oxidation state can not oxidise iodide to I2 ., , AgNO3  2 Ag   S 2   Ag2 S, (H 2 S ), [100×1 =100 millimole],  2 mole  1 mole,  100 miliimole  50 millimole H 2 S required, , a, b,  1000 ,  1000  350  0.1 .....(2), , 53, 40, From (1) and (2) b =1gm., (c) From solution of (31), From equation (1), , (b) (H 2 SO 4 ), , (b), , 2, , Beleongs to, alkene Cn H 2n, , 50, 1,  ., 100 2, , Ratio, 42., , (c) At room temperature 2 H 2(g) , , O2(g)  2 H 2 O(l), , t =0, t =t, , 50 ml, 50 – 2x, =0, In this case H 2 is limiting reagent, , 50 ml, 0, 50 – x 2x, 25 gases (50)liquid, , x = 25 ml, At 110°C, , t =t, , 2 H 2(g)  O2(g)  2 H 2O(g), 0, , 25 ml, , 50 ml, , Vgas  75 ml

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2, , 43., , (c), , 2, , 2, , CuSO 4  2 KI  K 2 SO 4  CuI 2 ;, , 1.12 mL is obtained from 4.12 mg, , 1, , 2 CuI 2  CuI 2  I2, ,  22400 mL will be obtained from, , I2  2 Na 2 S 2O3  2 NaI  Na 2 S 4 O6, , 4 .12,  22400 mg  84.2 g, 1.12, , Eq. wt. Of CuSO 4 .5 H 2O  Mol.wt.  250, 100 ml of 0.1 N hypo  100 ml of 0.1 N CuSO 4 .5 H 2O, , , 44., , (d), , (b), Element, , 250  0 .1  100,  2 .5 gm, 100, , X, Y, , HNO3  KOH  KNO 3  H 2O, , 0.2  56  11.2 gm ., , (a) Isobutane and n-butane, formula; C4 H10, , 46., , (b) n , , C4 H10 , , have same molecular, , 5  208, O2 ,  17.9 kg, 58, , (b) Cd 2 , , Ratio = 2 : 1, 5 6., , (b) Mg 2  H 2, , n, , 3ml (O)  1ml O 3, 5 7., , (a), , Turpentine oil absorb ozone., (a) 50% HCl itself means 50 gm HCl react with 100 gm sample, , 50,  100  50% ., 100, AgNO3  HCl  AgCl  HNO 3, , KMnO4, , 5 8., , 1, 5, , ROH  CH 3 MgI  CH 4  Mg, 1 mol., , (c), , CaCO 3  2 HCl  CaCl 2  CO 2  H 2 O, 2N, , 100 g, , 44 g, , 100 g CaCO 3 with 1 N HCl gives 22 g CO 2, , Critical Thinking Questions, , 3., , 2, 1, 1, ,  10 ,   0.2, 10, 10 5, , (c), , 1 .5,  3 mole, 0 .5, , 24  3  72 gm ., , 2., , M 1 V1, n, M 2 V2, =, ; M1 V1  1 M 2 V2, n1, n2, n2, , 5 1., , 1, O 2  MgO, 2, 0 .5 mole, 0.5 mole of oxygen react with 1 mole of Mg, Mg , 1mole, , 1.5 mole of oxygen react with, , =14.345 gm, 0.1mole, , FeSO 4, , For (d), M1V1  0 .02  10 , , 12 gm 1,  mole of H 2, 24 gm 2, , 100 g CaCO 3 with 2 N HCl gives 44 g CO 2, , 30, 500  0 .2, 170, 1000, t =0 0.176 mole 0.1 mole limiting, t =t 0.076 mole 0, , (d), , S 2   CdS, , Cu 2  S 2   CuS, 20×0.5 =10, , O2, H2, , 0 .25, 0 .25, , 9, mole, 100, , 20×1= 20, , % Purity , , 50., , 1, , At NTP Vol. CO 2  0.09  22.4  2.016 L ., , V of O2  V of O3  135  10  145ml, , (a), , 2.5, , CaCO 3  CO 2  0.09 mole, , 55., , 150  10, x,  15ml, 100, , 49., , 20, , A3 (BC4 )2  3  2  [5  (2  4 )]2  0 ., , 90% pure 9gm , , 30ml (O)  10 ml O 3, , 48., , 50, , (b) CaCO 3  CaO  CO 2, 10 gm, , 16.8,  0 .75 mole of H 2 and O 2, 22.4, , (a) , , Ratio, 2, , 5 4., , 0.5 mole H 2 – 0.5 mole H 2 O = 9gm., 47., , a/b, 5, , (a), , 13, , O2  4 CO 2  5 H 2 O, 2, , 2 H 2 O  2 H 2  O2 0 .75, 2:1, 2 mole H 2 – 2 mole H 2 O, , At.wt.(b), 10, , 5 3., , For 58gm of C4 H10 208 gm O 2 is required then for 5 kg of, , C4 H10, , %(a), 50, , Simplest formula  X 2 Y, , 12 . 6,  0.2 mole; HNO3  KOH, 63, 0.2 mole  0.2 mole, , 45 ., , 5 2., , (b) H 2 O contains H and O in a fixed ratio by mass. It illustrates, the law of constant composition., (b) 100 g of ZnSO crystals are obtained from =22.65 g Zn, 4, , 1g of ZnSO crystals will be obtained from , 4, , 20 g of ZnSO crystals obtained from , 4, , 4., , OR, I, , 1 mol  22400 cc, , 22.65, g Zn, 100, , 22.65,  20  4 .53 g, 100, , (a) If same volume is occupied by the gas, the no. of molecules are, same, so no. of moles are same., 1 mole of N 2 gas  2  14  28 gm

