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Atoms and Molecules, Recap Notes, Laws of chemical combination : Lavoisier, and Joseph L. Proust established the two, laws of chemical combination :, X Law of conservation of mass : It states, that mass can neither be created nor, destroyed in a chemical reaction. In other, words, in any chemical reaction, the total, mass of the reactants is equal to the total, mass of the products., C +, O2 → CO2, Carbon, (12 g), , X, , X, , X, , Oxygen, (32 g), , Carbon dioxide, (44 g), , Law of constant proportions : It is also, known as the law of definite proportions., It states that in a chemical substance, the, elements are always present in definite, proportions by mass.In water, the ratio, of the mass of hydrogen to the mass of, oxygen is always 1 : 8., Dalton’s atomic theory : This theory was, based on the laws of chemical combination., According to this theory, all matter,, whether an element, a compound or a, mixture is composed of small particles, called atoms., The postulates of Dalton’s atomic theory, are stated as follows :, – All matter is made of very tiny particles, called atoms., – Atoms are indivisible particles, which, cannot be created or destroyed in a, chemical reaction., – Atoms of a given element are identical, in mass and chemical properties., – Atoms of different elements have, different, masses, and, chemical, properties., , – Atoms combine in the ratio of small, whole numbers to form compounds., – The relative number and kinds of atoms, are constant in a given compound., Atom :, X The building blocks of all matter are, atoms., X The smallest unit of an element, which, may or may not exist independently, but, always takes part in a chemical reaction,, is called an atom., X The size of an atom is indicated by its, radius which is called atomic radius. It is, measured in ‘nanometres’. One nanometre, is one billionth part of a metre., –9, , , 1 nanometre (nm) = 10 metre (m) or, 9, 1 metre (m) = 10 nanometre (nm), Modern day symbols, different elements :, X, , of, , atoms, , of, , IUPAC (International Union of Pure and, Applied Chemistry) approves names of, the different elements., – Many of the symbols are the first one, or two letters of the element’s name in, English., – The first letter of a symbol is always, written as a capital letter and the, second letter as a small letter., – Symbols of some elements are formed, from the first letter of the name and a, letter, appearing later in the name., – Other symbols have been taken from, the names of elements in Latin,, German or Greek.

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X, , X, , X, , X, , X, , X, , X, , X, , Atomic mass : Atoms are extremely, small particles. The actual masses of the, atoms are so small that it is difficult to, determine the actual masses of individual, atoms. It was found convenient to compare, the masses of atoms of different elements, with some reference atom. The masses, thus obtained are called relative atomic, masses. It has no units., IUPAC recommended the use of an, isotope of carbon with mass number 12, as the standard reference for measuring, atomic masses. It is called carbon–twelve, (C–12) and is represented as 12C., Atomic mass is defined as the number of, times one atom of an element is heavier, 1, than, th of the mass of an atom of, 12, carbon –12 isotope., Atomic mass =, Mass of one atom of an element, 1, mass of one atom of carbon − 12, 12th, The atomic mass of an element is now, defined as the average relative mass of, its atoms as compared with the mass of, an atom of carbon–12 isotope taken as, 12 units., Atomic mass of an element is expressed, in amu or u or it is written without units, because it is the relative mass of the atom, of that element., Atomic mass unit (amu) may be defined as, 1, th of the mass of an atom of carbon–12, 12, isotope on the atomic scale., 1, th of mass of C–12 isotope, 1 amu =, 12, Atomic mass expressed in grams is called, gram atomic mass of that element. It is the, quantity in grams which is numerically, equal to the atomic mass of an element on, a.m.u. scale., The amount of an element having mass, equal to gram atomic mass is called one, gram atom (or g atom) of the element., , Existence of atoms :, X Atoms of most elements are not able to, exist independently., X Atoms form molecules and ions. These, , molecules or ions aggregate in large, numbers to form the matter that we can, see, feel or touch., Gold atom, , , Molecules :, X A molecule is a group of two or more atoms, that are chemically bonded together, that, is, tightly held together by attractive, forces., X A, molecule can be, O, O, O, defined as the smallest, An, ozone, molecule, particle of an element, (O3), or a compound that is, capable of an independent existence and, shows all the properties of that substance., X Atoms of the same element or of different, elements can join together to form molecule., X Molecules of elements : The molecules, of an element are constituted by the same, type of atoms., X Metals and some, H, O, H, O, other, elements,, such as carbon, do Hydrogen, Oxygen, molecule, molecule, not have a simple, structure but consist of a very large and, indefinite number of atoms bonded together., X The number of atoms constituting a, molecule is known as its atomicity. On, the basis of their atomicities, elements, are classified as monatomic, diatomic,, triatomic, tetratomic, etc., Monatomic, element, , Diatomic, molecule, , Tetratomic, molecule, , Krypton (Kr) Nitrogen (N2), X, , Phosphorus (P4), , Molecules of compounds : Atoms, of different elements join together in, definite proportions to form molecules of, compounds., H, O, H, H, Water, molecule, , H, C, H, H, Methane, molecule

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Ions :, X Compounds composed of metals and, non-metals contain charged species., The charged species are known as ions., Loss of, electron(s), Cation, (+), X, , X, , Neutral, atom, , [Relative atomic mass : C = 12, O = 16], Element, , Mass, ratio, , C, , 3, , 3 1, =, 12 4, , 1, ×4=1, 4, , O, , 8, , 8 1, =, 16 2, , 1, ×4=2, 2, , Gain of, electron(s), Anion, (–), , An ion is a charged particle and can be, negatively or positively charged., – A negatively charged ion is called an, anion and the positively charged ion is, called a cation., – Ions may consist of a single charged, atom or a group of atoms that have a, net charge on them., – A group of atoms carrying a charge is, known as a polyatomic ion., +, A radical is an atom, H, or a group of atoms, carrying positive, H, H, N, or negative charge, that behaves as, H, a single unit in a, Ammonium ion, chemical reaction., +, For example, NH4, , Writing chemical formulae :, The chemical formula of a compound is a, symbolic representation of its composition., X The chemical formulae of a compound can, be written by the following methods :, – Mass ratio method, – Valency method, X Writing formula using mass ratio, method : The ratio of number of atoms in a, molecule of a compound can be calculated,, if the ratio of their masses and the atomic, masses of the elements, constituting the, molecule of the compound are known., – Collect the mass ratio of each element., – Find the atomic ratio of each element., Atomic ratio of an element =, Mass ratio of the element, , Relative atomic mass of the elemen, nt, – Change the atomic ratio into whole, number to get a simple ratio., For example, if a sample of carbon, dioxide contains carbon and oxygen in, the mass ratio of 3 : 8, then, , Atomic, ratio, , Simple, ratio, , The formula of carbon dioxide = CO2, X, , X, , X, , X, , Formulae of simple compounds : In, case of simple molecular Symbol H, O, compounds, which are, Valency 1, 2, made up of two different, Formula : H2O, elements (also called, binary compounds), the symbols of two, elements are written side by side and, their respective valencies are written, below their symbols., In case of simple ionic compounds, made, up of monatomic ions, the symbol of the, O, metal atom (forming the Symbol Al, cation) is written first Charge 3+, 2−, followed by the symbol Formula : Al O, 2 3, of the non-metal atom, (forming the anion) and their respective, valencies are written below their symbols., In case of ionic Symbol Al, (SO4), compounds containing, 2−, polyatomic ions, the Charge 3+, formula of the polyatomic Formula : Al2(SO4)3, ion is written in brackets and the valencies, are written below., Then we must cross-over the valencies of, the combining atoms or the charges., , Molecular mass and mole concept :, X Molecular mass : Molecular mass of a, substance is the average relative mass, of its molecules as compared with that, of an atom of C–12 isotope taken as 12., In other words, molecular mass of a, substance represents the number of times, the molecule of that substance is heavier, than 1/12th of the mass of an atom of C–12, isotope., Molecular mass =, Mass of one molecule of a substance, 1, mass of one atom of carbon − 12, 12th

