LEMO NC E CO PT MOLE CONCEPT SIGNIFICANT FIGURES : (A) Every scientific observation involves some degree of uncertainity depending upon the limitation of instrument. To represent scientific data, role of significant figures has its own importance. (B) Significant figures are equal to the number of digits in numbers with last digit uncertain and rest all are certain digits i.e. all the digits of datum including the uncertain one , are called significant figures. (C) Rules for determination significant figure: (i) All non zero digits are significant. Example : 3.14 has three significant figures (ii) The zeros to the right of the decimal point are significant. Example : 3.0 has two significant figures. (iii) The zeros to the left of the first non zero digit in a number are not significant. Example : 0.02 has one significant figure. (iv) The zeros between two non zero digits are also significant. Example : 6.01 has three significant figures. (v) Exponential form : N × 10n. Where N show the significant figure. Example : 1.86 × 104 has three significant figure. (vi) Rounding off the uncertain digit : (a) If the left most digit to be rounded off is more than 5 , the preceding number is increased by one. Example : 2.16 is rounded to 2.2 (b) If the left most digit to be rounded off is less than 5 , the preceding number is retained. Example : 2.14 is rounded off to 2.1 (c) If the left most digit to be rounded off is equal to 5 , the preceding number is not changed if it is even and increased by one if it is odd. Example : 3.25 is rounded off to 3.2 2.35 is round off to 2.4 LAWS OF CHEMICAL COMBINATION : Law of definite proportion [Proust, 1799] (a) According to the law , the composition of a compound always remains a constant i.e. the ratio of weights of different elements in a compound ; no matter by whatever method , it is prepared or obtained from different sources, remains always a constant Example : In H2O ratio of weight = 1 : 8 In CO2 ratio of weight = 3 : 8 Law of multiple proportion [John Dalton, 1804] According to this law, when two elements A and B combine to form more than one chemical compound then different weights of A , which combine with a fixed weight of B , are in a proportion of simple whole number 12 The atomic mass unit (amu) is equal to 1 amu = 1 × mass of one C-12 atom 12 ~ mass of one nucleon in C-12 atom. = 1. The ratio of weights of H and O in Water is 1 : 8 Gay-Lussac’s [1808] law of combining volumes : This law states that under similar conditions of pressure and temperature.A.66 × 10–24 gm or 1. Therefore when H combines with O they should combine in the ratio of 4 : 32 (i. 12 grams of C reacts with 4 grams of H whereas in CO2 12 gram of C reacts with 32 grams of O. The same is found to be true in H2O molecule. In CH4 .A. = 1 : 8) or in simple multiple of it. their combining ratio is directly reciprocated if they combine directly Example : C with H form methane and with O form CO2. ratio 1 : 3 : 2 AVOGADRO'S HYPOTHESIS : or Vapour density = Molecular weight 2 On hydrogen scale : Relative atomic mass (R. 1792-94] When two elements combines separately with third element and form different types of molecules.Example : CO & CO2 12 : 16 & 12 : 32 ratio = 16 : 32 = 1 : 2 Law of reciprocal proportions [Ritche.e.66 × 10–27 kg Atomic mass = R. N2(g) + 3H2(g) 2NH3(g) vol.M × 1 amu . Ex.A.M) = Mass of one atom of an element mass of one hydrogen atom Mass of one atom of an element 1 mass of one C 12 atom 12 Atomic mass unit (or amu) : th 1 mass of one atom of carbon-12 isotope. volume ratio of gases is always in terms of simple integers.M) = Relative atomic mass (R. Mass Relative density or Vapour density : Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature and pressure.r.4 2 Volume at STP × mol. wt. Mole-mole analysis : This analysis is very much important for quantitative analysis point of view.D. lt 2.Y-map : Interconversion of mole . Molarity of solution = number of moles of solute volume of solution in litre . × At. Now consider again the decomposition of KClO3 .4 ×2 lt 2. i. mass and number of particles : Number ×N A N A Mole mol. and from the stoichiometry of reaction we can write Moles of O 2 Moles of KClO 3 Moles of KCl = = 3 2 2 Now for any general balance chemical equation like a A + b B c C + d D you can write. moles of B reacted moles of C produced moles of D produced Moles of A reacted = = = a b c d Molarity (M) : The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity of the solution. Students are advised to clearly understand this analysis. wt. d gas Vapour density = d H2 Mgas Mgas V.. At.t.D. Relative density can be calculated w. 2KClO3 2KCl + 3O2 In very first step of mole-mole analysis you should read the balanced chemical equation like 2 moles KClO3 on decomposition gives you 2 moles KCl and 3 moles O2.volume.e. wt. to other gases also. wt. = M = H2 2 Mgas = 2 V. .. a n x n Mathematically... % weight by volume (w/v) : It is given as mass of solute present in per 100 ml of solution. i. a1x1 a 2 x 2 .e... and x1.. .e.. % weight by weight (w/w) : It is given as mass of solute present in per 100 gm of solution... i.. average atomic mass of X (Ax) = 100 Where : a1.. atomic mass of isotopes.. mass of solute in gm % w/w = mass of solution in gm 100 % w/v = mass of solute in gm 100 volume of solution in ml % volume by volume (v/v) : It is given as volume of solute