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1 mole of CO gas  12  16  28 gm, 5., , 1000  100  4.17  t, , 6., , 3 .92  1000 20,  W  1000 ,  58  1000   18  392  1000  5 W=3.476gm/L., , , , 14., , (d) Volume m of HCl neutralised by NaOH = (Caustic soda) = V1, , 1000, t,  2 .4 K ., 100  4 .17, (b)  8gm sulphur is present in 100 gm of substance, , , 7., , M1 V1 M 2 V2  W,  V, , ,  1000   2, 1, 5, M V,  5, , 75, (c) Heat capacity of water per gram ,  4 .17, 18, Q = mST, , 32gm sulphur will present =, , (b) (a) 6.023  10, , 23, , N1V1  N 2 V2 ; 0.1  V1  0.2  30 ; V1  60ml, V total (HCl ) = 100 ml, 40 ml, 40ml 0.1N HCl is now neutralised by KOH (0.25 N) , , molecules of CO 2, , (HCl ) N1V1  N 2 V2 (KOH), , No. of atoms  3  6.023  10 21 = 18.069  10 21 atoms, , 0.1  40  0.25  V2 ;, , (b) 22.4L of CO 2, No. of atoms = 6.023  10 23  3  18.069  10 23 atoms, , 15 ., , (c), , (c) 0.44gm of CO 2, No. of moles , , 0.44, 1, ,  6.023  10 23 moles, 44, 100, , B, ,  6.023  10 moles  3  6.023  10 atoms, 21, , 8., 9., , (a) 200mg of CO 2 = 200  10 3  0.2 gm, = 6  10 23 molecules, , 0.2gm of CO 2 =, , 6  10 23,  0 .2  0 .0272  10 23, 44, , 17., , (a) %C , , 2 WH 2 O, 2, ,  100 , 18, W, 18, Element, %(a), At.wt.(b), C, 83.6, 12, H, 16.4, 1, , 18., , 2 K 2 Cr2 O 4  2 HCl  K 2 Cr2 O7  2 KCl  H 2 O, , W, 24, 1000  Mg 2 ; EW ,  12, E, 2, , 12  10, 12, 12., , (c), , 3, , Mol.wt., Eq. wt. , 6, , 13., , (a), , KMnO4 = Mohr salt, , M 64, ,  16 ;, 4, 4, , Twice 16  2  32, , Assertion & Reason, 1., , (e) We know that from the reaction H 2  Cl2  2 HCl that the, ratio of the volume of gaseous reactants and products is in, agreement with their molar ratio. The ratio of, H 2 : Cl2 : HCl volumes is 1 : 1 : 2 which is the same as, ,  1000  1 ., , K 2 Cr2 O7  4 H 2 SO 4  K 2 SO 4 Cr2 (SO 4 )3,  6/two atom,  12/two atom, Change by 6, 4 H 2 O  3[O], , 7, , (b) SO 2  2 H 2 O  S  2 H 2 O2, 0, 4, EW , ,  2.85  10 3 ., , (a) Meq of Mg 2  Meq of washing soda, , 1 .28,  100  16.4%, .858, a/b, Ratio, 6.96, 1, ×3, 16.4, 2.3, , , , C3 H7  12  3  7  43 gm ., , Now, 6.023  10 23 molecules = 1mole, , 11., , 12 WCO 2, 12, 2.63, ,  100 , ,  100  83.6%, 44, W, 44 0.858, , %H , ,  10 21 (2.72  1) = 1.72  10 21 molecules, , (d), , W, V, W, 112, , , ;, ; W  0.08 gm ., M 22400 16 22400, , (c), , So remaining molecules  2.72  10 21  10 21, , 10 ., , n, , 16., , Now 10 21 molecule are removed., , 1  1 .72  10 21,  0 .285  10  2, 6 .023  10 23, , 3, mole ; 2 mole – 3 mole, 2, , V  3  22.4  67.2 L ., ,  2.72  10 21 molecule, , 1.72  10 21 molecules , , 3, 21.6, H 2  B  3 HCl ; B ,  2 mole, 2, 10.8, , 3, H2, 2, , 1mole , , 18.069  10 21 atoms, (b) It is about 22.4L., , V2  16ml ., , BCl3  3[H ]  B  3 HCl, BCl3 , , 21, , 44gm of CO 2, , = 60 ml, , V1, , 100,  32  400 ., 8, , their molar ratio. Thus volume of gas is directly related to the, number of moles. Therefore, the assertion is false but reason is, true., 2., , (e) We know that molecular weight of substance is calculated by, adding the atomic weight of atoms present in one molecules., We also know that molecular weight of oxygen (O2 ) =2x, (Atomic weight of oxygen)  2  16  32 a.m.u. Atomic

