1 Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry 1 1 Some Basic Concepts of Chemistry

Matter –– The three states of matter can be inter-converted by ‹‹ Anything which occupies space, has mass and which can be changing the conditions of temperature and pressure as felt by our senses is called matter. follows :

‹‹ Ev Liq Classification of Matter : Co GasGVa ap as uefacti n po orat tion ndensa – tion – Physical Classification : io riza at io on n o tion tion Matter lim ndensa or r Deposi Co Sub or Solid Liquid Solid Liquid Solids Liquids Gases Definite shape Definite No definite Melting or Fusion Freezing or Crystallization and volume volume but no shape and Endothermic state changes Exothermic state changes definite shape volume –– Chemical Classification : Matter

Pure Substances Mixtures Fixed ratio of masses No fixed ratio of of constituents masses of constituents

Elements Compounds Homogeneous Heterogeneous Consists of only one Composed of two Uniform composition Composition is not kind of atoms or more atoms of throughout uniform throughout different elements Solids at room temperature, high density, possess lustre, Metals good conductors of heat and Organic Inorganic electricity Compounds Compounds Originally obtained from Usually obtained from Brittle, do not possess animals and plants, contain minerals and rocks, do not lustre, poor conductors Non-metals carbon and few other elements contain C–H bonds of heat and electricity like H, O, S, N, X (halogens), Possess characteristics etc. having C–H bonds Metalloids of both metals and non- metals

Units and Measurements Mass m kilogram kg ‹‹ Fundamental Units : The units which can neither be derived Time t second s from one another nor can be further resolved into any other Temperature T kelvin K units are called fundamental units or basic units. Electric current I ampere A The SI system has seven basic units : Luminous intensity Iv candela cd Physical quantity Symbol for Name of Symbol quantity unit for unit Amount of substance n mole mol ‹‹ Derived Units : The units for other quantities which can be Length l metre m derived from fundamental units are called derived units. 2 Chemistry – Physical Symbol Unit (S.I.) Symbol – This can be done by method called factor label method or Quantity unit factor method. – Area A Square metre m2 – This is done by using conversion factor (C.F.), which is a factor equal to one that converts a quantity in one unit Volume V Cubic metre m3 to the same quantity in another unit. Some conversion –3 Density r Kilogram per kg m factors are as follows : cubic metre Velocity v Metre per second ms–1 Conversion Factors –8 –10 –1 2 Force F Newton N = kg m s–2 1 angstrom (Å) = 10 cm = 10 m = 10 nm = 10 pm 1 inch = 2.54 cm Pressure P Pascal Pa = N m–2 1 cm = 0.3937 inch Energy E Joule J = N m 1 metre = 39.37 inch = kg m2s–2 1 km = 0.621 mile Frequency u Hertz Hz or s–1 1kg = 2.20 pounds (lb) Electric q Coulomb C = A s 1g = 0.0353 ounce charge –1 1 pound = 453.6 g Potential E° Volt V = J C –24 –27 –1 –1 1 atomic mass unit (amu) = 1.6605 × 10 g = 1.6605 × 10 kg difference = J A s –3 –10 2 –3 –1 ≡ 1.492 × 10 erg = 1.492 × 10 J = kg m s A = 3.564 × 1011 cal = 9.310 × 108 eV ‹‹ Precision : It refers to the closeness of various measurements for the same quantity. = 931.48 MeV 1 atmosphere (atm) = 760 torr = 760 mm Hg = 76 cm Hg ‹‹ Accuracy : It is the agreement of a particular value to the 5 true value of the result. = 1.01325 × 10 Pa 7 ‹‹ Significant figures : Significant figures are those digits in 1 calorie (cal) = 4.18 × 10 erg = 4.18 J a measured number that include all certain digits and one = 2.613 × 1019 eV doubtful digit. 1 erg = 10–7 J = 2.389 × 10–8 cal = 6.242 × 1011 eV 4 ‹‹ Rules for determining the number of significant figures : 1 faraday (F) = 9.6487 × 10 coulomb –– All non-zero digits are significant. 1 dyne (dyne) = 10–5 N –– A zero becomes significant when it comes in between 1 joule = 107 erg = 0.2390 cal two non-zero numbers. 1 litre = 1000 cc = 1000 mL = 1 dm3 –– Zeros at the beginning of a number are not significant. = 10–3 m3 –– All zeros to the right of a number are significant. 1 coulomb (coul) = 2.9979 × 109 esu ‹‹ Scientific Notation : Numbers are represented in the term 1 curie (Ci) = 3.7 × 1010 disintegrations sec–1 N × 10n (where N lies between 1 to 10 and n is the exponent 1 electron volt (eV) = 1.6021 × 10–12 erg = 1.6021 × 10–19 J having positive or negative value). = 3.827 × 10–20 cal ‹ ‹ Dimensional Analysis : The expression of any particular –1 quantity in terms of fundamental quantity is known as = 23.06 kcal mol –10 dimensional analysis. 1 electrostatic unit (esu) = 3.33564 × 10 coul Illustration 1 (ii) 15.15 × 10–12 m or 1.515 × 10–11 m (a) Convert the following into basic units : –6 –6 (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg (iii) 25365 mg = 25365 × 10 kg [ 1 mg = 10 kg] –2 (NCERT Exemplar) = 2.5365 × 10 kg (b) How many significant figures are there in each of the (b) (i) two (ii) three following numbers? (iii) four (iv) two –3 (i) 17 (ii) 103 (iii) 1.035 (iv) 0.0010 (c) (i) 0.0048 = 4.8 × 10 5 (c) Express the following in the scientific notation. (ii) 234200 = 2.342 × 10 3 (i) 0.0048 (ii) 234200 (iii) 8008 (iv) 500.0 (iii) 8008 = 8.008 × 10 2 Soln.: (a) (i) 28.7 × 10–12 m or 2.87 × 10–11 m (iv) 500.0 = 5.000 × 10 Some Basic Concepts of Chemistry 3 Laws of Chemical Combination with the fixed weight of the other, bear a simple whole number ratio to one another. ‹‹ Law of Conservation of Mass : It states that during any physical or chemical change, total mass of products is equal ‹‹ Law of Reciprocal Proportions : It states that when two elements combine separately with a fixed mass of third to the total mass of reactants. element, then the ratio between their masses in which they ‹‹ Law of Definite or Constant Proportions : It states that a combine will be either same or simple multiple of the ratio compound always contains the same elements combined in in which they combine with each other. the same definite proportion by weight. ‹‹ Gay Lussac’s Law of Combining Volumes : It states that ‹‹ Law of Multiple Proportions : It states that when two or under similar conditions of temperature and pressure, more elements combine to form two or more compounds, whenever gases react together, the volumes of the reacting the different weights of one of the elements which combine gases as well as products bear a simple whole number ratio. Illustrations

2 A and B combines with each other to form X, Y, and Z. The 3 One volume of nitrogen combines with three volumes of following reactions take place in the formation of X, Y and Z. hydrogen to form two volumes of ammonia. 0.6 g A + 0.8 g B → 1.4 g X (a) What volumes of nitrogen and hydrogen are required to 9.0 g A + 24.0 g B → 33.0 g Y form 50 L of ammonia? 40.0 g A + 160.0 g B → 200.0 g Z (b) What volume of nitrogen will react completely with 30 L of Which law is illustrated by these data? hydrogen? What volume of ammonia will be formed? Soln.: On the basis of reactions given, 1 g of A reacts with Soln.: (a) N2 + 3H2 → 2NH3 different amounts of B, which are as follows : x L 3x L 2x L B 08. B 24.0 Given, 2x L = 50 L ⇒ x = 25 For X ⇒ = = 1.33 g; For Y ⇒ = = 2.66 g \ required volume of nitrogen = 25 L A 06. A 90. and volume of hydrogen required = 3 × 25 L = 75 L B 160 For Z ⇒ = = 4.0 g (b) Given, A 40 30 Thus, ratio of different amounts ofB , which react with 1 g of A. 3xx L = 30 L ⇒= = 10 = 1.33 : 2.66 : 4.0 = 1 : 2 : 3 3 Since, this is a simple ratio, so the above results illustrates law \ Required volume of nitrogen = x L = 10 L of multiple proportions. and volume of ammonia formed = 2x L = 2 × 10 L = 20 L

Dalton's Atomic Theory ‹‹ It does not explain Gay-Lussac's law of combining gaseous Main postulates of Dalton's theory are as follows : volumes. ‹‹ Matter is made up of small indivisible particles called atoms. ‹‹ It does not give an idea about isotopes and isobars. ‹‹ Atoms can neither be created nor destroyed. ‹‹ It fails to explain why atoms of different elements show ‹‹ Atoms of a given element are identical in properties. different properties like mass, size, etc. ‹‹ Atoms of different elements differ in properties. ‹‹ It does not explain the difference between an atom and a ‹‹ Atoms of different elements combine in a fixed ratio to form molecule. molecule of a compound. ‹‹ Avogadro's Law : It states that equal volume of all gases Limitations of Dalton's Atomic Theory contain equal number of molecules at same temperature and pressure. ‹‹ It fails to explain the cause of chemical combination. Self Test 1

1. Which one of the following represents smallest quantity? B and C is 1 : 3 : 5. If 32 parts by mass of X combines with –4 (a) 1850 ng (b) 1.85 × 10 g 84 parts by mass of Y in B, then in C, 16 parts by mass of X 3 –6 (c) 1.85 × 10 µg (d) 1.85 × 10 kg will combine with 2. Two elements X (at. mass = 16) and Y (at. mass = 14) combine (a) 28 parts by mass of Y (b) 70 parts by mass of Y to form compounds A, B and C. The ratio of different (c) 32 parts by mass of Y (d) 56 parts by mass of Y. masses of Y which combines with a fixed mass of X in A, 4 Chemistry 3. Bar is unit of (a) law of multiple proportions (a) temperature (b) pressure (b) law of gaseous volumes (c) volume (d) density. (c) law of definite proportion (d) none of the above. (.30 52−×02.)(.199105) 4. In the final answer of the expression 10. The number of significant figures in the three numbers 13. 7 the number of significant figures is 252 cm, 0.252 cm and 0.0252 cm are (a) 1 (b) 2 (a) 3, 4 and 5 respectively (b) 3, 4 and 4 respectively (c) 3 (d) 4 (c) 3, 3 and 4 respectively (d) 3, 3 and 3 respectively. 5. Two students X and Y report the weights of the same 11. The average speed (with significant digit) of a ball that substance as 4.0 g and 4.00 g respectively. Which of the covers a distance of 9.856 m in 5.4 s is following statements is correct? (a) 1.8251 m/sec (b) 1.8 m/sec (a) Both are equally accurate. (c) Both of these (d) None of these (b) X is more accurate than Y. 12. When 2 L mixture of CO and CO2 is passed over red hot (c) Y is more accurate than X. charcoal, its volume increases by 1.4 times. If all the volume (d) Both are inaccurate scientifically. measurements are made under similar conditions of P and 6. A sample of 1.375 g cupric oxide, when reduced in a stream T, then the % composition of mixture is of hydrogen gave 1.098 g Cu. On the other hand, a sample (a) 50% CO, 50% CO2 (b) 40% CO, 60% CO2 of 1.179 g pure Cu gave 1.476 g cupric oxide when Cu was (c) 60% CO, 40% CO2 (d) 10% CO, 90% CO2 in HNO and nitrate formed was heated strongly to get 3 13. 2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide, 1.42 g cupric oxide. These data prove of the same metal heated in steam gave 2.27 g of its oxide. (a) law of conservation of mass Which law is shown by this data? (b) law of definite proportions (a) Law of constant composition (c) law of multiple proportions (b) Law of multiple proportions (d) none of the above. (c) Law of reciprocal proportions 7. 5.8 g of sodium chloride was allowed to react with 9.8 g (d) None of these of sulphuric acid. 12.0 g of sodium hydrogen sulphate and 14. Which one of the following represents Avogadro’s hypothesis? gaseous hydrogen chloride were formed. Calculate the (a) Equal volumes of all gases under same conditions of mass of the gas formed. temperature and pressure contain equal number of (a) 36 g (b) 3.6 g atoms. (c) 3.4 g (d) 6.2 g (b) Equal volumes of all gases under same conditions of 8. Which is not a basic postulate of Dalton's atomic theory? temperature and pressure contain equal number of (a) Each element is composed of extremely small particles molecules. called atoms. (c) Gases react together in volumes which bear a simple (b) Atoms are neither created nor destroyed in a chemical ratio to one another. reaction. (d) The rates of diffusion of gases are inversely (c) Atoms of an element may be different due to presence proportional to the square root of their densities. of isotopes. 15. If vapour density is defined with respect to oxygen in place (d) Different elements have different types of atoms. of hydrogen, then the ratio of the vapour density with 9. The percentage of nitrogen in hydrides, ammonia and respect to oxygen and hydrogen is hydrazine is experimentally found to be 82.35% and 87.5% (a) 1 : 1 (b) 1 : 16 respectively. The data prove the (c) 16 : 1 (d) 1 : 8

Atomic Mass Atomic mass is expressed in amu or u. 1 amu = 1.66 × 10–24g ‹‹ Atomic Mass : It is the average relative mass of atom of 1 1 element as compared with times the mass of an atom of ‹‹ One atomic mass unit (amu) is equal to th of the mass of 12 12 carbon-12 isotope. an atom of carbon-12 isotope. Average mass of an atom ‹‹ Gram Atomic Mass (GAM) : Atomic mass of an element Atomic mass = 1/12 × Mass of an atom of 12C expressed in g is called gram atomic mass or gram atom or mole atom. ‹‹ Average Atomic Mass : If an element exists in two isotopes –– Number of g atoms or mole atoms having atomic masses 'a' and 'b' in the ratio x : y then average Mass of an element in g ()xa×+()yb× = atomic mass is atomic mass unit (a.m.u.) Gram at. mass xy+ Some Basic Concepts of Chemistry 5

