CHEMISTRYTARGET IIT JEE 2014 XI (P) APPLICATION OF MOLE CONCEPT It's a furry animal !! What is a mole? A mole is something that my brother has on his face. Idiots ! Its chemistry A mole is a counting unit. A-10, "GAURAV TOWER", Road No.-1, I.P.I.A., Kota-324005 (Raj.) INDIA Tel.: 0744-2423738, 2423739, 2421097, 2424097 Fax: 0744-2436779 E-mail: [email protected] Website : www.bansal.ac.in Application of Mole Concept [2] CHEMISTRY – STUDY OF MATTER Overview of Chemistry Special Highlights BRAIN TEASERS: GENERAL MISTAKE: TEACHER’S ADVICE: Application of Mole Concept [3] . 2. substances reacts in the ratio of their equivalent masses. 1. In all chemical reactions.1 BASIC DEFINITION INVOLVED Atomic mass Atomic mass of an element can be defined as the number which indicates how many times the mass of one atom of the element is heavier in comparison to Atomic mass = 2. "Wonder these laws are useful?" "These are no longer useful in chemical calculations now but gives an idea of earlier methods of analysing and relating compounds by mass.2 1 th part of the mass of one atom of Carbon-12.5 Gay Lussac law of combining volumes [Gay Lussac] When gases combined or produced in a chemical reaction they do so in a simple ratio by volume provided all the gases are at same temperature and pressure.3 Law of multiple proportions [Dalton] When two elements combine to form two or more compounds. 1.1 LAWS OF CHEMICAL COMBINATION Law of conservation of mass [Lavoisier] In any physical or chemical change mass can neither be created nor be destroyed." 2.KEY CONCEPTS 1. bear a simple ratio to one another. 1. 1. the ratio in which they do so will be same or simple multiple if both directly combined with each other. 1.2 Law of constant composition [Proust] A chemical compound always contains the same element combined together in fixed proportion by mass.4 Law of reciprocal proportions [Richter] When two different elements combine with the same mass of a third element.12] 12 Atomic mass unit (amu) The quantity [ 1 × mass of an atom of C–12] is known as atomic mass unit. 12 Application of Mole Concept [4] . 12 [ Mass of an atom of the element] Mass of an atom in amu = 1 1 amu [ Mass of an atom of carbon . the different masses of one element which combine with a fixed mass of the other element. 4 Average atomic weight = % of isotope X molar mass of isotope.3 Gram atomic mass The gram atomic mass can be defined as the mass of 1 mole atoms of an element.12] 12 Mass of one molecule of the substance (in amu) 1 amu Gram Molecular mass Gram molecular mass can be defined as the mass of 1 mole of molecules. of moles of any compound and mi = molecular mass of any compound. Make yourselves clear in the difference between mole% and mass% in question related to above. 2.7 Average molecular weight = niMi ni where ni = no.6 Mass of one molecule of the substance (in amu) 1 [ Mass of an atom of C . its a mole%) 2. 2. 2.e. comparison to 12 Molecular mass = = 2. The % obtained by above expression (used in above expression) is by number (i.Can you calculate value of 1 amu in gram? Be clear in the difference between 1 amu and 1 gm. Application of Mole Concept [5] .5 Molecular mass Molecular mass is the number which indicates how many times one molecule of a substance is heavier in 1 th of the mass of one atom of C-12. Shortcut for % determination if average atomic weight is given for X having isotopes XA & XB. 3.8 Mole : One mole is a collection of that many entities as there are number of atoms exactly in 12 gm of C-12 isotope. (for atoms). then no. or 1 mole = collection of 6.7 litre.0821 lit-atm/mol-K (when P is in atmosphere and V is in litre.02 × 1023 = NA = Avogadro's No. How would I calculate moles if volume of a solid is given? Application of Mole Concept [6] . Atomic wt.02 × 1023 species 6. 1 mole of ions is termed as 1 gm-ion and 1 mole of molecule termed as 1 gm-molecule. 1 mole of atoms is also termed as 1 gm-atom. NA (a) If no.% of XA Average atomic weight wt of X B = × 100 difference in weight of X A & X B Try working out of such a shortcut for XA. Given wt. METHODS OF CALCULATIONS OF MOLE : Given no. of moles = (b) If weight of a given species is given.) use n = 1 mole of any gas at STP occupies 22. If volume of a gas is given along with its temperature (T) and pressure (P) PV RT where R = 0. of some species is given. XB. Gases do not have volume. What is meant by "Volume of gas"? Do not use this expression (PV = nRT) for solids/liquids. then no of moles = or (c) = Given wt. XC 2. (for molecules) Molecular wt. Limiting Reagent : It is very important concept in chemical calculation.1 CONCEPT OF LIMITING REAGENT. MOLECULAR FORMULA : Empirical formula : Formula depicting constituent atom in their simplest ratio. Application of Mole Concept [7] . Molecular formula : Formula depicting actual number of atoms in one molecule of the compound 4. 