Problems in Physical Chemistry for JEE PDF

For More Material Join: @JEEAdvanced_2024 For More Material Join: @JEEAdvanced_2024 For More Material Join: @JEEAdvanced_2024  1 1 STOICHIOMETRY Laws of Chemical Combination...

For More Material Join: @JEEAdvanced_2024 For More Material Join: @JEEAdvanced_2024 For More Material Join: @JEEAdvanced_2024  1 1 STOICHIOMETRY Laws of Chemical Combination Chemical reactions take place according to certain laws. These laws are called the Laws of Chemical Combination. “These are no longer useful in chemical calculations now but gives an idea of earlier methods of analysing and relating compounds by mass.” v Law of Conservation of Mass [Lavoisier (1774)] During any physical or chemical change, the sum of masses of all substances present in reactions vessel remain conserved. v Law of Constant Composition or Definite Proportions [Proust (1799)] In a given chemical compound, the elements are always combined in the same proportions by mass. v Law of Multiple Proportions [Dalton (1803)] Whenever two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers. v Law of Reciprocal Proportions [Richter (1792)] When two elements combine separately with a fixed mass of a third elements then the ratio of their masses in which they do so is either same or some whole number multiple of the ratio in which they combine with each other. v Gay-Lussac’s Law of combining volumes According to Gay-Lussac’s law of combining volume, when gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the products, if these are also gases, provided all measurements of volumes are done under similar conditions of temperature and pressure. v Avogadro’s Law The volume of a gas (at fixed pressure and temperature) is proportional to the number of moles (or molecules of gas present). Mathematically we can write V µn Or Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. For More Material Join: @JEEAdvanced_2024 2 Dalton’s Atomic Theory v Matter consists of tiny particles called atoms. v Atoms are indestructible. In chemical reactions, the atoms rearrange but they do not themselves break apart. v In any sample of a pure element, all the atoms are identical in mass and other properties. v The atoms of different elements differ in mass and other properties. v When atoms of different elements combine to form compounds, new and more complex particles are formed. However, in a given compound the constituent atoms are always present in the same fixed numerical ratio. Modern Atomic Theory v Atom is no longer considered to be indivisible : It has been found that an atom has a complex structure and is composed of sub-atomic particles such as electrons, protons and neutrons. v Atoms of the same element may not be similar in all respects : 23 24 Ex. Isotopes ( 11 Na , 11 Na) v Atoms of different elements may be similar in one or more respects : Ex. Isobars. ( 40 40 20 Ca , 18 Ar ) v Atom is the smallest unit which takes part in chemical reactions : Although atom is composed of sub-atomic particles, yet it is the smallest particle which takes part in chemical reactions. v The ratio in which atoms unite may be fixed and integral but may not be simple : For example, in sugar molecule (C 12H 22O 11 ), the ratio of C, H and O atoms is 12 : 22 : 11 which is not simple. v Atoms of one element can be changed into atoms of other element : Ex. Artificial Nuclear Reactions v The mass of atom can be changed into energy : According to Einstein’s equation E = mc 2 (E = Energy, m = mass, c = the speed of light, i.e., 3 ´ 10 cm sec -1 ), mass and energy are inter-convertible. Mole Concept (a) Definition of one mole : One mole is a collection of that many entities as there are number of atoms exactly in 12 gm of C-12 isotope. æ 1 ö 1g (b) 1u = 1 amu = ç ÷ of mass of 1 atom of C 12 =. ´ 10 -24 g = 166 è 12 ø NA (c) For Elements : v 1 g atom = 1 mole of atoms = N A atoms. v g atomic mass (GAM) = mass of N A atoms in g. Mass ( g) v Mole of atoms = GAM or molar mass For More Material Join: @JEEAdvanced_2024  3 (d) For molecules : v 1 g molecule = 1 mole of molecule = N A molecule. v g molecular mass (GMM) = mass of N A molecule in g. Mass ( g) v Moles of molecules = GMM or molar mass (e) For ionic compounds : v 1 g formula unit = 1 mole of formula unit = N A formula unit. v g formula mass (GFM) = mass of N A formula unit in g. Mass ( g) v Mole of formula unit = GFM or molar mass Contains 6.022 × 1023 particles (f) 1 mole of a substance Mass as much as molecular mass/atomic mass/ionic mass in grams. If it is a gas, one mole occupies a volume of 22.4 L at 1 atm & 273 K (g) Average or mean atomic mass : Average atomic mass of element A x + A 2 x 2 +...... A Avg. = 1 1 x 1 + x 2 +...... Here A1 , A 2 are isotopic mass of element and x 1 , x 2 are natural abundance of isotopes. (h) Average or mean molar mass : The average molar mass of the different substances present in the container M n + M 2n 2 +...... M Avg. = 1 1 n 1 + n 2 +..... Here M 1 , M 2 are molar mass of substances and n 1 , n 2 are mole of substances present in the container. Empirical Formula, Molecular Formula (a) Empirical formula : Formula depicting constituent atom in their simplest ratio. Molecular formula : Formula depicting actual number of atoms in one molecule of the compound. (b) Relation between molecular formula and empirical formula : Molecular mass n= Empirical Formula mass (c) Densities : Mass v Density = , Unit: g/cc Volume Density of any substance v Relative density = Density of reference substance Density of any substance v Specific gravity = Density of water at 4° C v Vapour density : Ratio of density of vapours to the density of hydrogen at similar pressure and temperature. For More Material Join: @JEEAdvanced_2024 4 Density of vapours at some temperature and pressure Vapour density = Density of H 2 gas at same temperature and pressure Molecular mass Vapour density = 2 Stoichiometry Stoichiometry pronounced (“stoy – key – om – e – tree”) is the calculations of the quantities of reactants and products involved in a chemical reaction. Following methods can be used for solving problems. (a) Mole Method (Balancing is required) D For Ex. : 2KClO 3 ¾¾ ® 2KCl + 3O 2 Moles of KClO 3 Moles of KCl Moles of O 2 = = 2 2 3 (b) Principle of Atom Conservation (P.O.A.C.) method (Balancing is not required) D For Ex. : KClO 3 ¾¾ ® KCl + O 2 POAC for K : 1 ´ mole of KClO 3 = 1 ´ mole of KCl POAC for Cl : 1 ´ mole of KClO 3 = 1 ´ mole of KCl POAC for O : 3 ´ mole of KClO 3 = 2 ´ mole of O 2 Concept of Limiting Reagent (a) Limiting Reagent : It is very important concept in chemical calculation. It refers to reactant which is present in minimum stoichiometry quantity for a chemical reaction. It is reactant consumed fully in a chemical reaction. So all calculations related to various products or in sequence of reactions are made on the basis of limiting reagent. (b) Calculation of Limiting Reagent : Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. Percentage Yield Actual yield The percentage yield of product = ´ 100 theoretical maximum yield Concentration Terms (a) For solutions (i.e., homogeneous mixtures) : v If the mixture is not homogeneous, then none of them is applicable. æwö Mass of solute (i) % by mass ç ÷ = ´ 100 è W ø Mass of solution [ X % by mass means 100 gm solution contains X gm solute ; \ (100 - X ) gm solvent ] æwö Mass of solute (ii) % ç ÷= ´ 100 è V ø Volume of solution æwö [ X % ç ÷ means 100 mL solution contains X gm solute] èV ø For More Material Join: @JEEAdvanced_2024  5 ævö Volume of solute (iii) %ç ÷= ´ 100 è V ø Volume of solution v For gases % by volume is same as mole % Moles of solute (iv) Mole % = ´ 100 Total moles Moles of solute (v) Mole fraction ( X ) = Total moles Moles of solute (vi) Molarity ( M ) = Volume of solution (in litre) Moles of solute (vii) Molality (m ) = Mass of solvent (in kg) Mass of solute Mass of solute (viii) Parts per million (ppm) = ´ 10 6 ~ = ´ 10 6 Mass of solvent Mass of solution No. of formula unit (ix) Formality ( F ) = Volume of solution (in litre) (b) (i) On adding solvent in a solution (dilution) : Number of mole of solute remains constant M f V f = M i Vi (ii) Mixing of two solutions of same solute M f V f = M 1 V 1 + M 2V 2 +.... (c) Volume strength of H 2O 2: H 2O 2 (aq) solution labelled as ‘ x V ’ ‘volume H 2O 2 (for e.g., ‘20 V’ H 2O 2 ), it means x volume of O 2 (in litre) at 1 atm and 273K that can be obtained from 1 litre of such a sample when it decomposes according to 1 H 2O 2 ¾® H 2O + O 2 2 Volume strength of H 2O 2 Molarity of H 2O 2 = 11.2 Eudiometry (For reactions involving gaseous reactants and products) v Eudiometry or gas analysis involves the calculations based on gaseous reactions or the reactions in which at least two components are gaseous, in which the amounts of gases are represented by their volumes, measured at the same pressure and temperature. v Gay-Lussac’s law of volume combination holds good. v Problem may be solved directly is terms of volume, in place of mole. The volume of gases produced is often given by mentioning