present in per 100 ml solution.... volume of solute in ml % v/v = volume of solution in ml × 100 Miscellaneous : AVERAGE/ MEAN ATOMIC MASS : The weighted average of the isotopic masses of the element’s naturally occuring isotopes.e. a2.. % calculation : The concentration of a solution may also expressed in terms of percentage in the following way. molality = number of moles of solute 1000 mass of solvent in gram Mole fraction (x) : The ratio of number of moles of the solute or solvent present in the solution and the total number of moles present in the solution is known as the mole fraction of substances concerned. mole % of isotopes. i.. x2... x3 .Molality (m) : The number of moles of solute dissolved in1000 gm (1 kg) of a solvent is known as the molality of the solution...e..... a3 . i.. The number of sodium atoms in 2 moles of sodium ferrocyanide Na4[Fe(CN)6].022 × 1023 atoms (B) 4 molecules (D) 24. In which of the following pairs do 1 g of each have an equal number of molecules? (A) N2O and CO (B) N2 and C3O2 (C) N2 and CO (D) N2O and CO2 Section (B) : Density and vapour density B-1.I : OBJECTIVE QUESTIONS * Marked Questions are having more than one correct option. the number of moles in 46 g of sodium is : (A) 1 (B) 2 (C) 2.2 litre (B) 44.6 A-5.0 g (atomic) oxygen and 1. A-4. moles.0 g of atomic oxygen. A-6. (C) 1.02 × 1023 A-8. e = charge on one electron) (A) A-2. (B) 12g carbon (C) 8g oxygen gas (D) 16 g oxygen atom 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% A-10*.PART . One mole of P4 molecules contains (A) 1 molecule (C) 1/4 × 6.4 litre .0 g of ozone. 5. Which of the following will contain same number of atoms as 20g of calcium? (A) 24g magnesium A-9.P. A-3.3 (D) 4.0 g of oxygen gas. (D) All contain same number of atoms.T. is(A) 2 (B) 6.0 g of ozone. B-2.6 litre of a gas at N.023 × 1023 (C) 8 × 6. Section (A) : Atoms. 1. Out of 1. avogadro's hypothesis A-1. The charge on 1 gram ions of Al3+ is : (NA = Avogadro number.088 × 1023 atoms A-7. 1 N e coulomb 27 A (B) 1 × NAe coulomb 3 What is correct for 10 g of CaCO3 (A) It contains 1g-atom of carbon (C) It contains 12 g of calcium (C) 1 × NAe coulomb 9 (D) 3 × NAe coulomb (B) It contains 0.3 g-atoms of oxygen (D) None of these Which of the following contains the largest number of atoms (A) 11g of CO2 (B) 4g of H2 (C) 5g of NH3 (D) 8g of SO2 If the atomic mass of Sodium is 23.02 × 1023 (D) 4 × 6. (B) 1.8 litre (C) 2 litre (D) 22. 2 moles of H2 at NTP occupy a volume of (A) 11.0 g dioxygen. molecules. weighs equal to 8 gm the vapour density of gas is (A) 32 (B) 16 (C) 8 (D) 40. the maximum number of oxygen atoms are contained in (A) 1. (D) 0. Density of ozone relative to methane under the same temperature & pressure conditions is : (A) 1 (B) 3 (C) 1.04 (D) 55.8 gm (B) 18 gm (C) 3.P.. Mol.002 (B) 35.5 mole of methane (C) 3. The vapour density of the gas B is 20. the vapour density of the gas A is : (A) 30 (B) 40 (C) 50 (D) 60 B-12.48 litres of methane at N.2 x 1022 molecules of methane (B) 0.P. If the molecular weight of B is M. Vapour density of a gas if its density is 0.56 B-13.0 L oxygen gas both measured at S. the molecular weight of A will be : (A) M B-10.B-3....05 kg (C) 3605 kg (D) M 2 (D) 3.605 kg B-11. The molecular formula of the compound is : (A) C2H4O2 (B) C4H8O4 (C) C3H6O3 (D) all of these . is valid for (A) metals (B) non metals (C) solids (D) gases B-5. wt.(Cu = 63.089 (B) 2M (C) 3M Mass of H2O in 1000 kg CuSO4. is 2.2H2O are .. The ratio of the weight of one litre of a gas to the weight of 1.52 (C) 71.T. C-3. 4. A nugget of gold and quartz was found to contain x g of gold and y g of quartz and has density d.5) (A) 360.22.5 (D) 2.. If the densities of gold and quartz are d1 and d2 respectively then the correct relation is : x y xy (A) d + d = d 1 2 (B) xd1 + yd2 = (x + y) d x y xy (C) d + d = d 2 1 (D) x x xy + d + d =0 d 1 2 The vapour density of a gas A is twice that of a gas B. C-2.178 (B) 2 (C) 4 B-8. The molecular weight of the gas would be : (A) 14. The molar mass of normal water is .1 mole of methane B-4. B-9.6 (C) 59.(Ba = 137) (A) 2 moles (B* ) 4 moles (C) 3 moles (D) 5 moles B-6..T.5H2O is ..5 mole of K4[Fe(CN)6] is (A) 1.3 The empirical formula of a compound of molecular mass 120 is CH2O.6 gm (D) 36 gm The percentage of nitrogen in urea is about(A) 38. The relative density of a gas A with respect to another gas B is 2.4 (B) 46. (A) 10% less (B) 10% high (C) 2% less (D) zero% less Section (C) : Percentage composition and molecular formula C-1.. correspond to(A) 1. as compared to heavy water. Number of moles of water in 488 gm of BaCl2. = vapour density × 2. The mass of carbon present in 0.1 (D) 61.5 B-7.178 g/L at NTP is : (A) 0.5 kg (B) 36.2 gm of methane (D) 0. . The percent of N in 66% pure (NH4)2 SO4 sample is (A) 32 (B) 28 (C) 14 (D) None of these C-6. [Atomic mass Na = 23. The empirical formula of a compound is CH. (.6 kg Fe2O3 (at.7 (D) None C-7. An oxide of a metal (M) contains 40% by mass of oxygen...07 (B) 2. The empirical formula of the oxide is(A) M2O (B) MO (C) M2O3 (D) M3O4 C-10. O2 . The mass of oxygen that would be required to produce enough CO...) if molecular weight of compound is 200.. The molecular formula of the compound will be (A) C2H2 (B) C3H3 (C) C4H4 (D) C6H6 C-9. (D) 0.4% sulphur by mass... H2O) is (A) 5 (B) 2. Calculate the molecular formula of compound which contains 20% Ca and 