6. Vapour density = Molecular mass 2 Can you prove the above expression? Is the above parameter temperature dependent? 5. It refers to reactant which is present in minimum stoichiometry quantity for a chemical reaction. So all calculations related to various products or in sequence of reactions are made on the basis of limiting reagent.1 EMPIRICAL FORMULA. 4.2 Relation between the two : Molecular formula = Empirical formula × n Molecular mass n = Empirical Formula mass Check out the importance of each step involved in calculations of empirical formula. STOICHIOMETRY : Stoichiometry pronounced (“stoy – key – om – e – tree”) is the calculations of the quantities of reactants and products involved in a chemical reaction. 4. For solving any such reactions.4.first step is to calculate L. It comes into picture when reaction involves two or more reactants. Following methods can be used for solving problems.3 Vapour density : Vapour density : Ratio of density of vapour to the density of hydrogen at similar pressure and temperature. (a) Mole Method Balance reaction required (b) Factor Label Method (c) POAC method } Balancing not required (d) Equivalent concept } to be discussed later 6. It is reactant consumed fully in a chemical reaction.R. the reactants is converted to product (desired) and waste. PERCENTAGE YIELD actual yield The percentage yield of product = the theoretical maximum yield 100 For reversible reactions. 8.2 Quantitative Analysis 1. Estimation of Oxygen Application of Mole Concept [8] . (c) Divide given moles of each reactant by their stoichiometric coefficient. Estimation of Halogens 4.1 SOME EXPERIMENTAL METHODS : For determination of atomic mass : Dulong's and Petit's Law : Atomic weight × specific heat (cal/gm°C) 6. 8. Estimation of Nitrogen (i) Dumas method (ii) Kjeldahl’s method 3. 8. the one with least ratio is limiting reagent. Estimation of Phosphorus 6.4 Gives approximate atomic weight and is applicable for metals only. Estimation of Sulphur 5. [Useful when number of reactants are more than two. the actual amount of any limiting reagent consumed in such incomplete reactions is given by [% yield × given moles of limiting reagent] For irreversible reaction with % yield less than 100. Estimation of Carbon and Hydrogen 2.6.] 7. [Useful when only two reactant are there] (b) By calculating amount of any one product obtained taking each reactant one by one irrespective of other reactants. The one giving least product is limiting reagent.2 Calculation of Limiting Reagent : (a) By calculating the required amount by the equation and comparing it with given amount. Take care of units of specific heat. (c) Chloroplatinate salt method : (for organic bases) Lewis acid : electron pair acceptor Lewis base :electron pair donor Procedure : Some amount of organic base is reacted with H2PtCl6 and forms salt known as chloroplatinate. Can you guess why? (b) Silver salt method : (for organic acids) Basicity of an acid : No. This comes in picture when any gas is collected over water. If acidity of base is 'n' w2 1 M n (410) M salt w1 and M = salt then base 2 195 n Application of Mole Concept [9] . (w2 gm).8. The volume of air displaced over water is given (V) and the following expressions are used. If base is denoted by B then salt formed (i) with monoacidic base = B2H2PtCl6 (ii) with diacidic base = B2(H2PtCl6)2 (iii) with triacidic base = B2(H2PtCl6)3 The known amount (w1 gm) of salt is heated and Pt residue is measured. converted to vapour and collected over water.3 For molecular mass determination : (a) Victor Maeyer's process : (for volatile substance) Procedure : Some known weight of a volatile substance (w) is taken. Then if the basicity of acid is n. w w RT RT M= or M= (P P' )V PV If aq. of replacable H+ atoms in an acid (H contained to more electronegative atom is acidic) Procedure : Some known amount of silver salt (w1 gm) is heated to obtain w2 gm of while shining residue of silver. tension is not given If aq. tension is P' Aqueous tension : Pressure exerted due to water vapours at any given temperature. molecular weight of acid would be w2 1 Msalt w1 and molecular weight of acid = M – n(107) salt 108 n This is one good practical application of POAC. UF6. from an excess of uranium ? The molar mass of UF6 is 352 gm/mol. A sample of methane contains 5. in which nitrogen dioxide reacts with water.8 Substance A and B are colourless gases obtained by combining sulphur with oxygen. what volume of ethanol should be measured out? Q.47 gm of carbon atom and 0.4 1. having mass number M and (M + 0. Substance A results from combining 6 gm of sulphur with 6 gm of oxygen and substance B result from combining 8.88 gm of oxygen. Show that two sample obey law of multiple proportion. The density of liquid alcohol is 0.993 gm of H atoms