certain solvent which absorb certain gases. For More Material Join: @JEEAdvanced_2024 6 Solvent gas (es) absorbed KOH (aq.) CO 2 , SO 2 , Cl 2 Ammoniacal Cu 2Cl 2 CO Turpentine oil O3 Alkaline pyrogallol O2 Water NH 3 , HCl Anhydrous CuSO 4 /CaCl 2 H 2O Redox Processes (a) Oxidation Number : It is the charge (real or imaginary) which an atom appears to have when it is in combined state. It may be a whole no. or fractional. An element may have different values of oxidation number depending. It depends on nature of compound in which it is present. There are some operational rules to determine oxidation number. (b) Definition of Oxidation and Reduction : v Oxidation : Addition of oxygen , removal of hydrogen , addition of electronegative element, removal of electropositive element , loss of electrons , increase in oxidation number (de-electronation). v Reduction : Removal of oxygen, addition of hydrogen, removal of electronegative element, addition of electropositive element, gain of electrons, decrease in oxid. no. (electronation). v Redox Reaction : A reaction in which oxidation & reduction occur simultaneously. (c) Oxidising and Reducing Agents : v Oxidising Agents (Oxidants, Oxidisers) : They oxidise others, themselves are reduced & gain electrons. e.g., O 2 , O 3 , HNO 3 , MnO 2 , H 2O 2 , halogens, KMnO 4 , K 2Cr 2O 7 , KIO 3 , Cl(SO 4 ) 3 , FeCl 3 , NaOCl, hydrogen ions.(Atoms present in their higher oxidation state.) Compounds is which reacting atom is present in its highest oxidation state may act as oxidant only. v Reducing Agents (Reductants or Reducers) : They reduce others, themselves get oxidised & lose electrons. Also called reductants or reducers. H 2 molecular form is weak reducing agent but Nascent hydrogen is strong reducing agent e. g. , C, CO, H 2S, SO 2 , SnCl 2 , Sodium thio Sulphate (Na 2S 2O 3 ), Al, Na, CaH 2 , NaBH 4 , LiAlH 4 etc. Compounds in which reacting atom is present in its lowest oxidation state may act as reductant only. Both Oxidising & Reducing Agents : Compounds having reactive atom in its intermediate (between highest and lowest oxidation states) oxidation state may act as both oxidant or reductant. e. g. , SO 2 , H 2O 2 , O 3 , NO 2 , etc. (d) Balancing of redox reactions : v Ion - electron method v Oxidation number method [Concept involved is that, in any chemical reaction electrons ( e - ) cannot be produced so no. of e - s in L.H.S. & R.H.S. should be the same] For More Material Join: @JEEAdvanced_2024  7 Equivalent Concept (a) Law of Chemical equivalence : It states that in any chemical reaction the equivalents of all the reactants and products must be same. 2 A + 3 B ¾® 4C Equivalents of ‘ A’ = Equivalents of ‘ B ’ = Equivalents of ‘ C ’ (b) Terms used in equivalent concept : Molar mass of A v Equivalent mass of A = Valency factor or n factor Mass of ‘ A’ (in g) v Equivalents of ‘ A’ = Equivalent mass of ‘ A’ v Numbers of equivalents of ‘ A’ = no. of moles of ‘ A’´n-factor v Normality (N) : For solutions, concentration term normality (N) is used, which can be defined as “the number of equivalent of solute present in one litre (1000 mL) solution”. Number of g - equivalents of solute N= Volume of solution (in L ) W( g) ´ 1000 N= E ´ V (in mL ) N = Molarity × Valence factor milli-equivalents of solute = N ´ V (in mL) = M ´ V (in mL) ´n factor (c) Valence factor (n-factor) calculation : n-factor here we mean a conversion factor by which we divide molar mass of substance to get equivalent mass and it depends on nature of substance which vary from one condition to another condition. We can divide n-factor calculations in two category. v In case of non-redox reaction. n-factor = moles of charge displaced per mole of species v In case of redox reaction. n-factor of oxidizing or reducing agent = mole of electrons gain or lost per mol of species. (d) Volumetric analysis (Titration) : Titration is a procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a standard solution of another substance, whose concentration is known. v Primary standard : A substance available in a pure form or state of known purity which is used in standardizing a solution. v Standardization : The process by which the concentration of a solution is accurately ascertained. v Standard solution : A solution whose concentration has been accurately determined. v Titrant : The reagent (a standard solution) which is added from a burette standard term to react with the analyte. v Titrate : This mainly involves titrations based chemistry. It can be divided into two major category. (I) Non-redox reactions (II) Redox reactions For More Material Join: @JEEAdvanced_2024 8 (e) Type of reactions : (i) Non-redox reactions : This involve following kind of titrations : 1. Acid-Base titrations 2. Double indicator acid-base titration 3. Precipitation titration 4. Back titration (ii) Redox reactions : This involve following kind of titrations : 1. Iodimetry titrations 2. Iodometry titrations 3. Back titration (f) Titrations : (I) Non-redox titrations (1) Acid-Base titration : To find out strength or concentration of unknown acid or base, it is titrated against base or acid of known strength. At the equivalence point, equivalents of titrate and titrant solution are equal. So by using law of chemical equivalence strength of unknown solution in determined. At equivalence point, milliequivalents of acid = milliequivalents of base. (2) Double indicator acid-base titration : In the acid-base titration the equivalence point is known with the help of indicator which changes its colour at the end point. In the titration of polyacidic base or polybasic acid there are more than one end point. Some times one indicator is not able to give colour change at every end point. So to find out each end point we have to use more than one indicator. For example, in the titration of Na 2CO 3 against HCl there are two end points. Na 2CO 3 + HCl ¾® NaHCO 3 + NaCl At first equivalence point NaHCO 3 + HCl ¾® H 2CO 3 + NaCl At second equivalence point When we use phenolphthalein in the above titration it changes its colour at first end point when NaHCO 3 is formed and with it we can not determine second end point. Similarly with methyl orange it changes its colour at second end point only and we can not determine first end point. It is because all indicator changes colour on the basis of pH of titration mixture. So in titration of NaHCO 3 , KHCO 3 against acid, phenolphthalein can not be used. Titration Indicator pH Range n-factor (against acid) Na 2CO 3 Phenolphthalein 8.3 – 10 1 Na 2CO 3 Methyl orange 3.1 – 4.4 2 Note : When we carry out dilution of solution, milliequivalent, equivalent, milli mole or mole of substance does not change because they represent amount of substance, however molar concentration may change. (3) Precipitation titration : In precipitation titrations involving ionic reactions, we determine the strength of unknown solution of salt by titrating it against a reagent with which it can form precipitate. For example NaCl strength can be determine by titrating it against AgNO 3 solution with which it form white ppt. of AgCl. For More Material Join: @JEEAdvanced_2024  9 At equivalence point, Milliequivalent of NaCl = milliequivalent of AgNO 3 used = milliequivalent of AgCl formed (II) Redox Titrations At equivalence point, milliequivalent of Oxidizing agent used = milliequivalent of reducing agent reacted. S.No. Reagent Half Reaction n-factor of reagent 1. FAS (Mohr’s salt) Fe 2+ ® Fe 3+ + e- 1 [FeSO 4(NH 4 )2SO 4 × 6H 2O] 2. MnO -4 (Permanganate ion) MnO -4 + 8H + + 5 e- ® Mn 2+ + 4H 2O 5 (in acidic medium) 3. MnO -4 (in basic medium) Mn 7 + + e- ® Mn 6+ 1 7+ 4+ 4. MnO -4(in mild basic or Mn - + e ® Mn 3 neutral medium) 5. Cr2O72- (dichromate ion) Cr2O72- + 14H + + 6 e- ® 2Cr 3+ + 7H 2O 6 6. C 2O 2- 4 (Oxalate ion) C 2O 42- ® 2CO 2 + 2 e- 2 7. As 2O 3 As 2O 3 + 5H 2O ® 2 AsO 34- + + 10H + 4 e - 4 8. CaOCl 2 (Bleaching powder) CaOCl 2 + H 2O + 2KI ® Ca(OH )2 + I2 + 2KCl 2 D 9. MnO 2 MnO 2 + 4HCl (Conc. ) ¾¾ ® MnCl 2 + Cl 2 + 2H 2O 2 10. IO -3 IO -3 - + + 5I + 6H ® 3I2 + 3H 2O 5 11. H 2O 2 (act as oxidizing agent) H 2O 2 + 2 e- ® 2H 2O 2 - 12. H 2O 2 (act as reducing agent) H 2O 2 ® O 2 + 2 e 2 13. H 2O 2 (disproportion) H 2O 2 ® H 2O + 1 2 O 2 1 14. Cl 2 (disproportion) 3Cl 2 + 6OH - (strong) ® ClO -3 + 5Cl - + 3H 2O 5/3 - + 15. H 2S (in acidic medium) H 2S + I2 ® S + 2I + 2H 2 2+ 2+ 4+ - 16. Sn (in acidic medium) Sn + I2 ® Sn + 2I 2 17. N 2H 4 N 2H 4 ® N 2 + 4H + + 4 e- 4 18. SO 2- 3 (in acidic medium) SO 23- + H 2O ® SO 24- + 2 e + 2H- + 2 19. Na 2S2O 3 (Sodium thiosulphate 2Na 2S2O 3 + I2 ® Na 2S4O 6 + 2NaI 1 or Hypo) 20. I2 I2 + 2 e- ® 2I- 2 (1) Iodimetric Titration or Iodimetry : In such titrations iodine solution is used as an oxidant and iodine is directly titrated against a reducing agent. This type of titrations are used for the determination of strength of reducing agents like sulphides, arsenides, thiosulphates etc., by titrating them against a standard solution of iodine. For More Material Join: @JEEAdvanced_2024 10 This type of titration involves