–– Mass of an element in g = No. of gram atoms ‹‹ Equivalent Mass : × Gram at. mass –– The number of parts by mass of the substance which –– Number of atoms in 1 gram at. mass = 6.02 × 1023 combine or displace directly or indirectly 1.008 parts by ‹‹ Calculation of Atomic Weight : mass of hydrogen or 8 parts by mass of oxygen or 35.5 –– (Dulong and Petit’s law for solids) parts by mass of chlorine. Atomic mass × specific heat = 6.4 (approx.) ‹‹ Gram Equivalent Mass : –– The quantity of a substance whose mass (in grams) is 6.4 numerically equal to its equivalent mass is called its one ∴ Approx. atomic mass = sp. heat gram equivalent or gram equivalent mass (GEM). Thus, Number of gram equivalent Approx. at. mass Valency = Mass of the substance (in grams) Eq. mass = GEM of the substance Atomic mass = Eq. mass × Valency ‹‹ Calculation of Equivalent Mass : Molecular Mass Atomic mass –– For elements, Equivalent mass = ‹‹ Molecular Mass : Molecular mass of a molecule of an Valency element or a compound may be defined as a number which Molecular mass –– For acids, Equivalent mass = indicates how many times heavier is a molecule of that Basicity of acid 1 Molecular mass element or compound as compared with of the mass –– For bases, Equivalent mass = 12 Acidity of base of an atom of carbon-12. Molecular mass is also expressed –– For salts, Equivalent mass Formula mass in a.m.u. = Mass of one molecule of the substance Total positive or negative charge Molecular mass = 11/ 2 × MMass of one atom of C-12 –– For oxidising agents, Equivalent mass Actual mass of one molecule Formula mass –24 = = Mol. mass (in amu) × 1.66 × 10 g Number of electrons it gains ∑ MX –– For reducing agent, Equivalent mass ‹ ii ‹ Average Molecular Mass = Formula mass ∑ XTotal = Number of electrons it loses where M1, M2, M3 ... are molecular masses of species 1, 2, 3 .... etc. with % as X1, X2, X3. Note: In ionic compounds, the term molecular mass is replaced ‹‹ Gram Molecular Mass (GMM) : Molecular mass of an by Formula mass which is obtained by adding atomic masses of element or compound when expressed in g is called its gram various atoms present in a formula. molecular mass or gram molecule or mole molecule. Formula Unit Mass Mass of substance in g Number of gram molecules = ‹ Gram mol. mass ‹ It is the mass of a molecule of an ionic compound. Mass of substance in g = No. of gram molecules × Gram mol. ‹‹ It is also equal to the sum of atomic masses of all the elements mass present in the formula. ‹‹ Average Atomic Mass and Molecular Mass : Mole Concept ∑ AXii A (Average atomic mass) = ‹‹ One mole of any substance contains a fixed number ∑ Xtotal 23 (6.022 × 10 , known as Avogadro's number, NA) of any type ∑ MX M (Average molecular mass) = ii of particles (atoms or molecules or ions) and has a mass ∑ Xtotal equal to the atomic or molecular weight in grams. Where A1, A2, A3 .... are atomic masses and M1, M2, M3 ... Weight (grams) Number of moles = are molecular masses of species 1, 2, 3, ... etc. with % as X1, Weight of one mole (g/mole) X , X ..... etc. 2 3 Weight ‹‹ Calculation of Molecular Mass : = gram at. mass or gram mol. mass –– Molecular mass = 2 × Vapour density –– Molecular mass = Mass of 22.4 L of vapour at STP 1 Mole = 1 g-atom = 1 g-molecule = 1g-ion. Mass ()g r1 M2 ‹‹ Number of moles, n = –– Rates of diffusion, = − Molar mass (g mol 1) r2 M1 6 Chemistry

V × 23 ‹‹ Number of moles, n = (V = volume of gas in litres at 6.022 10 19 22.4 = 2.69 × 10 molecules/cc NTP or STP) 22400 N ÷ Molecular weight ‹‹ Number of moles, n = (N = Number of particles) NA ÷ Atomic weight ÷ 22.4 L 1 1 Mass Mole Volume at STP ‹‹ 1 amu (or 1 u) = th of mass of an atom of C-12 = g. × Molecular weight × 22.4 L 12 N × Atomic weight For this reason, 1 amu (or 1 u) is also called 1 Avogram. ÷ NA × NA ‹‹ The number of molecules present in 1 cc of an ideal gas at STP is called Loschmidt number. Its value is Number of molecules Illustrations 99..985 ×+10015 × 2 100.015 4 Calculate the average atomic mass of hydrogen using the = ==1.00015 u following data : 100 100 21 Isotope % Natural Molar mass 5 Find out the mass of 10 molecules of Zn. abundance Soln.: For Zn (i.e. monoatomic substance) number of atoms = number of molecules 1H 99.985 1 Number of moles of Zn 2 H 0.015 2 N 1021 weight weight = == = 23 (NCERT Exemplar) NA 6.023 ×10 atomic weight 65.38 Soln.: Average atomic mass = Abundance of 1H × Molar mass + 1021 2 Weight of Zn = × 65.38 Abundance of H × Molar mass 6.023 ×1023 100 = 0.108 g Percentage Composition number of atoms present in one molecule of the compound. ‹‹ Mass percentage of an element in a compound Relationship between Empirical Formula and Mass of that element in the compound = ×100 Molecular Formula Molecular mass of the compound ‹‹ Molecular formula = n × Empirical Formula Percentage Yield Determination of Empirical Formula

Actual yield ‹‹ The % by weight of each element present in one molecule of ‹‹ Percentage yield = ×100 Theoretical yield the compound is determined. ‹‹ Then % by weight of each element is divided by its atomic Empirical Formula and Molecular Formula weight. It gives atomic ratio of the elements present in the compound. Empirical Formula ‹‹ Atomic ratio of each element is divided by the minimum ‹‹ The empirical formula of a compound expresses the simplest value of atomic ratio so as to get simplest ratio of atoms. whole number ratio of atoms of various elements present in one molecule of the compound. ‹‹ If the simplest ratio is fractional then it is multiplied with suitable coefficient to convert it into nearest whole number.

Molecular Formula ‹‹ Write the empirical formula as we get the simplest ratio of ‹‹ The molecular formula of a compound represents the actual atoms. Illustration The molecular weight of the compound is 118 g. Calculate 6 An organic substance containing carbon, hydrogen and the molecular formula of the compound. oxygen gave the following percentage composition. Soln.: To calculate the empirical formula of the compound. C = 40.687%; H = 5.085% and O = 54.228% Some Basic Concepts of Chemistry 7

Element Symbol Percentage of At. mass of Relative no. of Simplest Simplest whole element element atoms atomic ratio no. atomic ratio Carbon C 40.687 12 40.687 3.390 2 = 3.390 = 1 12 3.389

Hydrogen H 5.085 1 5.085 5.085 3 = 5.085 = 15. 1 3.389

Oxygen O 54.228 16 54.228 3.389 2 = 3.389 = 1 16 3.389

\ Empirical Formula is C2H3O2 Molecular formula = n × (Empirical formula) \ Empirical formula mass = (2 × 12) + (3 × 1) = 2 × C H O + (2 × 16) = 59 2 3 2 Molecular mass 118 = C4H6O4 n === 2 Empirical formula mass 59 Thus, the molecular formula is 4C H6O4.

Self Test 2

1. The percentage of Se in peroxidase enzyme is 0.5% by 6. 1 g of hydrogen contains 6 × 1023 atoms. The atomic weight mass (atomic mass of Se = 78.4 amu). Then, the minimum of helium is 4 u. Then the number of atoms in 1 g of He is 1 23 molecular mass of enzyme which contains not more than (a) ××61023 (b) 6 × 10 one Se atom is 4 (a) 1.568 × 104 amu (b) 1.568 × 107 amu (c) 4 × 6 × 1023 (d) 1 × 1023 3 6 (c) 1.568 × 10 amu (d) 1.568 × 10 amu 7. The maximum number of molecules are present in 2. Three flasks each of volume one litre are separately filled (a) 15 L of H2 gas at S.T.P. (b) 5 L of N2 gas at S.T.P. with the gases H2, He and O2 at the same temperature and (c) 0.5 g of H2 gas (d) 10 g of O2 gas pressure. The ratio of total number of atoms of these gases 8. 22.4 litre of water vapour at NTP, when condensed to water present in different flasks would be occupies an approximate volume of (a) 1 : 1 : 1 (b) 1 : 2 : 2 (a) 1 litre (b) 18 litre (c) 2 : 1 : 2 (d) 3 : 2 : 2 (c) 1 mL (d) 18 mL 3. 0.188 g of an organic compound having an empirical 9. 1.020 g of metallic oxide contains 0.540 g of the metal. formula, CH2Br displaced 24.2 cc. of air at 14°C and Taking the symbol for the metal as M, find the molecular 752 mm pressure. Calculate the molecular formula of the formula of the oxide. The specific heat of the metal is compound. (Aqueous tension at 14°C is 12 mm.) 0.216 cal deg–1 g–1. (a) CH2Br (b) C2H2Br2 (a) MO2 (b) M2O3 (c) C2H4Br4 (d) C2H6Br4 (c) M2O (d) MO 4. Rearrange the following (I to IV) in the order of increasing 10. 3.011 × 1022 molecules are removed from a vessel containing masses and choose the correct answer. 1680 cc of nitrogen at STP. How many moles are remaining? (Atomic masses : N = 14, O = 16, Cu = 63) (a) 0.025 (b) 0.05 I. 1 atom of oxygen II. 1 atom of nitrogen (c) 0.5 (d) 1 III. 1 × 10–10 × (gram molecular weight of oxygen) 11. 10 g of hydrofluoric acid gas occupies 5.6 litre of volume IV. 1 × 10–10 × (gram atomic weight of zinc) at NTP. If the empirical formula of the gas is HF, then its (a) II < I < III < IV (b) IV < III < II < I molecular formula will be (At. mass of F = 19) (c) II < II < I < IV (d) III < IV < I < II (a) HF (b) H3F3 5. The equivalent weight of phosphoric acid, H PO in the 3 4 (c) H2F2 (d) H4F4 reaction, 12. The number of moles of carbon dioxide which contain 8 g NaOH + H PO → NaH PO + H O is 3 4 2 4 2 of oxygen is (a) 59 (b) 49 (a) 0.1 mole (b) 0.20 mole (c) 25 (d) 98 (c) 0.50 mole (d) 0.25 mole 8 Chemistry

13. A sample of ammonium phosphate (NH4)3PO4 contains (a) 3.5 g (b) 2.9 g 3.18 mole of H-atoms. The number of moles of O atoms in (c) 15.2 g (d) 12.9 g the sample is 15. If 1021 molecules are removed from 100 mg CO , then (a) 2.12 (b) 0.58 2 number of moles of CO left are (c) 1.06 (d) 3.18 2 (a) 6.10 × 10–4 (b) 2.8 × 10–3 14. How much NaNO3 must be weighed out to make 50 mL of –3 –2 an aqueous solution containing 70 mg Na+ ions per mL? (c) 2.28 × 10 (d) 1.36 × 10

Stoichiometry and therefore, controls amount of product. The remaining

‹‹ One of the most important aspects of a chemical equation or left out reactant is called the excess reagent. is that when it is written in the balanced form, it gives ‹‹ In case where there is a limiting reagent, the initial amount quantitative relationship between the various reactants and of the limiting reagent must be used to calculate the amount products in terms of moles, molecules, masses and volumes. of product formed. This is called stoichiometry (Greek word meaning 'to –– For a reaction, X + Y → Z measure an element'). nX ()initial  X  –– If < S   , then X is limiting Limiting Reagent nY ()initial  Y  ‹‹ The reactant which is completely consumed in the reaction – nX ()initial  X  – If > S   , then Y is limiting and hence limits the amount of product formed is called nY ()initial  Y  limiting reagent. where, nX and nY are the moles of X and Y. S is the ‹‹ Limiting reagent is present in least stoichiometric amount stoichiometric ratio of X and Y. Illustrations

7 Calculate the amount of carbon dioxide that could be 88 16 g of O2 reacts with C to produce × 16= 22 g of CO2. produced when 64 (i) 1 mole of carbon is burnt in air. 8 On heating 9.8 g KClO3, its mass is reduced by 0.384 g. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. Calculate the percentage of original KClO3 sample, that has been decomposed. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. (NCERT Exemplar) Soln.: Thermal decomposition of KClO3 occurs as : Soln.: (i) CO+2 → CO2 ∆ 1 mole 1 mole 1 mole KClO32()ss→+KClO() 32/ ()s ()12 gg()32 ()44 g Thus, loss in mass is due to O2 escaped and so mass of O2 Hence, 1 mole of C produces 44 g of CO2. formed is 0.384 g. (ii) CO+2 →CO2 12 gg32 44 g 0.384 or Moles of O2 = 32 Here, O2 is the limiting reagent. = 1.2 × 10–2 32 g of O2 reacts with C to produce 44 g of CO2 . . 3 44 . 1 × Mole of KClO3 ≡× Moles of O2 16 g of O2 reacts with C to produce × 16 = 22 g of CO2 2 32 −2 12. ×10 −3 (iii) 22CO+→2CO \ Moles of KClO3 = =×810 mole 2 2 15. 24 gg 64 88 g –3 and mass of KClO3 decomposed = 8 × 10 × 122.5 = 0.980 g Here, O2 is the limiting reagent. 0.980 \ % of decomposition = ×=100 10% 64 g of O2 reacts with C to produce 88 g of CO2 98.