80% Br (by wt. 12 g of alkaline earth metal gives 14. If the formula of first oxide is M3O4 then formula of second oxide is (A) MO (B) M2O (C) M2O3 (D) MO2 C-11.37% nitrogen by mass.5 (C) 1 D-3.607 (C) 260.. What will be the minimum molecular weight of insulin (A) 94.117 (B) 1884 (C) 941 (D) 976 C-5.6% and 30% oxygen respectively. Section (D) : Balanced chemical equation analysis & Limiting Reagent D-1.2 kg of Na2CO3 completely into NaHCO3. A compound was found to contain 5. Br = 80) (A) Ca1/2Br (B) CaBr2 (C) CaBr (D) Ca2Br C-8... Ca = 40] CaCO3 CaO + CO2 Na2 CO3 + CO2 + H2O 2NaHCO3 (A) 100 Kg (B) 20 Kg (C) 120 Kg (D) 30 Kg D-4. The moles of O2 required for reacting with 6.. Insulin contains 3. Two oxides of Metal contain 27. (Take at wt of M = 56. Atomic weight of metal is (A) 12 (B) 20 (C) 40 (D) 14. NX is produced by the following step of reactions M + X2 M X2 3MX2 + X2 M3X8 M3 X8 + N2CO3 NX + CO2 + M3O4 How much M (metal) is consumed to produce 206 gm of NX. NO + .8 g of its nitride.8 gm of ammonia. X = 80) (A) 42 gm (B) 56 gm (C) 14 gm 3 (D) 7 gm 4 . Metal (M) has atomic mass of 24.. NH3 +.. which completely reduces 1.. The simplest formula of a compound containing 50% of element X (atomic mass = 10) and 50% of the element Y (atomic mass = 20) by weight is (A) XY (B) X2Y (C) XY2 (D) X2Y3. Its molecular weight is 78. N=23. (Atomic wt. Ca = 40. mass Fe = 56) is : (A) 240 gm (B) 480 gm (C) 720 gm (D) 960 gm D-2..C-4.5 What weight of CaCO3 must be decomposed to produce the sufficient quantity of carbon dioxide to convert 21.8 D-5. What is the minimum molecular weight of compound(A) 26. (A) 1 mol 2 (B) 1 mol 3 (C) 1 mol 4 (D) 2 mol 3 .8 litre H2 & 22.4 litre HCl (D) 22. Then : (A) X is the limiting reagent (B) Y is the limiting reagent (C) No reactant is left over and mass of X2Y3 formed is double the mass of ‘X’ taken (D) none of these D-14*.4 litre HCl D-11. (A) 2 mole (B) 3 mole (C) 4 mole (D) 5 mole D-12.4 (B) 0. Then which option is/are correct ? (A) 1 mole of A3B2C2 is formed (B) 1 2 mole of A3B2C2 is formed (C) 1/2 mole of A3B2 is formed (D) 1 2 mole of A3B2 is left finally D. then : (A) Oxygen is the limiting reagent.T.4 L. How many mol Fe2+ ions are formed. 2H2 + O2 2H2O. when excess of iron is treated with 50mL of 4.5 D-9. 8 mol of P and 5 mol of Q will produce (A) 8 mol of R (B) 5 mol of R (C) 4 mol of R D-13. A + B A3B2 (unbalanced) A3B2 + C A3B2C2 (unbalanced) Above two reactions are carried out by taking 3 moles each of A and B and one mole of C. 12 litre of H2 and 11. What will be the change in volume when 100 mL of phosphine is decomposed ? PH3 P4 + H2 (A) + 50 mL (B) 500 mL (C) + 75 mL (D) – 500 mL D-17.5 mole of H2SO4 is mixed with 0. (D) 13 mol of R Equal weight of 'X' (At.15* If 27 g of Carbon is mixed with 88 g of Oxygen and is allowed to burn to produce CO2 . (C) 0.5 (C) 0. (B) Volume of CO2 gas produced at NTP is 50. wt. A mixture containing 100 gm H2 and 100 gm O2 is ignited so that water is formed according to the reaction. How much water will be formed (A) 113 gm (B) 50 gm (C) 25 gm (D) 200 gm D-8. wt. = 36) and 'Y' (At. 0.2 (D) 0. For the reaction : A + 2B C 5 mole of A and 8 mole of B will produce (A) 5 mole of C (B) 4 mole of C (C) 8 mole of C (D) 13 mole of C D-10. = 24) are reacted to form the compound X2Y3.2 mole of Ca (OH)2.1 (C) 0. For the reaction 2P + Q R. (D) Volume of unreacted O2 at STP is 11.2 L. How many mole of Zn(FeS2) can be made from 2 mole zinc. The composition by volume of mixture is (A) 24 litre of HCl (B) 0. (C) C and O combine in mass ratio 3 : 8.8 D-7. 3 mole iron and 5 mole sulphur.0M HCl under inert atmosphere ? Assume no change in volume (A) 0.2 litre oxygen at N.4 (D) 1.P.D-6. How many moles of potassium chlorate need to be heated to produce 11.2 (B) 0. The maximum number of moles of CaSO4 formed is (A) 0.8 lit HCl. D-16.2 litre of Cl2 are mixed and exploded.8 litre Cl2 and 20. Phosphine (PH3) decomposes to produce vapours of phosphorus and H2 gas. 18 1 moles of oxygen combine with Al to form Al2O3. If 500 ml of 1 M solution of glucose is mixed with 500 mlof 1 M solution of glucose final molarity of solution will be : (A) 1 M (B) 0.2g of chlorine are made to react completely to yield a mixture of Cl and Cl3. If 1 gm of HCl and 1 gm of MnO2 heated together the maximum weight of Cl2 gas evolved will be [MnO2 + 4HCl MnCl2 + Cl2 + 2H2O] [Atomic mass of Mn=55] (A) 2gm (B) 0.4 (D) 1.1 mole.10 M NaOH solution to make a solution in which the molarity of the H2SO4 is 0. the weight of Al used in the reaction is : (Al = 27) 2 (A) 27g (B) 54g (C) 40.6 g of silver coin is dissolved in HNO3.6g (B) 85.2 M HCl to obtain 0.24 litre E-2.5 mole of H2SO4 is mixed with 0. (A) 0. The molar ratio of Fe++ to Fe+++ in a mixture of FeSO4 and Fe2(SO4)3 having equal number of sulphate ion in both ferrous and ferric sulphate is (A) 1 : 2 (B) 3 : 2 (C) 2 : 3 (D) can't be determined D-19.18 mol of H atoms. Calculate the number of moles of Cl and Cl3 formed.06 (D) 3. 71g (D) 17.486 gm (D) 0. (Ag) (A) 74. How many grams of this ore would have to be processed in order to obtain 1.5g E-5. An ore contains 1.265 (B) 0. 25g of this sample is treated with excess of HCl.D-18.6 M HCl and 750 ml of 0. 219.5 mole. 0. by weight.5g. 0.050 M ? (A) 400 mL (B) 50 mL (C) 100 mL (D) 150 mL . 25. What volume of a 0.5 mL F-2.972 gm D-21.5 D-22.975 gm (C) 0.2 litre (D) 2.2 mole.25 M solution of HCl is : (A) 750 ml (B) 100 ml (C) 200 m (D) 300 m F-4.2 mole (C) 0. 