free iodine, here iodine solution is treated with known sodium thiosulphate solution. I 2 + 2Na 2S 2O 3 ¾® 2NaI + Na 2S 4 O 6 Equivalents of I 2 = Equivalents of Na 2S 2O 3 used (2) Iodometric Titration or Iodometry: It is an indirect method of estimation of iodine. In this titration an oxidizing agent is used to liberate iodine from solution containing iodide ion ( e. g. , KI) and the liberated iodine is treated with a standard solution of a reducing agent added from a burette. Here a neutral or an acidic solution of an oxidizing agent is used and the amount of liberated I 2 is equal to the equivalents of this oxidizing agent. These titrations are used to determine the concentration of K 2Cr 2O 7 , KMnO 4 , CuSO 4 , Ferric ions, H 2O 2 etc. These titrations are carried out in following two steps : v Step-1 : Oxidizing agent ( X ) + KI (excess)¾® I 2 + reduced state of oxidant Equivalents of ( X ) = Equivalents of I 2 v Step-2 : Liberated I 2 + 2Na 2S 2O 3 ¾® 2NaI + Na 2S 4 O 6 Equivalents of I 2 = Equivalents of Na 2S 2O 3 used (g) Back titration : Back titration is used in volumetric analysis to find out excess amount of reagent added by titrating it with suitable reagent. It is also used to find out percentage purity of sample. For example in an acid-base titration suppose we have added excess base [B(OH) q ] in acid (H n A) solution. To find excess base it is back titrated with another acid (H m B ) of known concentration. Equivalents of base = equivalents of (H n A + H m B ) Þ q ´ moles of base taken = n ´ moles of H n A reacted + m ´ moles of H m B reacted Hardness of Water (a) Definition of hard water : Hard water is having soluble salts of calcium and magnessium ions. (b) Degree of hardness : Degree of hardness defined as number of parts by mass of CaCO 3 (or its equivalent quantities of other substance) present in million parts of mass of water. é Mass of CaCO 3 ù Hardness of water = ê ´ 10 6 ú ppm. ë Mass of water û For More Material Join: @JEEAdvanced_2024  11 ` Level 1 1. Calculate number of neutrons present in 12 ´ 10 25 atoms of oxygen ( 8 O 17 ) : (Given: N A = 6 ´ 10 23 ) (a) 1800 (b) 1600 (c) 1800 N A (d) 3200 N A -23 2. If mass of one atom is 3.32 ´ 10 g, then calculate number of nucleons (neutrons and protons) present in 2 atoms of the element : (a) 40 (b) 20 (c) 10 (d) 40 N A 3. Calculate number of electrons present in 9.5 g of PO 3- 4 : (a) 6 (b) 5 N A (c) 0.1 N A (d) 4.7 N A 4. What is the number of moles of O-atom in 126 amu of HNO 3 ? 2 6 (a) 2 (b) (c) 6 (d) NA NA 5. What is the charge of 96 amu of S 2- ? (a) 2 C (b) 3.2 ´ 10 -19 C (c) 9.6 ´ 10 -19 C (d) 6 C 6. A sample of sodium has a mass of 46 g. What is the mass of the same number of calcium atoms as sodium atoms present in given sample ? (a) 46 g (b) 20 g (c) 40 g (d) 80 g 7. The total number of neutrons present in 54 mL H 2O (l) are: (a) 3 N A (b) 30 N A (c) 24 N A (d) none of these 8. Total number of electrons present in 48 g Mg 2+ are: (a) 24 N A (b) 2 N A (c) 20 N A (d) none of these 9. The number of neutrons in 5 g of D 2O (D is 12H) are: (a) 0.25 N A (b) 2 × 5 N A (c) 1 × 1 N A (d) none of these 10. Cisplatin, an anticancer drug, has the molecular formula Pt(NH 3 ) 2 Cl 2. What is the mass (in gram) of one molecule ? (Atomic masses : Pt = 195, H = 1.0, N = 14, Cl = 35.5) (a) 4.98 ´ 10 -21 (b) 4.98 ´ 10 -22 (c) 6.55 ´ 10 -21 (d) 3.85 ´ 10 -22 11. Aspirin has the formula C 9 H 8 O 4. How many atoms of oxygen are there in a tablet weighing 360 mg? (a) 1.204 ´ 10 23 (b) 1.08 ´ 10 22 (c) 1.204 ´ 10 24 (d) 4.81 ´ 10 21 12. 20 g of an ideal gas contains only atoms of S and O occupies 5.6 L at 1 atm and 273 K. What is the molecular mass of gas ? (a) 64 (b) 80 (c) 96 (d) None of these 13. A sample of ammonium phosphate, (NH 4 ) 3 PO 4 , contains 6 moles of hydrogen atoms. The number of moles of oxygen atoms in the sample is: (a) 1 (b) 2 (c) 4 (d) 6 14. Total number of moles of oxygen atoms in 3 litre O 3( g) at 27°C and 8.21 atm are: (a) 3 (b) 1 (c) 1 (d) none of these 15. 3.011 ´ 10 22 atoms of an element weighs 1.15 gm. The atomic mass of the element is: (a) 10 amu (b) 2.3 amu (c) 35.5 amu (d) 23 amu For More Material Join: @JEEAdvanced_2024 12 16. One atom of an element x weighs 6.643 ´ 10 -23 g. Number of moles of atoms in its 20 kg is: (a) 4 (b) 40 (c) 100 (d) 500 -23 17. Mass of one atom of the element A is 3.9854 ´ 10 g. How many atoms are contained in 1 g of the element A? (a) 2.509 ´ 10 23 (b) 6.022 ´ 10 23 (c) 12.044 ´ 10 23 (d) None of these 18. Which of the following contains the largest mass of hydrogen atoms? (a) 5.0 moles C 2H 2O 4 (b) 1.1 moles C 3H 8 O 3 (c) 1.5 moles C 6 H 8 O 6 (d) 4.0 moles C 2H 4 O 2 19. Which has minimum number of oxygen atoms ? (a) 10 mL H 2O (l) (b) 0.1 mole of V 2O 5 ( s) (c) 12 gm O 3(g) (d) 12.044 ´ 10 22 molecules of CO 2 20. Rearrange the following (I to IV) in the order of increasing masses: (I) 0.5 mole of O 3 (II) 0.5 gm atom of oxygen (III) 3.011 ´ 10 23 molecules of O 2 (IV) 5.6 litre of CO 2 at 1 atm and 273 K (a) II< IV < III< I (b) II< I< IV < III (c) IV < II< III< I (d) I< II< III< IV 21. If the volume of a drop of water is 0.0018 mL then the number of water molecules present in two drop of water at room temperature is: (a) 12.046 ´ 10 19 (b) 1084. ´ 10 18 (c) 4.84 ´ 10 17 (d) 6.023 ´ 10 23 22. It is known that atom contain protons, neutrons and electrons. If the mass of neutron is assumed to half of its original value whereas that of proton is assumed to be twice of its original value then the atomic mass of 146 C will be : (a) same (b) 14.28% less (c) 14.28% more (d) 28.56% less 23. Common salt obtained from sea-water contains 8.775% NaCl by mass. The number of formula units of NaCl present in 25 g of this salt is : (a) 3.367 ´ 10 23 formula units (b) 2.258 ´ 10 22 formula units 23 (c) 3.176 ´ 10 formula units (d) 4.73 ´ 10 25 formula units 24. The number of hydrogen atoms present in 25.6 g of sucrose (C 12H 22O 11 ) which has a molar mass of 342.3 g is : (a) 22 ´ 10 23 (b) 9.91 ´ 10 23 (c) 11 ´ 10 23 (d) 44 ´ 10 23 25. Caffiene has a molecular mass 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 26. The density of water is 1g/mL. What is the volume occupied by 1 molecule of water ? (a) 1.44 ´ 10 -23 mL (b) 1mL (c) 18 mL (d) 2.88 ´ 10 -23 mL 27. A 25.0 mm ´ 40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is 19.32 g /cm 3. How many gold atoms are in the sheet ? (Atomic weight : Au = 197.0) (a) 7.7 ´ 10 23 (b) 1.5 ´ 10 23 (c) 4.3 ´ 10 21 (d) 1.47 ´ 10 22 28. If average molecular mass of air is 29, then assuming N 2 and O 2 gases are there, which options are correct regarding composition of air ? (i) 75% by mass of N 2 (ii) 75% by moles N 2 (iii) 72.41% by mass of N 2 (a) only (i) is correct (b) only (ii) is correct (c) both (ii) and (iii) are correct (d) both (i) and (ii) are correct For More Material Join: @JEEAdvanced_2024  13 29. Density of dry air containing only N 2 and O 2 is 1.15 g/L at 740 mm of Hg and 300 K. What is % composition of N 2 by mass in the air ? (a) 78% (b) 85.5% (c) 70.02% (d) 62.75% 30. A gaseous mixture of H 2 and CO 2 gases contains 66 mass % of CO 2. The vapour density of the mixture is: (a) 6.1 (b) 5.4 (c) 2.7 (d) 10.8 31. The vapour density of a mixture containing NO 2 and N 2O 4 is 27.6. The mole fraction of N 2O 4 in the mixture is: (a) 0.1 (b) 0.2 (c) 0.5 (d) 0.8 32. Density of an ideal gas at 2 atm and 600 K is 2 g/L. Calculate relative density of this gas with respect to Ne(g) under similar conditions : (Given : R = 1/12 atm L/mol.K) (a) 2.5 (b) 2 (c) 3 (d) 5 33. Average atomic mass of magnesium is 24.31 amu. This magnesium is composed of 79 mole % of 24 Mg and remaining 21 mole % of 25 Mg and 26 Mg. Calculate mole % of 26 Mg. (a) 10 (b) 11 (c) 15 (d) 16 34. Indium (atomic mass = 114.82) has two naturally occurring isotopes, the predominant one form has isotopic mass 114.9041 and abundance of 95.72%. Which of the following isotopic mass is the most likely for the other isotope ? (a) 112.94 (b) 115.90 (c) 113.90 (d) 114.90 35. Calculate density of a gaseous mixture which consist of 3.01 ´ 10 24 molecules of N 2 and 32 g of O 2 gas at 3 atm pressure and 860 K temperature (Given : R = 1/12 atm L/mole.K) (a) 0.6 g/L (b) 1.2 g/L (c) 0.3 g/L (d) 12 g/L 36. A mixture of O 2 and gas “Y” (mol. mass 80) in the mole ratio a : b has a mean molecular mass 40. What would be mean molecular mass, if the gases are mixed in the ratio b : a under identical conditions ? (Assume that gases are non-reacting): (a) 40 (b) 48 (c) 62 (d) 72 37. If water sample are taken from sea, rivers or lake, they will be found to contain hydrogen and oxygen in the approximate ratio of 1 : 8. This indicates the law of: (a) multiple proportion (b) definite proportion (c) reciprocal proportions (d) none of these 38. Hydrogen and oxygen combine to form H 2O 2 and H 2O containing 5.93% and 11.2% hydrogen respectively. The data illustrates : (a) law of conservation of mass (b) law of constant proportion (c) law of reciprocal proportion (d) law of multiple proportion 39. Which one of the following combinations illustrate law of reciprocal proportions ? (a) N 2O 3 , N 2O 4 , N 2O 5 (b) NaCl, NaBr, NaI (c) CS 2 , CO 2 , SO 2 (d) PH 3 , P2O 3 , P2O 5 40. Carbon and oxygen combine to form two oxides, carbon monoxide and carbon dioxide in which the ratio of the masses 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 For More Material Join: @JEEAdvanced_2024 14 41. 