combined in different way.12 What total volume. Q. (NA = 6 × 1023) Q. If 1.7 gm of C atom and 1.6 What time in sec.2 Find the number of neutrons in 0.10H2O contains same number of oxygen atoms as present in x gm of H2SO4.10 Nitric acid is manufactured by the Ostwald process.5 Ethanol. Q.61 gm of Na2SO4.3 A gaseous mixture contains CO2 (g) and N2O (g) in a 2 : 5 ratio by mass.04 mg of uranium hexafluoride. Show that the mass ratio in two substances are in simple multiple of each other.9 MOLE How many gm of HCl is needed for complete reaction with 69. Q.7893 g/ml at 293 K.. as represented by this equation U(s) + 3F2(g) UF6(g) How many fluorine molecules are required to produce 7.5) respectively. Q. UF6.9 gm of H atom combined in certain way.1 2 isotopes of an element are present in 1 : 2 ratio of number. it would take to spend Avogadro’s number of rupees at the rate of 10 lac rupees per second? Q.EXERCISE # I LAWS OF CHEMICAL COMBINATION AND AVERAGE MOLAR MASS Q.45 g water. in litre at 627°C and 1 atm. Application of Mole Concept [10] . Find the mean mass number of element. Find the ratio of the number of molecules of CO2 (g) and N2O (g) Q.2 gm HNO3 ? Q. assuming that all the hydrogen atoms are H1 atoms and all the oxygen atoms are O16 atoms Q. C2H5OH.6 gm of sulphur with 12. could be formed by the decomposition of 16 gm of NH4NO3 ? Reaction : 2 NH 4NO3 2N2 + O2 + 4H2O(g).6 gm MnO2 ? HCl + MnO2 MnCl2 + H2O + Cl2 Q. Whereas a sample of propane contain 4. 3 NO2 (g) + H2O (l) 2 HNO3 (aq) + NO (g) How many grams of nitrogen dioxide are required in this reaction to produce 25.7 Methane and propane are both constituent of natural gas. Calculate x.2 mole of ethanol are needed for a particular experiment. is the substance commonly called alcohol.11 Flourine reacts with uranium to produce uranium hexafluoride. Q. which in turn is obtained from titanium oxide by the following process : 3 TiO2(s) + 4C (s) + 6Cl2 (g) 3TiCl4(g) + 2CO2(g) + 2CO (g) A vessel contains 4. The evolved hydrogen collected at 0°C has a volume of 44.20 A sample containing only CaCO3 and MgCO3 is ignited to CaO and MgO.17 Sulphuric acid is produced when sulphur dioxide reacts with oxygen and water in the presence of a catalyst : 2SO2(g) + O2 (g) + 2 H2O(l) 2 H2SO4 . HCl forms magnesium chloride.21 Determine the percentage composition of a mixture (by mass) of anhydrous sodium carbonate and sodium bicarbonate from the following data: wt.0 moles each of LiH & BF3.55 gm was treated to precipitate all the Ca as CaCO3 which was then heated and quantitatively converted to 1.Q.158 mol KO2 and 0.13 Q. Calculate the mass percentages of CaCO3 and MgCO3 in the sample.6 mol of SO2 reacts with 4. of the mixture taken = 2 gm Loss in weight on heating = 0.18 Potassium superoxide.66 gm CO2 gas is evolved.124 gm. suppose the reaction goes to completion as written. Q. 36 gm of carbon was mixed with 142 gm of Cl2. Q. 7. KO2. Q. how many gram of TiCl4 can be produced ? (Ti = 48) Q. 0. LIMITING REACTANT Titanium. Determine the percentage composition of the original mixture by mass. Application of Mole Concept [11] . what is the maximum number of moles of H2SO4 that can be obtained ? Q.32 gm TiO2. 5.19 PROBLEMS RELATED WITH MIXTURE 39 gm of an alloy of aluminium and magnesium when heated with excess of dil.8 litres at 1 atm pressure. is used in rebreathing gas masks to generate oxygen : KO2(s) + H2O(l) KOH (s) + O2 (g) If a reaction vessel contains 0.15 A chemist wants to prepare diborane by the reaction 6 LiH + 8BF3 6Li BF4 + B2H6 If he starts with 2.1 gm Cl2.23 When 4 gm of a mixture of NaHCO3 and NaCl is heated. how many moles of O2 can be produced? Q. Q. The mixture of oxides produced weight exactly half as much as the original sample.68 gm of CaO.10 mol H2O.8 mol of O2 and a large excess of water. How many moles of B2H6 can be prepared.14 Calculate mass of phosphoric acid required to obtain 53. Calculate the composition of the aluminium by moles.16 Carbon reacts with chlorine to form CCl4.76 gm C and. which is used to make air plane engines and frames. Calculate the mass percentage of CaCl2 in the mixture. aluminium chloride and hydrogen. can be obtained from titanium tetrachloride. Q. If 5.22 A sample of mixture of CaCl2 and NaCl weighing 5.4 gm pyrophosphoric acid. Calculate mass of CCl4 produced and the remaining mass of reactant. Exhaled air contains CO2 and H2O. is utilised in closed system breathing apparatus. NaClO3. how much cyclohexene will be obtained from 100 gm of cyclohexanol ? H 2SO 4 C6H12O con . both of which are removed and the removal of water generates oxygen for breathing by the reaction 4KO2(s) + 2H2O(l) 3O2(g) + 4KOH(s) The potassium hydroxide removes CO2 from the apparatus by the reaction : KOH (s) + CO2 (g) KHCO3(s) (a) What mass of KO2 generates 48 gm of oxygen ? (b) What mass of CO2 can be