35. all silver is precipitated as AgCl.8 M solution contains 100 milli moles of the solute? (A) 100 mL (B) 125 mL (C) 500 mL (D) 62.7g (C) 107.795 (C) 1.1 mole (B) 0.10 M H2SO4 must be added to 50 mL of a 0. What weights of P4O6 and P4O10 will be produced by the combustion of 31g of P4 in 32g of oxygen leaving no P4 and O2.5g (C) 55g.5 M F-3. A sample of calcium carbonate is 80% pure.5 mole (D) 0.2 mole of Ca (OH)2. The number of mol of O atoms in the sample is : (A) 0. How much volume of CO2 will be obtained at NTP.00g of pure silver. The maximum number of moles of CaSO4 formed is (A) 0.4 g of iodine and 14. 0.75g.9g (D) 134.1 mole. 190. Ag2S. (A) 4. The volume of water that must be added to a mixture of 250 ml of 0.48 litre (B) 5.5g (B) 27.5g. 21. What volume of 0.0g E-3. (A) 2. 0.34% of the mineral argentite.2 mole E-4.2 (B) 0.5g (D) 81 g Section (E) : Principle of Atom Conservation (POAC) E-1. 0. The weight of AgCl is found to be 14.35 g then % of silver in coin is : (A) 50% (B) 75% (C) 100% (D) 15% Section (F) : Concentration terms F-1.5 (C) 0. When NaCl is added to this solution.5 M (C) 2 M (D) 1. A sample of ammonium phosphate (NH4)3PO4 contains 3.6 litre (C) 11. If 1 D-20. 6g F-10.4 M (D) 18 M F-9. the molarity is (A) 18 M (B) 36 M (C) 54 M (D) 0.8% w/v) and 100 mL solution of H2SO4 (9.1 M (B) 0.18 M (B) 8. Mole fraction of A in H2O is 0. (A) 1/2 (B) 2 (C) 1.5. If the density is 1. H3PO4 (98 g mol–1) is 98% by mass of solution.5 M solution is (A) 40 ml (B) 16 ml (C) 50 ml (D) 24ml F-16. What is the molarity of H2SO4 solution that has a density of 1. 50 mL solution of BaCl2 (20.6 g (C) 29.8%( mass / volume) solution of KOH is : (Given atomic mass of K = 39 ) is : (A) 0.8 g (D) 11.5 (D) 1 F-12. Cl = 35.) (B) 1 M H2SO4 (56 Rs per lt.5 (C) 14.2 M (D) 2.9 (B) 15. Which sample of acid is the best choice for you.F-5.02 × 1022 molecules.2.8% w/v) are mixed (Ba = 137.0 M F-6. Suppose you want an acidic solution to carry out a chemical reaction with 2 moles of NaOH.84 g/cc and contains 98% by mass of H2SO4? (Given atomic mass of S = 32) (A) 4.5) (A) 1 M H2SO4 (50 Rs per lt. (A) 1000 g (B) 1098. The volume of water is required to make 0.4 g (D) 1019. S = 32) BaCl2 + H2SO4 BaSO4 + 2HCl BaSO4 formed is : (A) 23. wt.8 F-7.6 g (C) 980.8 g/ml.1 M (B) 1.18 M F-13. A solution of glucose received from some research laboratory has been marked mole fraction x and molality (m) at 10ºC. Calculate the total weight of the solution having 1000 gm of solvent. The concentration of the solution is (A) 0. : S = 32. Cl = 35. (At.3 g (B) 46. The molarity of the solution containing 2. Then the ratio of the concentration of cation and anion.) (D) 1 M HCl (27 Rs per lt. 500 mL of a glucose solution contains 6. The molality of a sulphuric acid solution is 0. When you will calculate its molality and mole fraction in your laboratory at 24ºC you will find (A) mole fraction (x) and molality (m) (B) mole fraction (2x) and molality (2m) (C) mole fraction (x/2) and molality (m/2) (D) mole fraction (x) and (m ± dm) molality F-15.2 M (D) 1 M F-8.) F-11.5 M (C) 0.2.65 g F-14. 2M of 100 ml Na2 SO4 is mixed with 3M of 100 ml NaCl solution and 1M of 200 ml CaCl2 solution.) (C) 1 M HCl (30 Rs per lt.5 (D) 16.20 M solution from 16 mL of 0.0 M (C) 0. The molality of A in H2O is : (A) 13.14 M (C) 18. A solution of FeCl3 is (A) M 90 M its molarity for Cl– ion will be 30 (B) M 30 (C) M 10 (D) M 5 . The mixture is heated in presence of Pt catalyst.4 m What is the molarity of solution if density of solution is 1.67 M (C) 2.68% 5.5 M (B) 6. we can find other concentration terms also.6 gm/ml (A) 5. Select the correct statement (A) SO2 is the limiting reagent (C) both SO2 and O2 are limiting reagent (B) O2 is the limiting reagent (D) cannot be predicted Number of moles of SO3 formed in the reaction will be (A) 10 (B) 4 (C) 8 (D) 14 Number of moles of excess reactant remaining (A) 4 (B) 2 (C) 6 (D) 8 Comprehension # 2 Read the following comprehension carefully and answer the questions. The concentrations of solutions can be expressed in number of ways.II : MISCELLANEOUS QUESTIONS Comprehensions Type Comprehension # 1 Read the following comprehension carefully and answer the questions.4% (B) 78% (C) 48.8% (D) 19. (A) 24. = a Msolvent (a Msolvent 1000 ) where a = molality and Msolvent = Molar mass of solvent We can change : Mole fraction Molality Molarity 4. The definition of different concentration terms are given below : Molarity : It is number of moles of solute present in one litre of the solution. knowing one concentration term for the solution.59 M (D) none (D) none . Molar concentration (Molarity) and Molal concentration (molality). viz : mass fraction of solute (or mass percent).5 m (C) 24. 1. (A) 4. Determine the mass percent of sodium chloride in the final solution. 60 gm of solution containing 40% by mass of NaCl are mixed with 100 gm of a solution containing 15% by mass NaCl. 3. 2. Molality : It is the number of moles of solute present in one kg of the solvent moles of solute Mole Fraction = moles of solute moles of solvent If molality of the solution is given as 'a' then mole fraction of the solute can be calculated by Mole Fraction = a 1000 a Msolvent .e. What is the molality of the above solution. (B) 5. Following reaction takes place : 2SO2(g) + O2(g) 2SO3(g) Assuming the reaction proceeds to completion. These terms are known as concentration terms and also they are related with each other i.PART .4 m 6. 