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 mass 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 42. The law of multiple proportion is illustrated by the two compounds : (a) Sodium chloride and sodium bromide (b) Ordinary water and heavy water (c) Caustic soda and caustic potash (d) Sulphur dioxide and sulphur trioxide 43. All the substances listed below are fertilizers that contribute nitrogen to the soil. Which of these is the richest source of nitrogen on a mass percentage basis ? (a) Urea, (NH 2 ) 2 CO (b) Ammonium nitrate, NH 4 NO 3 (c) Nitric oxide, NO (d) Ammonia, NH 3 44. One mole of element X has 0.444 times the mass of one mole of element Y. One atom of element X has 2.96 times the mass of one atom of 12 C. What is the atomic mass of Y ? (a) 80 (b) 15.77 (c) 46.67 (d) 40.0 23 45. A given sample of pure compound contains 9.81 gm of Zn, 1.8 ´ 10 atoms of chromium and 0.60 mole of oxygen atoms. What is the simplest formula ? (a) ZnCr 2O 7 (b) ZnCr 2O 4 (c) ZnCrO 4 (d) ZnCrO 6 46. The formula of an acid is HXO 2. The mass of 0.0242 moles of the acid is 1.657 g. What is the atomic mass of X ? (a) 35.5 (b) 28.1 (c) 128 (d) 19.0 47. What is the empirical formula of vanadium oxide, if 2.74 g of the metal oxide contains 1.53 g of metal ? (a) V 2O 3 (b) VO (c) V 2O 5 (d) V 2O 7 48. Determine the empirical formula of Kelvar, used in making bullet proof vests, is 70.6% C, 4.2% H, 11.8% N and 13.4% O : (a) C 7 H 5NO 2 (b) C 7 H 5N 2O (c) C 7 H 9 NO (d) C 7 H 5NO 49. The hydrated salt Na 2CO 3×xH 2O undergoes 63% loss in mass on heating and becomes anhydrous. The value of x is: (a) 10 (b) 12 (c) 8 (d) 18 50. A 6.85 g sample of the hydrated Sr(OH) 2 ×xH 2O is dried in an oven to give 3.13 g of anhydrous Sr(OH) 2. What is the value of x ? (Atomic masses : Sr = 87.60, O = 16.0, H = 1.0) (a) 8 (b) 12 (c) 10 (d) 6 51. What percentage of oxygen is present in the compound CaCO 3 ×3Ca 3(PO 4 ) 2 ? (a) 23.3% (b) 45.36% (c) 41.94% (d) 17.08% 52. Dieldrin, an insecticide, contains C, H, Cl and O. Combustion of 29.72 mg of Dieldrin gave 41.21 mg CO 2 and 5.63 mg of H 2O. In a separate analysis 25.31 mg of Dieldrin was converted into 57.13 mg AgCl. What is the empirical formula of Dieldrin ? (a) C 6 H 4 Cl 3O (b) C 8 H 8 ClO (c) C 12H 8 Cl 6 O (d) C 6 H 4 Cl 3O 2 53. A gaseous compound is composed of 85.7% by mass carbon and 14.3% by mass hydrogen. Its density is 2.28 g/litre at 300 K and 1.0 atm pressure. Determine the molecular formula of the compound. (a) C 2H 2 (b) C 2H 4 (c) C 4 H 8 (d) C 4 H 10 For More Material Join: @JEEAdvanced_2024  15 54. Complete combustion of 0.86 g of compound X gives 2.64 g of CO 2 and 1.26 g of H 2O. The lowest molecular mass X can have : (a) 43 g (b) 86 g (c) 129 g (d) 172 g 55. The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO 4 × 7H 2O. The atomic mass of M is : (a) 40.3 (b) 36.3 (c) 24.3 (d) 11.3 -1 56. In an organic compound of molar mass 108 gm mol C, H and N atoms are present in 9 : 1 : 3.5 by mass. Molecular formula can be : (a) C 6 H 8 N 2 (b) C 7 H 10 N (c) C 5H 6 N 3 (d) C 4 H 18 N 3 57. On analysis, a certain compound was found to contain 254 g of iodine (at. mass 127) and 80 g oxygen (at. mass 16). What is the formula of the compound ? (a) IO (b) I 2O (c) I 5O 3 (d) I 2O 5 58. An element A is tetravalent and another element B is divalent. The formula of the compound formed from these elements will be : (a) A 2 B (b) AB (c) AB 2 (d) A 2 B 3 59. A compound used in making nylon, contains 43.8% oxygen. There are four oxygen atoms per molecule. What is the molecular mass of compound? (a) 36 (b) 116 (c) 292 (d) 146 60. Suppose two elements X and Y combine to form two compounds XY 2 and X 2Y 3. 0.05 mole of XY 2 weighs 5 g while 3.011 ´ 10 23 molecules of X 2Y 3 weigh 85 g. The atomic masses of X and Y are respectively : (a) 20, 30 (b) 30, 40 (c) 40, 30 (d) 80, 60 61. 44 g of a sample on complete combustion gives 88 g CO 2 and 36 g of H 2O. The molecular formula of the compound may be: (a) C 4 H 6 (b) C 2H 6 O (c) C 2H 4 O (d) C 3H 6 O 62. 40 milligram diatomic volatile substance ( X 2 ) is converted to vapour that displaced 4.92 mL of air at 1 atm and 300 K. Atomic mass of element X is nearly: (a) 400 (b) 240 (c) 200 (d) 100 63. Two elements X (at. mass = 75) and Y (at. mass = 16) combine to give a compound having 75.8% of X. The formula of the compound is: (a) XY (b) X 2Y (c) X 2Y 2 (d) X 2Y 3 64. A sample of phosphorus that weighs 12.4 g exerts a pressure 8 atm in a 0.821 litre closed vesel at 527°C. The molecular formula of the phosphorus vapour is: (a) P2 (b) P4 (c) P6 (d) P8 65. Manganese forms non-stoichiometric oxides having the general formula MnO x. The value of x for the compound that analyzed 64% by mass Mn: (a) 1.16 (b) 1.83 (c) 2 (d) 1.93 66. 1.44 gram of titanium (At. mass = 48) reacted with excess of O 2 and produce x gram of non-stoichiometric compound Ti 1.44 O. The value of x is: (a) 2 (b) 1.77 (c) 1.44 (d) none of these 67. Which statement is false for the balanced equation given below ? CS 2 + 3O 2 ¾® 2SO 2 + CO 2 For More Material Join: @JEEAdvanced_2024 16 (a) One mole of CS 2 will produce one mole of CO 2 (b) The reaction of 16 g of oxygen produces 7.33 g of CO 2 (c) The reaction of one mole of O 2 will produce 2/3 mole of SO 2 (d) Six molecules of oxygen requires three molecules of CS 2 68. Which of the following setups is correct to calculate the mass (in g) of KClO 3 produced from the reaction of 0.150 moles of Cl 2 ? 3Cl 2 + 6KOH ¾® 5KCl+ KClO 3 + 3H 2O (a) 0.150 moles Cl 2 ´ 1 mole KClO 3 /3 moles Cl 2 ´ 122.5 g/1 mole KClO 3 (b) 0.150 moles Cl 2 ´ 1 mole KClO 3 /3 moles Cl 2 ´ 1 mole KClO 3 /122.5 g (c) 0.150 moles Cl 2 ´ 3 moles Cl 2 /1 mole KClO 3 ´ 122.5 g/1 mole KClO 3 (d) 0.150 moles Cl 2 ´ 3 moles Cl 2 /1 mole KClO 3 ´ 1 mole KClO 3 /122.5 g 69. 2.0 g of a sample contains mixture of SiO 2 and Fe 2O 3. On very strong heating, it leaves a residue weighing 1.96 g. The reaction responsible for loss of mass is given below. Fe 2O 3 ( s) ¾® Fe 3O 4 ( s)+ O 2( g), (unbalance equation) What is the percentage by mass of SiO 2 in original sample ? (a) 10% (b) 20% (c) 40% (d) 60% 70. What volume of air at 1 atm and 273 K containing 21% of oxygen by volume is required to completely burn sulphur (S 8 ) present in 200 g of sample, which contains 20% inert material which does not burn. Sulphur burns according to the reaction 1 S 8 ( s)+ O 2( g) ¾® SO 2( g) 8 (a) 23.52 litre (b) 320 litre (c) 112 litre (d) 533.33 litre 71. For the reaction, 2Fe(NO 3 ) 3 + 3Na 2CO 3 ¾® Fe 2(CO 3 ) 3 + 6NaNO 3 initially 2.5 mole of Fe(NO 3 ) 3 and 3.6 mole of Na 2CO 3 are taken. If 6.3 mole of NaNO 3 is obtained then % yield of given reaction is: (a) 50 (b) 84 (c) 87.5 (d) 100 72. How many moles of P4 can be produced by reaction of 0.10 moles Ca 5(PO 4 ) 3 F, 0.36 moles SiO 2 and 0.90 moles C according to the following reaction ? 4 Ca 5(PO 4 ) 3 F + 18 SiO 2 + 30 C ¾® 3P4 + 2CaF2 + 18CaSiO 3 + 30 CO (a) 0.060 (b) 0.030 (c) 0.045 (d) 0.075 73. Some older emergency oxygen masks contains potassium superoxide, KO 2 which reacts with CO 2 and water present in exhaled air to produce oxygen according to the given equation. If a person exhales 0.667 g of CO 2 per minute, how many grams of KO 2 are consumed in 5.0 minutes? 4KO 2 + 2H 2O + 4CO 2 ¾® 4KHCO 3 + 3O 2 (a) 10.7 (b) 0.0757 (c) 1.07 (d) 5.38 74. The mass of N 2F4 produced by the reaction of 2.0 g of NH 3 and 80 g of F2 is 3.56 g. What is the per cent yield? 2NH 3 + 5F2 ¾® N 2F4 + 6HF (a) 79.0 (b) 71.2 (c) 84.6 (d) None of these For More Material Join: @JEEAdvanced_2024  17 75. Calculate the mass of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO 3 ) : (a) 104.4 kg (b) 105.4 kg (c) 212.8 kg (d) 106.4 kg 76. Phosphoric acid (H 3PO 4 ) prepared in a two step process. (1) P4 + 5O 2 ¾® P4 O 10 (2) P4 O 10 + 6H 2O ¾® 4H 3PO 4 We allow 62 g of phosphorus to react with excess oxygen which form P4 O 10 in 85% yield. In the step (2) reaction 90% yield of H 3PO 4 is obtained. Mass of H 3PO 4 produced is: (a) 37.485 g (b) 149.949 g (c) 125.47 g (d) 564.48 g 77. 9 moles of “D” and 14 moles of E are allowed to react in a closed vessel according to given reactions. Calculate number of moles of B formed in the end of reaction, if 4 moles of G are present in reaction vessel. (Percentage yield of reaction is mentioned in the reaction) Step-1 3 D + 4 E ¾ 80 ¾¾ % ® 5C + A Step-2 3C + 5G ¾ 50 ¾¾% ® 6B + F (a) 2.4 (b) 30 (c) 4.8 (d) 1 78. The chief ore of Zn is the sulphide, ZnS. The ore is concentrated by froth floatation process and then heated in air to convert ZnS to ZnO. 2ZnS + 3O 2 ¾ 80% ¾¾® 2ZnO + 2SO 2 ZnO + H 2SO 4 ¾100% ¾¾® ZnSO 4 + H 2O 2ZnSO 4 + 2H 2O ¾ 80% ¾¾® 2Zn + 2H 2SO 4 + O 2 The number of moles of ZnS required for producing 2 moles of Zn will be : (a) 3.125 (b) 2 (c) 2.125 (d) 4 79. 