Calculation of Concentration of a Solutions Mass of component A 6 ‹‹ Parts per million of A = ×10 Total mass of solution WA ‹‹ Mass % of A = ×100 Moles of solute WWAB+ ‹‹ Molarity ()M =×1000 Volume of solution (in L) VA ‹‹ Volume % of A = ×100 Moles of solute VV+ ‹‹ Molality ()m =×1000 AB Mass of solvent (in kg) Some Basic Concepts of Chemistry 9

Gram equiv. of solute increase in volume, so molarity, normality, etc. decrease as ‹‹ Normality ()N = ×1000 temperature is increased. Volume of solution (in L) ‹‹ Molality, mole fraction, % by mass, etc. are not temperature Molecular mass =×Molarity dependent. Equivalent mass ‹‹ Molarity of a solution after dilution ‹ ‹ All the units of concentration using volume terms e.g., M1V1 = M2V2 molarity, normality, formality, etc. are temperature ‹‹ Molarity of a solution after mixing dependent. As rise in temperature usually results in M1V1 + M2V2 = M3(V1 + V2) Illustrations 1.4924V 9 What would be the molarity of a solution obtained by \ Molarity of solution = = 76. 1 M –1 2V mixing equal volumes of 30% by mass H2SO4 (d = 1.218 g mL ) × –1 98 and 70% by mass H2SO4 (d = 1.610 g mL )? 1000 Soln.: Let V mL of each solution are mixed, then 10 If 4 g of NaOH dissolves in 36 g of H O, calculate For first solution : H SO is 30% by mass 2 2 4 the mole fraction of each component in the solution. Also, Mass of H2SO4 = 30 g determine the molarity of solution (specific gravity of and mass of solution = 100 g solution is 1 g mL–1). 100 \ Volume of solution = mL Soln.: Mole fraction of H O 1.218 2 No. of moles of HO 100 = 2 i.e., mL contains 30 g H2SO4 Total no. of moles ()H ON+ aOH 1.218 2 30 ××V 1.218 36 \ V mL contains g H SO No. of moles of H2O = = 2 moles 100 2 4 18 4 For second solution : H2SO4 is 70% by mass No. of moles of NaOH = = 01. mol Mass of H2SO4 = 70 g 40 Mass of solution = 100 g Total number of moles = 2 + 0.1 = 2.1 100 2 \ Volume of solution = mL Mole fraction of H2O = = 09. 5 1.610 21. 100 01. i.e., mL contains 70 g H2SO4 Mole fraction of NaOH = = 0.047 1.610 21. 70 ××V 1.610 Mass of solution \ V mL contains g H24SO 100 = Mass of H2O + Mass of NaOH = 36 + 4 = 40 g On mixing these two, total mass of H SO 40 2 4 Volume of solution = = 40 mL 30 ×1..218 70 ×1 610 1 = + Vg = 1.4924 Vg   No. of moles of solute 01.  100 100  Molarity = ==25. M Total volume of solution = 2V mL. (... equal volumes are mixed) Volume of solution in L 00. 4 L

Self Test 3

1. With increase of temperature, which of these changes? proceeds to completion. Which of the following statements (a) Molality is true?

(b) Weight fraction of solute (a) 'SO2' will remain in excess. (c) Fraction of solute present in water (b) Only 'S' and 'H2O' will remain in the reaction vessel. (d) Mole fraction (c) 'H2S' will remain in excess. 22 2. If 1.6 g of SO2 and 1.5 × 10 molecules of H2S are mixed (d) None of these. and allowed to remain in contact in a closed vessel until the 3. The density of 4% (w/v) NaOH solution is 1.02 g/mL. What reaction, is the molality of the solution? 2H2S + SO2 → 3S + 2H2O (a) 1.89 m (b) 1.02 m 10 Chemistry (c) 2.8 m (d) 5 m (a) 0.28 (b) 0.3 4. 200 g sample of hard water is passed through a cation (c) 0.24 (d) 2.32 exchanger in which H+ ions are exchanged by Ca2+ ions. 10. The change required to deposit 1 mmol of aluminium The water coming out of cation exchanger needed 75 mL metal by the passage of 9.65 A through an aqueous solution of 0.1 N NaOH for complete neutralisation. The hardness of aluminium ions is of water due to Ca2+ ion is (a) 3 × 96.5 c (b) 2 × 96.5 c (a) 250 ppm (b) 500 ppm (c) 96.5 c (d) 6 × 96.5 c (c) 750 ppm (d) 1000 ppm 11. 6 g magnesite mineral on heating liberated carbon dioxide

5. 0.037 g of an alcohol, R–OH, was added to C2H5MgI and which measures 1.12 L at STP. What is the percentage the gas evolved measured 11.2 cc at STP. The molecular purity of mineral? mass of the alcohol is (a) 50% (b) 60% (a) 42 u (b) 46 u (c) 70% (d) 80% (c) 58 u (d) 74 u 12. Calculate the molarity of a solution of ethanol in water in 6. A compound of iron and chlorine is soluble in water. An which the mole fraction of ethanol is 0.040. excess of silver nitrate was added to precipitate the chloride (a) 2.31 M (b) 0.04 M ion as silver chloride. If 134.8 mg of the compound (c) 5.55 M (d) 4 M gave 304.8 mg of AgCl, then what is the formula of the 13. 90 mL of pure dry O2 is subjected to silent electric compound? discharge. If only 10% of it is converted to O3, volume of (a) FeCl6 (b) FeCl3 the mixture of gases (O2 and O3) after the reaction will be (c) FeCl2 (d) FeCl4 ...... and after passing through turpentine oil will be ...... (a) 65 mL and 85 mL (b) 81 mL and 87 mL 7. CaCO3 is 90% pure. Volume of CO2 collected at STP when (c) 85 mL and 65 mL (d) 87 mL and 81 mL 10 g of CaCO3 is decomposed, is (a) 20.16 L (b) 1.008 L 14. What volume of carbon monoxide at 2 atm and 273 °C is (c) 2.96 L (d) 2.016 L required in order to produce 5.6 g of metal by the reduction 8. In a gas phase reaction, 50 kg of nitrogen and 10 kg of of ferric oxide? hydrogen are mixed to produce ammonia. The maximum (a) 5.6 L (b) 4.25 L amount of ammonia produced is (c) 6.72 L (d) 3.36 L (a) 46.67 kg (b) 50 kg 15. 2 g of a mixture of CO and CO2 on reaction with excess (c) 56.67 kg (d) 10 kg I2O5 produced 2.54 g of I2. What would be the mass % of CO in the original mixture? 9. In the reaction 4A + 2B + 3C → A4B2C3, what will be the 2 number of moles of product formed, starting from 1 mole (a) 60 (b) 30 of A, 0.6 mole of B and 0.72 mole of C? (c) 70 (d) 35 AT A GLANCE

Prefixes Used in the SI System 1 cm = 0.3937 inch –1 9 deci (d) 10 giga (G) 10 1 metre = 39.37 inch centi (c) 10–2 femto (f) 10–15 1 km = 0.621 mile milli (m) 10–3 atto (a) 10–18 1kg = 2.20 pounds (lb) micro (m) 10–6 zepto (z) 10–21 nano (n) 10–9 yocto (y) 10–24 1g = 0.0353 ounce pico (p) 10–12 tera (T) 1012 1 pound = 453.6 g deca (da) 101 peta (P) 1015 1 atomic mass = 1.6605 × 10–24 g = 1.6605 × 10–27 kg –3 –10 hecto (h) 102 exa (E) 1018 unit (amu) ≡ 1.492 × 10 erg = 1.492 × 10 J 11 8 kilo (k) 103 zeta (Z) 1021 = 3.564 × 10 cal = 9.310 × 10 eV mega (M) 106 yotta (Y) 1024 = 931.48 MeV 1 atmosphere = 760 torr = 760 mm Hg = 76 cm Hg Useful Conversion factors (atm) = 1.01325 × 105 Pa –8 –10 –1 2 1 angstrom (Å) = 10 cm = 10 m = 10 nm = 10 pm 1 calorie (cal) = 4.18 × 107 erg = 4.18 J 1 inch = 2.54 cm = 2.613 × 1019 eV Some Basic Concepts of Chemistry 11

1 erg = 10–7 J = 2.389 × 10–8 cal = 6.242 × 1011 eV ‹‹ Equivalent weight of a base Molecular weight of the salt 1 faraday (F) = 9.6487 × 104 coulomb = Total positive valency of the metal atoms 1 dyne (dyne) = 10–5 N 1 joule = 107 erg = 0.2390 cal ‹‹ Equivalent weight of oxidizing /reducing agent 3 Molecular weight of the substance 1 litre = 1000 cc = 1000 mL = 1 dm = No. of electrons gained/llost by one molecule 1 coulomb (coul) = 10–3 m3 9 1 curie (Ci) = 2.9979 × 10 esu Concentration of Solutions 1 electron volt = 3.7 × 1010 disintegrations s–1 –12 –19 No. of moles of the solute n2 (eV) = 1.6021 × 10 erg = 1.6021 × 10 J ‹‹ Molarity = = Vol. of solution in litre V = 3.827 × 10–20 cal = 23.06 kcal mol–1 where moles (n2) of solute 1 electrostatic = 3.33564 × 10–10 coul Mass of the solute in g w2 unit (esu) = = Molecular mass of the solute M2 w Mole Concept M = 2 MV× ‹‹ Molar mass = mass of 1 mole 2 23 ‹‹ No. of moles of solute n2 w2 1 Mole = 6.023 × 10 particles (atom/molecule) ‹‹ Molality = == × ‹‹ 1 Mole atoms = gram atomic mass Mass of the solvent in kg W1 M2 W1 23 ‹‹ 1 g atom = N atoms = 6.023 × 10 atoms Number of formula mass of solute ‹‹ Formality = ‹‹ 1 Mole molecules = Gram molecular mass (or 1 g molecule) Volume of the solution in litre 23 ‹‹ No. of g. eq. of the solute 1 g molecule = N molecules = 6.023 × 10 molecules ‹‹ Normality = = gram molecular mass = 22.4 L of any gas at STP Vol. of solution in litre ‹‹ 1 Mole ionic compound = Gram formula mass Mass of the solute in g w w 23 where g. eq. = =⇒N = = 6.023 × 10 formula units Eq. mass of the solute E EV× 23 ‹‹ 22.4 L of gas at S.T.P. = 6.023 × 10 molecules of gas. Mol. mass ‹‹ Normality of a solution = Molarity × Molecular Mass Eq. mass ‹‹ Normality of an acid = Molarity × Basicity ‹‹ Molecular mass = 2 × Vapour density Normality of a base = Molarity × Acidity ‹‹ Molecular mass = Mass of 22.4 L of vapour at STP ‹‹ Mole fraction of solute in solution (x2) ‹‹ Dulong and Petit law = At. weight × specific heat = 6.4 n2 wM22 (for metals only) = = nn12+ wM1212+ wM Equivalent Weight Mole fraction of solvent in solution (x1) n wM ‹‹ Equivalent weight of an element in a redox reaction = 1 = 11 Atomic weight nn12+ wM1212+ wM = No. of electrons lost or gained by one atom oof that element where w , M are mass and molecular mass of solvent and 1 1 w2, M2 for the solute. x1 + x2 = 1. Atomic weight of element ‹‹ Equivalent weight of an element = wA Valency of element ‹‹ Mass fraction of component Ax()A = ‹‹ Equivalent weight of a compound wwAB+ Molecular weight of compound wB = Mass fraction of component Bx()= (xA + xB = 1) Total charge on its cation or anion B + wwAB Wt. of the solute in g Formula weight of ion ‹‹ % by weight = ×100 ‹‹ Equivalent weight of anion = Wt. of the solution in g Charge on ion Vol. of solute in cc Molecular weight of salt ‹‹ % by volume = × 100 ‹‹ Equivalent weight of an acid salt = Vol. of solution in cc Replaceable H atom in it Wt. of solute in g Molecular weight of the acid ‹ ‹‹ Equivalent weight of an acid = ‹ % by wt./vol = × 100 Basicity Vol. of solution in cc Molecular weight of the base Wt. of the solute in g ‹‹ Equivalent weight of a base = ‹‹ Strength of a solution = Acidity Vol. of solution in litres 12 Chemistry

‹‹ Parts per million (ppm) of substance A ‹‹ Molarity equation : M1V1 = M2V2. Mass of A 6 Vol. of A 6 ‹‹ If two non-reacting solutions of different normalities are ppm = ×10 or ×10 Mass of solution Vol. of solution mixed, the normality of the resulting solution can be calculated :