removed from the apparatus by 63.Q. KO2.9 gm of KO2 ? Q.How much volume of CO2 will be obtained at 1 atm & 273 K? Q. C6H10 Q.28 gm/ml and 36.28 Q. If the yield of this reaction is 75%.32 Sodium chlorate. how many tons SO2 are dumped into the atmosphere each day? Q. 25 gm of this sample is treated with excess of HCl.26 A sample of calcium carbonate is 80% pure. H2SO4.24 PERCENTAGE YIELD AND PERCENTAGE PURITY A power company burns approximately 475 tons of coal per day to produce electricity.2 % by weight. KClO3(s) KCl(s) + O2(g) Q. can be prepared by the following series of reactions: 2KMnO4 + 16 HCl 2 KCl + 2 MnCl2 + 8H2O + 5 Cl2 6Cl2 +6 Ca(OH)2 Ca(ClO3)2 + 5 CaCl2 + 6H2O Ca(ClO3)2 + Na2SO4 CaSO4 + 2 NaClO3 What mass of NaClO3 can be prepared from 100 ml of concentrated HCl (density 1. how much H2SO4 will 1200 gm of PbS produce? (Pb = 208) Q.29 How many grams of 80% pure Na2SO4 can be produced from 130 gm of 90% pure NaCl ? SEQUENTIAL REACTIONS Sulphur trioxide may be prepared by the following two reactions : S8 + 8O2(g) 8SO2(g) 2SO2(g) + O2(g) 2SO3(g) How many grams of SO3 will be produced from 1 mol of S8? Q.5% by mass)? Assume all other substances are present in excess amounts. Application of Mole Concept [12] .27 Cyclohexanol is dehydrated to cyclohexene on heating with conc.31 Potassium superoxide. If the sulphur content of the coal is 1.30 2PbS + 3O2 2PbO + 2SO2 3SO2 + 2HNO3 + 2H2O 3H2SO4 + 2NO According to the above sequence of reactions.25 Calculate the percent loss in weight after complete decomposition of a pure sample of potassium chlorate. Find out the percentage of nitrogen in the compound. 6H2O.45 10 gm of an organic compound on complete combustion gives 17.3g of an organic compound gave 24.34 In Dumas.18 g of water. Application of Mole Concept [13] . 0. The residual acid required 60 mL of 0.4 gm SO2. Q.40 gm of the chloroplatinate of a mono acid base on ignition gave 0.26 l SO3 at 1 atm.Q. H and O.75 gm of an organic compound gave on complete combustion 1.46 An organic compound contains C. Q.6 gm water and 6. 0. Q. Find the percentage composition of nitrogen in the compound.64 g of an organic compound gave 0. This is heated and decomposed to magnesium pyrophosphate. (Pt = 195) Q. the ammonia evolved from 0.57 ml of nitrogen at1 atm and 273 K. What weight of NaH2PO4 was present originally? Q.39 MISCELLANEOUS PROBLEM P4S3 + 8O2 P4O10 + 3SO2 Calculate minimum mass of P4S3 is required to produce at least 0. 0.1 g of an organic compound gave 0.33 QUANTITATIVE ANALYSIS On complete combustion. according to the following reaction ? H2O + SO3 H2SO4 Q.47 gm of water. Determine the percentage composition of carbon and hydrogen in the compound. The ammonia evolved was absorbed in 50 ml of 0. 0.93 mL of nitrogen collected at 300K temperature and 775mm pressure.9 gm/ml) reacts with 49. Q.42 Calculate the amount of H2SO4 produced (in gm) when 40 ml H2O (d = 0. Q. neutralized 10 mL of 1 M H2SO4. A given weight of the compound when heated with nitric acid and silver nitrate gave an equal weight of silver chloride. wt of the base.466 g of barium sulphate. If the weight of 1 mole of compound is 60.33 gm of Mg2P2O7. a mixture of CO and CO2 is obtained.5 g of the compound in Kjeldahl’s estimation of nitrogen.44 0.41 In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate Mg(NH4)PO4. Calculate the mol.5 M solution of NaOH for neutralization.176g of carbon dioxide and 0.75 gm of the compound gave 58.37 In sulphur estimation. Find out the percentage of bromine in the compound. 1 g of an organic compound gave 0.18 gm of H2O.6 gm of carbon dioxide and 0. What is the percentage of sulphur in the compound? Q. (Aqueous tension at 300K=15 mm) Q. and 300 K. method for estimation of nitrogen. What is the mass ratio of the mixture obtained when 20 grams of O2 reacts completely with 12 grams of carbon ? Q.43 2. Mg2P2O7 which is weighed.35 During estimation of nitrogen present in an organic compound by Kjeldahl’s method.094 g of AgBr.Calculate the percentage composition of nitrogen in the compound. Deduce the empirical formula of compound. Q.6 gm CO2.44 gm CO2 and 0.38 A sample of 0. find molecular formula.8 gm of Pt.50 g of an organic compound was treated according to Kjeldahl's method. 3. Deduce the empirical formula of the compound? Q.3 gm of this compound on combustion yielded 0.40 By the reaction of carbon and oxygen.96 gm of each product.36 In Carius method of estimation of halogen. 0.5 M H2SO4. A solution of NaH2PO4 yielded 3. 6 gm. in that order.6 If the yield of chloroform obtainable from acetone and bleaching powder is 58%. Q.70 m.gr. 1.46 g per litre at 27°C and 1. Calculate % by mass of NaCl in the original mixture.8 gm/ml] for complete neutralisation. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1. (iii) What would be energy released if 16 kg of Fe2O3 reacts with 2. What would be the mass of gas in 22. with a film 300 layers thick ? The density of the film is 1. 