0.8 mole of a mixture of CO and CO 2 requires exactly 40 gram of NaOH in solution for complete conversion of all the CO 2 into Na 2CO 3. How many moles more of NaOH would it require for conversion into Na 2CO 3 , if the mixture (0.8 mole) is completely oxidised to CO 2 ? (a) 0.2 (b) 0.6 (c) 1 (d) 1.5 80. Silver oxide (Ag 2O) decomposes at temperature 300°C yielding metallic silver and oxygen gas. A 1.60 g sample of impure silver oxide yields 0.104 g of oxygen gas. What is the per cent by mass of the silver oxide in the sample ? (a) 5.9 (b) 47.125 (c) 94.25 (d) 88.2 81. 342 g of 20% by mass of Ba(OH) 2 solution (sp. gr. 0.57) is reacted with 1200 mL of 2 M HNO 3. If the final density of solution is same as pure water then molarity of the ion in resulting solution which decides the nature of the above solution is: (a) 0.25 (b) 0.5 M (c) 0.888 M (d) None of these 82. 100 mL of H 2SO 4 solution having molarity 1 M and density 1.5 g/mL is mixed with 400 mL of water. Calculate final molarity of H 2SO 4 solution, if final density is 1.25 g/mL : (a) 4.4 M (b) 0.145 M (c) 0.52 M (d) 0.227 M 3 83. What volume of HCl solution of density 1.2 g/cm and containing 36.5% by mass HCl, must be allowed to react wtih zinc (Zn) in order to liberate 4.0 g of hydrogen ? (a) 333.33 mL (b) 500 mL (c) 614.66 mL (d) None of these 84. An ideal gaseous mixture of ethane (C 2H 6 ) and ethene (C 2H 4 ) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O 2 to produce CO 2 and H 2O. Mole fraction at C 2H 6 in the mixture is: (a) 0.6 (b) 0.4 (c) 0.5 (d) 0.8 For More Material Join: @JEEAdvanced_2024 18 85. Wood’s metal contains 50.0% bismuth, 25.0% lead, 12.5% tin and 12.5% cadmium by mass. What is the mole fraction of tin ? (Atomic mass : Bi = 209, Pb = 207, Sn = 119, Cd = 112) (a) 0.202 (b) 0.158 (c) 0.176 (d) 0.221 86. The density of a 56.0% by mass aqueous solution of 1-propanol (CH 3CH 2CH 2OH) is 0.8975 g/cm 3. What is the mole fraction of the 1-propanol ? (a) 0.292 (b) 0.227 (c) 0.241 (d) 0.276 2- 87. What is the molartiy of SO 4 ion in aqueous solution that contain 34.2 ppm of Al 2(SO 4 ) 3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 ´ 10 -4 M (b) 2 ´ 10 -4 M (c) 10 -4 M (d) None of these 88. The relation between molarity (M) and molality (m) is given by : (r = density of solution (g mL ), M 1 = molecular mass of solute) 1000 M 1000 r M (a) m = (b) m = 1000 r - M 1 1000 r - MM 1 1000 MM 1000 M (c) m = (d) m = 1000 r - MM 1 1000 r - MM 1 89. Molarity and molality of a solution of an liquid (mol. mass = 50) in aqueous solution is 9 and 10 respectively. What is the density of solution? (a) 1 g/cc (b) 0.95 g/cc (c) 1.05 g/cc (d) 1.35 g/cc 90. An aqueous solution of ethanol has density 1.025 g/mL and it is 2 M. What is the molality of this solution ? (a) 1.79 (b) 2.143 (c) 1.951 (d) None of these 91. 0.2 mole of HCl and 0.2 mole of barium chloride were dissolved in water to produce a 500 mL solution. The molarity of the Cl - ions is : (a) 0.06 M (b) 0.09 M (c) 1.2 M (d) 0.80 M 92. Calculate the mass of anhydrous HCl in 10 mL of concentrated HCl (density = 1.2 g / mL) solution having 37% HCl by mass. (a) 4.44 g (b) 4.44 mg (c) 4.44 ´ 10 -3 mg (d) 0.444 mg 93. Calculate the molality of 1 L solution of 80% H 2SO 4 (w/V), given that the density of the solution is 1.80 g mL -1. (a) 8.16 (b) 8.6 (c) 1.02 (d) 10.8 94. Fluoxymesterone, C 20 H 29 FO 3 , is an anabolic steroid. A 500 mL solution is prepared by dissolving 10.0 mg of the steroid in water. 1.0 mL portion of this solution is diluted to a final volume of 1.00 L. What is the resulting molarity ? (a) 1.19 ´ 10 -10 (b) 1.19 ´ 10 -7 (c) 5.95 ´ 10 -8 (d) 2.38 ´ 10 -11 95. The 25 mL of a 0.15 M solution of lead nitrate, Pb(NO 3 ) 2 reacts with all of the aluminium sulphate, Al 2(SO 4 ) 3 , present in 20 mL of a solution. What is the molar concentration of the Al 2(SO 4 ) 3 ? 3Pb(NO 3 ) 2( aq)+ Al 2(SO 4 ) 3( aq) ¾® 3PbSO 4 ( s)+ 2Al(NO 3 ) 3( aq) (a) 6.25 ´ 10 -2 M (b) 2.421 ´ 10 -2 M (c) 0.1875 M (d) None of these 96. Concentrated HNO 3 is 63% HNO 3 by mass and has a density of 1.4 g/mL. How many millilitres of this solution are required to prepare 250 mL of a 1.20 M HNO 3 solution? (a) 18.0 (b) 21.42 (c) 20.0 (d) 14.21 For More Material Join: @JEEAdvanced_2024  19 97. 50 mL of 20.8% (w/V) BaCl 2 (aq) and 100 mL of 9.8% (w/V) H 2SO 4 (aq) solutions are mixed. Molarity of Cl - ions in the resulting solution is: (At. mass of Ba = 137) (a) 0.333 M (b) 0.666 M (c) 0.1 M (d) 1.33 M 98. 100 mL of 10% NaOH (w/V) is added to 100 mL of 10% HCl (w/V). The nature of resultant solution is: (a) alkaline (b) strongly alkaline (c) acidic (d) neutral 99. How many millilitres of 0.1 M H 2SO 4 must be added to 50 mL of 0.1 M NaOH to give a solution that has a concentration of 0.05 M in H 2SO 4 ? (a) 400 mL (b) 200 mL (c) 100 mL (d) None of these 100. 1 M HCl and 2 M HCl are mixed in volume ratio of 4 : 1. What is the final molarity of HCl solution? (a) 1.5 (b) 1 (c) 1.2 (d) 1.8 101. Three solutions X, Y, Z of HCl are mixed to produce 100 mL of 0.1 M solution. The molarities of X, Y and Z are 0.07 M, 0.12 M and 0.15 M respectively. What respective volumes of X, Y and Z should be mixed ? (a) 50 mL, 25 mL, 25 mL (b) 20 mL, 60 mL, 20 mL (c) 40 mL, 30 mL, 30 mL (d) 55 mL, 20 mL, 25 mL 102. A bottle of an aqueous H 2O 2 solution is labelled as ‘28.375 V’ H 2O 2 and the density of the solution (in g/mL) is 1.25. Choose the correct option: (a) Molality of H 2O 2 solution is 2 (b) Molarity of H 2O 2 solution is 5 (c) Molality of H 2O 2 solution is 2.15 (d) None of these 103. The impure 6 g of NaCl is dissolved in water and then treated with excess of silver nitrate solution. The mass of precipitate of silver chloride is found to be 14 g. The % purity of NaCl solution would be: (a) 95% (b) 85% (c) 75% (d) 65% 104. Al 2(SO 4 ) 3 solution of 1 molal concentration is present in 1 litre solution of density 2.684 g/cc. How many moles of BaSO 4 would be precipitated on adding excess of BaCl 2 in it ? (a) 2 moles (b) 3 moles (c) 6 moles (d) 12 moles 105. A certain public water supply contains 0.10 ppb (part per billion) of chloroform (CHCl 3 ). How many molecules of CHCl 3 would be obtained in 0.478 mL drop of this water ? (assumed d = 1 g/mL) (a) 4 ´ 10 -13 ´ N A (b) 10 -3 ´ N A (c) 4 ´ 10 -10 ´ N A (d) None of these 106. Decreasing order (first having highest and then others following it) of mass of pure NaOH in each of the aqueous solution: (i) 50 g of 40% (w/W) NaOH (ii) 50 mL of 50% (w/V) NaOH [d soln. = 1.2 g / mL] (iii) 50 g of 15 M NaOH [d soln. = 1 g / mL] (a) i, ii, iii (b) iii, ii, i (c) ii, iii, i (d) ii, i, iii 107. What is the molar mass of diacidic organic Lewis base ( B ), if 12 g of its chloroplatinate salt ( B H 2PtCl 6 ) on ignition produced 5 g residue of Pt? (a) 52 (b) 58 (c) 88 (d) None of these For More Material Join: @JEEAdvanced_2024 20 108. On strong heating, one gram of the silver salt of an organic dibasic acid yields 0.5934 g of silver. If the mass percentage of carbon in it 8 times the mass percentage of hydrogen and one-half the mass percentage of oxygen, determine the molecular formula of the acid. (a) C 4 H 6 O 4 (b) C 4 H 6 O 6 (c) C 2H 6 O 2 (d) C 5H 10 O 5 109. 0.607 g of a silver salt of tribasic organic acid was quantitatively reduced to 0.37 g of pure Ag. What is the molecular mass of the acid? (a) 207 (b) 210 (c) 531 (d) 324 110. A sample of peanut oil weighing 2 g is added to 25 mL of 0.40 M KOH. After saponification is complete, 8.5 mL of 0.28 M H 2SO 4 is needed to neutralize excess of KOH. The saponification number of peanut oil is : (saponification number is defined as the milligrams of KOH consumed by 1 g of oil) (a) 146.72 (b) 223.44 (c) 98.9 (d) None of these 111. 20 mL of a mixture of CO and H 2 were mixed with excess of O 2 and exploded and cooled. There was a volume contraction of 18 mL. All volume measurements corresponds to room temperature (27°C) and one atmospheric pressure. Determine the volume ratio V 1 : V 2 of CO and H 2 in the original mixture. (a) 1 : 2 (b) 3 : 2 (c) 2 : 3 (d) 4 : 1 112. In the reaction, 2 Al( s) + 6HCl( aq) ¾® 2 Al 3+ ( aq) + 6Cl - ( aq) + 3H 2( g) (a) 6 L HCl( aq) is consumed for every 3 L H 2( g) produced (b) 33.6 L H 2( g) is produced regardless of temperature and pressure for every mole Al that reacts (c) 67.2 L H 2( g) at 1 atm and 273 K is produced for every mole Al that reacts (d) 11.2 L H 2( g) at 1 atm and 273 K is produced for every mole HCl(aq) consumed 113. The percentage by volume of C 3H 8 in a gaseous mixture of C 3H 8 , CH 4 and CO is 20. When 100 mL of the mixture is burnt in excess of O 2 , the volume of CO 2 produced is: (a) 90 mL (b) 160 mL (c) 140 mL (d) none of these 114. 