‹‹ Normality equation : N1V1 = N2V2. N1V1 + N2V2 = N3V3. EXTRA EDGE

‹‹ Relation between molality and molarity Now mole fraction of solute x2 If M = Molarity = Moles of solute dissolved in 1000 mL of moles of solute n2 == solution and molecular mass of solute = M2 moles of solvent + moles of solute nn12+ Then mass of solute =M M2 M MM1 \ Mass of solution = 1000 r – MM = = 2 1000ρ − MM 1000ρ −+MM MM As molality m = moles of solute per 1000 g of solvent 2 + M 21 M M 1 = ×1000 ρ − MM 1000 MM2 = 1 1000ρ −−MM()M M mρ 21 or Molality ()m = or Molarity (M) = MM2  mM2  MM1 ρ − 1+  or x2 = 1000  1000  1000ρ −−MM()21M ‹ ‹ Relation between molarity and density of solution ‹‹ Relation between molality and mole fraction no. of moles of solute Molarity of the solution = Molality = moles of solute per 1000 g of solvent volume of solution Moles of solute = n2 n M = 1 1000 ()nM + nM / ρ Moles of solvent = n1 = 11 22 M1 Here n1M1 = mass of solute, n2M2 = mass of solvent, i.e. where M1 = Molecular mass of solvent n1M1 + n2M2 = mass of solution n2 n2 Now, x2(mole fraction of solute) = = , ‹ 1000 ‹ Relation between molarity and mole fraction nn12+ n2 + If molarity = M = Moles of solute dissolved in 1000 mL of M1 solution nM21 x2 = M2 = Molecular mass of solution nM21+1000 Volume of solution = 1000 mL, density of solution = r nM21 Mass of solute = MM , Mass of solution = 1000r 2 1000 mM1 Where M = Molarity. or we can write x = = 2 nM +1000 mM +1 Thus mass of solvent = Mass of solution – mass of solute 21 1 1000 = 1000r – MM2 1000ρ − MM because moles of solute dissolved in 1000 g of solvent is Now number of moles of solvent = 2 n M called molality i.e. 2 = m 1 1000 where M1 = Molecular mass of solvent Some Basic Concepts of Chemistry 13 Exercise

LEVEL 1 36.85% of nitrogen. Which law of chemical combination is illustrated by this data? ‰‰ General Introduction (a) Law of conservation of mass (b) Law of definite proportions 1. Which one of the following forms part of seven basic (c) Law of reciprocal proportions SI units? (d) Law of multiple proportions (a) Watt (b) Candela (c) Newton (d) Pascal 10. Which among the following combinations illustrates the law of reciprocal proportions? 2. Which of the following is an example of a homogeneous mixture? (a) N2O3, N2O4, N2O5 (b) NaCl, NaBr, NaI (a) Water + Alcohol (b) Water + Sand (c) H2O, H2S, SO2 (d) PH3, P2O3, P2O5 (c) Water + Oil (d) Petrol 11. For the gaseous reaction : HC22()gg+→lH() 2 Cl()g . If 3. Match the following : 40 mL of hydrogen completely reacts with chlorine then find out the required volume of chlorine and volume of Column I Column II –5 produced HCl(g). A. 1 faraday (i) 10 N (a) 20 mL, 40 mL (b) 40 mL, 80 mL B. 1 dyne (ii) 0.2390 cal (c) 40 mL, 20 mL (d) 80 mL, 40 mL –8 C. 1 joule (iii) 2.389 × 10 cal 12. Which one of the following pairs of compounds illustrates 4 D. 1 litre (iv) 9.6487 × 10 coulomb the law of multiple proportions? –3 3 E. 1 erg (v) 10 m (a) H2O, Na2O (b) MgO, Na2O (a) (A)-(iv), (B)-(i), (C)-(ii), (D)-(v), (E)-(iii) (c) Na2O, CaO (d) SO2, SO3 (b) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii), (E)-(v) 13. The law of conservation of mass holds good for all of the (c) (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv), (E)-(v) following except (d) (A)-(v), (B)-(iii), (C)-(iv), (D)-(ii), (E)-(i) (a) chemical reactions (b) nuclear reactions 4. Which of the following is an example of matter according (c) exothermic reactions (d) endothermic reactions. to physical state at room temperature and pressure? 14. Statement 1 : Pure water obtained from different sources (a) Solid (b) Liquid such as river, well, spring, sea, etc. always contains hydrogen (c) Gas (d) All of these and oxygen combined in the ratio 1 : 8 by mass. 5. Given the numbers : 461 cm, 0.461 cm, 0.0461 cm. The Statement 2 : A chemical compound always contains number of significant figures for the three numbers is elements combined together in same proportion by mass. (a) 3, 3 and 4 respectively (b) 3, 4 and 4 respectively (a) If both statement 1 and statement 2 are true and (c) 3, 4 and 5 respectively (d) 3, 3 and 3 respectively. statement 2 is the correct explanation of statement 1. 6. A pure substance can only be (b) If both statement 1 and statement 2 are true but (a) a compound statement 2 is not the correct explanation of statement 1. (b) an element (c) If statement 1 is true but statement 2 is false. (c) an element or a compound (d) If both statement 1 and statement 2 are false. (d) a homogeneous mixture. 15. Which of the following equations does not obey the law of 7. Which of the following is a compound? conservation of mass? (a) Graphite (b) Brass (a) 4H + O2 2H2O (b) H2 + O 2H2O (c) Cement (d) Marble (c) 2H + O2 H2O (d) Both (b) and (c) ‰‰ Laws of Chemical Combination 16. Two gaseous samples were analysed. One contained 1.2 g of carbon and 3.2 g of oxygen. The other contained 27.3% 8. The law of definite proportion was proposed by carbon and 72.7 % oxygen. The experimental data are in (a) Lavoisier (b) Dalton (c) Proust (d) Gay-Lussac. accordance with (a) law of conservation of mass 9. Ammonia contains 82.35% of nitrogen and 17.65% of (b) law of definite proportions hydrogen. Water contains 88.90% of oxygen and 11.10% of (c) law of reciprocal proportions hydrogen. Nitrogen trioxide contains 63.15% of oxygen and (d) law of multiple proportions. 14 Chemistry 17. Different proportions of oxygen in the various oxides of (c) Equal volumes of gases contain equal number of nitrogen prove the law of molecules under the same conditions of temperature (a) conservation of mass (b) constant proportions and pressure. (c) multiple proportions (d) reciprocal proportions. (d) Atom is the smallest particle that takes part in a chemical reaction. 18. In the reaction N2 + 3H2 → 2NH3, ratio by volume of N2, H2 and NH3 is 1 : 3 : 2. This illustrates law of 25. Dalton’s atomic theory could not explain (a) conservation of mass (a) law of conservation of mass (b) definite proportions (b) law of definite proportions (c) multiple proportions (c) Gay Lussac’s law of combining volume (d) gaseous volumes. (d) all of these. 19. There are two common oxides of sulphur, one of which ‰‰ Atomic and Molecular Masses contains 50% O by weight, the other almost exactly 60%. 2 26. The modern atomic weight scale is based on The weights of sulphur which combine with 1 g of O 12 16 2 (a) C (b) O (fixed) are in the ratio of (c) 18O (d) 1H (a) 2 : 3 (b) 2 : 1 (c) 1 : 3 (d) 3 : 2 27. Chlorine occurs as two isotopes Cl-35 (atomic mass 34.9689 u) and Cl-37 (atomic mass 36.9659 u). 75.77% of 20. H2S contains 94.11% sulphur; SO2 contains 50% oxygen natural chlorine is Cl-35. The atomic weight of chlorine is and H2O contains 11.11% hydrogen. Thus, (a) 35 (b) 34.96 (a) law of conservation of mass is followed (c) 37 (d) 35.45 (b) law of reciprocal proportions is followed 28. If the mass of neutron is assumed to half of its original (c) law of multiple proportions is followed value whereas that of proton is assumed to be twice of its (d) all of the above. 14 original value, then the atomic mass of 6C will be ‰‰ Dalton’s Atomic Theory (a) same (b) 14.28 % less (c) 14.28% more (d) 28.56% more. 21. Which of the following postulates of Dalton’s atomic theory explains the law of multiple proportion? 29. Bromine has two naturally occuring isotopes, Br-79 and (a) Atoms of two elements may combine with one another Br-81. The atomic mass and per cent natural abundance of to form more than one compound. Br-79 are 78.9183 u and 50.54%. What is the atomic mass (b) Atoms combine in the ratio of small whole numbers of the second isotope, Br-81? (Br = 79.90) (a) 81.50 (b) 80.90 to form compounds. (c) 49.46 (d) 79.05 (c) The relative number and kinds of atoms are constant in a given compound. 30. 1 amu is equal to 1 (d) All of the above. (a) of mass of 1 atom of C-12 22. Which of the following facts went against the Dalton’s 12 1 theory? (b) of mass of 1 atom of O-16 (a) Discovery of radioactivity 14

(b) Discharge tube experiment (c) 1 g of H2 (c) Discovery of isotopes (d) 1.66 × 10–23 kg. (d) All of these 31. An element A has the following isotopic composition: 23. The main drawback(s) of Dalton’s atomic theory is 200 199 202 A: 90%, A : 8.0% and A : 2.0%. The weighted average (a) it could not explain the law of gaseous volumes atomic mass of the naturally occurring element A is (b) it could not explain why atoms of different elements (a) 205 u (b) 210 u have different masses, sizes, etc (c) 190 u (d) 200 u (c) it could not explain how and why atoms combine to form molecules 32. How many grams are contained in 1 gram-atom of Mg? (a) 12 g (b) 24 g (d) all of the above 1 24. Which of the following postulates is not a part of Dalton’s (c) 1 g (d) g 24 atomic theory? (a) Compounds are formed by the combination of atoms 33. At NTP, the density of a gas is 0.00445 g/mL. Find out the molecular mass of the gas. of more than one element in fixed whole number ratios. (a) 50 (b) 75 (b) Chemical reactions involve only the rearrangement of (c) 60 (d) 100 atoms. Some Basic Concepts of Chemistry 15 54 56 57 34. Given that the abundances of isotopes Fe, Fe and Fe 46. Calculate the mass of 4 NA molecules of CO2 . are 5%, 90% and 5% respectively, the atomic mass of Fe is (a) 22 g (b) 44 g (a) 55.75 (b) 55.95 (c) 88 g (d) 176 g (c) 55.85 (d) 56.05 47. Calculate the effective molecular weight of air. 35. Specific heat of a solid element is 0.1 cal g–1 °C and its (a) 28 (b) 32 equivalent weight is 31.8. Its exact atomic weight is (c) 28.84 (d) 32.21 (a) 35.8 (b) 63.6 48. The volume occupied by one molecule of water 3 (c) 31.8 (d) 95.4 (density = 1 g/cm ) is –23 3 –23 3 36. Which property of an element is not variable? (a) 9.0 × 10 cm (b) 6.023 × 10 cm (c) 3.0 × 10–23 cm3 (d) 5.5 × 10–23 cm3 (a) Valency (b) Atomic weight (c) Equivalent weight (d) All of these 49. How many moles of magnesium phosphate Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? 37. Relative density of a volatile substance with respect to CH 4 (a) 0.02 (b) 3.125 × 10–2 is 4. Its molecular weight would be –2 –2 (c) 1.25 × 10 (d) 2.5 × 10 (a) 16 (b) 32 (c) 64 (d) 128 50. How many years it would take to spend Avogadro’s number of rupees at the rate of 1 million rupees in one second? 38. A compound possesses 8% sulphur by mass. The least (a) 19.098 × 1019 years (b) 19.098 × 1015 years molecular mass is (c) 19.098 × 109 years (d) 19.098 × 1023 years (a) 200 (b) 400 –1 51. Atomic mass of mercury is 200 and density is 13.6 g cc . (c) 155 (d) 355 How many moles of metal are present in one litre? 35 37 39. In chlorine gas, ratio of Cl and Cl is (a) 55 (b) 68 (a) 1 : 3 (b) 3 : 1 (c) 74 (d) 86 (c) 1 : 1 (d) 1 : 4 52. 0.44 g of a colourless oxide of nitrogen occupies 224 mL at 40. Dulong and Petit’s law is valid only for STP. The compound is (a) metals (b) non-metals (a) N2O (b) NO (c) gaseous elements (d) solid elements. (c) N2O2 (d) NO2 ‰‰ Mole Concept and Molar Masses 53. Haemoglobin contains 0.25% iron by mass. The molecular mass of haemoglobin is 89600. The number of iron atoms 41. How many carbon atoms are present in 0.35 mol of per molecule of haemoglobin is C6H12O6? (a) 2 (b) 4 (a) 1.26 × 1023 carbon atoms (c) 6 (d) 8 23 (b) 6.023 × 10 carbon atoms 54. What is the molecular mass of a substance, each molecule 24 (c) 1.26 × 10 carbon atoms of which contains 9 carbon atoms, 13 hydrogen atoms and (d) 6.023 × 1024 carbon atoms 2.33 × 10–23 g of other component? 42. How many atoms are present in one cc of helium gas at (a) 108 amu (b) 125.08 amu STP? (c) 124.04 amu (d) 135.04 amu (a) 6.023 × 1023 (b) 2.25 × 1010 55. The alloy aluminium-bronze contains 90% Cu and 10% Al (c) 2.69 × 1019 (d) 4.08 × 1015 by mass. In what atomic ratio are the two metals present in 43. The density (in g/mL) of a 3.60 M sulphuric acid solution the alloy? (a) 63.5 : 27 (b) 142 : 37 that is 29% H SO (molar mass = 98 g/mol) by mass is 2 4 (c) 90 : 10 (d) 50 : 50 (a) 2.645 (b) 1.882 24 (c) 1.216 (d) 1.458 56. From 160 g of SO2(g) sample, 1.2046 × 10 molecules of SO are removed. Then find out the volume of SO left 44. 14 g of nitrogen gas and 22 g of CO gas are mixed together. 2 2(g) 2 over at STP. Find the volume of gaseous mixture at STP. (a) 5.6 L (b) 11.2 L (a) 22.4 L (b) 44.8 L (c) 22.4 L (d) 44.8 L (c) 11.2 L (d) 5.6 L 57. The total number of atoms of all elements present in mole 45. How many moles of electron weigh one kilogram? of ammonium dichromate is 1 23 23 (a) 6.023 × 1023 (b) ×1023 (a) 6.023 × 10 (b) 19.05 × 10 9.108 (c) 114.473 × 1023 (d) 84.322 × 1023 54 6.023 ×10 1 58. If N is Avogadro number, then the number of valence (c) (d) ×108 A 9.108 9..108 × 6 023 electrons in 4.2 g of N3– ions is 16 Chemistry