10 moles of SO2 and 4 moles of O2 are mixed in a closed vessel of volume 2 litres. 8 .4 L.2 M AlCl3 solution + 400 ml of 0.24 L 10 9 (B) 200 9 (C) 700 9 (D) 350 9 (B) 9 g/L 7 (C) 2 g/L 7 (D) can’t be determined Density of air at NTP is : (A) 1 g/L Match the Column Type : 10. 11.5% (w/v) H2SO4 (s) [Cl¯] = 0.25 g HCl (Ag = 108) (q) 22.4 M KCl + 50 ml H2O (q) [SO42–] = 0.2 L If air is treated as a solution of O2 and N2 then % W/W of oxygen is : (A) 9. Column-I Column-II (A) 100 ml of 0. According to the Avogadro’s law.1 M HCl solution (p) Total concentration of cation(s) = 0.2 7.Comprehension # 3 Read the following comprehension carefully and answer the questions. Now let us assume air to consist of 80% by volume of Nitrogen (N2) and 20% by volume of oxygen (O2).5 M (D) 200 ml 24.12 M (B) 50 ml of 0. X O2 0.2 mol of O2 hence the mole fractions of N2 and O2 are given by XN2 0. If air is taken at STP then its 1 mol would occupy 22. 1 mol of air would contain 0. Volume occupied by air at NTP containing exactly 11.2 M Column I Column II (A) Zn(s) + 2HCl(aq) ZnCl2(s) + H2(g) above reaction is carried out by taking 2 moles each of Zn and HCl (p) 50% of excess reagent left (B) AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(g) above reaction is carried out by taking 170 g AgNO3 and 18. Even if we have a mixture of non-reacting gases then Avogadro’s law is still obeyed by assuming mixture as a new gas. equal number of moles of gases occupy the same volume at identical condition of temperature and pressure.8 mol of N2 and 0.96 L (C) 11. (D) 2KClO3(s) 2KCl(s) + 3O2(g) 2/3 moles of KClO3 decomposed (s) HCl is the limiting reagent .4 L of gas at STP is liberated (C) CaCO3(s) CaO(s) + CO2(g) 100 g CaCO3 is decomposed (r) 1 moles of solid (product) obtained.2 gm of Nitrogen : (A) 22.2 M K2SO4 + 70 ml H2O (r) [SO42–] = 2. 8. (D) 2.06 M (C) 30 ml of 0.4 L (B) 8. C4 H10 has a mass of 58. it is necessary to know the density of the solution. Statement-1 : The average mass of one Mg atom is 24. Statement-2 is True. Statement-2 : Three isotopes. which is not the actual mass of one Mg atom. 15.0 g cm–3 at room temperature. Statement-2 is NOT a correct explanation for Statement-1. (A) Statement-1 is True. (C) Statement-1 is True. Statement-2 : Gram atomic mass of an element contains Avogadro’s number of atoms.022 × 1023 molecules and has a mass of 58. molarity and the mole fraction of solute can be calculated from the weight percentage and the density of the solution 16. Statement-1 : A molecule of butane. Statement-1 : A one molal solution prepared at 20°C will retain the same molality at 100°C. 25Mg and 26Mg. 19. (E) Statement-1 and Statement-2 both are False. 24Mg. provided there is no loss of solute or solvent on heating. Statement-2 : One mole of butane contains 6. Statement-1 : Both 12 g. Statement-2 is False. (B) Statement-1 is True. if the molarity is known. Statement-2 : These units are not defined in terms of any volume.305 amu. (D) Statement-1 is False. Statement-2 : The density of water is about 1. Statement-1 : For calculating the molality or the mole fraction of solute. 17. Statement-1 : Molality and mole fraction concentration units do not change with temperature. 207 Statement-2 : Atomic weights are relative masses.02 × 1023 atoms. Statement-2 is True. of Mg are found in nature. 13. of aluminium will have 6.12 amu. Statement-2 : Molality is independent of temperature. Statement-1 : The ratio of the mass of 100 billion atoms of magnesium to the mass of 100 billion atoms of lead can be expressed as 24 . of carbon and 27 g. . (B). Statement-2 is a correct explanation for Statement-1. (D) and (E) out of which ONLY ONE is correct.Assertion / Reason Type Direction : Each question has 5 choices (A). 18. Statement-2 : Molality. Statement-2 is True.12 g. 12. 14. Statement-1 : The molality and molarity of very dilute aqueous solutions differ very little. (C). 5 (C) 2 (D) 3 . (A) (n/2)g (B) 100g (C) (100/n)g (D) 100ng 3. In an organic compound of molar mass greater than 100 containing only C.5) will be: (A) 12.4. the weight of polyethene formed as per the equation n(C2H4) (CH2CH2)n is. One mole of potassium chlorate (KClO3) is thermally decomposed and excess of aluminium is burnt in the gaseous product. When 100g of ethylene polymerises entirely to polyethene.4 (B) 0.25 (B) 24.6 : 0. 122. if the compound contains 12.2 11. what is a : b.8% sulphur how many nitrogen and sulphur atoms are present per atom of sodium? (A) 2 and 1 (B) 1 and 3 (C) 1 and 2 (D) 3 and 1 9.00 2. The amount of polyethylene possibly obtainable from 64.65 mol O2 to give a mixture of only FeO and Fe2O3.6. the mean molar mass at which all the N2O4 may be presumed to have dissociated : N2O4 tends to the lower value of 39.17 (D) 49.3 (C) 0. The sodium salt of methyl orange has 7% sodium. In the preceding problem. How many mol of aluminium oxide (Al2O3) are formed ? (A) 1 (B) 1.2 : 0. What is the minimum molecular weight of the compound? (A) 420 (B) 375 (C) 329 (D) 295 8.I : MIXED OBJECTIVE Single choice type 1.1 : 0.6 : 0.5 grams of oxygen for every 7. What is the least molar mass? (A) 175 (B) 140 (C) 105 (D) 210 5.0 grams of iron. XeF6 fluorinates 2 to F7 and liberates Xenon(g).3 (D) 0.5 : 0.PART . If it