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X, , X, , X, , X, , X, , X, , X, , The molecular mass of a substance is, the sum of the atomic masses of all the, atoms in a molecule of the substance. It is, therefore the relative mass of a molecule, expressed in atomic mass units (u)., Like atomic mass, molecular mass of a, substance (element or compound) is also, expressed in amu or u or it is written, without units because it is the relative, mass of one molecule of the substance., Formula unit mass : The term formula, unit is used for those substances whose, constituent particles are ions., The formula unit mass of a substance is a, sum of the atomic masses of all atoms in a, formula unit of a compound., – Formula unit mass is calculated in, the same manner as we calculate the, molecular mass., Mole concept : It is more convenient, to refer to the quantity of a substance in, terms of the number of its molecules or, atoms, rather than its mass. So, a new, unit mole was introduced., – One mole of any species (atoms,, molecules, ions or particles) is that, quantity in number having a mass equal, to its atomic or molecular mass in grams., The number of particles (atoms, molecules, or ions) present in 1 mole of any substance, is fixed, with a value of 6.022 × 1023. This, is an experimentally obtained value. This, number is called the Avogadro constant, or Avogadro number (represented by N0),, named in honour of the Italian scientist,, Amedeo Avogadro., 1 mole (of anything) = 6.022 × 1023 (in number)., Mass of 1 mole of a particular substance, is also fixed., , The mass of 1 mole of a substance is equal, to its relative atomic or molecular mass, in grams. The atomic mass of an element, gives us the mass of one atom of that, element in atomic mass units (u)., 1 mole = 6.022 × 1023 number = Relative, mass in grams, Thus, a mole is the chemist’s counting unit., No. of moles = n, Given mass = m, Molar mass, = M, Given number of particles = N, Avogadro, number of particles = N0, N, m n= N, ;, ; m = M × n; m = M ×, ;, n=, N, N, M, 0, 0, X, , m, × N0 ; N = n × N0 ;, M, X Avogadro’s law states that equal volumes, of all gases under the same conditions of, temperature and pressure contain the, same number of particles., X The volume occupied by one mole of a gas, is called its molar volume., X Mole concept for ionic compounds: A, mole of an ionic compound is that amount, of the substance which has mass equal to, gram formula unit mass, i.e., formula unit, mass of the ionic compound expressed in, grams., X A mole is defined as that amount of the, substance which contains Avogadro’s, number of formula units., 1 mole = Gram formula unit mass, = 6.022 × 1023 formula units, X Percentage composition : Percentage, composition is the percentage by mass of, each element in a compound., % of an element in a compound, Mass of that element, × 100, =, Mass of the compound, N=

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Practice Time, OBJECTIVE TYPE QUESTIONS, , 1. If 12 g of carbon burns completely in 40 g, oxygen in a closed container, the product has, (a) 44 g of carbon dioxide, (b) 8 g of oxygen, (c) both (a) and (b), (d) none of these., 2. Elements belonging to different groups, of the periodic table are given below. If the, element X forms a chloride whose formula is, ‘XCl 2 ’ then element ‘X’ belongs to the group, whose representative element is, (a) Al, (b) Na, (c) Mg, (d) Si, 3. Identify the incorrect statement., (a) The building blocks of all matter are atoms., (b) Atoms are very small. They cannot be seen, by the naked eye., (c) The size of an atom is expressed in metres., (d) An atom of hydrogen has the radius of the, order of 10–10 m., 4. The number of atoms in 0.1 mole of a, triatomic gas is, (a) 6.026 × 1022, (b) 1.806 × 1023, (c) 3.6 × 1023, (d) 1.8 × 1022, 5. Which of the following statements is not, true about an atom?, (a) Atoms are not able to exist independently., (b) Atoms are the basic units from which, molecules and ions are formed., (c) Atoms are always neutral in nature., (d) Atoms aggregate in large numbers to form, the matter that we can see, feel or touch., 6. Match List-I with List-II and mark the, correct option., , List-I, List-II, (P) 0.25 mole oxygen 1. 6.022 × 1023 , molecules, (Q) 18 g water, 2. 1.505 × 1023 , molecules, (R) 46 g Na atom, 3. 6.022 × 1023 atoms, (S) 1 mole C atom, 4. 12.044 × 1023 atoms, (a) P-1, Q-2, R-3, S-4 (b) P-2, Q-1, R-4, S-3, (c) P-4, Q-1, R-3, S-2 (d) P-1, Q-4, R-3, S-2, 7. Which one of the following pair of gases, contains the same number of moles?, (a) 16 g of O2 and 14 g of N2, (b) 8 g of O2 and 22 g of CO2, (c) 28 g of N2 and 22 g of CO2, (d) 32 g of O2 and 32 g of N2, 8. Observe the given experimental set-up in, which an ignition tube containing solution of X,, is dipped in a conical flask containing solution, of Y. Tilt and swirl the flask, so that both the, solutions get mixed., Cork, Thread, Conical flask, Small ignition tube, Solution of X, Solution of Y, , According to the law of conservation of mass,, what could be solutions X and Y?,    X  , Y, (a) Copper sulphate, Sodium carbonate, (b) Barium chloride, Sodium sulphate, (c) Lead nitrate, Sodium chloride, (d) All of these

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9. A c o m p o u n d , P Q 2 h a s t h e f o l l o w i n g, arrangement of electrons :, The elements P and Q are respectively, ××, , Q, , ××, , (a) N, Cl, (c) O, F, , , , ×, , , , P, , , , , ×, , ××, , Q, , ××, , ××, , ××, , (b) Cl, S, (d) Na, F, , 10. In a science project, Aditya has to make a, chart, illustrating various elements and their, atomicity., Aditya decided to show elements of different, atomicity by different shapes., Monoatomic →, Diatomic , , →, , Triatomic , , →, , Tetratomic, , →, , Polyatomic →, Pick the element which is shown incorrectly., (a), , Cl, , (b), , (c), , P, , (d), , Al, , S, , 11. The mass of a molecule of water is, (a) 3 × 10–26 kg, (b) 3 × 10–25 kg, –26, (c) 1.5 × 10, kg, (d) 2.5 × 10–26 kg, 12., (a), (b), (c), (d), , 52 g of He contains, 4 × 6.022 × 1023 atoms, 13 atoms, 13 × 6.022 × 1023 atoms, 4 atoms, , 13. How many grams of H2SO4 are present in, 0.25 mole of H2SO4?, (a) 2.45, (b) 24.5, (c) 0.245, (d) 25.4, 14. Identify the correct statements., (i) In a compound such as water, the ratio of, the mass of hydrogen to the mass of oxygen, is always 8 : 1., (ii) If 9 g of water is decomposed, 1 g of hydrogen, and 8 g of oxygen are always obtained., (iii) In ammonia, nitrogen and hydrogen are, always present in the ratio 3 : 14 by mass., (iv) Many compounds are composed of two or, more elements and each such compound has, the same elements in the same proportions., , (a) (i) and (iii), (c) (ii) and (iv), , (b) (i), (ii) and (iii), (d) All of these, , 15. A sample of pure water, irrespective of its, source contains 11.1% hydrogen and 88.9%, oxygen. The data supports, (a) law of constant proportions, (b) law of conservation of mass, (c) law of reciprocal proportions, (d) law of multiple proportions., 16. How many atoms in total are present in, CoCl3.6H2O?, (a) 17, (b) 22, (c) 8, (d) 18, 17. Match the columns by choosing the correct, option., Column I , Column II, (A) 52 g of He, (i) 2 moles, (B) 8 g of O2, (ii) 1 mole, (C) 2 g of H2, (iii) 0.25 mole, (D) 56 g of N2, (iv) 13 moles, (a) (A)-(i), (B)-(iv), (C)-(iii), (D)-(ii), (b) (A)-(iii), (B)-(ii), (C)-(i), (D)-(iv), (c) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i), (d) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii), 18. Which of the following pair of elements, represents a mole ratio of 1 : 1?, (a) 10 g of calcium and 12 g of magnesium, (b) 12 g of magnesium and 6 g of carbon, (c) 12 g of carbon and 20 g of calcium, (d) 20 g of sodium and 20 g of calcium, 19. Which of the following has the smallest, mass?, (a) 4 g of He, (b) 6.022 × 1023 atoms of He, (c) 1 atom of He, (d) 1 mole atoms of He, 20. Which of the following molecules are, diatomic?, (i) Nitrogen (ii) Neon , (iii) Oxygen, (iv) Sulphur (v) Phosphorus (vi) Ozone, (vii) Fluorine (viii)Hydrogen (ix) Fullerene, (a) (ii), (iv), (v) and (vi), (b) (iv), (v) and (ix), (c) (ii) and (vi), (d) (i), (iii), (vii) and (viii)