40 mL gaseous mixture of CO, CH 4 and Ne was exploded with 15 mL of oxygen. After treatment with KOH the volume reduced by 9 mL and again on treatment with alkaline pyrogallol, the volume further reduced by 1.5 mL. Percentage of CH 4 in the original mixture is: (a) 22.5 (b) 77.5 (c) 7.5 (d) 15 115. A gaseous mixture of propane and butane of volume 3 litre on complete combustion produces 11.0 litre CO 2 under standard conditions of temperature and pressure. The ratio of volume of butane to propane is: (a) 1 : 2 (b) 2 : 1 (c) 3 : 2 (d) 3 : 1 116. Phosphorous has the oxidation state of +1 in: (a) Orthophosphoric acid (b) Phosphorous acid (c) Hypophosphorous acid (d) Metaphosphoric acid 117. The oxidation state(s) of Cl in CaOCl 2 (bleaching powder) is/are: (a) +1 only (b) –1 only (c) +1 and -1 (d) none of these 118. The oxidation number of sulphur in S 8 , S 2F2 , H 2S and H 2SO 4 respectively are: (a) 0 , + 1, - 2 and 6 (b) + 2 , 0 , + 2 and 6 (c) 0 , + 1, + 2 and 4 (d) - 2 , 0 , + 2 and 6 For More Material Join: @JEEAdvanced_2024  21 119. Fe shows an oxidation state of +1 in: (a) Fe(CO) 5 (b) [Fe(H 2O) 5 NO] SO 4 (c) Fe 4 [Fe(CN) 6 ]3 (d) FeCl -4 120. When SO 2 is passed into an acidified potassium dichromate solution, the oxidation numbers of sulphur and chromium in the final products respectively are: (a) +6, +6 (b) +6, +3 (c) 0, +3 (d) +2, +3 121. The oxidation number of nitrogen atoms in NH 4 NO 3 are: (a) +3, +3 (b) +3, - 3 (c) - 3 , + 5 (d) - 5, + 3 122. The oxidation states of S-atoms in Caro’s and Marshall’s acids are: (a) + 6 , + 6 (b) + 6 , + 4 (c) + 6 , - 6 (d) + 4 , + 6 123. In which of the following the oxidation number of oxygen has been arranged in increasing order : (a) OF2 < KO 2 < BaO 2 < O 3 (b) BaO 2 < KO 2 < O 3 < OF2 (c) BaO 2 < O 3 < OF2 < KO 2 (d) KO 2 < OF2 < O 3 < BaO 2 124. The oxidation numbers of oxygen in KO 3 , Na 2O 2 respectively are : (a) 3, 2 (b) 1, 0 (c) 0, 1 (d) –0.33, –1 125. The oxidation number of phosphorus in Ba(H 2PO 2 ) 2 is : (a) –1 (b) +1 (c) +2 (d) +3 126. If it is known that in Fe 0. 96 O, Fe is present in +2 and +3 oxidation state, what is the mole fraction of Fe 2+ in the compound ? (a) 12/25 (b) 25/12 (c) 1/12 (d) 11/12 127. Which of the following sequence of compounds is according to the decreasing order of the oxidation state of nitrogen ? (a) HNO 3 , NO , NH 4 Cl , N 2 (b) HNO 3 , NO , N 2 , NH 4 Cl (c) HNO 3 , NH 4 Cl , NO , N 2 (d) NO,HNO 3 ,NH 4 Cl,N 2 128. 2 mole of N 2H 4 loses 16 mole of electron is being converted to a new compound X. Assuming that all of the N appears in the new compound. What is the oxidation state of ‘N’ in X ? (a) - 1 (b) - 2 (c) + 2 (d) + 4 129. When K 2Cr 2O 7 is converted to K 2CrO 4 , the change in the oxidation state of chromium is : (a) 0 (b) 6 (c) 4 (d) 3 130. When a manganous salt is fused with a mixture of KNO 3 and solid NaOH, the oxidation number of Mn changes from +2 to : (a) +4 (b) +3 (c) +6 (d) +7 - 131. In Fe(II)–MnO 4 titration, HNO 3 , is not used because: (a) it oxidises Mn 2+ (b) it reduces MnO -4 (c) it oxidises Fe 2+ (d) it reduces Fe 3+ formed 132. Which species are oxidized and reduced in the reaction ? FeC 2O 4 + KMnO 4 ¾® Fe 3+ + CO 2 + Mn 2+ (a) Oxidised : Fe, C ; Reduced : Mn (b) Oxidised : Fe ; Reduced : Mn (c) Reduced : Fe, Mn ; Oxidised : C (d) Reduced : C ; Oxidised : Mn, Fe For More Material Join: @JEEAdvanced_2024 22 133. In which of the following reactions, H 2O 2 is acting as a reducing agent ? (a) SO 2 + H 2O 2 ¾® H 2SO 4 (b) 2KI + H 2O 2 ¾® 2KOH + I 2 (c) PbS + 4H 2O 2 ¾® PbSO 4 + 4H 2O (d) Ag 2O + H 2O 2 ¾® 2Ag + H 2O + O 2 134. Following reaction describes the rusting of iron 4Fe + 3O 2 ¾® 4Fe 3+ + 6O 2- Which one of the following statement is incorrect ? (a) This is an example of a redox reaction (b) Metallic iron is reduced to Fe 3+ (c) Fe 3+ is an oxidising agent (d) Metallic iron is a reducing agent 135. Which reaction does not represent auto redox or disproportionation ? (a) Cl 2 + OH - ¾® Cl - + ClO 3- + H 2O (b) 2H 2O 2 ¾® H 2O + O 2 (c) 2Cu + ¾® Cu 2+ + Cu (d) (NH 4 ) 2 Cr 2O 7 ¾® N 2 + Cr 2O 3 + 4H 2O 136. Which of the following is redox reaction ? (a) H 2SO 4 reacts with NaOH (b) In atmosphere, O 3 is formed from O 2 by lightning (c) Evaporation of H 2O (d) Oxides of nitrogen are formed form nitrogen & oxygen by lightning 137. Which of the following is a redox-reaction ? (a) 2Na[Ag(CN) 2 ] + Zn ¾® Na 2[Zn(CN) 4 ] + 2Ag (b) BaO 2 + H 2SO 4 ¾® BaSO 4 + H 2O 2 (c) N 2O 5 + H 2O ¾® 2HNO 3 (d) AgNO 3 + KI ¾® AgI + KNO 3 138. For the redox reaction, MnO -4 + C 2O 24- + H+ ¾® Mn 2+ + CO 2 + H 2O the correct coefficients of the reactants for the balanced reaction are respectively MnO -4 , C 2O -4 , H+ : (a) 2, 5, 16 (b) 16, 3, 12 (c) 15, 16, 12 (d) 2, 16, 5 139. In a chemical reaction, K 2Cr 2O 7 + xH 2SO 4 + ySO 2 ¾® K 2SO 4 + Cr 2(SO 4 ) 3 + zH 2O; the value of x, y and z respectively are : (a) x = 1, y = 3, z = 1 (b) x = 4, y = 1, z = 4 (c) x = 3, y = 2, z = 1 (d) x = 2, y = 2, z = 1 140. Balance the following equation and choose the quantity which is the sum of the coefficients of the products :......CS 2 +....... Cl 2 ¾® CCl 4 +...... S 2Cl 2 (a) 5 (b) 3 (c) 6 (d) 2 141. Balance the following equation and choose the quantity which is the sum of the coefficients of reactants and products:....PtCl 4 +.... XeF2 ¾® PtF6 +...ClF +... Xe (a) 16 (b) 13 (c) 18 (d) 12 142. It 0.1 mole H 3PO x is completely neutralised by 5.6 g KOH then select the true statement. (a) x = 3 and given acid is dibasic (b) x = 4 and given acid has no P–H linkage (c) x = 2 and given acid does not form acid salt (d) all of these For More Material Join: @JEEAdvanced_2024  23 143. When potassium permanganate is titrated against ferrous ammonium sulphate in acidic medium, the equivalent mass of potassium permanganate is: molecular mass molecular mass (a) (b) 3 5 molecular mass molecular mass (c) (d) 2 10 144. Equivalent mass of FeS 2 in the half reaction, FeS 2 ¾® Fe 2O 3 + SO 2 is : (a) M/10 (b) M/11 (c) M/6 (d) M/1 145. The equivalent mass of HCl in the given reaction is: K 2Cr 2O 7 + 14HCl ¾® 2KCl+ 2CrCl 3 + 3Cl 2 + H 2O (a) 16.25 (b) 36.5 (c) 73 (d) 85.1 146. Equivalent mass of H 3PO 2 when it disproportionate into PH 3 and H 3PO 3 is: (a) M (b) M/2 (c) M/4 (d) 3M/4 147. In the following reaction, As 2S 3 + H + NO 3 ¾® NO + H 2O + AsO 4 + SO 24- + - 3- the equivalent mass of As 2S 3 is related to its molecular mass by : (a) M/2 (b) M/4 (c) M/24 (d) M/28 148. Sulphur forms the chlorides S 2Cl 2 and SCl 2. The equivalent mass of sulphur in SCl 2 is : (a) 8 g/mol (b) 16 g/mol (c) 64.8 g/mol (d) 32 g/mol 149. The equivalent mass of an element is 4. Its chloride has a vapour density 59.25. Then, the valency of the elements is : (a) 4 (b) 3 (c) 2 (d) 1 150. 6 ´ 10 -3 mole K 2Cr 2O 7 reacts completely with 9 ´ 10 -3 mole X n+ to give XO -3 and Cr 3+. The value of n is : (a) 1 (b) 2 (c) 3 (d) None of these 151. What mass of H 2C 2O 4 × 2H 2O (mol. mass = 126) should be dissolved in water to prepare 250 mL of centinormal solution which act as a reducing agent ? (a) 0.63 g (b) 0.1575 g (c) 0.126 g (d) 0.875 g 152. The equivalent mass of the salt, KHC 2O 4 × H 2C 2O 4 × 4H 2O when it act as reducing agent is : Mol.mass Mol.mass Mol.mass Mol.mass (a) (b) (c) (d) 1 2 3 4 153. The equivalent mass of divalent metal is W. The molecular mass of its chloride is : (a) W + 35.5 (b) W + 71 (c) 2W + 71 (d) 2W + 35.5 154. When BrO -3 ion reacts with Br - in acid medium, Br 2 is liberated. The equivalent mass of Br 2 in this reaction is: 5M 5M 3M 4M (a) (b) (c) (d) 8 3 5 6 155. If m A gram of a metal A displaces m B gram of another metal B from its salt solution and if the equivalent mass are E A and E B respectively then equivalent mass of A can be expressed as: mA m A ´ mB mB mA (a) E A = ´ EB (b) E A = (c) E A = ´ EB (d) E A = ´ EB mB EB mA mB For More Material Join: @JEEAdvanced_2024 24 156. Hydrazine reacts with KIO 3 in presence of HCl as N 2H 4 + IO -3 + 2H+ + Cl - ¾® ICl+ N 2 + 3H 2O The equivalent masses of N 2H 4 and KIO 3 respectively are: (a) 8 and 53.5 (b) 16 and 53.5 (c) 8 and 35.6 (d) 8 and 87 157. What will be the normality of a solution obtained by mixing 0.45 N and 0.60 N NaOH in the ratio 2 : 1 by volume ? (a) 0.4 N (b) 0.5 N (c) 1.05 N (d) 0.15 N -3 2+ -3 158. A solution containing 2.7 ´ 10 mol of A ions required 16. ´ 10 mole of MnO -4 for the 2+ - oxidation of A to AO 3 the medium used is : (a) neutral (b) acidic (c) strong basic (d) none of these 159. H 2O 2 is used as bleaching reagent because on dissociation it gives oxygen 1 (H 2O 2 ¾® H 2O + O 2 ). 2 “Chachi 420" used H 2O 2 solution to bleach her hair and she required 2.24 L O 2 gas at 1 atm and 273 K. She has a H 2O 2 solution labelled as ‘5.6 V’ then what volume of such solution must she required to bleach her hair ? (a) 200 mL (b) 300 mL (c) 400 mL (d) 500 mL 160. 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 : (a) 100 (b) 150 (c) 120 (d) 200 161. 10 mL of N-HCl, 20 mL of N/2 H 2SO 4 and 30 mL of N 3 HNO 3 are mixed together and volume made to one litre. The normality of H + in the resulting solution is : (a) 3 N 100 (b) N 10 (c) N 20 (d) N 40 162. 