(a) 2.4 NA (b) 1.6 NA (a) 22.4 L (b) 11.2 L (c) 4.2 NA (d) 3.2 NA (c) 5.6 L (d) 33.6 L 59. The number of potassium ions in two moles of potassium 69. The number of moles of oxygen in 1 L of air containing ferrocyanide is 21% oxygen by volume, under standard conditions, is (a) 6.023 × 1023 (b) 2 × 6.023 × 1023 (a) 0.0093 mol (b) 2.10 mol (c) 4 × 6.023 × 1023 (d) 8 × 6.023 × 1023 (c) 0.186 mol (d) 0.21 mol 60. What is the ratio of number of molecules, if the mass ratio 70. A sample of ethane has the same mass as 107 molecules of N2 and O2 is 4 : 1 ? of methane. How many ethane molecules does the sample (a) 28 : 32 (b) 14 : 16 contain? (c) 32 : 7 (d) 16 : 32 (a) 6.023 × 1023 (b) 3.826 × 108 6 4 61. Statement 1 : Under identical conditions 1 L of O2 and 1 L (c) 5.34 × 10 (d) 1.25 × 10 of O3 gas contain the same number of moles. ‰‰ Percentage Composition Statement 2 : 1 L of O2 and 1 L of O3 gas contain the same number of oxygen atoms. 71. An oxide of nitrogen gave the following percentage (a) If both statement 1 and statement 2 are true and composition : statement 2 is the correct explanation of statement 1. N = 25.94 and O = 74.06 (b) If both statement 1 and statement 2 are true but The empirical formula of the compound is statement 2 is not the correct explanation of statement 1. (a) NO (b) N2O (c) If statement 1 is true but statement 2 is false. (c) N2O3 (d) N2O5 (d) If both statement 1 and statement 2 are false. 72. In a compound 52.2% carbon, 13% hydrogen, 34.8% – 2+ 62. Calculate the number of Cl and Ca ions in 222 g oxygen are present and molecular mass of the compound anhydrous CaCl2. is 92. The molecular formula of the compound is 2+ – (a) 2NA ions of Ca and 2NA ions of Cl (a) C H O (b) C H O – 2+ 6 12 6 4 12 2 (b) 2NA ions of Cl and 4NA ions of Ca (c) C3H8O (d) C14H10O (c) 2N ions of Ca2+ and 4N ions of Cl– A A 73. Percentage composition of an organic compound is as (d) None of these follows : . 63. A hydrate of iron (III) thiocyanate [Fe(SCN)3 xH2O], was C = 10.06, H = 0.84, Cl = 89.10 found to contain 19% H2O. The value ofx is Which of the following corresponds to its molecular (a) 2 (b) 3 formula if the vapour density is 60.0? (c) 4 (d) 6 (a) CH3Cl (b) CHCl3 64. The vapour density of a mixture containing NO2 and N2O4 (c) CH2Cl2 (d) CHCl ° –1 is 38.3 at 27 C. Calculate the mole of NO2 in 100 mole 74. In an organic compound of molar mass 108 g mol C, H mixture. and N atoms are present in 9 : 1 : 3.5 by weight. Molecular (a) 76.60 (b) 33.48 formula can be (c) 38.32 (d) None of these. (a) C6H8N2 (b) C7H10N 65. When gypsum is totally dehydrated, what is the percentage (c) C5H6N3 (d) C4H18N3 weight loss? 75. A giant molecule contains 0.25% of a metal whose atomic (a) 20.93 (b) 31.80 weight is 59. Its molecule contains one atom of that metal. (c) 26.20 (d) 36.40 Its minimum molecular weight is 66. Calculate the total number of N2 and O2 molecules in a (a) 11800 (b) 23600 breath in which a person inhales 224 mL air (at STP) and 100 × 59 in air N and O are 78% and 21% respectively. (c) 5900 (d) 2 2 04. (a) 1.67 × 1024 (b) 6.023 × 1023 (c) 5.96 × 1021 (d) 6.89 × 1020 76. A compound has the composition : Na 29.0%, S 40.5% and O 30.5%. If the formula mass is 158, what is the formula 67. In a solution of C H and C H CH , x is 0.5. Calculate 6 6 6 5 3 C6H6 unit of the compound? the mass % composition of mixture. (a) Na SO (b) Na SO (a) 50, 50 (b) 45.9, 54.1 2 4 2 3 (c) Na S O (d) Na S O (c) 52.59, 47.5 (d) 60.2, 39.8 2 2 3 2 2 8 77. Calculate the molecular formula of compound which 68. The density of O at 0°C and 1 atm is 1.429 g/L. The molar 2 contains 20% Ca and 80% Br (by wt.) if molecular weight volume of gas is of compound is 200. (Atomic wt. of Ca = 40, Br = 80) Some Basic Concepts of Chemistry 17

(a) Ca3Br2 (b) CaBr2 (a) 0.4 mole (b) 0.1 mole (c) CaBr (d) Ca2Br (c) 0.2 mole (d) 0.9 mole.

78. Calculate the empirical formula for a compound that 88. How many grams of pure KMnO4 should be heated with contains 26.6% potassium, 35.4% chromium and 38.1% H2SO4 to produce 48 g of O2? oxygen by mass. (Atomic wt : K = 39, Cr = 52, O = 16) (a) 180.6 g (b) 189.6 g (a) KCrO2 (b) K2CrO4 (c) 150.5 g (d) 310 g. (c) K Cr O (d) None of these. 2 2 7 89. 1000 g aqueous solution of CaCO3 contains 10 g of calcium 79. Caffeine has a molecular weight of 194. It contains 28.9% carbonate. Concentration of solution is by mass of nitrogen. Number of atoms of nitrogen in one (a) 10 ppm (b) 100 ppm molecule of it (c) 1000 ppm (d) 10000 ppm. (a) 6 (b) 3 90. What weight of magnesia is obtained by complete (c) 4 (d) 5 combustion of two grams of metal? 80. An organic compound has the composition : C = 58.5%, (a) 2.4 g (b) 3.3 g H = 4.06%, N = 11.4% and O = 26.0%. What is the empirical (c) 4.8 g (d) 6.6 g formula of the compound? 91. How many moles of lead (II) chloride will be formed in a (a) CH3NO2 (b) C6H5NO reaction between 6.5 g of PbO and 3.2 g of HCl? (c) C6H5NO2 (d) C6H4N2O4 (a) 0.088 (b) 0.033

81. Calculate the mass percent of oxygen present in Na2SO4. (c) 0.011 (d) 0.029 (a) 32.4 (b) 22.5 92. Given that 10 g of a dibasic acid (mol. mass =100) are present (c) 45 (d) 43.2 in 500 mL of the solution. The density of the solution is –1 82. A compound contains 21 atoms of carbon per molecule. 1.02 g mL . Match the entries of column I with appropriate The mass percentage of carbon in the compound is 69.98%. entries of column II and choose the correct option. Its molar mass is Column I Column II (a) 176.5 (b) 252.2 A. Normality of the solution p. 0.98 (c) 287.6 (d) 360.1 B. Molality of the solution q. 0.996 83. Insulin contains 3.4% sulphur the minimum molecular C. Mole fraction of solvent r. 0.2 weight of insulin is (a) 941.176 (b) 944.328 D. Mass fraction of solvent s. 0.4 (c) 946.278 (d) 943.265 (a) A-p; B-q; C-r; D-s (b) A-s; B-r; C-q; D-p (c) A-s; B-r; C-p; D-q (d) A-r; B-s; C-q; D-p ‰‰ Stoichiometry and Stoichiometric Calculations 3 93. 2Al +→OAlO 84. The molarity of a solution prepared by dissolving 4 g NaOH 2 232 in enough water to form 250 mL of the solution is 9 g of Al will react with (a) 0.2 M (b) 0.4 M (a) 6 g O2 (b) 8 g O2 (c) 0.6 M (d) 1 M (c) 9 g O2 (d) 4 g O2 85. 3 g of magnesium is burnt in a closed vessel containing 3 g 94. The hydrated salt, Na2SO4.xH2O undergoes 55.9% loss in of oxgygen. The weight of excess reactant left is weight on heating and becomes anhydrous. The value of x (a) 0.5 g of oxygen (b) 1.0 g of magnesium will be (c) 1.0 g of oxygen (d) 0.5 g of magnesium (a) 6 (b) 5 86. In a gaseous reaction of the type (c) 7 (d) 10 aA + bB → cC + dD 95. How many grams of 80% pure marble stone on calcination which statement is wrong? can give 14 grams of quick lime? (a) a mole of A combines with b moles of B to give C and D. (a) 25 g (b) 31.25 g (b) a litre of A combines with b litre of B to give C and D. (c) 26.85 g (d) 35.15 g (c) a g of A combines with b g of B to give C and D. 96. In the reaction (d) a molecules of A combines with b molecules of B to 2 Al + 6HCl → 2Al3+ + 6Cl– + 3H give C and D. (s) (aq) (aq) (aq) 2(g) (a) 6 L HCl is consumed for every 3 L H produced 87. 0.2 mole of orthophosphoric acid and 1.0 mole of calcium (aq) 2(g) hydroxide were allowed to react. Calculate the maximum (b) 33.6 L H2(g) is produced regardless of temperature and number of moles of calcium phosphate formed. pressure for every mole Al that reacts 18 Chemistry

(c) 67.2 L H2(g) at STP is produced for every mole Al that 107. 26 mL of CO2 are passed over red hot coke. The volume of reacts CO evolved is

(d) 11.2 L H2(g) at STP is produced for every mole HCl(aq) (a) 26 mL (b) 10 mL consumed. (c) 20 mL (d) 52 mL. 97. For the reaction A + 2B → C, 6 mol of A and 8 mol of B will 108. In the aqueous solution of sulphuric acid the mole fraction produce of water is 0.85. The molality of the solution is (a) 5 mol of C (b) 4 mol of C (a) 1 m (b) 0.19 m (c) 8 mol of C (d) 13 mol of C (c) 9.8 m (d) 2.5 m 98. A solution contains one mole of alcohol and four moles of 109. Two solutions of a substance (non-electrolyte) are mixed in water. What are the fractions of water and alcohol? the following manner: (a) 4/1, 1/4 (b) 1/4, 4/1 500 mL of 1.5 M of first solution with 500 mL of 1.2 M of (c) 4/5, 1/5 (d) 1/5, 4/5 second solution. The molarity of final solution is 99. 100 mL of 30% (w/v) NaOH solution is mixed with (a) 1.20 M (b) 1.50 M 100 mL 90% (w/v) NaOH solution. Find the molarity of (c) 1.35 M (d) 2.70 M final solution. 110. How many moles of potassium chlorate need to be heated (a) 1.5 M (b) 13 M to produce 11.2 litres oxygen at NTP? 1 1 1 (c) M (d) 15 M (a) mol (b) mol 15 3 2 1 100. Volume of O2 at STP required to burn completely 70 mL of (c) mol (d) 1 mol acetylene is 4 (a) 100 mL (b) 150 mL 111. Which of the following relations is incorrect for solutions?

(c) 175 mL (d) 200 mL (a) 1 M Al2(SO4)3 = 6 N Al2(SO4)3

101. 1 L of CO2 is passed over hot coke. When the volume (b) 2 M H2SO4 = 4 N H2SO4 of reaction mixture becomes 1.4 L, the composition of (c) 1 M H3PO4 = 1/3 N H3PO4 reaction mixture is (d) 3 N Al2(SO4)3 = 0.5 M Al2(SO4)3

(a) 0.6 L CO (b) 0.8 L CO2 112. 4 g of each pure hydrochloric acid and pure caustic soda are (c) 0.6 L CO2 and 0.8 L CO (d) 0.8 L CO2 and 0.6 L CO together dissolved in water. The weight of sodium chloride obtained is 102. In the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l), when (a) 58.5 g (b) 5.85 g 1 mol of ammonia and 1 mol of O2 are made to react to completion (c) 4 g (d) None of these.