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21. If three gases X, Y and Z are arranged in, increasing order of their relative molecular, mass and the mass of each gas is 10 g at S.T.P, state, which gas will contain the least number, of molecules and which will contain the most?, (a) X least and Y maximum, (b) X maximum and Z least, (c) Y maximum and Z least, (d) Y least and Z maximum, 22. Which of the following would weigh the, highest?, (a) 0.2 mole of sucrose (C12H22O11), (b) 2 moles of CO2, (c) 2 moles of CaCO3, (d) 10 moles of H2O, 23. Match the columns by choosing the correct, option., Column I, Column II, (Molecule), (Mass ratio of elements), (A) Water (H : O), (i), 14 : 3, (B) Ammonia (N : H), (ii) 1 : 8, (C) Carbon dioxide (C : O) (iii) 1 : 1, (D) Sulphur dioxide (S : O) (iv) 3 : 8, (a) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii), (b) (A) - (iii), (B) - (ii), (C) - (i), (D) - (iv), (c) (A) - (i), (B) - (iv), (C) - (iii), (D) - (ii), (d) (A) - (iv), (B) - (iii), (C) - (ii), (D) - (i), 24. Arrange the following in the order of, increasing mass. (Atomic mass of O = 16 u,, Cu = 63 u, N = 14 u), I. one atom of oxygen, II. one atom of nitrogen, III. 1 × 10–10 mole of oxygen gas, IV. 1 × 10–10 mole of copper, (a) II < I < III < IV, (b) I < II < III < IV, (c) III < II < IV < I, (d) IV < II < III < I, 25. The total number of electrons present in, 16 g of methane gas is, (a) 96.352 × 1023, (b) 48.176 × 1023, 23, (c) 60.22 × 10, (d) 30.110 × 1023, 26. The formula of chloride of a metal M is MCl3,, then the formula of the phosphate of metal M, will be, (a) MPO4, (b) M2PO4, (c) M3PO4, (d) M2(PO4)3, , 27. The given figure shows the set-up to study, the reaction between gas X and copper (II) oxide:, CuO, Gas X, , N2 (g), heat, Water, Ice, , Water, , Which of the following statements is/are correct?, (Given : Atomic mass of N = 14 u, H = 1 u, O = 16 u,, C = 12 u), I. Gas X is a compound of two elements,, nitrogen and hydrogen., II. The number of atoms present in 0.5 mole of, N2 is 6.023 ×1023., III. 1 mole of H 2O contains 1 mole of oxygen, molecules and 2 moles of hydrogen atoms., (a) I and II only, (b) I and III only, (c) II only, (d) III only, 28. Which of the following has highest number, of molecules?, (a) 8 g of CH4, (b) 4.4 g of CO2, (c) 34.2 g of C12H22O11 (d) 2 g of H2, 29. All samples of carbon dioxide contain carbon, and oxygen in the mass ratio 3 : 8. This is in, agreement with the law of, (a) conservation of mass, (b) constant proportions, (c) multiple proportions, (d) gaseous volumes., 30. Two gaseous samples were analysed. One, contained 1.2 g of carbon and 3.2 g of oxygen. The, other contained 27.3% carbon and 72.7% oxygen., The experimental data are in accordance with, (a) law of conservation of mass, (b) law of definite proportions, (c) law of reciprocal proportions, (d) law of multiple proportions., 31., (a), (b), (c), (d), , Which of the following weighs most?, 2 g atoms of nitrogen, 25 g iron, 2×× 1023 atoms of carbon, 1 mole of SO2, , 32. Which of the following correctly represents, 360 g of water?, (i) 2 moles of H2O, (ii) 20 moles of water

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(iii) 6.022 × 1023 molecules of water, (iv) 1.2044 × 1025 molecules of water, (a) (i), (b) (i) and (iv), (c) (ii) and (iii), (d) (ii) and (iv), 33. Arrange the following in the increasing, order of mass in grams :, (i) One atom of silver, (ii) Two grams atom of nitrogen, (iii) One mole of calcium, (iv) Two grams of sodium, [At. masses : Ag = 108 u, N = 14 u, Ca = 40 u, Na, = 23 u], (a) (i) < (ii) < (iii) < (iv) (b) (iv) < (iii) < (ii) < (i), (c) (i) < (iv) < (ii) < (iii) (d) (iii) < (ii) < (i) < (iv), , Case I : Read the passage given below and, answer the following questions from 37 to 39.,  he knowledge of valencies of various radicals, T, helps us to write the formulae of chemical, compounds. The total positive charge on positive, ions (cations) is equal to the total negative, charge on negative ions (anions) in a molecule., Therefore, in writing the formula of a compound,, the positive and negative ions are adjusted in, such a way that the total number of positive, charges of positive ions (cations) becomes equal, to the total number of negative charges of, negative ions (anions)., There is another simple method for writing the, formulae of ionic compounds. In this method, the, valencies (or positive or negative charges) of the, ions can be ‘crossed over’ to give subscripts. The, purpose of crossing over of charges is to find the, number of ions required to equalise the number, of positive and negative charges., 37. Element X has two valencies 5 and 3 and Y, has valency 2. The elements X and Y are most, likely to be respectively, (a) copper and sulphur, (b) sulphur and iron, (c) phosphorus and fluorine, (d) nitrogen and iron., 38. The formula of the sulphate of an element, X is X2(SO4)3. The formula of nitride of element, X will be, , 34. 3.42 g of sucrose are dissolved in 18 g of, water in a beaker. The number of oxygen atoms, in the solution are, (a) 6.68 × 1023, (b) 6.09 × 1022, 23, (c) 6.022 × 10, (d) 6.022 × 1021, 35. It was found that 0.10 mole of MSO 4, combines with 9.0 g of water to form the hydrate, salt MSO4.nH2O. What is the value of n?, (a) 2, (b) 3, (c) 4, (d) 5, 36. The molecular mass of X is 106. X among the, following is, (a) CaCO3, (b) SO3, (c) Na2CO3, (d) NaCl, , (a) X2N, (c) XN, , (b) XN2, (d) X2N3, , 39. The formula of a compound is X 3 Y . The, valencies of elements X and Y will be respectively, (a) 1 and 3, (b) 3 and 1, (c) 2 and 3, (d) 3 and 2, Case II : Read the passage given below and, answer the following questions from 40 to 43., A mole of an atom is a collection of atoms whose, total mass is the number of grams equal to the, atomic mass. Since equal number of moles of, different elements contain an equal number, of atoms it becomes convenient to express the, amounts of the elements in terms of moles. A, mole represents a definite number of particles, viz, atoms, molecules, ions or electrons. This, definite number is called Avogadro number, or Avogadro constant which is equal to, 6.022 × 1023. Hence a mole represents 6.022 × 1023, particles of the substance. One mole of substance, represents one gram-formula of the substance., One mole of a gas at standard temperature and, pressure occupies 22.4 litres., 40. How many grams of sodium must be taken, to get 1 mole of the element ?, (a) 23 g, (b) 35.5 g, (c) 63.5 g, (d) 46 g