is regarded as a mixture of FeO and Fe2O3 in the weight ratio a : b. 1 mol of iron (Fe) reacts completely with 0.0 kg CaC2 can be (A) 28Kg (B) 14kg (C) 21kg (D) 42 kg 4. One mole of a mixture of N2. On heating to a temperature 2NO2.2 : 0.5 times the percentage of N. A certain mixed oxide of iron contains 2. H and N. The % loss in weight after heating a pure sample of potassium chlorate KClO3 (M.50 (C) 39. Formation of polyethene from calcium carbide takes place as follows : CaC2+H2O Ca(OH)2 + C2H2 C2H4. the percentage of C is 6 times the percentage of H while the sum of the percentages of C and H is 1. What is the mole ratio of N2 : NO2 : N2O4 in the original mixture ? (A) 0.5 : 0.8% nitrogen and 9.1 : 0. wt. (atomic weight of iron = 56) (A) 9 : 10 (B) 9 : 20 (C) 14 : 5 (D) 1 : 1 10. NO2 and N2O4 has a mean molar mass of 55. 210 mmol of XeF6 can yield a maximum of_____ mmol of IF7 (A) 420 (B) 180 (C) 210 (D) 245 7. Mole ratio of ferrous oxide to ferric oxide is : (A) 3 : 2 (B) 4 : 3 (C) 20 : 13 (D) none of these 6. n(C2H4) (CH2CH2)n. 2 (C) 287.50 M? (A) 33 mL (B) 66 mL (C) 133 mL (D) 100 mL . if 107 grams NH4Cl is produced.721 (C) 1.0 mL of 0. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. Then the moles of MgCl2 formed is : (At. The mass of CaCO3 produced when carbon dioxide is passed in excess through 500 ml of 0.3 (D) 25.924 g ml 1. 14.72g X. Cl = 35. If 10g of X4O6 has 5. no solid residue is left behind. What is the least possible molar mass of the compound ? (A) 86 (B) 63 (C) 94 (D) 78 13. If a piece of iron gains 10% of its weight due to partial rusting into Fe2O3 the percentage of total iron that has rusted is : (A) 23 (B) 13 (C) 23.7 litres (B) 1. Hence percentage of ethyl alcohol by weight is (A) 54% (B) 25% (C) 75% (D) 46% 17. Its molar mass is : (A) 176.67 18.12 (D) 2.33. Its density is 1. hydrogen. A certain organic substance used as a solvent in many reactions contains carbon.33 and 2. 4% CaCI2 and 6% NH4Cl will be (A) 0. 25. wt. The percentage contraction in volume is : (A) 8 % (B) 2 % (C) 3 % (D) 4 % 24. A certain compound has the molecular formula X4O6. The mole of dissolved ammonia gas in one litre water bottle is (dwater 1 gm/ml) (A) 5.5 (B) 252.67.40 M Ba(OH)2 must be added to 50. Mole fraction of ethyl alcohol in aqueous ethyl alcohol (C2H5OH) solution is 0.9 litres (C) 1. The molarity of Cl¯ in an aqueous solution which was (w/V) 2% NaCl.8.8 g/cm3.8 ×10–4 mol (B) 1 × 10–2 mol (C) 0. (C) y/x must be less than or equal 1.5 M Ca(OH)2 will be(A) 10 gm (B) 20 gm (C) 50 gm (D) 25 gm. Mg = 24.85 litres 22.5) (A) 3 moles (B) 6 moles (C) 5 moles (D) 10 moles 23. Which of the following statements is correct ? (A) y/x must lie between 1. N = 14.12. oxygen and sulphur. the resulting solution is found to have a density of 0. When x grams of carbon are heated with y grams of oxygen in a closed vessel.58 ×10–2 mol (D) same as w/w 21.342 (B) 0. Molarity of H2SO4 is 18 M.0 ml of ethyl alcohol of density 0.67 (B) y/x must be greater than or equal 2.7.18 16.6 (D) 360. atomic mass of X is : (A) 32 amu (B) 37 amu (C) 42 amu (D) 98 amu 15.5 litres (D) 1. What volume of nitrogen measured at STP could be obtained when the mixture has been passed over red hot copper? (A) 1.25. What approximate volume of 0. hence molality is (A) 18 (B) 100 (C) 36 (D) 500 19.792 g ml 1 with 15 ml of pure water at 40 C. The mixture is now dissolved in HCl to form MgCl2 and NH4Cl. 2 litres of a mixture of nitrous and nitric oxides at STP have a mean molecular weight of 39.1 20. On mixing 15. A mineral water sample was analysed and found to contain 1 × 10–3 % ammonia (w/w). The weight ratio C : O : S = 3 : 2 : 4.33 (D) y/x must be greater than or equal 1. 120 g Mg was burnt in air to give a mixture of MgO and Mg3N2. The mass percentage of carbon in cortisone is 69.30 M NaOH to get a solution in which the molarity of the OH– ions is 0. Weight % of hydrogen in the compound is 7.98%. 1 gm/ml) then : (A) concentration of solution remains same (B) volume of solution become 200 ml (C) mass of H2SO4 in the solution is 98 gm (D) mass of H2SO4 in the solution is 19. When 4 gm of a mixture of NaHCO3 and NaCl is heated. Determine the empirical formula of the compound. how much H2SO4 will 1146 gm of PbS produce? 9.More than one choice type 26. was prepared by heating styrene with to contain 10. Calculate the percentages of CaCO3 and MgCO3 in the sample. Calculate % by mass of NaCl in the original mixture.8 litre. produced after combustion 44. which was then heated and quantitatively converted to 1.6 gm 27. A substance used as a water softener has the following mass percentage composition : 42.9% P.4 g of water.5) (A) Mixture contains 50% NaCl (B) Mixture contains 60% CaCl2 (C) Mass of CaCl2 is 2. A plystyrene. and 39. Show that the results are in accordance with the law of conservation of mass.8 g of CO2 and 5. (Na = 23.T. If the fixed C atoms are all stored after photosynthesis as starch (C6H10O5). 8. A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. how long will it take for the algae to double their own weight? [Mw of (C6H10O5)n = 162]. The left over gas is combusted in the presence of excess oxygen then (NA = 6 × 1023) (Density of water = 1gm/ml) (A) 2 Moles of C2H6 left for combustion (B) Volume of CO2 at S. Ca = 40. Na = 23.P. 18. g of green algae absorbs 6 × 10–3 mol CO2 per hour by photosynthesis. The mixture of oxides produced weight exactly half as much as the original sample. 