(a) 1.5 mol of H2O is produced 113. If 0.4 mol of BaCl2 is mixed with 0.2 mol of Na3PO4. The (b) 1.5 mol of NO is produced maximum number of moles of Ba3(PO4)2 that can be (c) all the oxygen is consumed formed is (d) all the ammonia is consumed. 3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl

103. 10 mL of concentrated H2SO4 (18 M) is diluted to one litre. (a) 0.7 (b) 0.5 The approximate strength of the dilute acid is (c) 0.3 (d) 0.1 (a) 180 M (b) 18 M 114. How many gram of KCl would have to be dissolved in 40 g

(c) 0.18 M (d) 1.8 M H2O to give 20% by weight of solution? 104. If 8 mL of uncombined O2 remain after exploding O2 with (a) 15 g (b) 10 g 4 mL of hydrogen, the number of mL of O2 originally were (c) 11.5 g (d) 31.5 g (a) 10 (b) 15 115. The percent loss in mass after heating a pure sample of (c) 12 (d) 20 potassium chlorate (mol. mass = 122.5) will be 105. Density of 2.05 M solution of acetic acid in water is (a) 24.5 (b) 28.50 1.02 g/mL. The molality of same solution is (c) 39.18 (d) 49.0 –1 –1 (a) 5.56 mol kg (b) 3.28 mol kg 116. Liquid benzene (C6H6) burns in oxygen according to the –1 –1 (c) 2.28 mol kg (d) 1.28 mol kg equation 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(g). How 106. The production cost of hydrogen from a mineral acid using many litres of O2 at STP will be needed for the complete zinc is rupees 10 per mole. How many electrons is worth of combustion of 39 g of liquid benzene? one rupee? (Mol. wt. of O2 = 32, C6H6 = 78) (a) 6.022 × 1023 (b) 1.2 × 1023 (a) 100 L (b) 11.2 L (c) 2.4 × 1022 (d) None of these (c) 22.4 L (d) 84 L Some Basic Concepts of Chemistry 19

117. 500 mL of gaseous hydrocarbon when burnt in excess of O2 (a) 14.75 g (b) 12 g gave 2.5 L of CO2 and 3.0 L of water vapours under same (c) 2.49 g (d) 4.95 g conditions. Molecular formula of the hydrocarbon is 2. The volume of ethyl alcohol (density 1.15 g/cc) that has to (a) C3H8 (b) C4H10 be added to prepare 100 cc of 0.5 M ethyl alcohol solution (c) C6H12 (d) C5H12 in water is 118. Equal weights of ethane and hydrogen are mixed in an (a) 1.15 cc (b) 2 cc empty vessel at 25 °C. The fraction of the total pressure (c) 2.15 cc (d) 2.30 cc exerted by hydrogen is 3. Statement 1 : The conversion factor of changing the (a) 1/2 (b) 1 volume of a gaseous substance into moles is the same for (c) 1/16 (d) 15/16 all gases. 119. 2 g of metal carbonate is neutralized completely by 100 mL Statement 2 : The molar volumes of gases do not change of 0.1 N HCl. The equivalent weight of metal carbonate is with their molar masses. (a) 50 (b) 100 (a) If both statement 1 and statement 2 are true and (c) 150 (d) 200 statement 2 is the correct explanation of statement 1. (b) If both statement 1 and statement 2 are true but 120. The minimum quantity in gram of H2S needed to statement 2 is not the correct explanation of statement 1. precipitate 63.5 g of Cu2+ will be 2+ + (c) If statement 1 is true but statement 2 is false. Cu + H2S → Cu2S + 2H (d) If both statement 1 and statement 2 are false. (a) 31.75 g (b) 63.5 g 4. A certain oxide of iodine has been found to contain iodine (c) 34 g (d) 68 g and oxygen. The ratio of iodine : oxygen is 254 : 112. On 121. 40% w/V NaCl solution (specific gravity = 1.12) is being dissolved in water this oxide can produce equivalent to (a) HIO2 (b) HIO4 5 6 (a) 3.57 × 10 ppm (b) 3.57 × 10 ppm (c) H5IO6 (d) Both (b) and (c) 6 5 (c) 1 × 10 ppm (d) 4 × 10 ppm 5. Which of the following is not a mixture? 122. If in a reaction HNO3 is reduced to NO, the mass of HNO3 (a) Wax (b) Brass absorbing one mole of electrons would be (c) Iodized salt (d) Marble

(a) 21.0 (b) 63.0 6. A solution containing Na2CO3 and NaOH requires (c) 42.0 (d) 31.5 300 mL of 0.1 N HCl using phenolphthalein as an indicator.

123. Equivalent weight of I2 in the reaction, Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 N HCl is required. The amount I2 + 5Cl2 + 6H2O → 2HIO3 + 10HCl is (a) M (b) M/2 of NaOH present in the original solution is (c) M/5 (d) M/10 (a) 0.5 g (b) 1 g (c) 2 g (d) 4 g 124. An oleum is labelled as 109%. Calculate mass percent of 7. The number of significant figures present in 35000 can be free SO3 and H2SO4. (a) 50%, 50% (b) 60%, 40% (a) 2 (b) 3 (c) 40%, 60% (d) 10%, 90% (c) 5 (d) all of these. 8. Number of atoms in 558.5 g Fe is 125. Statement 1 : Molality is preferred over molarity in –1 expressing the concentration of a solution. (at. wt. of Fe = 55.85 g mol ) (a) half that in 60 g carbon (b) 6.023 × 1022 Statement 2 : The S.I. unit of molarity is mol dm–3. (c) half that in 8 g He (d) 6.023 × 1024 (a) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1. 9. Among the following which does not have the same mass? (b) If both statement 1 and statement 2 are true but (a) 0.1 mole of SO2 gas statement 2 is not the correct explanation of statement 1. (b) 0.1 mole of O2 gas 22 (c) If statement 1 is true but statement 2 is false. (c) 6.02 × 10 molecules of SO2 gas 23 (d) If both statement 1 and statement 2 are false. (d) 1.204 × 10 molecules of O2 gas. 10. 16 g of oxygen has same number of molecules as in LEVEL 2 (a) 16 g of CO (b) 28 g of N2 (c) 14 g of N (d) 10 g of H 1. What mass of barium chloride would be decomposed 2 2 by 9.8 g of sulphuric acid, if 12 g of barium sulphate and 11. Select the incorrect statement. (a) One mole of electrons weighs 0.55 mg. 2.75 g of hydrogen chloride were produced in the reaction (b) Mole correlates mass and number of species present assuming that the law of conservation of mass is true? in a definite amount of matter. 20 Chemistry (c) At STP, 2 mole of a gas occupies 44.8 L. (c) 1.5 mole of NO is formed (d) None of these. (d) both (a) and (c). 12. 10 g of ammonium chloride was mixed with 10 g of slaked 21. Statement 1 : If 100 cc of 0.1 N HCl is mixed with 100 cc of lime and the mixture was heated. Then 0.2 N HCl, the normality of the final solution will be 0.15.

(a) 3.1 g Ca(OH)2 will be left unreacted Statement 2 : Normality of a solution is equal to the (b) 10.4 g of CaCl2 will be formed number of moles of solute present in one litre solution. (c) 3.2 g of NH3 will be formed (a) If both statement 1 and statement 2 are true and (d) all of these are correct. statement 2 is the correct explanation of statement 1. 13. Two solutions are taken separately and both solutions are (b) If both statement 1 and statement 2 are true but found to be at the same temperature. The two solutions statement 2 is not the correct explanation of statement 1. are mixed and a thermometer shows that the temperature (c) If statement 1 is true but statement 2 is false. of the mixture has decreased. Which of the following (d) If both statement 1 and statement 2 are false. statements is incorrect? 22. 100 mL mixture of CO and CO2 is mixed with 30 mL of (a) The chemical reaction is absorbing energy. oxygen and sparked in a eudiometer tube. The residual (b) The chemical reaction is releasing energy. gas after treatment with aqueous KOH has a volume of (c) The chemical reaction is endothermic. 10 mL which remains unchanged when treated with (d) None of these. alkaline pyrogallol. If all the volumes are under the same conditions, point out the incorrect option. 14. Law of multiple proportions is illustrated by which of the (a) The volume of CO that reacts, is 60 mL. following pairs? (b) The volume of CO that remains unreacted, is 10 mL. (a) H2S and SO2 (b) SO2 and SO3 (c) The volume of O2 that remains unreacted, is 10 mL. (c) Na2S and Na2O (d) All of these (d) The volume of CO2 that gets absorbed by aqueous 15. 10 litres of O2 gas is reacted with 30 litres of CO at STP. The KOH, is 90 mL. volumes of gas present at the end of the reaction is 23. Mass of one mole molecules of CO is (a) CO = 10 L (b) CO = 20 L 2 2 (a) 44 g (b) 44 u (c) O2 = 10 L (d) both (a) and (b). (c) 2.65 × 1025 u (d) both (a) and (c). 16. Molecular mass = vapour density × 2, this relation is/are 24. Match the column I with column II and select the correct valid for answer using the code given below the lists : (a) metals (b) non-metals (c) solids (d) gases. Column I Column II (Reactant) (Product) 17. Given below are few statements. Mark the statements which are incorrect. A 1 g hydrogen + 4 g p 1.125 g (a) Gram atomic mass of an element may be defined as carbon the mass of Avogadro’s number of atoms. B 1 g hydrogen + 1g q 4 g (b) The molecular mass of a diatomic elementary gas is oxygen twice its atomic mass. C 1 g nitrogen + r 1.21 g (c) Gay Lussac’s law of chemical combination is valid for 1 g hydrogen all substances. (d) A pure compound has always a fixed proportion of D 1 g carbon + s 3.66 g masses of its constituents. 4 g oxygen (a) A → p; B → q; C → r; D → s 18. 8 g of O has the same number of molecules as in 2 (b) A → q; B → p; C → r; D → s (a) 11 g of CO (b) 22 g of CO 2 2 (c) A → s; B → r; C → p; D → q (c) 7 g of CO (d) both (b) and (c). (d) A → r; B → s; C → q; D → p 19. Among FeSO4. 7H2O (A), CuSO4.5H2O (B), ZnSO4.7H2O 25. Which of the following have same number of significant (C), MnSO4.4H2O (D), isomorphous salts are figures? (a) A and C (b) A and D (a) 50 (b) 5.0 (c) B and C (d) B and D. (c) 0.50 (d) All of these 20. In the reaction, 4 NH3 + 5O2 → 4NO + 6H2O, when one 26. One mole electron does not mean mole of ammonia and one mole of oxygen are made to (a) NA electrons react to completion, then the incorrect option is (b) 6.023 × 1023 electrons (a) 1.0 mole of H2O is produced (c) electron present in 1F charge (b) all the oxygen is consumed (d) 2NA electrons. Some Basic Concepts of Chemistry 21 27. Sulphuric acid reacts with sodium hydroxide as follows : The empirical formula of the welding gas is

H2SO4 + 2NaOH → Na2SO4 + 2H2O (a) CH (b) CH2 When 1 L of 0.1 M sulphuric acid solution is allowed to (c) CH3 (d) None of these. react with 1 L of 0.1 M sodium hydroxide solution, the 34. The molar mass of the above welding gas is amount of sodium sulphate formed and its molarity in the (a) 20 g (b) 22 g solution obtained is (c) 24 g (d) 26 g –1 –1 (a) 7.10 g and 0.1 mol L (b) 7.10 g and 0.025 mol L 35. The molecular formula of the above welding gas is (c) 5.40 g and 0.1 mol L–1 (d) 3.55 g and 0.025 mol L–1 (a) C2H2 (b) C2H4 28. 10 g of hydrogen and 64 g of oxygen were filled in a steel (c) C2H6 (d) none of these. vessel and exploded. Amount of water produced in this 36. A metal oxide has the formula Z2O3. It can be reduced by reaction will be hydrogen to give free metal and water. 0.1596 g of the metal (a) 3 mol (b) 5 mol oxide requires 6 mg of hydrogen for complete reduction. (c) 72 g (d) 36 g The atomic weight of the metal is 29. Which of the following have the same mass? (a) 27.9 (b) 159.6

(i) 0.1 mol of Fe2O3 (c) 79.8 (d) 55.8 23 (ii) 3.01 × 10 molecules of SO2 37. One litre of a mixture of O2 and O3 at NTP was allowed to (iii) 0.5 mol of O2 gas react with an excess of acidified solution of KI. The iodine (iv) 11200 mL of ozone gas at STP. liberated required 40 mL of M/10 sodium thiosulphate

(a) (i) and (ii) solution for titration. Moles of O2 in the mixture was (b) (i) and (iii) (a) 0.044 (b) 0.042 (c) (ii) and (iii) (c) 0.956 (d) 0.958

(d) (i), (iii), (iii) and (iv) 38. Weight of O3 in the above mixture was 3 30. 50 cm of 0.04 M K2Cr2O7 in acidic medium oxidizes a (a) 0.042 g (b) 0.096 g sample of H2S gas to sulphur. Volume of 0.03 M KMnO4 (c) 1.344 g (d) 0.044 g

required to oxidize the same amount of H2S gas to sulphur, 39. Percentage of O3 in the above mixture was in acidic medium is (a) 67% (b) 33% (a) 60 cm3 (b) 80 cm3 (c) 58% (d) 25% 3 3 (c) 90 cm (d) 120 cm 40. 1 cc N2O at NTP contains 31. Which of the following statements is/are correct? 18. (a) ×1022 atoms I. One gram atom of carbon contains Avogadro’s 224 number of atoms. 60. 2 II. One mole of oxygen gas contains Avogadro’s number (b) ×1023 molecules of molecules. 22400 III. One mole of hydrogen molecule contains Avogadro’s 13. 2 23 number of atoms. (c) ×10 electrons 23 224 IV. One mole of electrons stands for 6.02 × 10 electrons. (d) all of the above. (a) II and IV only (b) I only 41. From a mixture which makes up crude oil, a particular (c) III only hydrocarbon ingredient (that is one containing hydrogen and carbon atoms only) has been isolated. 10 g of this (d) I, II and IV only liquid are burned in excess of oxygen and the products are 32. Ethanol-water mixture has 46 g ethanol in 100 g mixture. 31.4 g of carbon dioxide and 12.9 g of water. The molar By a suitable technique volatile component goes off. Thus, ratio of carbon dioxide and water present is (a) 3 mol of non-volatile component are left (a) 1 : 2 (b) 1 : 1 (b) 9NA atoms of non-volatile component are left (c) 2 : 1 (d) 1 : 4 (c) 9N atoms of volatile component are separated A 42. If we burn an equimolar mixture of the above hydrocarbon (d) all of these are correct. and oxygen in a closed vessel, then after the reaction the 33. A welding fuel gas contains only carbon and hydrogen. gaseous mixture present in the vessel will consist of Burning a small sample of it in oxygen gives 3.38 g of (a) CO2 and H2O carbon dioxide, 0.690 g of water and no other products. A (b) CO2, H2O and O2 volume of 10.0 L (measured at STP) of this welding gas is (c) CO2, H2O and hydrocarbon found to weigh 11.6 g. (d) CO2, H2O, hydrocarbon and O2. 22 Chemistry