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41. What is the mass in grams of a single atom, of chlorine? (Atomic mass of chlorine = 35.5), (a) 6.54 × 1023 g, (b) 5.9 × 10–23 g, (c) 0.0025 g, (d) 35.5 g, 42. How many number of moles are there in, 5.75 g of sodium ?, (Atomic mass of sodium = 23), (a) 0.25, (b) 0.5, (c) 1, (d) 2.5, 43. What is the mass in grams of 2.42 mol of, zinc? (Atomic mass of Zn = 65.41), (a) 200 g, (b) 25 g, (c) 85 g, (d) 158 g, Case III : Read the passage given below and, answer the following questions from 44 to 46., The molecular mass of a substance is the relative, mass of its molecule as compared with the mass, of a carbon–12 atom taken as 12 units. The, molecular mass of a substance indicates the, number of times one molecule of a substance is, 1, heavier than, of C-12 atom. It is equal to the, 12, sum of atomic masses of all the atoms present, in a molecule. Depending on the number of, atoms of same or different elements present in, the molecule, it can be monoatomic, diatomic,, triatomic, tetratomic or a polyatomic molecule., 44. Which is an example of a polyatomic molecule?, (a) S8, (b) HNO3, (c) C2H5OH, (d) All of these, 45. Total number of atoms in 44 g of CO2 is, (a) 6.02 × 1023, (b) 6.02 × 1024, (c) 1.806 × 1024, (d) 18.06 × 1022, 46. Carbon dioxide, hydrogen sulphide, calcium, chloride and sodium oxide are examples of, (a) triatomic molecules, (b) triatomic and tetratomic molecules, (c) diatomic and triatomic molecules, (d) tetratomic molecules., Case IV : Read the passage given below and, answer the following questions from 47 to 50., According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is, composed of small particles called atoms which, can neither be created nor destroyed during, , a chemical reaction. Dalton’s theory provides, a simple explanation for the laws of chemical, combination. He used his theory to explain, law of conservation of masses, law of constant, proportions and law of multiple proportions,, based on various postulates of the theory. Dalton, was the first scientist to use the symbols for, the elements in a very specific sense. When he, used a symbol for an element he also meant a, definite quantity of that element, that is one, atom of that element., 47. Which postulate of Dalton’s atomic theory, is the result of the law of conservation of mass?, (a) Atoms can neither be created nor destroyed., (b) Each element is composed of extremely, small particles called atoms., (c) All the atoms of a given element are identical., (d) During chemical combination, atoms of, different elements combine in simple ratios., 48. Which postulate of Dalton’s atomic theory, explains law of definite proportions?, (a) Atoms of an element do not change during, a chemical reaction., (b) An element consists of atoms having fixed, mass and the number and kind of atoms in, a given compound is fixed., (c) Different elements have different kind of atoms., (d) Atoms are of various kinds., 49. “If 100 g of calcium carbonate (whether in, the form of marble or chalk) is decomposed,, 56 g of calcium oxide and 44 g of carbon dioxide, are formed.” Which law of chemical combination, is illustrated by this statement?, (a) Law of constant proportions, (b) Law of conservation of mass, (c) Law of multiple proportions, (d) Law of conservation of energy, 50. When 5 g calcium is burnt in 2 g oxygen,, 7 g of calcium oxide is produced. When 5 g of, calcium is burnt in 20 g of oxygen, then also, 7 g of calcium oxide is produced. Which law of, chemical combination is being followed?, (a) Law of conservation of mass, (b) Law of multiple proportions, (c) Law of constant proportions, (d) No law is being followed.

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For question numbers 51-60, a statement of assertion followed by a statement of reason is given. Choose the correct, answer out of the following choices., (a) Both assertion and reason are true, and reason is correct explanation of the assertion., (b) Both assertion and reason are true, but reason is not the correct explanation of the assertion., (c) Assertion is true, but reason is false., (d) Assertion is false, but reason is true., 51. Assertion : Both 44 g CO2 and 16 g CH4, have same number of carbon atoms., Reason : Both contain 1 g atom of carbon which, contains 6.023 × 1023 carbon atoms., 52. Assertion : The molecular mass and formula, unit mass of a substance is the sum of atomic, masses of all the atoms in the molecular formula, or formula unit of a compound respectively., Reason : The only difference between the, molecular mass and formula unit mass is that,, former is for molecular compounds (covalent, compounds) and latter is for ionic compounds., However, their numerical value is the same., 53. Assertion : 1 amu equals to 1.66 × 10–24 g., 1, –24, th of mass, Reason : 1.66 × 10 g equals to, 12, of a C–12 atom., , 56. Assertion : Pure water obtained from, different sources such as river, well, spring,, sea etc. always contains hydrogen and oxygen, combined in the ratio of 1 : 8 by mass., Reason : A chemical compound always contains, same elements combined in same fixed proportion, by mass., 57. Assertion : The balancing of chemical, equations is based on law of conservation of, mass., Reason : Total mass of reactants is equal to, total mass of products., 58. Assertion : Molecular weight of SO 2 is, double to that of O2., Reason : One mole of SO2 contains double the, number of molecules present in one mole of O2., , 54. Assertion : When 10 g of CaCO 3 is, decomposed, 5.6 g of residue is left and 4.4 g of, CO2 escapes., Reason : Law of conservation of mass is followed., , 59. Assertion : One atomic mass unit (amu), is mass of an atom equal to exactly one-twelfth, the mass of a carbon–12 atom., Reason : Carbon–12 isotope was selected as, standard., , 55. Assertion : Law of conservation of mass, holds good for nuclear reactions., Reason : It states that mass can neither be, created nor destroyed in a chemical reaction., , 60. Assertion : Atomic mass of aluminium is 27., Reason : An atom of aluminium is 27 times, heavier than 1/12th of the mass of carbon-12, atom., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1. How many H atoms are in 0.80 moles of, hexane, C6H14?, 2. Formula of the carbonate of a metal M is, M2CO3. Write the formula of its chloride., 3., , Define law of conservation of mass., , 4., , Define the atomicity of a molecule of an element?, , 5., , Calculate formula mass of sodium carbonate., , 6. Calculate the mass of 1.12 moles of sulphur, trioxide molecules., 7. Write the formulae for the following :, (a) Calcium phosphate, (b) Magnesium hydroxide