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weight of oxygen is 16, because it is 16 times heavier than1/12 of, carbon atom. Therefore assertion is false but reason is true., , 2 HCl  H 2  I2 (Bimolecular), , th, , 3., , 4., , (c) According to Dalton's atomic theory atoms can neither be, created nor destroyed and according to berzelius hypothesis,, under similar condition of temperature and pressure equal, volumes of all gases contain equal no. of atom. Therefore, assertion is true but reason is false., , , 11., , (a) Both assertion and reason are true and reason is the correct, explanation of assertion., , 12., , (e) Equivalent wt. of Cu in CuO =, , (e) One mole of any substance corresponding to 6.023  10 23, entities is respective of its weight., , , 5., , 6., , Molecular weight of SO 2 is double to that of O 2 ., , (d) 1.231 has four significant figures all no. from left to right are, counted, starting with the first digit that is not zero for, calculating the no. of significant figure., , 13., , (e) Mass spectrometer is the instrument used for the, determination of accurate atomic mass and the relative, abundance of the isotopes., , 14., , (a) Both assertion and reason are true and reason is the correct, explanation of assertion., , 15 ., , (a) Example, , 16., , 17., , 6 .023  10 23  5 .6, 22.4, , 1,  6 .023  10 23, 4, , According to avagadro's hypothesis equal volume of all gases, contain equal no. of molecules under similar condition of, temperature and pressure., 7., , (a) For universally accepted atomic mass unit in 1961,, C-12, was selected as standard. However the new symbol used is ' v', (unified mass) in place of amu., , 8., , (c) Vapour density of B , , M, ,, 2, , Vapour density of A  4 , , M,  2M, 4, , Molecular mass of A  2  2 M  4 M ., 9., , (a) Pure water always contains hydrogen and oxygen in the ratio, 1 : 8 by mass. This is in accordance with the law of constant, composition., , 10 ., , (b) The number of moles of a solute present in litre of solution is, known is as molarity ( M)., The total no. of molecules of reactants present in a balanced, chemical equation is known as molecularity. For example,, , PCl5  PCl3  Cl2 (Unimolecular), , compounds, , are, , (b) No. of atoms present in a molecules of a gaseous element is, called atomicity., For example, O 2 has two atoms and hence its atomicity is 2., , 32 gm  6.023  10 23 molecules = 22.4L, , , , isomorphous, , ZnSO 4 . 7 H 2O, MgSO4 . 7 H 2O, FeSO4 . 7 H 2O (valency, of Zn, Mg, Fe =2)., , Similarly, O2  2  16  32 gm ,, , 22.4 L  6.023  10 23 or 5 .6 L , , of, , K2 SO 3 , K2CrO4 , K2 SeO 4 (valency of S, Cr, Se = 6) and, , Now 22.4L of N 2  volume occupied by one mole of, , , , 63.6, =63.6, 1, , (Valency of Cu =1)., , (d) Molar volume (at NTP) = 22.4 L, , N 2  28 gm  6.023  10 23 molecules., , 63.6, At.wt., =31.8, , 2, Valency, , Equivalent wt. of Cu in Cu 2 O , , Molecular weight of SO 2  32  2  16  64 gm ., Molecular weight of O2  16  2  32 gm ., , Molarity and molecularity are used in different sense., , (a) 12gm of C-12 contain 6.023  10 23 atom, , , , 12,  10  23  1.66  10  24 ., 6.023