43. A student performs a titration with different burettes Column I Column II and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. (A) Mass of H produced when (i) 3.01 × 1023 The number of significant figures in the average titre 2 0.5 mole of zinc reacts with molecules value is excess of HCl (a) 2 (b) 4 23 (c) 3 (d) 1 (B) Mass of all atoms of a (ii) 6.023 × 10 compound with formula molecules 44. A sample of 3.55 g of bleaching powder was dissolved in C70H22 100 mL of water. 25 mL of this solution, on treatment with –21 KI and dilute acid, required 10 mL of 0.125 N sodium (C) Number of molecules in (iii) 1.43 × 10 g thiosulphate solution. The percentage of available chlorine 35.5 g of Cl2 in the sample is (D) Number of molecules in (iv) 1 g (a) 5 % (b) 95 % 64 g of SO2 (c) 10 % (d) 70 % (a) (A) → (ii), (B) → (i), (C) → (iv), (D) → (iii) 45. Match the column I with column II and select the correct (b) (A) → (i), (B) → (ii), (C) → (iii), (D) → (iv) answer using the code given below the lists : (c) (A) → (iv), (B) → (iii), (C) → (i), (D) → (ii) Column I Column II (d) (A) → (iv), (B) → (iii), (C) → (ii), (D) → (i) A 46 g Na p 0.5 mol 48. If 20.0 g of CaCO3 is treated with 20.0 g of HCl, then grams B 5.6 L O q 6.023 × 1023 molecules 2 of CO2 produced according to the reaction C 22 g CO r 7.83 × 1024 atoms 2 CaCO3(s) + 2HI(aq) → CaCl2(aq) + H2O(l) + CO2(g) is D 100 g CaCO3 s 2 mol 23 (a) 12 g (b) 9 g E 52 g He t 3.01 × 10 atoms (c) 44 g (d) 6 g (a) A → p; B → r; C → t; D → q; E → s 49. 3 g of activated charcoal was added to 50 mL of acetic acid (b) A → q; B → s; C → r; D → p; E → t solution (0.06 N) in a flask. After an hour it was filtered (c) A → s; B → t; C → p; D → q; E → r and the strength of the filtrate was found to be 0.042 N. The (d) A → r; B → p; C → t; D → q; E → s amount of acetic acid adsorbed (per gram of charcoal) is 46. A solution containing 0.1 mol of a metal chloride MClx (a) 42 mg (b) 54 mg requires 500 mL of 0.8 M AgNO3 solution for complete (c) 18 mg (d) 36 mg reaction MClx + xAgNO3 → xAgCl + M(NO3)x. The value 50. When 3.2 g S is vaporized at 450 °C and 723 mm pressure, of x is the vapours occupy a volume of 780 mL. If the molecular (a) 3 (b) 1 formula of S vapour under these conditions is Sx, then the (c) 5 (d) 4 value of x is 47. Match the column I with column II and mark the (a) 6 (b) 4 appropriate choice. (c) 2 (d) 8

SELF TEST-ANSWER KEYS SELF TEST 1 1. (a) 2. (b) 3. (b) 4. (c) 5. (c) 6. (b) 7. (b) 8. (c) 9. (a) 10. (d) 11. (b) 12. (c) 13. (a) 14. (b) 15. (b) SELF TEST 2 1. (a) 2. (c) 3. (c) 4. (a) 5. (d) 6. (a) 7. (a) 8. (d) 9. (b) 10. (a) 11. (c) 12. (c) 13. (d) 14. (d) 15. (a) SELF TEST 3 1. (c) 2. (a) 3. (b) 4. (c) 5. (d) 6. (c) 7. (d) 8. (c) 9. (c) 10. (a) 11. (c) 12. (a) 13. (d) 14. (d) 15. (b) Some Basic Concepts of Chemistry 23 Hints & Explanations

LEVEL 1 chemical changes the total mass of the system remains constant or in a chemical reaction, mass is neither created nor destroyed. 1. (b) In (b) and (c), mass of reactants is not equal to the mass of 2. (a) : Water and alcohol are completely mixed and form products. uniform solution. 16. (b) : Percentage of carbon in sample 1 12. 3. (a) = ×=100 27.%3 12..+ 32 4. (d) : According to the physical state at room temperature 32. and pressure, the matter is present in three states solid, liquid Percentage of oxygen in sample 1 = ×=100 72.%7 12..+ 32 and gas. Both samples have same percentage composition, hence this is 5. (d) in accordance with law of definite proportions. 6. (c) 17. (c)

7. (d) : Marble (CaCO3) is a compound. 18. (d) 8. (c) 19. (d) : First oxide Second oxide 9. (c) : In ammonia, % of N = 82.35%; % of H = 17.65% Sulphur 50% 40% Now, the wt. of nitrogen that combines with 1 part by wt. of H Oxygen 50% 60% 82.35 46. 7 H 50 in NH3 ==17.65 In first oxide, 1 part oxygen combines with sulphur = = 1 part. 50 In water, % of H = 11.10%; % of O = 88.90% 40 Similarly for second oxide = = 06. 7 parts Now, the wt. of oxygen that combines NH3 H2O 60 with 1 part by wt. of H in H2O So the ratio is 1 : 0.67 or 3 : 2 88.90 ==80. 1 20. (b) : In H S, 5.89 g H combines with 94.11 g S 11.10 N N O O 2 2 3 Hence, 1 g H combines with 16 g S In the two compounds NH3 and H2O, weights of nitrogen and oxygen that combine with the same weight (1 part) of hydrogen, In SO2, 50 g O combines with 50 g S are in the ratio of 4.67 : 8.01 or 1 : 1.71 ...(i) Hence, 1 g O combines with 1 g S In nitrogen trioxide, In H2O, 11.11 g H combines with 88.89 g O % of N = 36.85 ; % of O = 63.15 Hence, 1 g H combines with 8 g O In this compound, the ratio of the weights of nitrogen to oxygen Thus, law of reciprocal proportions is followed. is 36.85 : 63.15 or 1 : 1.71 ...(ii) 21. (b) As the ratio of N and O in both the cases i.e. in eq. (i) and (ii) 22. (d) are same, these data illustrate the law of reciprocal proportions. 23. (d) 10. (c) : H2O, H2S and SO2 illustrate the law of reciprocal proportions. 24. (c) : It is Avogadro’s hypothesis. 11. (b) : According to Gay Lussac’s Law : 25. (c) : Dalton failed to explain Gay-Lussac’s law of combining volume. HC22()gg+→l2() HCl()g . . 26. (a) . 1 mL of H2(g) will react with 1 mL of Cl2(g) and 2 mL of HCl(g) will produce. 27. (d) : The natural abundance of 75.77% for Cl-35 means that out of 100 chlorine atoms in nature, 75.77 are of the isotope \ 40 mL of H2(g) will react with 40 mL of Cl2(g) and 80 mL of Cl-35. HCl(g) will produce \ The natural abundance of Cl-37 = 100 – 75.77 = 24.23% \ Required volume of Cl2(g) = 40 mL \ The relative atomic mass of chlorine = Produced volume of HCl(g) = 80 mL 75.77 24.23 12. (d) 34.9689 ×+36.9659 ×=35.45 u 100 100 13. (b) \ The atomic weight of chlorine is 35.45. 14 14. (a) 28. (c) : Number of protons in 6C = 6 14 15. (d) : According to law of conservation of mass in all Number of neutrons in 6C = 8 24 Chemistry 14 As per given new atomic mass of 6C = 12 + 4 = 16 35 ×+xx37()1− = 35.5 (As the mass of electron is negligible as compared to neutron 1 and proton) x = 0.75 ⇒ 1 – x = 0.25 16 −14 35 37 35 37 % increase in mass = ×=100 14.28 \ Ratio of Cl and Cl is Cl : Cl = 3 : 1 14 40. (d) : This law is applicable only to solid elements except Be, 29. (b) : The natural abundance of Br-81 = 100 – 50.54 = 49.46% B, C and Si. According to this law, If x is the atomic mass of Br-81 then the relative atomic mass 6.4 50..54 49 46 Atomic weight = =×78.9183 +×xx=+39..8853 0 4946 Specific heat 100 100 ... \ 39.8853 + 0.4946x = 79.90 41. (c) : 1 mol of C6H12O6 = 6 NA atoms of C ⇒ 0.4946x = 79.90 – 39.8853 =40.0147 \ 0.35 mol of C6H12O6 = 6 × 0.35 NA atoms of C ⇒ = = 2.1 NA atoms x 80.9031 23 The atomic mass of Br-81 is 80.90 u. = 2.1 × 6.022 × 10 = 1.26 × 1024 carbon atoms 30. (a) 42. (c) : At STP, 22400 cc = 6.023 × 1023 atoms 31. (d) : Average atomic mass Number of atoms present in one cc at STP 90 8 2 = 200 ×+199 ×+202 × 6.022 ×1023 100 100 100 = =×26. 91019 4 = 180 + 15.92 + 4.04 = 199.96 ≈ 200 u 22. 41× 0 32. (b) 43. (c) : The mass of 3.60 mol of H2SO4 = 3.60 mol × 98 g/ mol Density of gas 0.004450 = 352.8 g 33. (d) : V.D. === 50 Density of H 0.000089 100 2 The mass of 1 L of the acid solution = 352.8 g × Molecular mass = 2 × V.D. = 2 × 50 = 100 29 = 1216.5 g 34. (b) : 54Fe → 5% 56Fe → 90% 1216.5 g 57 \ Density = =1.2165 g/mL Fe → 5% 1000 mL Av. atomic mass = x1A1 + x2A2 + x3A3 14 44. (a) : Mole of N2 = = 05. = 54 × 0.05 + 56 × 0.9 + 57 × 0.05 = 55.95 28 35. (b) : According to Dulong and Petit’s rule 22 Mole of CO2 = = 05. Atomic weight × sp. heat  6.4 44 64. (Approx.) atomic weight  = 64 So, total moles = 0.5 + 0.5 = 1 01. Volume at STP = 1 × 22.4 = 22.4 L Equivalent weight = 31.8 –31 64 45. (d) : 9.108 × 10 kg is the weight of one electron = Valency of element (n) = 2 1 31.8 \ 1 kg is the weight of electron Atomic weight = n × E = 2 × 31.8 = 63.6 9.108 ×10−31 1 1 36. (b) = × mole electron × −31 × 23 mass of 1 mol gas 9..108 10 6 023 10 37. (c) : Relative vapour density = 108 mass of 1 mol CH4 = m × 4 = \ m = 64 9..108 6 023 m 64 16 So, V.D == = 32 46. (d) : NA molecules of CO2 has molecular mass = 44 g 2 2 4 NA molecules of CO2 has mass = 44 × 4 = 176 g \ \ V.D. of substance = 32 Mol wt. = 2 × 32 = 64 47. (c) : Air is a homogenous mixture. It has 79% nitrogen and 38. (b) : The compound must possess at least one mole of 21% of oxygen by volume. If we take one mole of air, it would sulphur atoms, i.e. 32 g sulphur, i.e. 8% of molecular mass = 32 g contain 0.79 mole of nitrogen and 0.21 mole of oxygen. 32 \ Molecular mass = ×100 = 400 The weight of one mole of air = 0.79 × 28 + 0.21 × 32 = 28.84 8 Thus, the effective molecular weight of air is 28.84. 39. (b) : Suppose x fraction of the gas is Cl35 then there will be (1 – x) fraction of Cl37 48. (c) : The molar mass of water = 18 g Some Basic Concepts of Chemistry 25