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8. Give one example of (i) a polyatomic cation,, (ii) a polyatomic anion., , (a) Fe2O3, (b) BaCl2, , 9. What are the valencies of metal and nonmetal in the following compounds?, , 10. Write the chemical formulae of the sulphates, of the following cations : K+, Ba2+, Al3+, , Short Answer Type Questions (SA-I), 11. Why does the atomic mass of an element, not represent the actual mass of its atom?, , 15. C a l c u l a t e t h e n u m b e r o f m o l e s i n, 12.044 ××1023 helium atoms., , 12. 8.4 g of sodium bicarbonate on reaction with, 20 g of acetic acid liberated 4.4 g of carbon, dioxide gas into the atmosphere. What is the, mass of the residue left?, , 16. How many particles are represented by, 0.25 mole of an element?, , 13. How many ions are there in 80 g of, magnesium oxide?, 14. Carbon and oxygen react with each other in, the ratio 3 : 8 by mass. What weight of carbon, should be used to react completely with 40 g of, oxygen?, , 17. What are the postulates of Dalton’s atomic, theory. Give at least four points., 18. (a) Define atomic mass unit., (b) What is meant by saying that “the atomic, mass of oxygen is 16”?, 19. How many moles are present in 11.5 g of, sodium?, 20. A piece of copper weighs 0.635 g. How many, atoms of copper does it contain?, , Short Answer Type Questions (SA-II), 21. An element X shows a variable valency of, 3 and 5. What are the formulae of the oxide, formed by it?, 22., (a), (b), (c), , Convert into mole., 12 g of oxygen gas, 20 g of water, 22 g of carbon dioxide, , 23. What is Avogadro number? How many atoms, of each element are present in 6.3 g of nitric, acid (HNO3)?, 24. 4.9 g of sulphuric acid contains 0.1 g of, hydrogen, 1.6 g of sulphur and rest oxygen., Calculate the mass percentage composition of, all the elements of sulphuric acid., 25. Calculate the formula unit masses of ZnO,, Na2O, K2CO3. Given atomic masses of Zn = 65, u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u., 26. What mass of silver nitrate will react with, 5.85 g of sodium chloride to produce 14.35 g of, silver chloride and 8.5 g of sodium nitrate if the, law of conservation of mass is true?, , 27. Calculate the number of aluminium ions, present in 0.051 g of aluminium oxide., (Hint : The mass of an ion is the same as that, of an atom of the same element. Atomic mass, of Al = 27 u), 28. 1022 atoms of an element ‘X’ are found to, have a mass of 930 mg. Calculate the molar, mass of the element ‘X’., 29. A 0.24 g sample of compound of oxygen and, boron was found by analysis to contain 0.096 g, of boron and 0.144 g of oxygen. Calculate the, percentage composition of the compound by, weight., 30. The molecular formula of ferric sulphate is, Fe2(SO4)3. (Atomic mass : Fe = 56 u, S = 32 u,, O = 16 u), (a) Calculate the molar mass of Fe2(SO4)3., (b) How many moles of each element are there, in 40 g of ferric sulphate?, 31. Calculate the number of atoms of each, element present in 122.5 g of KClO3.

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32. (a) An element shows variable valencies 4, and 6. Write the formulae of its two oxides., (b) An element forms an oxide A2O5., (i) What is the valency of the element A?, (ii) What will be the formula of the chloride of, the element?, , (c) 10 moles of sodium sulphite (Na2SO3)?, , 33. What is the mass of, (a) 1 mole of nitrogen atoms?, (b) 4 moles of aluminium atoms (atomic mass, of aluminium = 27)?, , 35. A flask P contains 0.5 mole of oxygen gas., Another flask Q contains 0.4 mole of ozone gas., Which of the two flasks contains greater number, of oxygen atoms?, , 34. Which amongst the following has more, number of atoms?, (i) 11.5 g of sodium or, (ii) 15 g of calcium [Na = 23; Ca = 40], , Long Answer Type Question (LA), 36. Calculate the number of moles of phosphorus, (P) atoms in 100 g of phosphorus. If phosphorus, is considered to contain P4 molecules, then how, many moles of P4 molecules are there?, 37. Weight of copper oxide obtained by heating, 4.32 g of metallic copper with nitric acid was, 5.40 g. In another experiment, 2.30 g of copper, oxide on reduction yielded 1.84 g of copper. Show, that these findings are in accordance with the, law of constant proportions., 38. A sample of ammonia (NH3) weighs 2.00 g., , OBJECTIVE TYPE QUESTIONS, 1., , (c) : C, , + O2, , 1 mole 1 mole, 12 g, , 32 g, , CO2, 1 mole, 44 g, , Remaining oxygen = (40 – 32)g = 8 g, 2. (c) : Mg has valency of +2. So, it will form MgCl 2, while Al, Na and Si will form chlorides AlCl3, NaCl and SiCl4, respectively., 3. (c) : The size of an atom is expressed in nanometres., 1 nm = 10–9 m, 4. (b) : 1 mole of triatomic gas contains, 3 × 6.022 × 1023 atoms, 0.1 mole of triatomic gas contains, 3 × 6.022 × 1023 × 0.1 = 1.806 × 1023 atoms, 5. (a) : Atoms of inert gases exist in monoatomic or, independent form., 6. (b), , What mass of sulphur dioxide (SO2) contains the, same number of molecules as are in 2.00 g of, ammonia?, 39. 25.4 g of iodine and 14.2 g of chlorine are, made to react completely to yield a mixture of, ICl and ICl3. Calculate the ratio of the moles of, ICl and ICl3., 40. (i) What is the mass of one atom of, hydrogen? (atomic mass of hydrogen = 1 u), +, (ii) How many NH4 ions are present in 1.5 moles, of (NH4)3PO4?, , 16 1, =, (a) : 16 g O2 has number of moles =, 32 2, 14 1, 14 g N2 has number of moles = =, 28 2, 7., , No. of moles are same, so no. of molecules are also same., 8., , (d), , 9. (c) : K, L, M, Valency, N (at. no. 7): 2, 5, – 3, Cl (at. no. 17) : 2, 8, 7 1, Thus, N and Cl will form compound, NCl3., K, L, M, Valency, Cl (at. no. 17) : 2, 8, 7 1, S (at. no. 16) : 2, 8, 6 2, Thus, S and Cl will form compound, SCl2., K, L, M, Valency, O (at. no. 8) : 2, 6, – 2, F (at. no. 9) : 2, 7, – 1, Thus, O and F will form compound, OF2.

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K, L, M, Valency, Na (at. no. = 11): 2, 8, 1 1, F (at. no. = 9) : 2, 7, – 1, Thus, Na and F will form compound, NaF., 10. (c) : Phosphorus (P) is a tetratomic element (written as, P4). Hence, it should indicated as  P ., 11. (a) : 6.022 × 1023 molecules have mass = 18 g, 18, 1 molecule of water has mass =, 6.022 × 1023, –23, –26, = 3 × 10 g = 3 × 10 kg, 12. (c) : No. of atoms = No. of moles × Avogadro’s, number, 52, = × 6.022 × 1023 atoms, 4, = 13 × 6.022 × 1023 atoms, 13. (b) : Mass = Number of moles × Molar mass, , = 0.25 × 98 = 24.5 g, 14. (c) : In a compound such as water, the ratio of the, mass of hydrogen to the mass of oxygen is always 1 : 8. In, ammonia, nitrogen and hydrogen are always present in the, ratio 14 : 3 by mass., 15. (a) : Water obtained from any source contains hydrogen, and oxygen in the same proportion by mass. Hence, the data, supports the law of constant proportions., 16. (b) : Atom of Co present = 1, Atoms of H present = 12, Atoms of O present = 6, Atoms of Cl present = 3, Total no. of atoms = 22, , 17. (c) : (A) 4 g of He = 1 mole, 1, \ 52 g of He = × 52 = 13 moles, 4, (B) 32 g of O2 = 1 mole, 1, × 8 = 0.25 mole, \ 8 g of O2 =, 32, (C) 2 g H2 = 1 mole, (D) 28 g of N2 = 1 mole, 1, \ 56 g of N2 =, × 56 = 2 moles, 28, 18. (b) : 1 mole of Ca = 40 g, 1 mole of Na = 23 g, 1 mole of Mg = 24 g, 1, (\ 12 g Mg =, moles), 2, 1 mole of C = 12 g, 1, (\ 6 g C =, moles), 2, , Hence, mole ratio of Mg and C = 1 : 1, 19. (c) : 4 g of He = 6.022 × 1023 atoms of He, , = 1 mole atoms of He, 4, = 0.664 × 10 −23 g, \ Mass of 1 He atom =, 6.022 × 1023, , = 6.64 × 10–24 g, \ 1 atom of He has the smallest mass., 20. (d) : (i) Nitrogen – N2 (ii) Neon – Ne, (iii) Oxygen – O2   (iv) Sulphur – S8, (v) Phosphorus – P4   (vi) Ozone – O3, (vii) Fluorine – F2   (viii) Hydrogen – H2, (ix) Fullerene – C60, 21. (b) : Suppose molecular mass of X, Y and Z are, M, M + 1 and M + 2 respectively., Number of moles in X, Y, Z are, 10 10, 10, ,, ,, M M +1 M + 2, 10, 10, 10, >, >, M M +1 M + 2, Number of molecules = Number of moles × NA, X has the maximum number of molecules and Z has the least, number of molecules., 22. (c) : Weight = Number of moles × Molar mass, 0.2 mole of sucrose (C12H22O11) = 0.2 × 342 = 68.4 g, 2 moles of CO2 = 2 × 44 = 88 g, 2 moles of CaCO3 = 2 × 100 = 200 g, 10 moles of H2O = 10 × 18 = 180 g, 23. (a) : Molecule Ratio by mass of elements, H2O 2 : 16 = 1 : 8, NH3 14 : 3 = 14 : 3, CO2 12 : 32 = 3 : 8, SO2 32 : 32 = 1 : 1, 24. (a) : (I) Mass of one atom of oxygen =, = 2.65 × 10, , –23, , g, , (II) Mass of one atom of nitrogen =, = 2.32 × 10, , –23, , 14, 6.023 × 1023, , g, , (III) Mass of 1 × 10, , –10, , mole of oxygen gas, , –10, , –9, , –10, , –9, , 16, 6.023 × 1023, , = 32 × 1 × 10 g = 3.2 × 10 g, –10, (IV) Mass of 1 × 10 mole of copper, = 63 × 1 × 10 g = 6.3 × 10 g, So, the correct order is II < I < III < IV.