500 mL of 0. . 0 = 16) 5. 2.2 M NaCl sol. is added to 100 mL of 0.12g of CaO.04% of oxygen. having formula Br3C6H2(C8H8)n. A sample containing only CaCO3 and MgCO3 is ignited to CaO and MgO.22 g (D) Mass of CaCl2 1.44 gm was treated to precipitate all the Ca as CaCO3. find the value of n.* If 100 ml of 1M H2SO4 solution is mixed with 100 ml of 98%(w/w) H2SO4 solution (d = 0.5 AgNO3 solution resulting in the formation of white precipitate of AgCl. Calculate the % of CaCl2 in the mixture.* 3 moles of the gas C2 H6 is mixed with 60 gm of this gas and 2. Determine the percentage composition of the original mixture. wt.12gm.46% bromine by weight. (C) Volume of liquid water produced is 54 ml (D) None PART .* A sample of a mixture of CaCl2 and NaCl weighing 4.II : SUBJECTIVE QUESTIONS 1. Cl = 35.4 × 1024 molecules of the gas is removed. How many moles and how many grams of AgCl are formed? Which is the limiting reagent? 4. 3.66 gm CO2 gas is evolved.07% Na. 6.11 g 28.(At . This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1. 0. P = 31. 2PbS + 3O2 2PbO + 2SO2 3SO2 + 2HNO3 + 2H2O 3H2SO4 + 2NO According to the above sequence of reactions. 3 g of ethane C2H6 on complete combustion gave 8. 7. I : IIT-JEE PROBLEMS (PREVIOUS YEARS) * Marked Questions are having more than one correct option. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23.610 g mL–1)? If the resulting solution has density 1. 1.2 M Al2(SO4)3 is mixed with 20 mL of 0. the mole concept was introduced.108 Which has maximum number of atoms : (A) 24 g of C (12) (B) 56 g of Fe (56) (C) [JEE-2002. and nitrogen contains the three elements in the respective ratio of 9 : 1 : 3 : 5. hydrogen. (B) 1 × 1031 9. Hg = 200 .023 × 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses.6 mL of 0. 1 Faraday = 96500 coulombs).425 g mL–1. A 4. A compound of carbon. and N on combustion gave 1. A large number of atoms/molecules (approximately 6. 13. calculate its molality.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. If the molecular weight of the compound is 108. biochemistry. which requires a clear understanding of the mole concept. What volume of 95% H2SO4 by weight (d = 1.20 M solution from 1600 mL of 0. What would be the molality of a solution obtained by mixing equal volumes of 30% by weight H2SO4 (d = 1.023 × 1054 9.10 g mL–1) 16. what is its molecular formula? 17.023 × 1023 2. 3/144] (D) 108 g Ag (108) Paragraph for Question Nos.6 M BaCl2.3 g of H2O. Calculate the concentrated of each ion in solution. electrochemistry and radiochemistry. The following example illustrates a typical case. What weight of Na2CO3 of 95% purity would be required to neutralise 45. H and N in the organic compound.2050 M solution? 12. To handle such large numbers conveniently.23 N acid? PART . Calculate molality of 1 litre solution of 93% H2SO4 by volume. 20 mL of 0. involving chemical / electrochemical reaction.108 6. 14.218 g mL–1) and 70% by weight H2SO4 (d = 1. (3) to (5) Chemical reactions involve interaction of atoms and molecules. 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. The density of solution is 1.85 g mL–1) and what mass of water must be taken to prepare 100 mL of 15% solution of H2SO4 (d = 1.10. H. 3/150] 6. This concept has implications in diverse areas such as analytical chemistry. How many moles of electron weigh one kilogram : (A) 6.108 (C) 27 g of Al (27) (D) 1 × 108 9. **[At the anode : 2Cl– Cl2 + 2e – At the cathode : Na+ + e– Na Na + Hg NaHg (sodium amalgam)] ** (These reactions were not present in IIT-JEE paper) .1 g of CO2 and 0. What is the percentage of C.84 g mL–1. Calculate the empirical formula. Calculate the ratio by weight of Hg2I2 and HgI2 formed (Hg = 200 .023 [JEE-2003. What volume of water is required to make 0. 18. I = 127) 11. 15. 0.45 g of an organic compound containing only C. In the reaction [AIEEE-2007. 4/162] (D) 3. (2) 22.50M (4) 1. the atomic mass of Fe is : [JEE-2009. What is the molarity of the final mixture ? [AIEEE 2005] (1) 2.125 × 10–2 (3) 1. What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.5 (B) 1.75 (D) 56.001 M (B) 0. 480 ml of 1.6 L H2(g) is produced regardless temperature and pressure for every moles that reacts.[AIEEE 2003] (1) 44.4 lit. 4/162] (A) 200 (B) 225 (C) 400 (D) 446 5.5 by weight.44 mol kg–1 9.02 (2) 3.02 g/ml.25 mole of oxygen atoms ? [AIEEE-2006] (1) 0.14 mol kg–1 (2) 3.64 (4) 1.05 PART . Which of the following concentration factor is affected by change in temperature ? [AIEEE 2002] (1) Molarity (2) Molality (3) Mole fraction (4) Weight fraction 2.II : AIEEE PROBLEMS [JEE-2007. In an organic compound of molar mass 108 g mol–1 C.0 [JEE-2007. [AIEEE 2002] (4) None of these . If we consider that 1/6. 3/120] (1) 1. The total charge (coulombs) required for complete electrolysis is : (A) 24125 (B) 48250 (C) 96500 6. the maximum weight (g) of amalgam formed from this solution is : [JEE-2007. (4) 11.02 M (D) 0.6 lit. 3/160] (A) 55. 3/120] 2AI(s) + 6HCl(aq) 2Al3+(aq) + 6Cl–(aq) + 3H2 (g) (1) 6L HCl(aq) is consumed for every 3L H2 produced. 5.5 × 10–2 10.88 11.85 (B) 55. The total number of moles of chlorine gas evolved is : (A) 0.5M first solution + 250 ml of 1.0 (C) 2.22 (2) 1. Given that the abundances of isotopes 54Fe.3.25 × 10–2 (4) 2.20M 8. 