200 units of energy is released.12gm. phosporus (P). The waterproofing film is deposited on the fabric layer upon layer.3 In one process for waterproofing. (ii) What should be the ratio of masses of Fe2O3 and Al taken so that maximum energy per unit mass of fuel is released.7 kg of Al. AlCl3 + 3NaOH Al(OH)3 + 3NaCl NaHCO3 + NaOH Na2CO3 + CO2 + H2O Q. Calculate % composition of mixture by moles. What minimum volume of 5 % by weight AgNO3 solution(sp. On heating same amount of mixture. and potassium (K) are the main nutrients in plant fertilizers. P2O5.4 litres at 1 atm and 273 K? Calculate the % composition of this gaseous mixture.7 fertilizer in terms of moles of each elements. According to an industry convention. The mixture was to have a density of 0. and express it as x : y : 1.2 A mixture of Ferric oxide (Fe2O3) and Al is used as a solid rocket fuel which reacts to give Al2O3 and Fe.4 A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate.5 124 gm of mixture containing NaHCO3.00 m by 3.00 atm. Application of Mole Concept [14] . 1.0. Each layer is 6. Q. 8% w/w NaOH solution [dNaOH = 1. by the reaction n(CH3)2SiCl2 + 2nOH– 2nCl– + nH2O + [(CH3)2SiO]n where n stands for a large integer. On complete reaction of 1 mole of Fe2O3. (i) Write a balance reaction representing the above change.2 : 4. Assume KNO3 does not decompose under given conditions. and K2O. How much (CH3)2 SiCl2 is needed to waterproof one side of a piece of fabric.0 Å thick [ the thickness of the (CH3)2SiO group]. it shows loss in weight of 18. No other reactants and products are involved. (Si = 28) Q. by volume.EXERCISE # II Q.8 A mixture of nitrogen and hydrogen. AlCl3 & KNO3 requires 500 ml.321 gm in order to ensure complete precipitation of chloride in every possible case? Q. Calculate the N : P : K ratio of a 28 : 14. KCl and NH4Cl separately or as mixture. a fabric is exposed to (CH3)2SiCl2 vapour.8 gm of chloroform ? 2CH3COCH3 + 6CaOCl2 Ca(CH3COO)2 + 2CHCl3 + 3CaCl2 + 2Ca(OH)2 Q. The vapour reacts with hydroxyl groups on the surface of the fabric or with traces of water to form the waterproofing film [(CH3)2SiO]n. was partially converted into NH3 so that the final product was a mixture of all these three gases. What is the weight of acetone required for producing 47. in the ratio of one mole of nitrogen to three moles of hydrogen. Weak base formed doesn't interfere in reaction. Q.7 Chloride samples are prepared for analysis by using NaCl.0 g/cm3.1 Nitrogen (N). the numbers on the label refer to the mass % of N.02 g ml–1) must be added to a sample of 0. of AgNO3 added (in ml) Q. To this solution AgNO3 solution of certain molarity was added gradually.9 gm sample of KClO3 was heated under such conditions that a part of it decomposed according to the equation (1) 2KClO3 2 KCl + 3O2 and remaining underwent change according to the equation.13 % 2ZnSO4 + 2H2O 80 2Zn + 2H2SO4 + O2 What mass of Zn will be obtained from a sample of ore containing 292. If it known that AgI is precipitated first. (2) 4KClO3 3 KClO4 + KCl If the amount of O2 evolved was 672 ml at 1 atm and 273 K. 24. Application of Mole Concept [15] .10 Equal weights of mercury and iodine are allowed to react completely to form a mixture of mercurous and mercuric iodide leaving none of the reactants.4 (in gm) 40 120 Vol.12 The chief ore of Zn is the sulphide.4) 4.2 kg of ZnS.11 Two substance P4 & O2 are allowed to react completely to form mixture of P4O6 & P4O10 leaving none of the reactants.. P4 + 3O2 P4O6 P4 + 5O2 P4O10 (i) If 1 mole P4 & 4 mole of O2 (ii) If 3 mole P4 & 11 mole of O2 (iii) If 3 mole P4 & 13 mole of O2 Q.9 In order to determine the composition of a mixture of halides containing MBr2 & NaI. Find out the value of ABCD where : AB = Atomic weight of metal 'M' CD = Mole percent of NaI in original mixture. along with zinc. formed 9. calculate the % by weight of KClO4 in the residue. The ore is concentrated by froth floatation process and then heated in air to convert ZnS to ZnO. 14 gm mixture was dissolved in water. Calculate the volume of O2 produced at 1 atm & 273 K. I = 127) Q. % 2ZnS + 3O2 75 2ZnO + 2SO2 % ZnO + H2SO4 100 ZnSO4 + H2O (a) (b) Q. ZnS.44 Total Mass of ppt. Precipitation of Br– does not start until the already precipitating I– precipitates completely. The mass of precipitate produced (in gm) were measured and it was plotted against volume of AgNO3 solution added (in ml). Calculate the ratio by mole of Hg2I2 and HgI2 formed (Hg = 200 .Q. Using this information calculate the composition of final mixture when mentioned amount of P4 & O2 are taken. (Zn = 65. (ii) At t =30 min.Q.2 M H2SO4 to produce BaSO4 and the acid. 2. (b) Moles of each specie at t = 30 min.16 The molecular mass of an organic acid was determined by the study of its barium salt.15 Consider the following set of reactions ( )n If 0. Information (i) At t = 0. of residue obtained is 54 gms then what will be the molecular mass of CH 3 — CH — CH 3 | Br n Q. The acid is mono basic. only 1 mole of A is present and the gas has V.1 moles of silver salt is taken & wt. the following informations are known.D. The barium salt contains two moles of water of hydration per Ba+2 ion.14 For a hypothetical chemical reaction represented by 3A(g) C(g) + D(g) . Q. the gaseous mixture consist of all three gases and has a vapour density = 75. (iii) Molecular Mass of C = 200 Calculate (a) Molecular weight of A and D. = 60. What is molecular mass of anhydrous acid ? 