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25. (c) : Total no. of electrons in 16 g i.e., 1 mole, of CH4 = 10 × 6.022 × 1023 electrons, = 60.22 × 1023 electrons, 26. (a) : Valency of M is 3., , Symbol M, PO4, , Valency 3, 3, Dividing by common factor, 1, 1, \ The formula of phosphate of metal M will be MPO4., 27. (a) : Gas X combines with CuO to give N2 and H2O. Thus,, it contains nitrogen and hydrogen. 1 mole of H2O contains 1, mole of oxygen atoms and 2 moles of hydrogen atoms., 28. (d) : Number of particles in 8 g of CH4, , =, , 6.022 × 1023, × 8 = 3.011 × 1023, 16, , No. of particles in 4.4 g of CO2, =, , 6.022 × 1023, × 4.4 = 0.6022 × 1023, 44, , No. of particles in 34.2 g of C12H22O11, =, , 6.022 × 1023, × 34.2 = 0.6022 × 1023, 342, , No. of particles in 2 g of H2, =, , 6.022 × 1023, × 2 = 6.022 × 1023, 2, , 29. (b) : Law of constant proportions states that a chemical, compound is always made up of the same elements combined, together in the same fixed proportion by mass., 30. (b) : Percentage of carbon in sample 1, 1.2, =, × 100 = 27.3%, 1.2 + 3.2, Percentage of oxygen in sample 1, 3.2, =, × 100 = 72.7%, 1.2 + 3.2, The two samples have same percentage composition which, is in accordance with the law of definite proportions., , (ii) 20 moles of water = 18 × 20 = 360 g, (iii) 6.022 × 1023 molecules of water = 1 mole = 18 g, (iv) 1 mole of water = 6.022 × 1023 molecules, 6.022 × 1023 molecules = 1 mole, 1, × 1.2044 × 1025, 1.2044 × 1025 molecules =, 23, 6.022 × 10, = 20 moles, = 20 × 18 = 360 g, 33. (c) : (i) 6.022 × 1023 atoms of silver weigh = 108 g, 108, \ 1 atom of silver weighs =, = 1.79 × 10 −22 g, 6.022 × 1023, (ii) 1 g atom of N = 14 g, 2 g atom of N = 2 × 14 = 28 g, (iii) Mass of 1 mole of calcium = 40 g, (iv) Mass of sodium = 2 g, Hence, increasing order of mass in gram is :, (i) < (iv) < (ii) < (iii)., 3.42, = 0.01 mol, 342, 1 mol of sucrose (C12H22O11) = 11 × NA` atoms of O, 34. (a) : Number of moles of sucrose =, , 0.01 mol of C12H22O11 = 0.01 × 11 × NA, , = 0.11 × NA atoms of O, 18, Number of moles of water =, = 1 mol, 18, 1 mol of water contains = 1 × NA atoms of O, Total number of oxygen atoms = 0.11 NA + 1.0 NA, , = 1.11 NA, Number of oxygen atoms in solution = 1.11 NA, = 1.11 × 6.022 × 1023 = 6.68 × 1023, 35. (d) : 18 g of H2O = 1 mole, \ 9 g of H2O = 0.5 mole, 0.1 mole of MSO4 combines with 0.5 mole of H2O, \ 1 mole of MSO4 will combine with, 0.5, =, × 1 mole of H2O = 5 moles of H2O, 0.1, , (d) : (a) 2 gram atom of nitrogen = 2 × 14 = 28 g, 25 g of iron, 6.022 × 1023 atoms of carbon weigh = 12 g, 2 × 1023 atoms of carbon will weigh, 12, =, × 2 × 1023 ≈ 4 g, 6.022 × 1023, , 36. (c) : Mass of CaCO3 = 40 + 12 + 3 × 16 = 100 g/mol, Mass of SO3 = 32 + 16 × 3 = 80 g/mol, Mass of Na2CO3 = 2 × 23 + 12 + 16 × 3 = 46 + 12 + 48, = 106 g/mol, Mass of NaCl = 23 + 35.5 = 58.5 g/mol, , (d) 1 mole of SO2 = 64 g, , 38. (c) : According to the given formula of sulphate, X is, trivalent and nitride, N3– is also trivalent. So formula of nitride, of X will be XN., , 31., (b), (c), \, , 32. (d) : (i) 1 mole of water = 18 g, 2 moles of water = 2 × 18 g = 36 g, , 37. (d) : N has valencies 3 and 5; Fe has valency 2.