4/162] (D) 193000 (PREVIOUS YEARS) 1.95 (C) 55.2 L H2(g) at STP is produced for every mole of Al that reacts . How many moles of magnesium phosphate.8 lit. respectively. 56Fe and 57Fe are 5%.1 M 6. (3) 67.0 If the cathode is a Hg electrode. (4) 67.28 mol kg–1 (3) 2.60 M sulphuric acid solution that is 29% (H2SO4 molar mass = 98 g mol–1) by mass will be [AIEEE-2007.02 × 1020 molecules of urea are present in 100 ml of its solution. (2) 33.344M (3) 1.6 gm of elemental boron (atomic mass = 10. Molecular formula can be : [AIEEE 2002] (1) C6H8N2 (2) C7H10N (3) C5H6N3 (4) C4H18N3 4.H and N atoms are present in 9 : 1 : 3. The density (in g mL–1) of a 3. mass of carbon atom is taken to be the relative atomic mass unit.05M solution of acetic acid in water is 1. (3) 89.2 L H2(g) at STP is produced for every mole of HCl(aq) consumed. the mass of one mole of a substance will [AIEEE 2005] (1) decrease twice (2) increase two fold (3) remain unchanged (4) be a function of the molecular mass of the substance 7. in place of 1/12.45 (3) 1. 4. 90% and 5%. 6.2M second solution.8) from the reduction of boron trichloride by hydrogen. Mg3(PO4)2 will contain 0. Two solution of a substance (non electrolyte) are mixed in the following manner. Number of atoms in 560g of Fe (atomic mass 56g mol–1) is : (1) Twice that of 70g N (2) Half that of 20g H (3) Both (1) and (2) 3.01 M (C) 0. Density of a 2. The concentration of urea solution is [AIEEE 2004] (A) 0. The molality of the solution is : [AIEEE-2006] (1) 1.70M (2) 1.2 lit.28 mol kg–1 (4) 0. Chlorine is prepared in the laboratory by treating manganese dioixde (MnO2) with aqueous hydrochloric acid according to the reaction. (ii) molar mass of the gas.9689 37 Cl 24.0215 8.0 g of manganese dioxide? . A volume of 10. what is its volume needed for making 2.15 0.000 (v) 500.364 (iii) 0. 9. If the density of methanol is 0.9% iron and 30.0 L (measured at STP) of this welding gas is found to weight 11.00 × 103 g dinitrogen reacts with 1. 2.77 34.41 g mL–1 and the mass per cent of nitric acid in it being 69%. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation : N2(g) + H2(g) 2NH3(g) (i) Calculate the mass of ammonia produced if 2.0125 + 0.1% dioxygen by mass.5785 (ii) 5 × 5.9659 4. Calculate the concentration of nitric acid in moles per litre in a sample which has a density.7864 + 0.02856 298. (ii) Will any of the two reactants remain unreacted? (iii) If yes.25 M solution? 5.NCERT QUESTIONS 1.38 g carbon dioxide. Burning a small sample of it in oxygen gives 3. which one and what would be its mass? 7.0025 (ii) 208 (iii) 5005 (iv) 126. Calculate the atomic mass (average) of chlorine using the following data : % Natural Abudance Molar Mass 35 Cl 75. CaCO3(s) + 2 HCl (aq) CaCl2 (aq) + CO2(g) + H2O() What mass of CaCO3 is required to react completely with 25 mL of 0.6 g.0 (vi) 2.112 0.00 × 103 g of dihydrogen.0034 6. How many significant figures are present in the following? (i) 0. 1. and (iii) molecular formula. How many significant figures should be present in the answer of the following calculations ? (i) 0.5 L of its 0.040 (assume the density of water to be one).690 g of water and no other products. 0. A welding fuel gas contains carbon and hydrogen only. Calculate : (i) empirical formula. 3. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction. Determine the empirical formula of an oxide of iron which has 69. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.75 M HCl? 10. 8.23 36. 4 HCl (aq) + MnO2(s) 2H2O () + MnCl(aq) + Cl2(g) How many grams of HCl react with 5.793 kg L–1. (C) E-1. (A) E-4. (B) F-2. (A) 13. (D) F-14. (A) E-2. (A) D-6. (B) p. (ABD) 27. (BC) . (D) 20. (C) 3. (C) D-11. (A) 15. (A) 17. (B) F-8. (B) B-2. (A) D-22. (A) D-12.I A-1. (C) D-17. (A) F-11. (B) D-7. (A) B-9. (D) 13. (A) D-9. (D) 16. (C) 24. (D) 17. (C) 1. (A) p. (C) 18.15* (BCD) D-16. (D) F-16. (AC) 28. (A) F-15. (B) 6.II 4. (C) p. (C) q. (C) C-11. (B) A-4. (A) F-3. (C) D-14*. (D) C-9. (B) 7. (B) Exercise # 2 PART . s . (C) C-6. (B) B-11. (C) F-5. (D) 14. (D) r 11. (B) B-13. (B) 3. (B) D-1. q . (D) F-12. A-8. (D) A-6. (B) B-10. (C) A-9. B-12. (C) PART . (A) F-1. (C) 23. (B) C-8. (A) C-1. r. (C) F-9. s . (C) F-4. (D) D-3. (C) D-13. (A) p. (B) 7. r . (B) C-4. (A) 5. (B) C-10. (CD) B-1. s . (D) 9. (B) D-2. (C) D-5. (D) 25. (C) 16. (B) 9. q. (D) 2. (B) D-20. (B) 8. (A) 26. (C) B-4. (B) E-3. (B) D-4. (A) 22. (A) 4. (B) A-5.I (B) 1. (A) B-5. (A) 15. (B) D-19. (A) 10. (B) D-10. (D) 19. (A) 18. 8. (C) C-5. (C) F-6. (D) C-2. (A) F-13. (A) 5. (B) s . (A) 11.Exercise # 1 PART . (D) A-7. (B) A-3. (BD) D. (C) (C) C-7. (A) F-7. (C) A-10*. (D) F-10. (B) C-3. (A) 19. (D) q 12. (B) 10. (A) 21. (D) A-2. (A) 12. (B) 6. (B) B-8. (D) (B) B-7. (B) 2. (A) D-8. (C) D-21. (B) D-18. (B) B-3. (B) B-6. (A) 14. (B) E-5. (i) 3 8. molar mass 26.93% 18. (B) 5. % of C = 66. (D) 5. 11.I 1.94 g CaCO3 10. 8. 90.00 11. Na3PO4 5. (1) 4. (D) 2. (3) 9.6 g 16. AgNO3 4. [Cl ] 13. (B) 6.6M 40 Exercise # 3 PART . (B) 4. 0. C6 H8 N2 17. (3) 3.40 g HCl (ii) 4 (iv) 3 (v) 4 (vi) 5 (iii) 4 7. (3) PART . (2) .4%. 63 % .42 14.PART . (i) 2. Empirical formula CH. 10. %NaCl = 77.72 × 102 g 8. CaCO3 = 28.66%. 40 mL 12. 37% 8.6% 7.05 mole.4 gm 9.22 15. 0.43 × 103 g (ii) yes (iii) Hydrogen will remain unreacted . (A) 3. 5. molecular formula C2H2 9. 0. (4) Exercise # 4 5. (1) 11. MgCO3 = 71. (2) 10.8% 10. (4) 8. (i) 2 (ii) 3 (iii) 4 6. n = 19 6. 10 3.0 g mol–1. 470. (1) 2. 0. % of N = 25.II 1.II 1.5978 g 24 0. (D) 6.532 : 1.