'x' M solution represent that there is 'x' mole of solute per litre of solution (Atomic mass of Ba = 137) Application of Mole Concept [16] .562 g of this salt reacts completely with 30 ml of 0. on strong heating. If the weight percentage of carbon in it 8 times the weight percentage of hydrogen and one-half the weight percentage of oxygen.16 gm of silver. is C(s) + H2(g) + O2 (g) C12H22O11(s) (A) 138. 2.5 % of NaI by weight .9 74 gm of a compoud on complete combustion gives 132 gm CO2 and 54 gm of H2O. Q. 3 mole of Y and 4 mole of Z is taken. Then mole % of gas B in the mixture would be ________. mass of 'B' = 120) 2AB2 + 4A2B 5A2 + 2B4 (A) 2120 gm (B) 1060 gm (C) 560 gm (D) 1660 gm Q. determine the molecular formula of the acid. Q. The molecular formula of the compound may be (A) C5H12 (B) C4H10O (C) C3H6O2 (D) C3H7O2 Q. If 1. The number of iodide ions going into his body everyday is [Atomic weight I =127] (A) 10–4 (B) 6. mass of 'A' = 10 . A person consumes 3 gm of salt everyday.7 The percentage by mole of NO2 in a mixture of NO2(g) and NO(g) having average molecular mass 34 is : (A) 25% (B) 20% (C) 40% (D) 75% Q.2 The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.6 40 gm of a carbonate of an alkali metal or alkaline earth metal containing some inert impurities was made to react with excess HCl solution.1 For the reaction. The liberated CO2 occupied 12. at 1 atm & 300 K. With 25% yield. Single Correct: Q.10 An iodized salt contains 0.5 Maximum mass of sucrose C12H22O11 produced by mixing 84 gm of carbon.5 (C) 172.8 The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is : (Given At.02 × 1023 Application of Mole Concept [17] . [Atomic weight of Ag = 108] (A) C4H6O4 (B) C4H6O6 (C) C2H6O2 (D) C5H10O5 Q.5 (D) 199. At.5 (B) 155.02 × 1019 (D) 6.02 ×10–4 (C) 6.4 3. O2 at 1 atm & 273 K according to given reaction.EXERCISE # III Fill in the blank: Q.3 For the reaction 2A + 3B + 5C 3D Initially if 2 mole of A.5 Q. The correct option is (A) Mass of impurity is 1 gm and metal is Be (B) Mass of impurity is 3 gm and metal is Li (C) Mass of impurity is 6 gm and metal is Li (D) Mass of impurity is 2 gm and metal is Mg Q.64 gram of the silver salt of an organic dibasic acid yields. moles of D which can be produced are _________.25 mole of W is obtained then % yield of this reaction is _______. 12 gm of hydrogen and 56 lit. 2X + 3Y + 4Z 5W Initially if 1 mole of X.315 lit. 4 mole of B and 6 mole of C is taken. 01 × 1019 (D) 6.17 The average atomic mass of a mixture containing 79 mole % of 24Mg and remaining 21 mole % of 25Mg and 26Mg .02 × 1020 Q. if PCl5 undergo 50% decomposition. How many iron atoms are there in a block of alloy measuring 10 cm × 20 cm × 15 cm? (A) 2.3%) has a density of 8.52 wt. nickel (45%) and manganese (0.442 × 1026 (C) 8. What is the molecular weight of aspartame? (A) 147 (B) 294 (C) 588 (D) 266 Application of Mole Concept [18] .12 In the quantitative determination of nitrogen. There are two nitrogen atoms per molecule.201 × 1026 (B)1.7%).13 The mass of P4O10 produced if 440 gm of P4S3 is mixed with 384 gm of O2 is P4S3 + O2 P4O10 + SO2 (A) 568 gm (B) 426 gm (C) 284 gm (D) 396 gm Q. N2 gas liberated from 0. All volumes are measured under the same conditions of temperature and pressure.42 gm of a sample of organic compound was collected over water.17 g/cm3.41 × 1025 (D) 6. (A) 2500 m3 (B) 500 m3 (C) 2000 m3 (D) 3000 m3 Q. tension at 250 K is 24 mm Hg and R = 0. calculate the theoretical volume of air which will be required for burning 200 m3 of acetylene gas completely. if a signature written by carbon pencil weights 1.19 All alloy of iron (54.31.04 × 1020 (B) 6. PCl5 PCl3 + Cl2 (A) 50% (B) 66. prepared to produce maximum energy is (Combustion reaction is exothermic) (A) 413.66 % (C) 33.02 × 1019 (C) 3. % by mass of nitrogen in the organic compound is [Aq.08 L atm mol–1 K–1 ] (A) 10 % 3 (B) 5 % 3 (C) 20 % 3 (D) 100 % 3 Q. % nitrogen. is 24.15 The mass of Mg3N2 produced if 48 gm of Mg metal is reacted with 34 gm NH3 gas is Mg + NH3 Mg3N2 + H2 (A) 200 3 (B) 100 3 (C) 400 3 (D) 150 3 Q. If the volume of N2 gas collected was 100 ml at