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39. (a) : According to the given formula, valency of X is 1, and valency of Y is 3., , 1 g atom of C = 12 g of C, 12 g of C contains 6.023 × 1023 carbon atoms., , 40. (a) : 1 mole of atom = gram atomic mass of atom, \ 1 mole of sodium = 23 g of sodium, , 52. (b), , 41. (b) : Mass of 6.023 × 10, , 23, , atoms of chlorine = 35.5 g, 35.5, Mass of 1 atom of chlorine =, 6.023 × 1023, = 5.9 × 10, , –23, , g, , Given mass in grams 5.75, =, = 0.25, 42. (a) : No. of moles =, 23, Molar mass, 43. (d) : Mass in grams = No. of moles × molar mass, = 2.42 × 65.41 = 158.2 g ≈ 158 g, , 53. (b) :, , 1, th of the mass of 1 atom of C–12 is, 12, , 1.66 × 10–24 g and atomic mass unit (1 amu) = 1.66 × 10–24 g., 54. (a), 55. (d) : Law of conservation of mass does not hold good, for nuclear reactions due to mass defect., 56. (a), , 44. (d) : All of these are polyatomic molecules., , 57. (a) : According to law of conservation of mass, in a, chemical reaction total mass of the products is equal to the, total mass of the reactants., , 45. (c) : Total no. of atoms in 44 g i.e., 1 mole of CO2, = 3 × 6.022 × 1023 = 1.806 × 1024, , 58. (c) : Both, 1 mole of SO2 and 1 mole of O2 contain same, number of molecules i.e., Avogadro’s number of molecules., , 46. (a) : Carbon dioxide (CO2), hydrogen sulphide (H2S),, calcium chloride (CaCl2) and sodium oxide (Na2O) are the, examples of triatomic molecules because all these contains, 3 atoms in their molecules., , 59. (a) : For universally accepted atomic mass unit in 1961,, C–12 was selected as standard. However,the new symbol, used is ‘u’(unified mass) in place of amu., , 47. (a) : Since an atom cannot be created or destroyed,, the number of various types of atoms in the products of a, chemical reaction is the same as the number of all the atoms, in the reactants., 48. (b) : According to Dalton’s atomic theory every element, has atoms having same mass and that, atoms of different, elements combine to form compounds and the number and, kind of atoms of each element in a compound is fixed. Since, number of atoms, kind of atoms and the mass of atoms of, each element in a given compound are fixed, a compound, will always have the same elements combined in the same, proportion by mass., 49. (b) : According to law of conservation of mass, total, mass of reactants = total mass of products. Total mass of, CaCO3 = 100 g (mass of reactant)., Total mass of products = mass of CaO + mass of CO2, , = 56 g + 44 g = 100 g, \ Total mass of reactants = Total mass of products, 50. (c) : A pure chemical compound always contains the, same elements combined together in the same proportion, by mass., 51. (a) : 44 g of CO2 = 1 mole = 1 g atom of C, 16 g of CH4 = 1 mole = 1 g atom of C, , 60. (a), , SUBJECTIVE TYPE QUESTIONS, 1. Number of hexane molecules, = Number of moles × 6.022 × 1023, = 0.8 × 6.022 × 1023 = 4.81 × 1023, One hexane molecule contains 14 H atoms, Hence, total number of H atoms in 0.8 moles, = 4.8 × 1023 × 14, = 67.2 × 1023 = 6.72 × 1024 atoms, 2., , In MCl valency of M is 1., , 3. Law of conservation of mass states that mass can neither, be created nor be destroyed in a chemical reaction., 4. The number of atoms present in one molecule of an, element is called its atomicity., 5. The formula of sodium carbonate is Na2CO3., \ Formula mass of sodium carbonate, = 23 × 2 + 12 + 3 × 16 = 46 + 12 + 48 = 106, 6. Relative molecular mass of SO3 = 1(32) + 3(16) = 80, Molar mass of SO3 = 80 g/mol, Mass of 1.12 moles of SO3, = Number of moles × Molar mass, = 1.12 mol × 80 g/mol = 89.6 g

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7., , Symbol, Charge, Formula:, 8., , (b) Magnesium hydroxide, , (a) Calcium phosphate, , +, , PO4, Ca, 2+, 3, Ca3(PO4)2, , OH, Symbol Mg, Charge 2+, 1, Formula: Mg(OH)2, , 2–, , (i) NH4 , (ii) SO4, , 9. , (a) Fe2O3, (b) BaCl2, , Metal, Non-metal, Fe(3) O(2), Ba(2) Cl(1), , 10. K2SO4, BaSO4, Al2(SO4)3, 11. Atomic mass of an element is the mass of its atom on, the atomic scale, on the other hand, the actual mass of an, atom is obtained by dividing the atomic mass by Avogadro’s, number., 12. Total mass of reactants = 8.4 g + 20 g = 28.4 g, Total mass of reactants = Total mass of products, 28.4 = 4.4 + x (mass of residue left), 28.4 – 4.4 = x ⇒ x = 24 g, 13. Number of moles of magnesium oxide, Mass of magnesium oxide, 80, =, =, =2, Molar mass of magnesium oxide, 40, MgO, Mg2+ + O2–, From the equation, we can see that 1 mol of MgO contains, 1 mol of Mg2+ ions and 1 mol of O2– ions. 2.0 mol of MgO, will contain 2.0 mol of Mg2+ ions and 2.0 mol of O2– ions., Hence, 2.0 mol of MgO contains 4.0 mol of ions., Number of ions = 4 × 6.022 × 1023 = 2.4088 × 1024, 14. Ratio by mass C : O, , 3, 8, , ?, 40, Mass of carbon required =, , 3 × 40, = 15 g, 8, , 15. Given : N = 12.044 × 1023 (of Helium), N0 = 6.022 × 1023, n = ?, We know that,, N, No.of particles(N ), No. of moles (n) =, or n =, N0, Avogadro’s number (N 0 ), \, , n=, , 12.044 × 1023, =2, 6.022 × 1023, , Thus, number of moles in 12.044 × 1023 helium atoms is 2., 16. 1 mole of an element contains 6.022 × 1023 particles, 0.25 mole of an element contains = 6.022 × 1023 × 0.25, , = 1.5 × 1023 particles, , 17. The postulates of Dalton’s atomic theory may be stated, as follows :, (i) All matter is made of very tiny particles called atoms., (ii) Atoms are indivisible particles, which cannot be created, or destroyed in a chemical reaction., (iii) Atoms of a given element are identical in mass and, chemical properties., (iv) Atoms of different elements have different masses and, chemical properties., (v) Atoms combine in the ratio of simple whole number to, form compounds., (vi) The relative number and kinds of atoms are constant in, a given compound., 1, 18. (a) The atomic mass unit is defined as, th of the mass, 12, of one atom of a carbon (C–12) isotope., (b) It means that one atom of oxygen is 16 times heavier, 1, than, th of the mass of a 12C atom., 12, Given mass (in g), Molar mass, 11.5, =, = 0.5, 23, 20. Gram atomic mass of copper = 63.5 g, 0.635, Number of moles in 0.635 g of copper =, = 0.01 moles, 63.5, Number of copper atoms in one mole = 6.022 × 1023, Number of copper atoms in 0.01 moles, , = 0.01 × 6.022 × 1023 = 6.022 × 1021, 19. No. of moles in 11.5 g of sodium =, , 21. (i) Formula of oxide when X has valency = 3, Symbol, , X, , O, , Formula = X2O3, Valency 3, 2, (ii) Formula of the oxide when X has valency = 5, Symbol, , X, , O, , Valency, , 5, , 2, , Formula = X2O5, , 22. (a) Molar mass of O2 = 2 × 16 = 32 g, \ Number of moles in 12 g of O2, Given mass 12, = = 0.375 mole, =, Molar mass 32, (b) Molar mass of H2O = 2 × 1 + 16 = 18 g, 20, = 1.11 moles, \ Number of moles in 20 g of H2O =, 18, (c) Molar mass of CO2 = 12 + 16 × 2 = 12 + 32 = 44 g, 22, = 0.5 mole, \ Number of moles in 22 g of CO2 =, 44