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1., , A m ixture of sand and iodine can be separated by, , 8., , [Kerala CEE 2002], , 2., , 2., , (a) Cry stallisation, , (b) Sublim ation, , (c) Distillation, , (d) Fractional distillation, , The elem ent sim ilar to carbon is, (a) Mg, , (b) Mn, , (c) Sn, , (d) Po, , The law of m ultiple proportions was proposed by, (a) Lav oisier, , (b) Dalton, , (c) Proust, , (d) Gay -Lussac, , 10., , 11., , (d) PbSO4 , NiSO4, , M is the m olecular weight of KMnO4 . The equiv alent, (a) M, , (b) M / 3, , (c) M / 5, , (d) M / 7, , An aqueous solution of 6.3 g of oxalic acid dihy drate is, m ade up of to 2 50 ml. The v olum e of 0.1, N NaOH required to com pletely neutralise 1 0 ml of this, [IIT 2001], , (a) 4 0 ml, , (b) 2 0 ml, , (c) 1 0 ml, , (d) 4 ml, , (b) Law of m ultiple proportions, (c) Law of reciprocal proportions, , (a) 1 1 N, , (b) 2 2 N, , (c) 3 3 N, , (d) 4 4 N, , (d) Gay -Lussac’s law of gaseous v olum es, One sam ple of atmospheric air is found to hav e 0.03 % of, carbon dioxide and another sam ple 0.04 %. This is, ev idence that, , 12., , The equivalent weight of phosphoric acid (H 3 PO4 ) in the, reaction, NaOH  H 3 PO4  NaH 2 PO4  H 2O is, [A IIMS 1999; BHU 2005], , (a) The law of constant com position is not alway s true, , (a) 2 5, , (b) 49, , (b) The law of m ultiple proportions is true, , (c) 59, , (d) 9 8, , (c) Air is a com pound, , 13., , (d) Air is a m ixture, , Volume of 0.6 M NaOH required to neutralize 30 cm 3 of, 0.4 M HCl is, , One part of an elem ent A com bines with two parts of, another B. Six parts of the element C com bine with four, parts of the element B. if A and C com bine together the, ratio of their weights will be governed by, [A MU 1984], , (a) 30 cm, , 3, , (c) 50 cm 3, 14., , (a) Law of definite proportion, (b) Law of m ultiple proportion, , 7., , (c) ZnSO 4 , MgSO4, , The normality of orthophosphoric acid hav ing purity of, 70% by weight and specific gravity 1.54 would be[CPMT 1992], , (a) Law of constant composition, , 6., , (b) MgSO4 , CaSO 4, , solution is, , 1 L of N 2 com bines with 3 L of H 2 to form 2 L of, , NH 3 under the sam e conditions. This illustrates the, , 5., , (a) ZnSO 4 , SnSO 4, , weight of KMnO 4 when it is conv erted into K 2 MnO4 is, , [IIT 1992], , 4., , 9., , Crystals of which pair are isom orphous [MP PMT 1985], , [KCET 1995], , (b) 20 cm, , (d) 45 cm 3, , One m ole of potassium dichrom ate com pletely oxidises, the following num ber of m oles of ferrous sulphate in, acidic m edium, [MP PET 1998], , (c) Law of reciprocal proportion, , (a) 1, , (b) 3, , (d) Law of conserv ation of m ass, , (c) 5, , (d) 6, , The m aximum amount of BaSO 4 precipitated on m ixing, equal v olum es of BaCl2 (0.5 M) with H 2 SO 4 (1 M) will, correspond to, , 15., , 3, , The number of equivalents of Na 2 S 2O3 required for the, volumetric estimation of one equivalent of Cu 2  is, , [A IIMS 1997], , [Kerala MEE 2000], , (a) 0.5 M, , (b) 1 .0 M, , (a) 1, , (b) 2, , (c) 1 .5 M, , (d) 2 .0 M, , (c) 3 /2, , (d) 3, , (SET -1)