total pressure 11 860 mm Hg at 250 K.11 The mass of CO2 produced from 620 gm mixture of C2H4O2 & O2.20 Aspartame.18 Air contains 20% oxygen by volume.19 × 1026 Q.04 gm (C) 440 gm (D) 320 gm Q.2 × 10–3 gm is (A) 12.14 Calculate percentage change in Mavg of the mixture.33 gm (B) 593. % mole of 26Mg is (A) 5 (B) 20 (C) 10 (D) 15 Q.16 The number of carbon atoms present in a signature. an artificial sweetener contains 9.Q.33 % (D) Zero Q. (A) Statement-1 is true.82 × 1022 (D) None of these Q. How many iron atoms are present in each ball of diameter 1 cm if the balls contain 84% iron.6 milligrams of vitamin C (ascorbic acid) having formula C6H8O6. Given : NA = 6 × 1023 Column I Column II (A) O-atoms present (P) 10–4 mole (B) Moles of vitamin C in 1 gm of vitamin C (Q) 5.26 The recommended daily dose is 17. (A) 4.6 times of produced moles of Na2SO4 (in Ist reaction. Match the following. More than one correct: Q.7 gm NH4Cl (D) Produced moles of NH4Cl (in IInd reaction) are 1.) if 4 gm of NaOH is taken Match the column: Q.Q. Statement -2 : 'B' is the liming reactant for the given data. statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is false.[Y = 89.32% (C) O (R) 44. Find the minimum possible molecular weight of compound? (A) 400 (B) 200 (C) 800 (D) 1600 Assertion Reason: Q.2.73% (B) Al (Q) 32.25 gm (NH4)2SO4 is required to produce 10. (B) Statement-1 is true.68 × 10–3 (C) Moles of vitamin C in recommended daily dose (R) 3. statement-2 is false. by mass ? The atomic mass of iron is 56.21 The specific gravity of the stainless steel spherical balls used in ball-bearings are 10.23 Statement -1 : 2A + 3B C 4/3 moles of 'C' are always produced when 3 moles of 'A' & 4 moles of 'B' are added.24 For the following reactions 40% Na2SO4 + H2O + NH3 (NH4)2SO4 + NaOH Yield = 40% 80% NH4Cl NH3 + HCl Yield = 80% Choose correct option(s).71 gm if 4 gm of NaOH is taken (B) 41.25 One type of artifical diamond (commonly called YAG for yttrium aluminium garnet) can be represented by the formula Y3Al5O12.82 × 1022 (C) 3. statement-2 is true and statement-2 is correct explanation for statement-1.6 × 1020 Application of Mole Concept [19] .22 An organic compound contains 8 % Oxygen and 4 % Sulphur by mass. Al =27] Column I Column II Element Weight percentage (A) Y (P) 22.7 gm NH4Cl (C) 25 gm of NaOH is required to produce 10. statement-2 is true. (D) Statement-1 is true.95% Q.12 × 1021 (B) 4. [Assuming all other reactants required are in sufficient amount] (A) Produced mass of NH4Cl is 2. . mole (Br2) 2 2 1 mole (Br2O).29 Calculate the amount of Mg(OH)2 (in gm) produced in above reaction.2 × 108 gm Q.32 If the yield of (ii) is 60% & (iii) reaction is 70% then mass of iron required to produce 2.30 Calculate the mass % of Mg converted into Mg3N2.(ii) (not balanced) Fe3Br8 + Na2CO3 NaBr + CO2 + Fe3O4 .4 gm of gaseous NH3 is generated according to given reactions. (A) 11.Q. 3.06 × 103 gm NaBr is formed (A) 20 (B) 10 (C) 40 (D) None Application of Mole Concept [20] . Br2 + 1 O Br2O 2 2 3 O Br2O3 2 2 Match composition of the final mixture for initial amount of reactants.6 (B) 17.31 (D) 40 Paragraph for question nos. 1 mole (Br2O3) 1 1 mole (Br2O3) .2 (D) 15. Mg + O2 MgO Mg + N2 Mg3N2 Mg3N2 + 6H2O 3Mg (OH)2 + 2NH3 Calculate the amount of magnesium oxide (in gm) in products..(i) FeBr2 + Br2 Fe3Br8 ... 31 to 33 NaBr..27 Br2 reacts with O2 in either of the following ways depending upon supply of O2. When products are treated with excess of H2O.4 (C) 23. mole (O2) 2 4 Comprehension: Q.2 × 105 kg (D) 4. Column I Column II (Initial reactants) (Final product) (A) 320 gm Br2 is mixed with 64 gm of O2 (P) 1 mole Br2O3 Br2 + (B) 160 gm Br2 is mixed with 8 gm of O2 (Q) (C) 80 gm Br2 is mixed with 32 gm O2 (R) (D) 160 gm Br2 is mixed with 48 gm O2 (S) 1 1 mole (Br2O). Fe = 56 ] (A) 420 gm (B) 420 kg (C) 4.8 Q.(iii) (not balanced) Mass of iron required to produce 2.33 If yield of (iii) reaction is 90% then mole of CO2 formed when 2. (A) 20 (B) 30 (C) 35 Q.. used to produce AgBr for use in photography can be self prepared as follows : Fe + Br2 FeBr2 .28 Paragraph for question nos.06 × 103 kg NaBr (A) 105 kg (B) 105 gm (C) 103 kg (D) None Q.. 28 to 30 24 gm pure sample of magnesium is burned in air to form magnesium oxide and magnesium nitride...8 (D) 32 Q.06 × 103 kg NaBr [Atomic weigth Br = 80. (A) 28 (B) 20 (C) 16. 34 Q. What is percentage of 'N' in the compound? (A) 7. If the vapour pressure at 27°C is 21 mm of Hg.4 % (B) 11. O and F] and 0.Q.27 mL of moist nitrogen at 27°C and at P mm Hg pressure.4 % (D) 21. [Assume that the empirical formula is same as molecular formula. 3 H 2 2 Mg + HCl MgCl2 + H2 Al + 3HCl AlCl3 + Q.66 % (B) 33.0 g of an alloy of Al and Mg when treated with excess of dil HCl forms MgCl2. The ammonia evolved was absorbed in 50 mL of 0.218 litre at 0.8 gram of a gas only.] (U = 238. 