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23. The number of particles (atoms, molecules or ions), present in 1 mole of any substance is fixed and is called, Avogadro number (NA). Its value is 6.022 × 1023., Molar mass of nitric acid (HNO3) = 1 + 14 + 3 × 16, , = 63 g mol–1, 63 g of HNO3 contains = 1 × 6.022 ×1023 atoms of H, , Mass of AgNO3 = (14.35 + 8.5) – 5.85 g, , = 22.85 – 5.85 = 17.0 g., , ∴, , \ 0.051 g of Al2O3 contains =, , 1 × 6.022 × 1023, 6.3 g of HNO3 contains =, × 6.3, 63, , , , = 6.022 × 1022 atoms of H, , 63 g of HNO3 contains = 1 × 6022 × 1023 atoms of N, , 27. Molar mass of Al2O3 = 2 × 27 + 3 × 16, , = 54 + 48 = 102 g, 102 g of Al2O3 contains = 2 × 6.022 × 1023 Al3+ ions, , , , 2 × 6.022 × 1023, × 0.051, 102, , = 6.022 × 1020 Al3+ ions, , , = 6.022 × 1022 atoms of N, 63 g of HNO3 contains = 3 × 6.022 × 1023 atoms of O, , 28. Molar mass of an element is the mass of Avogadro’s, number of atoms., 1022 atoms of the element have mass, 930, g = 0.930 g, = 930 mg =, 1000, , ∴, , 3 × 6.022 × 1023 × 6.3, 63, = 1.8066 × 1023 atoms of O, , \, , 24. Mass of sulphuric acid (W) = 4.9 g, Mass of hydrogen (w1) = 0.1 g, Mass of sulphur (w2) = 1.6 g, w1, 0.1, × 100, ∴ Mass percentage of hydrogen = × 100 =, W, 4.9, = 2.04%, , \, , ∴, , 6.3 g of HNO3 contains =, , 1 × 6.022 × 1023, × 6.3, 63, , 6.3 g of HNO3 contains =, , , , w2, 1.6, × 100 =, × 100 = 32.65%, W, 4.9, Mass percentage of oxygen = 100 – (2.04 + 32.65), = 65.31%, , Mass percentage of sulphur =, ∴, , 25. Formula unit mass of ZnO = atomic mass of Zn, + atomic mass of O = (65 + 16) u = 81 u, Formula unit mass of Na2O = 2 × atomic mass of Na, + atomic mass of O = (2 × 23 + 16) u = 62 u, Formula unit mass of K 2 CO 3 = 2 × atomic mass of, K + atomic mass of C + 3 × atomic mass of O, = (2 × 39 + 12 + 3 × 16) u = (78 + 12 + 48) u = 138 u, 26. The reaction is, Silver nitrate + Sodium chloride → Silver chloride +, , Sodium nitrate, If law of conservation of mass is true,, Total mass of reactants = Total mass of products, i.e., Mass of AgNO3 + Mass of NaCl = Mass of AgCl +, , Mass of NaNO3, We have : Mass of NaCl = 5.85 g, Mass of AgCl = 14.35 g,, Mass of NaNO3 = 8.5 g, Substituting these values in the above equation, we get, Mass of AgNO3 + 5.85 g = 14.35 g + 8.5 g, , 6.022 × 1023 atoms will have mass, , =, , 0.930, × 6.022 × 1023 g = 56.0 g, 1022, , Molar mass of the element = 56 g mol–1, N, Alternatively, n =, N0, m, Also, n =, M, N, 6.022 × 1023 atoms mol−1, m, or M = = m × 0 = 0.930 g ×, N, n, 1022 atoms, = 56.0 g mol–1, 29. We know that, % of any element in a compound , Mass of element, =, ×100, Mass of compound, 0.096, × 100 = 40%, % of boron =, 0.24, 0.144, × 100 = 60%, % of oxygen =, 0.24, 30. (a) The molar mass of Fe2(SO4)3, = 2 × 56 + 3 × 32 + 12 × 16 = 400, Given mass, 40, (b) No. of moles =, =, = 0.1, Molar mass 400, 1 mole of ferric sulphate contains 2 moles of iron, 0.1 mole of ferric sulphate contains = 2 × 0.1, = 0.2 moles of Fe, 1 mole of ferric sulphate contains 3 moles of sulphur, 0.1 mole of ferric sulphate contains = 3 × 0.1, = 0.3 moles of S, 1 mole of ferric sulphate contains 12 moles of oxygen, 0.1 mole of ferric sulphate contains = 12 × 0.1, = 1.2 moles of O

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31. No. of moles of KClO3 =, , 122.5, =1, 122.5, , (mol. wt. of KClO3 = 122.5), From the formula KClO3, we know that 1 mole of KClO 3, contains 1 mole of K atoms, 1 mole of Cl atoms and 3 moles, of O atoms., \ No. of atoms of K = 6.022 × 1023, No. of atoms of Cl = 6.022 × 1023, 23, No. of atoms of O = 3 × 6.022 × 1023 = 18.066 × 10, 32. (a) Let the element be represented by the symbol E., Formula of oxide in which valency of E = 4 are E2O4 or EO2, Formula of oxide in which valency of E = 6 are E2O6 or EO3, (b) Formula of oxide of the element = A2O5, (i) The valency of the element A in the oxide = 5, (ii) The formula of the chloride of the element A = ACl5, 33. (a) Mass of 1 mole nitrogen atoms, = number of moles × atomic mass = 1 × 14 = 14 g, (b) Mass of 4 moles aluminium atoms = 4 × atomic mass, , = 4 × 27 = 108 g, (c) Mass of Na2SO3 = 2 × 23 + 32 + 16 × 3, , = 46 + 32 + 48 = 126 g, 1 mole of Na2SO3 = 126 g, \ 10 moles of Na2SO3 = (10 × 126) g = 1260 g, , In flask Q :, 1 molecule of ozone (O3) = 3 atoms of oxygen, 1 mole of ozone gas = 6.022 × 1023 molecules, 0.4 mole of ozone gas = 6.022 × 1023 × 0.4 molecules, = 6.022 × 1023 × 0.4 × 3 atoms = 7.23 × 1023 atoms, \ Flask Q has a greater number of oxygen atoms as compared, to flask P., 36. Mass of phosphorus = 100 g, Atomic mass of P = 31.0 u, So, molar mass of P atom = 31 g, Therefore,, Number of moles of P atoms in 100 g of phosphorus, 100, =, = 3.22 mol, 31.0, If phosphorus is considered to be present as P4 molecules,, then, No. of moles of P4 molecules in 100 g of phosphorus, No. of molesof P atoms in100 g of phosphorus, 4, 3.22 mol, =, = 0.805 mol, 4, Thus, 100 g of phosphorus contains 3.22 moles, of P atoms and 0.805 moles of P4 molecules., =, , 34. We know that equal number of moles of different, elements contain equal number of atoms. Thus, we shall, convert masses of sodium and calcium to find which has, more number of moles., (i) For sodium, Gram atomic mass of sodium = 23 g, Mass of sodium = 11.5 g, Mass of sodium ingrams, \ No. of moles of sodium =, Gram atomic mass of sodium, 11.5 g, = 0.5 moles, =, 23 g, (ii) For calcium, Gram atomic mass of calcium = 40 g, Mass of calcium = 15 g, \ No. of moles of calcium, Mass of calcium ingrams, 15 g, =, = 0.375 moles, =, Gram atomic mass of calcium, 40 g, Therefore, sodium has more number of atoms than calcium., , 37. Experiment I, Weight of copper = 4.32 g, Weight of copper oxide = 5.40 g, \ Weight of oxygen = Weight of copper oxide –, , Weight of copper, = 5.40 – 4.32 = 1.08 g, , 35. In flask P :, 1 molecule of oxygen (O2) = 2 atoms of oxygen, 1 mole of oxygen gas = 6.022 × 1023 molecules, 0.5 mole of oxygen gas = 6.022 × 1023 × 0.5 molecules, = 6.022 × 1023 × 0.5 × 2 atoms = 6.022 × 1023 atoms, , As the ratio of copper and oxygen (by mass) is same in the, two experiments, law of constant proportions is verified., , Experiment II, Weight of copper = 1.84 g, Weight of copper oxide = 2.30 g, \ Weight of oxygen = Weight of copper oxide –, , Weight of copper, , = 2.30 – 1.84 = 0.46 g, Now, from experiment I, from experiment II, Ratio of copper and oxygen Ratio of copper and oxygen, Mass of copper 4.32 4, =, =, Mass of oxygen 1.08 1, , Mass of copper 1.84 4, =, =, Mass of oxygen 0.46 1, , 38. Mass of ammonia = 2.00 g, Molar mass of ammonia (NH3) = 1×14 +3 × 1, , = 17 g mol–1