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1., , (b) Iodine shows sublimation and hence v olatalizes on, heating, the vapour condenses on cooling to give pure, iodine., , 2., , (c) Carbon and tin both are same group elements so have, sim ilarities in properties., , 3., , (b) Law of m ultiple proportions was proposed by Dalton, and v erified by Berzelius., , 4., , (d) Gay - Lussac's law : The volumes of the reacting gases, and those of the gaseous products bear the sim ple, ratio (also called the law of gaseous v olum es)., , 5., , (d), , 6., , (c) The weights of two elements combining with a fixed, am ount of the third elem ent will bear the sam e, ratio(or sim ple m ultiple of it) in which they, them selv es react., , 7., , 10., ,  W 1000 ,  E  V   V1  N 2 V2, , , , 6.3 1000, ,  10  0 .1  V V= 4 0ml., 63 250, , 11., , solution/sam ple, V, , 12., , 13., , Hence 0.5 mole will react with 0.5 mole of H 2 SO 4, 8., , (c) Isom orphous substance m olecules contain the sam e, num ber of atom s bonded in sim ilar fashion., , 9., , (a) KMnO4  K 2 MnO4, , , , 14., , (d) NaOH  H 3 PO4  NaH 2 PO4, (PO43 ), ( NaPO42 ), MW, 98, , ., 1, no. of ionisableH , , (b) NaOH HCl, , (d) Cr2 O7  Cr 3  ; Fe   Fe  , n 1, n6, eq. of K2Cr2O7 = eq. of FeSO 4, , 6, ,  Equivalent weight of KMnO4, , W 100, 70  1000, , N,  11 N ., d 1 .54, 98  100 / 1.54, , N1 V1  N 2 V2 ; 0.6  V1  0.4  30 ; V1  20ml ., , i.e. BaCl2 is the limiting reagent., , Change in 0.5 per atom  7  6  1, , (a) 7 0% by weight 70 gm H 3 PO4  100 gm, , EW , , One m ole of BaCl2 reacts with one m ole of H 2 SO 4 ., , NaOH, , N 1 V1  N 2 V2, , (a) BaCl2  H 2 SO 4  BaSO4  2 HCl, , 7, , (a) Oxalic acid, , 1  6  x 1, 15., , (b) Cu 2   2 I   CuI 2 2CuI 2  Cu 2 I2  I2, , I2  2 Na 2S 2 O3  2 NaI  Na 2 S 4 O6, , M, Molecular weight of KMnO4, , M., 1, Change of 0.5 per atom, , Cu 2   2 Na 2 S 2O3, , ***

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1., , A m ixture of sand and iodine can be separated by, , 8., , [Kerala CEE 2002], , 2., , 2., , (a) Cry stallisation, , (b) Sublim ation, , (c) Distillation, , (d) Fractional distillation, , The elem ent sim ilar to carbon is, (a) Mg, , (b) Mn, , (c) Sn, , (d) Po, , The law of m ultiple proportions was proposed by, (a) Lav oisier, , (b) Dalton, , (c) Proust, , (d) Gay -Lussac, , 10., , 11., , (d) PbSO4 , NiSO4, , M is the m olecular weight of KMnO4 . The equiv alent, (a) M, , (b) M / 3, , (c) M / 5, , (d) M / 7, , An aqueous solution of 6.3 g of oxalic acid dihy drate is, m ade up of to 2 50 ml. The v olum e of 0.1, N NaOH required to com pletely neutralise 1 0 ml of this, [IIT 2001], , (a) 4 0 ml, , (b) 2 0 ml, , (c) 1 0 ml, , (d) 4 ml, , (b) Law of m ultiple proportions, (c) Law of reciprocal proportions, , (a) 1 1 N, , (b) 2 2 N, , (c) 3 3 N, , (d) 4 4 N, , (d) Gay -Lussac’s law of gaseous v olum es, One sam ple of atmospheric air is found to hav e 0.03 % of, carbon dioxide and another sam ple 0.04 %. This is, ev idence that, , 12., , The equivalent weight of phosphoric acid (H 3 PO4 ) in the, reaction, NaOH  H 3 PO4  NaH 2 PO4  H 2O is, [A IIMS 1999; BHU 2005], , (a) The law of constant com position is not alway s true, , (a) 2 5, , (b) 49, , (b) The law of m ultiple proportions is true, , (c) 59, , (d) 9 8, , (c) Air is a com pound, , 13., , (d) Air is a m ixture, , Volume of 0.6 M NaOH required to neutralize 30 cm 3 of, 0.4 M HCl is, , One part of an elem ent A com bines with two parts of, another B. Six parts of the element C com bine with four, parts of the element