37 to 39 An organic compound contains 69.92 atm pressure. 34 to 36 Water is added to 3.303 g of the organic compound. The products are 3. F = 19) The empirical formula of the gas is (A) HF2 (B) H2F (C) HF (D) HF3 The empirical formula of the solid product is (A) UF2O2 (B) UFO2 (C) UF2O (D) UFO The percentage of fluorine of the original compound which is converted into gaseous compound is (A) 66.4 % C.9 % Paragraph for question nos.41 Weight percentage of Al in the mixture is: (A) 60% (B) 40% (C) 55% (D) 45% Moles of MgCl2 produced will be? (A) 0.8 % H.38 Number of hydrogen atoms in the molecular formula of the compound? (A) 7 (B) 10 (C) 5 (D) 12 Q.303 g of this compound is analysed for nitrogen by Kieldahl's method.0125 (D) 0. A sample of 0.37 Paragraph for question nos. in Duma's method on combustion gave 32.025 (B) 0.52 grams of UF6.33 % (C) 50 % (D) 89.08 grams of a solid [containing only U. AlCl3 and hydrogen.40 Q. contains 95 % by mass fluorine.05 M H2SO4.55% (C) 19.0114 Application of Mole Concept [21] .39 0.1 M NaOH for neutralization.0167 (C) 0.36 Q. The excess acid required 25 mL of 0.35 Q. The gas [containing fluorine and hydrogen only]. 40 to 41 1. Find the value of 'P'? (A) 725 mm (B) 746 mm (C) 710 mm (D) 760 mm Paragraph for question nos. Molecular weight of organic compound is 121.6 % Q. 5. collected over Hg at 0°C has a volume of 1. The evolved hydrogen. 023×1014/cm2 and surface area is 1000 cm2. 1] Q.108 6.3] Q. [JEE 2005] Q.2 A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The number of glycine units present in the decapeptide is [JEE 2011] [Given : Decapeptide consumes nine molecules of H2O for hydrolysis] Application of Mole Concept [22] .3 At 100º C and 1 atmp . if the density of liquid water is 1.06 cm3 [JEE '2001 (Scr). Wt. find out the no.8 A decapeptide (Mol.6 cm3 (D) 0.023 Q.108 [JEE'2002 (Scr).0006 g cm 3 .7 Silver (atomic weight = 108 g mol–1) has a density of 10. If the virus is considered to be a sinlge particle. Wt.0 g when completely filled with liquid of density 0.023 ×1023 (B) 1 ×1031 9. Density of surface sites is 6.= 75).4 How many moles of e— weigh one Kg (A) 6. 1] (C) 6.46 cm3.0 % to the total weight of the hydrolysed products. The specific volume of the virus is 0.108 (D) 1 × 108 9.75 cm3 / g. of surface sites occupied per molecule of N2.001 atm and 298 K in a container of volume is 2.EXERCISE # IV Q.5 g cm–3. then the volume occupied by water molecules in 1 L of steam at that temperature is : (A) 6 cm3 (B) 60 cm3 (C) 0. The value of x is [JEE 2010] Q.5 Calculate the amount of Calcium oxide required when it reacts with 852 gm of P4O10. Determine the molecular weight of the gas. [JEE 2005] 6CaO + P4O10 2Ca3(PO4)2 Q. 148. On heating N2 gas evolved from sites and were collected at 0.6 20% surface sites have adsorbed N2.98 gml1 and 50. = 796) on complete hydrolysis gives glycine (Mol.0 g cm 3 and that of water vapour is 0. [JEE '98. alanine and phenylalanine.5 g when filled with an ideal gas at 760 mm at 300 K .023 × 1054 9. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y × 10x. Glycine contributes 47. find its molar mass.1 An evacuated glass vessel weighs 50 gm when empty . [JEE 1999] Q. 1185 Q.37 10 % Q.1 M+ Q.5 70 ml Q.1 20 : 2 : 1 Q.2 × 1023 2:5 Q.11 3.43 87. N2 = 16. NaCl = 37% Q.70 Application of Mole Concept [23] .2 1.ANSWER KEY EXERCISE # I 1 3 Q.6 gm Q.25 39.5 gm Q.5 AlCl3 = 33.9 116.8 gm Q. NaHCO3 = 50%.27 61.4 Q.9 4050 Q. (ii) 80 : 27.18% Q.39 1.21 NaHCO3 = 16.023 × 1017 Q.16 w CCl = 154 gm .35 56% Q.3 EXERCISE # II Q.48 litre Q.3 % Q.14 9.1 gm Q.44 C7H10NCl Q.2 (i) Fe2O3 + 2 Al Al2O3 + 2Fe.32 14.38 56% Q. 20 % Q.10 %NaCl = 77. Na2CO3 = 83.26 4.30 490 gm Q.0 gm Q. w = 24 gm c 4 Q. MgCO3 = 71.723 litre Q. NH3 = 33.6% Q.5 g/mol Q.10 27.29 640.33 42 %.19 Al = 66.2 gm Q.78 Q.34 9.8% 20 ml 1 : 2.6 ×1019 Q.6 40 gm Q.17 5.8 11.41 3.5 gm Q.8 gm Q.4 Q.7 Q.12 51.23 NaHCO3 = 63 % .8 %. (iii)10.36 4% Q.40 mco : m co 2 = 21 : 111 Q.4 1.22 60% Q.33%.33gm.4%.3 1.33% Q. KNO3 = 16.20 CaCO3 = 28.28 177.6% Q.45 C4H4SO Q.42 196 gm Q.25 mole Q.6 6.6 gm Q.31 (a) 142 gm (b) 39.161 gram Q.6 Q.18 0.000 units Q.2 % Q.6 gm Q.13 58.67% Q.15 0.5 gm Q.67 % H2 = 50%.46 C 2H 4O 2 Q.24 11. (C) P Q.6 2 Q.34 C Q.2 .20 B Q.75 Q.32 C Q.5 .41 B EXERCISE # IV Q.30 B Q. (B) Q. (C) Q Q. (ii) 2.16 × 103 lit.19 B Q.5 .4 B Q.22 C Q. nD =0.31 B Q.3 C Q.1 123 g/mol Q.12 A Q. (b)nA = 0.5 1008 gm Q.7 A Q.Q. (D) P Q.5 B Q.2 25% Q.7 7 Q.36 A Q. Q.27 (A) R. MD = 160.40 A Q.2 7.15 495 Q.14 (a) MA = 120.25 (A) R.33 B Q.29 B Q.2 Q.092 × 107 g Q. (b) 20.9 C Q.37 B Q.11 (i) 0.23 C Q.72 kg .11 C Q.39 B Q.26 (A) R.3 0.12 (a) 117. (B) P.72% Q.35 A Q.38 A Q.8 A Q.4 D Q.10 C Q. 1 (iii) 1.6 A Q.21 B Q.13 B Q.15 A Q.13 52.4.14 C Q.28 A Q. 0.24 BCD Q.17 C Q.16 B Q. 2 Q. nC= 0.1 50% Q. (C) S .8 6 Application of Mole Concept [24] .16 128 EXERCISE # III Q. (B) Q .18 A Q.