0.45 g of an acid of mol. mass 90 was neutralised by 20 mL of 0.54 N caustic potash (KOH). The basicity of acid is : (a) 1 (b) 2 (c) 3 (d) 4 163. A 3.4 g sample of H 2O 2 solution containing x% H 2O 2 by mass requires x mL of a KMnO 4 solution for complete oxidation under acidic condition. The molarity of KMnO 4 solution is : (a) 1 (b) 0.5 (c) 0.4 (d) 0.2 164. What volume of O 2( g) measured at 1 atm and 273 K will be formed by action of 100 mL of 0.5 N KMnO 4 on hydrogen peroxide in an acid solution ? The skeleton equation for the reaction is KMnO 4 + H 2SO 4 + H 2O 2 ¾® K 2SO 4 + MnSO 4 + O 2 + H 2O (a) 0.12 litre (b) 0.028 litre (c) 0.56 litre (d) 1.12 litre 165. A sample of 1.0 g of solid Fe 2O 3 of 80% purity is dissolved in a moderately concentrated HCl solution which is reduced by zinc dust. The resulting solution required 16.7 mL of a 0.1 M solution of the oxidant. Calculate the mole of electrons taken up by the oxidant. (a) 2 (b) 4 (c) 6 (d) 5 166. KMnO 4 reacts with oxalic acid according to the equation 2MnO -4 + 5C 2O 24- + 16H+ ¾® 2Mn 2++ 10CO 2 + 8H 2O Here, 20 mL of 0.1 M KMnO 4 is equivalent to : (a) 120 mL of 0.25 M H 2C 2O 4 (b) 150 mL of 0.10 M H 2C 2O 4 (c) 25 mL of 0.20 M H 2C 2O 4 (d) 50 mL of 0.20 M H 2C 2O 4 For More Material Join: @JEEAdvanced_2024  25 167. Ratio of moles of Fe (II) oxidised by equal volumes of equimolar KMnO 4 and K 2Cr 2O 7 solutions in acidic medium will be: (a) 5 : 3 (b) 1 : 1 (c) 1 : 2 (d) 5 : 6 168. The mass of a mixture containing HCl and H 2SO 4 is 0.1 g. On treatment with an excess of an AgNO 3 solution, reacted with this acid mixture gives 0.1435 g of AgCl. Mass % of the H 2SO 4 mixture is : (a) 36.5 (b) 63.5 (c) 50 (d) None of these 169. A solution of Na 2S 2O 3 is standardized iodometrically against 0.167 g of KBrO 3. This process requires 50 mL of the Na 2S 2O 3 solution. What is the normality of the Na 2S 2O 3 ? (a) 0.2 N (b) 0.12 N (c) 0.72 N (d) 0.02 N 170. 0.80 g of impure (NH 4 ) 2 SO 4 was boiled with 100 mL of a 0.2 N NaOH solution till all the NH 3( g) evolved. The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 5 mL of a 0.2 N H 2SO 4 solution. The percentage purity of the (NH 4 ) 2 SO 4 sample is: (a) 82.5 (b) 72.5 (c) 62.5 (d) 17.5 171. The NH 3 evolved due to complete conversion of N from 1.12 g sample of protein was absorbed in 45 mL of 0.4 N HNO 3. The excess acid required 20 mL of 0.1 N NaOH. The % N in the sample is: (a) 8 (b) 16 (c) 20 (d) 25 172. Find out % of oxalate ion in a given sample of an alkali metal oxalate salt, 0.30 g of it is dissolved in 100 mL water and its required 90 mL of centimolar KMnO 4 solution in acidic medium : (a) 66% (b) 55% (c) 44% (d) 6.6% 173. 320 mg of a sample of magnesium having a coating of its oxide required 20 mL of 0.1 M hydrochloric acid for the complete neutralisation of the latter. The composition of the sample is: (a) 87.5% Mg and 12.5% MgO (b) 12.5% Mg and 87.5% MgO (c) 80% Mg and 20% MgO (d) 20% Mg and 80% MgO 174. The concentration of bivalent lead ions in a sample of polluted water that also contains nitrate ions is determined by adding solid sodium sulphate (M = 142) to exactly 500 mL water. Calculate the molarity of lead ions if 0.355 g of sodium sulphate was needed for complete precipitation of lead ions as sulphate. (a) 1.25 ´ 10 -3 M (b) 2.5 ´ 10 -3 M (c) 5 ´ 10 -3 M (d) None of these 175. What volume of HNO 3 (sp. gravity 1.05 g mL -1 containing 12.6 (w/W) of HNO 3 ) that reduce into NO is required to oxidise iron 1 g FeSO 4 ×7H 2O in acid medium is: (a) 70 mL (b) 0.57 mL (c) 80 mL (d) 0.65 mL 176. The total volume of 0.1 M KMnO 4 solution that are needed to oxidize 100 mg each of ferrous oxalate and ferrous sulphate in a mixture in acidic medium is: (a) 1.096 mL (b) 1.32 mL (c) 5.48 mL (d) none of these N 177. When 2.5 g of a sample of Mohr’s salt reacts completely with 50 mL of KMnO 4 solution. 10 The % purity of the sample of Mohr’s salt is : (a) 78.4 (b) 70 (c) 37 (d) 40 178. 4 mole of a mixture of Mohr’s salt and Fe 2(SO 4 ) 3 requires 500 mL of 1 M K 2Cr 2O 7 for complete oxidation in acidic medium. The mole % of the Mohr’s salt in the mixture is: (a) 25 (b) 50 (c) 60 (d) 75 For More Material Join: @JEEAdvanced_2024 26 179. The equivalent mass of a metal is twice to that of oxygen. How many times is the equivalent mass of it’s oxide than the equivalent mass of the metal ? (a) 1.5 (b) 2 (c) 3 (d) 4 180. A metal oxide has the formula X 2O 3. It can be reduced by hydrogen to give free metal and water. 0.1596 g of metal oxide requires 6 mg of hydrogen for complete reduction. The atomic mass of the metal (in amu) is: (a) 15.58 (b) 155.8 (c) 5.58 (d) 55.8 181. Calculate the mass of anhydrous oxalic acid, which can be oxidised to CO 2( g) by 100 mL of an MnO -4 solution, 10 mL of which is capable of oxidising 50 mL of 1N I - to I 2. (a) 45 g (b) 22.5 g (c) 30 g (d) 12.25 g 182. A mixture of NaHC 2O 4 and KHC 2O 4 ×H 2C 2O 4 required equal volumes of 0.2 N KMnO 4 and 0.12 N NaOH separately. What is the molar ratio of NaHC 2O 4 and KHC 2O 4 ×H 2C 2O 4 in the mixture ? (a) 6 : 1 (b) 1 : 6 (c) 1 : 3 (d) 3 : 1 183. Stannous sulphate (SnSO 4 ) and potassium permanganate are used as oxidising agents in acidic medium for oxidation of ferrous ammonium sulphate to ferric sulphate. The ratio of number of moles of stannous sulphate required per mole of ferrous ammonium sulphate to the number of moles of KMnO 4 required per mole of ferrous ammonium sulphate, is : (a) 2.5 (b) 0.2 (c) 0.4 (d) 2.0 184. If a g is the mass of NaHC 2O 4 required to neutralize 100 mL of 0.2 M NaOH and b g that required to reduce 100 mL of 0.02 M KMnO 4 in acidic medium, then: (a) a = b (b) 2 a = b (c) a = 2 b (d) None of these 185. 2 mole, equimolar mixture of Na 2C 2O 4 and H 2C 2O 4 required V 1 L of 0.1 M KMnO 4 in acidic medium for complete oxidation. The same amount of the mixture required V 2L of 0.2 M NaOH for neutralization. The ratio of V 1 to V 2 is: (a) 1 : 2 (b) 2 : 1 (c) 4 : 5 (d) 5 : 4 186. A mixture containing 0.05 mole of K 2Cr 2O 7 and 0.02 mole of KMnO 4 was treated with excess of KI in acidic medium. The liberated iodine required 1.0 L of Na 2S 2O 3 solution for titration. Concentration of Na 2S 2O 3 solution was: (a) 0.40 mol L -1 (b) 0.20 mol L -1 (c) 0.25 mol L -1 (d) 0.30 mol L -1 187. 25 mL of 2 N HCl, 50 mL of 4 N HNO 3 and x mL of 2 M H 2SO 4 are mixed together and the total volume is made up to 1 L after dilution. 50 mL of this acid mixture completely reacted with 25 mL of a 1 N Na 2CO 3 solution. The value of x is: (a) 250 mL (b) 62.5 mL (c) 100 mL (d) None of these 188. In an iodometric estimation, the following reactions occur 2Cu 2+ + 4I - ¾® Cu 2I 2 + I 2 ; I 2 + 2Na 2S 2O 3 ¾® 2NaI+ Na 2S 4 O 6 0.12 mole of CuSO 4 was added to excess of KI solution and the liberated iodine required 120 mL of hypo. The molarity of hypo solution was: (a) 2 (b) 0.20 (c) 0.1 (d) 1.0 189. 1 g mixture of equal number of mole of Li 2CO 3 and other metal carbonate ( M 2CO 3 ) required 21.6 mL of 0.5 N HCl for complete neutralisation reaction. What is the approximate atomic mass of the other metal ? (a) 25 (b) 23 (c) 51 (d) 118 For More Material Join: @JEEAdvanced_2024  27 190. 32 g of a sample of FeSO 4 ×7H 2O were dissolved in dilute sulphuric acid and water and its volume was made up to 1 litre. 25 mL of this solution required 20 mL of 0.02 M KMnO 4 solution for complete oxidation. Calculate the mass % of FeSO 4 ×7H 2O in the sample. (a) 34.75 (b) 69.5 (c) 89.5 (d) None of these 191. In the mixture of (NaHCO 3 + Na 2CO 3 ), volume of HCl required is x mL with phenolphthalein indicator and y mL with methyl orange indicator in the same titration. Hence, volume of HCl for complete reaction of Na 2CO 3 is: (a) 2x (b) y (c) x/2 (d) ( y - x) 192. 0.1 g of a solution containing Na 2CO 3 and NaHCO 3 requires 10 mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator. Mass % of Na 2CO 3 in solution is: (a) 25 (b) 32 (c) 50 (d) None of these 193. A mixture of NaOH and Na 2CO 3 required 25 mL of 0.1 M HCl using phenolphthalein as the indicator. However, the same amount of the mixture required 30 mL of 0.1 M HCl when methyl orange was used as the indicator. The molar ratio of NaOH and Na 2CO 3 in the mixture was: (a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4 194. When 100 mL solution of NaOH and Na 2CO 3 was first titrated with N/10 HCl in presence of HPh, 17.5 mL were used till end point is obtained. After this end point MeOH was added and 2.5 mL of same HCl were required to attain new end point. The amount of NaOH in mixture is: (a) 0.06 g per 100 mL (b) 0.06 g per 200 mL (c) 0.05 g per 100 mL (d) 0.012 g per 200 mL 195. 1 gram of a sample of CaCO 3 was strongly heated and the CO 2 liberated was absorbed in 100 mL of 0.5 M NaOH solution. Assuming 90% purity for the sample, how many mL of 0.5 M HCl would be required to react with the resulting solution to reach the end point in presence of phenolphthalein? (a) 73 mL (b) 41 mL (c) 82 mL (d) 100 mL 196. A sample of pure sodium carbonate 0.318 g is dissolved in water and titrated with HCl solution. A volume of 60 mL is required to reach the methyl orange end point. Calculate the molarity of the acid. (a) 0.1 M (b) 0.2 M (c) 0.4 M (d) None of these 197. 