B. if A and C com bine together the, ratio of their weights will be governed by, [A MU 1984], , (a) 30 cm, , 3, , (c) 50 cm 3, 14., , (a) Law of definite proportion, (b) Law of m ultiple proportion, , 7., , (c) ZnSO 4 , MgSO4, , The normality of orthophosphoric acid hav ing purity of, 70% by weight and specific gravity 1.54 would be[CPMT 1992], , (a) Law of constant composition, , 6., , (b) MgSO4 , CaSO 4, , solution is, , 1 L of N 2 com bines with 3 L of H 2 to form 2 L of, , NH 3 under the sam e conditions. This illustrates the, , 5., , (a) ZnSO 4 , SnSO 4, , weight of KMnO 4 when it is conv erted into K 2 MnO4 is, , [IIT 1992], , 4., , 9., , Crystals of which pair are isom orphous [MP PMT 1985], , [KCET 1995], , (b) 20 cm, , (d) 45 cm 3, , One m ole of potassium dichrom ate com pletely oxidises, the following num ber of m oles of ferrous sulphate in, acidic m edium, [MP PET 1998], , (c) Law of reciprocal proportion, , (a) 1, , (b) 3, , (d) Law of conserv ation of m ass, , (c) 5, , (d) 6, , The m aximum amount of BaSO 4 precipitated on m ixing, equal v olum es of BaCl2 (0.5 M) with H 2 SO 4 (1 M) will, correspond to, , 15., , 3, , The number of equivalents of Na 2 S 2O3 required for the, volumetric estimation of one equivalent of Cu 2  is, , [A IIMS 1997], , [Kerala MEE 2000], , (a) 0.5 M, , (b) 1 .0 M, , (a) 1, , (b) 2, , (c) 1 .5 M, , (d) 2 .0 M, , (c) 3 /2, , (d) 3, , (SET -1)

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1., , (b) Iodine shows sublimation and hence v olatalizes on, heating, the vapour condenses on cooling to give pure, iodine., , 2., , (c) Carbon and tin both are same group elements so have, sim ilarities in properties., , 3., , (b) Law of m ultiple proportions was proposed by Dalton, and v erified by Berzelius., , 4., , (d) Gay - Lussac's law : The volumes of the reacting gases, and those of the gaseous products bear the sim ple, ratio (also called the law of gaseous v olum es)., , 5., , (d), , 6., , (c) The weights of two elements combining with a fixed, am ount of the third elem ent will bear the sam e, ratio(or sim ple m ultiple of it) in which they, them selv es react., , 7., , (a) BaCl2  H 2 SO 4  BaSO4  2 HCl, , 14., , (d) Cr2 O7  Cr 3  ; Fe   Fe  , n 1, n6, eq. of K2Cr2O7 = eq. of FeSO 4, , 1  6  x 1, 15., , (b) Cu, , i.e. BaCl2 is the limiting reagent., , 9., , (a) KMnO4  K 2 MnO4, , 7, , 6, , Change in 0.5 per atom  7  6  1, ,  Equivalent weight of KMnO4, , , , 10., , M, Molecular weight of KMnO4, , M., 1, Change of 0.5 per atom, , (a) Oxalic acid, , NaOH, , N 1 V1  N 2 V2,  W 1000 ,  E  V   V1  N 2 V2, , , , 6.3 1000, ,  10  0 .1  V V= 4 0ml., 63 250, , 11., , (a) 7 0% by weight 70 gm H 3 PO4  100 gm, solution/sam ple, V, , 12., , W 100, 70  1000, , N,  11 N ., d 1 .54, 98  100 / 1.54, , (d) NaOH  H 3 PO4  NaH 2 PO4, (PO43 ), ( NaPO42 ), EW , , 13., ,  2 I  CuI 2 2CuI 2  Cu 2 I2  I2, , Cu 2   2 Na 2 S 2O3, , Hence 0.5 mole will react with 0.5 mole of H 2 SO 4, (c) Isom orphous substance m olecules contain the sam e, num ber of atom s bonded in sim ilar fashion., , , , I2  2 Na 2S 2 O3  2 NaI  Na 2 S 4 O6, , One m ole of BaCl2 reacts with one m ole of H 2 SO 4 ., , 8., , 2, , MW, 98, , ., 1, no. of ionisableH , , (b) NaOH HCl, , N1 V1  N 2 V2 ; 0.6  V1  0.4  30 ; V1  20ml ., , ***