10 L of hard water required 5.6 g of lime for removing hardness. Hence temporary hardness in ppm of CaCO 3 is: (a) 1000 (b) 2000 (c) 100 (d) 1 198. 1 L of pond water contains 20 mg of Ca 2+ and 12 mg of Mg 2+ ions. What is the volume of a 2 N Na 2CO 3 solution required to soften 5000 L of pond water? (a) 500 L (b) 50 L (c) 5 L (d) None of these For More Material Join: @JEEAdvanced_2024 28 199. One litre of a sample of hard water contain 4.44 mg CaCl 2 and 1.9 mg of MgCl 2. What is the total hardness in terms of ppm of CaCO 3 ? (a) 2 ppm (b) 3 ppm (c) 4 ppm (d) 6 ppm 200. If hardness of water sample is 200 ppm, then select the incorrect statement : 0.02 (a) Mass ratio of CaCO 3 to H 2O is 100 (b) Mole ratio of CaCO 3 to H 2O is 3.6 ´ 10 -5 (c) Mass of CaCO 3 present in hard water is 0.2 g/L (d) 1 miliequivalent of CaCO 3 present in 1 kg of hard water For More Material Join: @JEEAdvanced_2024  29 Level 2 1. A mixture of NH 4 NO 3 and (NH 4 ) 2 HPO 4 contain 30.40% mass per cent of nitrogen. What is the mass ratio of the two components in the mixture ? (a) 2 : 1 (b) 1 : 2 (c) 3 : 4 (d) 4 : 1 2. What volume of 75% alcohol by weight ( d = 0.80 g /cm 3 ) must be used to prepare 150 cm 3 of 30% alcohol by mass ( d = 0.90 g /cm 3 )? (a) 67.5 mL (b) 56.25 mL (c) 44.44 mL (d) None of these 3. Calculate the number of millilitres of NH 3 (aq) solution ( d = 0.986 g /mL) contain 2.5% by mass NH 3 , which will be required to precipitate iron as Fe(OH) 3 in a 0.8 g sample that contains 50% Fe 2O 3. (a) 0.344 mL (b) 3.44 mL (c) 17.24 mL (d) 10.34 mL 4. In the preparation of iron from haematite (Fe 2O 3 ) by the reduction with carbon Fe 2O 3 + C ¾® Fe + CO 2 how much 80% pure iron may be produced from 120 kg of 90% pure Fe 2O 3 ? (a) 94.5 kg (b) 60.48 kg (c) 116.66 kg (d) 120 kg 5. A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of 12.5% by mass. 2.8 g of the mineral on heating lost 1.32 g of CO 2. What is the % by mass of the other metal ? (a) 87.5 (b) 35.71 (c) 65.11 (d) 23.21 6. 6.2 g of a sample containing Na 2CO 3 , NaHCO 3 and non-volatile inert impurity on gentle heating loses 5% of its mass due to reaction 2NaHCO 3 ¾® Na 2CO 3 + H 2O + CO 2. Residue is dissolved in water and formed 100 mL solution and its 10 mL portion requires 7.5 mL of 0.2 M aqueous solution of BaCl 2 for complete precipitation of carbonates. Determine mass (in gram) of Na 2CO 3 in the original sample. (a) 1.59 (b) 1.06 (c) 0.53 (d) None of these 7. Nitric acid can be produced from NH 3 in three steps process given below (I) 4NH 3( g)+ 5O 2( g) ¾® 4NO( g)+ 6H 2O( g) (II) 2NO( g)+ O 2( g) ¾® 2NO 2( g) (III) 3NO 2( g)+ H 2O(l) ¾® 2HNO 3( aq) + NO( g) percent yield of 1st , 2 nd and 3 rd steps are respectively 50%, 60% and 80% respectively then what volume of NH 3( g) at 1 atm and 0°C required to produced 1575 g of HNO 3. (a) 156.25 (b) 350 L (c) 3500 L (d) None of these 8. 1 M NaOH solution was slowly added into 1000 mL of 183.75 g impure H 2SO 4 solution and the following plot was obtained. The percentage purity of H 2SO 4 sample and slope of the curve respectively are: For More Material Join: @JEEAdvanced_2024 30 3 [H+ ] 2 (mol/L) 1 1 2 3 Volume of NaOH added (L) 1 1 (a) 75%, - (b) 80%, - (c) 80%, - 1 (d) None of these 3 2 9. MnO 2 on ignition converts into Mn 3O 4. A sample of pyrolusite having 75% MnO 2 , 20% inert impurities and rest water is ignited in air to constant mass. What is the percentage of Mn in the ignited sample ? (a) 24.6% (b) 37% (c) 55.24% (d) 74.05% 10. A 1.0 g sample of a pure organic compound containing chlorine is fused with Na 2O 2 to convert chlorine to NaCl. The sample is then dissolved in water, and the chloride precipitated with AgNO 3 , giving 1.96 g of AgCl. If the molecular mass of organic compound is 147, how many chlorine atoms does each molecule contain ? (a) 1 (b) 2 (c) 3 (d) 4 11. A 0.60 g sample consisting of only CaC 2O 4 and MgC 2O 4 is heated at 500°C, converting the two salts of CaCO 3 and MgCO 3. The sample then weighs 0.465 g. If the sample had been heated to 900°C, where the products are CaO and MgO. What would the mixtures of oxides have weighed ? (a) 0.12 g (b) 0.21 g (c) 0.252 g (d) 0.3 g 12. A metal M forms the sulphate M 2(SO 4 ) 3. A 0.596 gram sample of the sulphate reacts with excess BaCl 2 to give 1.220 g BaSO 4. What is the atomic mass of M ? (Atomic mass : S = 32, Ba = 137.3) (a) 26.9 (b) 69.7 (c) 55.8 (d) 23 13. Urea (H 2NCONH 2 ) is manufactured by passing CO 2( g) through ammonia solution followed by crystallization. CO 2 for the above reaction is prepared by combustion of hydrocarbons. If combustion of 236 kg of a saturated hydrocarbon (C n H 2n + 2 ) produces as much CO 2 as required for production of 999.6 kg urea then molecular formula of hydrocarbon is: (a) C 10 H 22 (b) C 12H 26 (c) C 13H 28 (d) C 8 H 18 14. 11.6 g of an organic compound having formula C n H 2n + 2 is burnt in excess of O 2( g) initially taken in a 22.41 litre steel vessel. Before reaction the gaseous mixture was at 273 K with pressure reading 2 atm. After complete combustion and loss of considerable amount of heat, the mixture of product and excess of O 2 had a temperature of 546 K and 4.6 atm pressure. The formula of organic compound is: (a) C 2H 6 (b) C 3H 8 (c) C 5H 12 (d) C 4 H 10 For More Material Join: @JEEAdvanced_2024  31 40% yield 15. H 2O 2 + 2KI ¾¾¾¾® I 2 + 2KOH 50% yield H 2O 2 + 2KMnO 4 + 3H 2SO 4 ¾¾¾® K 2SO 4 + 2MnSO 4 + 3O 2 + 4H 2O 150 mL of H 2O 2 sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M/2 H 2SO 4 for neutralisation. Other part was treated with KMnO 4 yielding 6.74 litre of O 2 at 1 atm. and 273 K. Using % yield indicated find volume strength of H 2O 2 sample used. (a) 5.04 (b) 10.08 (c) 3.36 (d) 33.6 16. SO 2Cl 2 (sulphuryl chloride) reacts with water to given a mixture of H 2SO 4 and HCl. What volume of 0.2 M Ba(OH) 2 is needed to completely neutralize 25 mL of 0.2 M SO 2Cl 2 solution: (a) 25 mL (b) 50 mL (c) 100 mL (d) 200 mL 17. 5 g sample contain only Na 2CO 3 and Na 2SO 4. This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M H 2SO 4. Calculate the % of Na 2SO 4 in the sample. (a) 42.4 (b) 57.6 (c) 36.2 (d) none of these 18. 20 mL of 0.2 M NaOH ( aq) solution is mixed with 35 mL of 0.1 M NaOH ( aq) solution and the resultant solution is diluted to 100 mL. 40 mL of this diluted solution reacted with 10% impure sample of oxalic acid (H 2C 2O 4 ). The mass of impure sample is: (a) 0.15 gram (b) 0.135 gram (c) 0.59 gram (d) None of these 19. A silver coin weighing 11.34 g was dissolved in nitric acid. When sodium chloride was added to the solution all the silver (present as AgNO 3 ) was precipitated as silver chloride. The mass of the precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin. (a) 4.8% (b) 95.2% (c) 90% (d) 80% 20. Two elements X (at. mass 16) and Y (at. mass 14) combine to form compounds A , B and C. The ratio of different masses of Y which combine with a fixed mass of X in A , B and C is 1 : 3 : 5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then in C , 16 parts by mass of X will combine with : (a) 14 parts by mass of Y (b) 42 parts by mass of Y (c) 70 parts by mass of Y (d) 84 parts by mass of Y 21. The conversion of oxygen to ozone occurs to the extent of 15% only. The mass of ozone that can be prepared from 67.2 L of oxygen at 1 atm and 273 K will be : (a) 14.4 g (b) 96 g (c) 640 g (d) 64 g 22. RH 2 (ion exchange resin) can replace Ca 2+ ions in hard water as RH 2 + Ca 2+ ¾® RCa+ 2H+. If 1 L of hard water after passing through RH 2 has pH = 3 then hardness in parts per million of Ca 2+ is: (a) 20 (b) 10 (c) 40 (d) 100 3 23. 100 cm of a solution of an acid (Molar mass = 98) containing 29.4 g of the acid per litre were completely neutralized by 90.0 cm 3 of aq. NaOH containing 20 g of NaOH per 500 cm 3. The basicity of the acid is: (a) 3 (b) 2 (c) 1 (d) data insufficient For More Material Join: @JEEAdvanced_2024 32 24. 20 mL of 0.1 M solution of compound Na 2CO 3 ×NaHCO 3 ×2H 2O is titrated against 0.05 M HCl. x mL of HCl is used when phenolphthalein is used as an indicator and y mL of HCl is used when methyl orange is the indicator in two separate titrations. Hence ( y - x) is: (a) 40 mL (b) 80 mL (c) 120 mL (d) None of these 25. A sample containing HAsO 2 (mol. mass = 108) and weighing 3.78 g is dissolved and diluted to 250 mL in a volumetric flask. A 50 mL sample (aliquot) is withdrawn with a pipet and titrated with 35 mL of 0.05 M solution of I 2. Calculate the percentage HAsO 2 in the sample: (a) 25% (b) 20% (c) 10% (d) none of these 26. A mixture of FeO and Fe 2O 3 is completely reacted with 100 mL of 0.25 M acidified KMnO 4 solution. The resultant solution was then treated with Zn dust which converted Fe 3+ of the solution to Fe 2+. The Fe 2+ required 1000 mL of 0.10 M K 2Cr 2O 7 solution. Find out the weight % Fe 2O 3 in the mixture. (a) 80.85 (b) 19.15 (c) 50 (d) 89.41 27. To a 10 mL, 1 M aqueous solution of Br 2 , excess of NaOH is added so that all Br 2 is disproportionated to Br - and BrO -3. The resulting solution is free from Br - , by extraction and excess of OH - neutralised by acidi