18 g 3 160 The molar volume of liquid water = = 18 cm 56. (b) : Given moles = = 25. 1 g/cm3 64 24 1 mol of water contains the Avogadro number (6.023 × 1023) of 1.2046 ×10 Removed moles = = 2 molecules. 6.023 ×1023 The volume occupied by one molecule So, left moles = 0.5 18 cm3 Volume left at STP = 0.5 × 22.4 = 11.2 L = =×29. 910−23 cm3 23 6.023 ×10 57. (c) : (NH4)2Cr2O7 contains 2 mole of N-atom, 8 mole of H-atom, 2 mole of Cr-atom and 7 mole of O-atom 49. (b) : 8 mole of O-atom ≡ 1 mole Mg3(PO4)2 10× .25 Total atoms of all element = 2 + 8 + 2 + 7 = 19 0.25 mole of O-atom = = 19 × 6.023 × 1023 atoms 8 23 –2 = 114.437 × 10 atoms = 3.125 × 10 mole of Mg3(PO4)2 6 42. 50. (c) : 10 rupees are spent in 1 sec 58. (a) : No. of valence electrons = 8 ××NNAA= 24. 23 14 23 16××.022 10 23 6.022 × 10 rupees are spent in = sec 59. (d) : 2 moles of K4[Fe(CN)6] contains 2 × 6.023 × 10 106 molecule = 4 × 2 × 6.023 × 1023 atoms of K. 23 16××.023 10 9 = years = 19.098 × 10 years 60. (c) : Ratio of molecules of N2 and O2 106 ×××60 60 24 × 365 = Ratio of number of moles of N2 and O2 51. (b) : Weight of Hg = Volume × Density 4 1 1 1 = ::==32 : 7 = 1000 × 13.6 = 13600 g 28 32 7 32 Weight 61. (c) : 1 L of O and 1 L of O contain the same number of Number of moles of Hg in 1 L = 2 3 Gram atomic weight moles and hence same number of molecules but not atoms. 13600 Number of atoms in O will be 1.5 times than that of O . ==68 3 2 200 62. (c) : Mol. wt. of CaCl2 = 111 g 52. (a) : ... 0.224 L has mass 0.44 g 2+ 111 g CaCl2 = NA ions of Ca 04..42× 24 22.4 L has mass = = 44 g (i.e., N2O) NA × 222 0.224 222 g of CaCl = = 2N ions of Ca2+ 2 111 A 53. (b) : Iron present in 100 g haemoglobin = 0.25 g – 02. 5 × 89600 Also, 111 g CaCl2 = 2NA ions of Cl \ Iron present in 89600 g haemoglobin = 2NA × 222 – – 100 222 g of CaCl2 = ions of Cl = 4 NA ions of Cl = 224 g 111 . 1 mole or NA molecules of haemoglobin has 224 g Fe or g-atoms 63. (b) : The formula of hydrate is Fe(SCN)3 xH2O of Fe = (224/56) = 4 Molecular mass of hydrate = 56 + 3 × (32 + 12 + 14) + 18 x = 4 g-atom of Fe = 4 NA atom of Fe = 230 + 18 x 18x ×100 \ 1 molecule of haemoglobin = 4 atoms of Fe % of H O = =⇒19 x = 3 2 + 54. (d) : The molecule has C, H and other component. 230 18x Mass of 9 C atoms = 12 × 9 = 108 amu 64. (b) : Mol. mass of mixutre (NO2 + N2O4) = 2 × V.D. Mass of 13 H atoms = 13 × 1 = 13 amu = 38.3 × 2 = 76.6 − Let x mole of NO is present in 100 mole mixture 23. 31× 0 23 2 Mass of other component = = 14.04 amu −24 \ Mass of NO2 + Mass of N2O4 = Mass of mixture 16. 61× 0 or, x × 46 + (100 – x) × 92 = 100 × 76.6 Total mass of one molecule = 108 + 13 + 14.04 = 135.04 x = 33.48 mole Mol. mass of substance = 135.04 amu 65. (a) : Molar mass of gypsum, CaSO .2H O = 172 g 55. (b) : 100 g of the alloy contains 90 g of Cu and 10 g of Al 4 2 Weight of water of hydration = 2 × 18 = 36 g 90 g Amount of Cu in 100 g of the alloy = = 14. 2 mol 172 g loses 36 g of water on dehydration. −1 63.5 g mol 100 \ Percentage weight loss =×36 = 20.93 10 g = 0.370 mol 172 Amount of Al in 100 g of the alloy = −1 27.0 g mol 224 −2 66. (c) : Moles of air ==10 mole Number of atoms of Cu 14. 2 mol × NA atomm/mol 14. 2 22400 = = (... 1 mole air at STP = 22400 mL) Number of atoms of Al 0.370 mol × NA atom/mol 0.370 \ No. of gas molecules = 10–2 × 6.023 × 1023 Thus, the ratio of Cu atoms to Al atoms in the alloy = 142 : 37 = 6.023 × 1021 air molecules 26 Chemistry

21 6.023 ××10 78 21 H 13 1 13 13 6 or No. of N2 molecules = =×47. 010 = 13 = 59. 7 100 1 2.175 21 6.023 ××10 21 21 and No. of O2 molecules = =×12. 610 O 24.8 16 34.8 2.175 1 100 = = 21 2.175 1 \ Total no. of N2 and O2 molecules = (4.70 + 1.26) × 10 16 2.175 = 5.96 × 1021 Empirical formula = C2H6O 67. (b) : Given that xC6H6 = 0.5 and so xC6H5CH3 = 0.5 Empirical formula mass = 12 × 2 + 6 + 16 = 46 Thus mixture contains equal moles of C6H6 and C6H5CH3. Molecular formula mass 92 Let one mole of each is mixed, then mixture will contain 78 g n === 2 Empirical formula mass 46 C H and 92 g C H CH with 170 g mass in total. 6 6 6 5 3 Molecular formula = 2 × (C H O) = C H O 78 2 6 4 12 2 \ %. C H =×100 = 45 9% 66 170 73. (b) : 92 Elements Relative no. Simple Simplest and %. C HCH =×100 = 54 1% 65 3 170 of atoms atomic ratio whole no. ratio 68. (a) : 1.429 g of O2 gas occupies volume = 1 L C 1 1 32 10.06 32 g of O gas occupies = = 22.4 litre/mol = 08. 4 2 1.429 12 21 69. (a) : 1 L of air contains 1 L ×=02. 1 L of oxygen H 08. 4 1 1 100 = 08. 4 1 Under standard conditions, 22.4 L of oxygen = 1 mol of oxygen 1 mol Cl 89.1 3 3 \ 0.21 L of oxygen = 02. 1 L × = 25. 1 22.4 L 35.5 = 0.009375 mol \ Empirical formula = CHCl3 107 70. (c) : Moles of CH4 = Empirical formula weight = 119.5 NA Molecular weight = 2 × V.D. = 2 × 60 = 120 107 \ Empirical formula ≡ Molecular formula Mass of CH4 = ×16 = Mass of C2H6 NA 74. (a) : Molar mass = 108 g/mol 7 10 ×16 Elements Wt. Wt. ratio/ Simplest Simplest So, moles of CH26= NA × 30 ratio Atomic mass ratio whole no. 107 ×16 ratio \ No. of molecules of CH = ×=N 53..41× 06 26 N × 30 A C 9 x 9xx3 3 3 A = 71. (d) : 12 4 Elements Relative no. of Simplest ratio Simplest whole H 1 x x 4 4 atoms no. ratio N 3.5 x 35. xx 1 1 = N 25.94 1 2 14 4 = 18. 5 14 \ Empirical formula is C3H4N O 74.06 2.5 5 Empirical formula mass = 12 × 3 + 4 + 14 = 54 = 46. 3 16 108 n ==2 So, empirical formula is N2O5. 54 \ Molecular Formula = C H N 72. (b) : 6 8 2 1× at wt. of metal % com- Atomic Relative Simplest Simplest 75. (b) : % of metal = ×100 position weight no. of ratio whole no. Minimum molecular weight atoms ratio 15××9 100 Elements 02. 5 = M C 52.2 12 52.2 43. 5 2 59 ×100 = 43. 5 = 2 M = = 23600 12 2.175 02. 5 Some Basic Concepts of Chemistry 27

76. (c) : N 11.4 11.4/14.0 = 0.814/0.814 1 % Relative no. Simplest Simplest 0.814 = 1.00 composition of atoms atomic ratio whole O 26.0 26.0/16.0 = 1.62/0.814 2 number ratio

Elements 1.62 = 1.99

Na 29.0 29.0/23.0 1.26/1.26 = 1 2 The empirical formula of the compound is 6C H5NO2. = 1.26 81. (c) : The molar mass of Na2SO4 S 40.5 40.5/32.0 1.26/1.26 = 1 2 = 2 × gram atomic weight of Na + 1 × gram atomic weight of S = 1.26 + 4 × gram atomic weight of O O 30.5 30.5/16.0 1.90/1.26 = 3 = 2 × 23.0 g + 1 × 32.0 g + 4 × 16.0 g = 142.0 g = 1.9 1.5 41× 60. g Hence, % of O = ×=100 45.0 The empirical formula of the compound is Na2S2O3. 142.0 g The empirical mass = 2 × 23 + 2 × 32 + 3 × 16 = 158 mass of C 82. (d) : % of C = ×100 Since the empirical mass is equal to the formula mass, the molar mass formula of the compound is Na2S2O3. 21×12 69.98 = ×100 77. (b) : M Elements % %/Atomic Simplest Simplest M = 360.1 mass ratio whole no. 1×× at. wt. of S 100 83. (a) : % of S = Ca 20 20/40 = 0.5 1 1 Minimum molecular mass Br 80 80/80 = 1 2 2 13××2 100 34. = Hence, empirical formula = CaBr2 M 200 32 ×100 n ==1 Molecular mass (M) = = 941.176 200 34. Hence, molecular formula = CaBr Number of moles of solute ()n 2 84. (b) : Molarity (M) = 78. (c) : Volume of solution in litres ()v 44 g / 0 g −1 Elements Relative no. of atoms Simplest Simplest ==04.. mol L = 04 M 0.250 L ratio whole no. 1 ratio 85. (c) : Mg + O → MgO 2 2 K 26.6 06. 8 2 = 06. 8 = 1 3 g 3 g 39 06. 8 = 0.125 mol = 0.1875 mol Here limiting reactant is Mg. Cr 35.4 06. 8 2 = 06. 8 = 1 So, oxygen left unreacted = 0.1875 – 0.125 = 0.0625 mol 52 06. 8 Hence, weight of oxygen = 0.0625 × 16 = 1.0 g O 38.1 23. 8 7 86. (c) = 23. 8 = 35. 16 06. 8 87. (b) : The stoichiometric equation is 3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 3H2O \ Empirical formula is K2Cr2O7 ≡ 194 × 28.9 3 moles of Ca(OH)2 2 moles of H3PO4 79. (c) : Wt. of nitrogen in caffeine = 1 mole of Ca(OH) ≡ 0.67 mole of H PO 100 2 3 4 ≡ 194 × 28.9 0.3 mole of Ca(OH)2 0.2 mole of H3PO4 No. of atoms in one molecule = × 2 = 4 H PO is the limiting reagent. Hence, salt formed is dependent 100 × 28 3 4 on the availability of acid only. 80. (c) : 2 moles of H3PO4 ≡ 1 mole of Ca3(PO4)2 % com- Relative no. of Simplest Simplest Number of moles of calcium phosphate formed position atoms atomic ratio whole no. 1 = 02..×=01 mole ratio 2 Elements 88. (b) : KMnO + 3H SO → K SO + 2MnSO + 3H O + 5/2 O C 58.5 58.5/12 = 4.88 4.88/0.814 6 4 2 4 2 4 4 2 2 2.5 moles of O is produced by 2 moles of KMnO = 5.99 2 4 2 H 4.06 4.06/1.01 = 4.02/0.814 5 48 g or 1.5 moles of O2 is produced by =×15..= 12 25. 4.02 = 4.94 Mass of KMnO4 = 1.2 × 158 = 189.6 g 28 Chemistry 6 89. (d) : ppm = mass of solute in 10 g H2O 80 g of calcium carbonate ≡ 100 g of marble stone 103 g H O contains 10 g CaCO 25 2 3 Weight of marble stone to be calcinated =×100 = 31.25 g 6 6 10 ×10 4 80 \ 10 g H O contains = = 10 g CaCO 3+ – 2 3 3 96. (d) : 2 Al + 6HCl → 2Al + 6Cl + 3H 10 (s) (aq) (aq) (aq) 2(g) For each mole of HCl reacted, 0.5 mole of H gas is formed 90. (b) : 2 Mg + O → 2MgO 2 2 at STP. 1 mole of an ideal gas occupies 22.4 L at STP. 2 moles of Mg ≡ 2 moles of MgO Volume of H2 gas formed at STP per mole of HCl reacted is 1 mole of Mg ≡ 1 mole of MgO 22.4 × 0.5 L = 11.2 L 24 g of Mg ≡ 40 g of MgO 97. (b) : A + 2B → C The weight of magnesia (MgO) produced by burning 2 g of Given 6 mol 8 mol 0 40 metal =×23= .333 g (reaction will occur according to B) 24 After reaction (6 – 4) (8– 8) 4 mol 91. (d) : PbO + 2HCl → PbCl2 + H2O 1 1 65. g 98. (c) : Xalcohol = = ; Xwater = 4/5 65. g of PbO ==0.029 mol 14+ 5 223.2 g/mol 99. (d) : Total wt. of NaOH = 30 + 90 = 120 32. g Total vol. of solution = 100 + 100 = 200 32. g of HCl ==0.088 mol 120 1000 36.5 g/mol M =× = 15 40 200 0.088 mol of HCl will react with 0.044 mol of PbO, so the 100. (c) : Combustion of acetylene is given as supplied PbO is less than the required amount. PbO is, therefore, 2C H + 5O → 4CO + 2H O the limiting reactant. 0.029 mol of PbO will produce 0.029 mol 2 2 2 2 2 2 moles of C2H2 ≡ 5 moles of O2 of PbCl2. 2 volumes of C2H2 = 5 volumes of O2 10 1 92. (b) : Normality =× ×=1000 04. 2 mL of C2H2 = 5 mL of O2 50 500 Volume of oxygen required to burn 70 mL of acetylene Mass of solution = 500 × 1.02 = 510 g 5 =×70 = 175 mL Mass of solvent = 510 – 10 g = 500 g 2 10 1 → ∴=Molality ××1000 = 02. 101. (c) : CO2(g) + C(s) 2CO(g) 100 500 Initially 1 L After reaction (1 –x )L 2x L Mole fraction : 500 g H2O = 500/18 mol = 27.78 mol 10 Given, 1– x + 2x = 1.4 10 g acid = mol = 0.1 mole \ 100 x = 0.4 L Hence, vol. of CO = 1 – 0.4 = 0.6 L 27.78 2 χ = = 0.996 and vol. of CO = 2 × 0.4 = 0.8 L HO2 27..78 + 01 5 Mass fraction = 500/510 = 0.98 102. (c) : 1 mol of ammonia requires = 12. 5 mol of oxygen 4 9 1 for complete reaction. 93. (b) : Moles of Al == Hence, the oxygen supplied is less than the required amount and 3 1 1 27 3 Moles of O2 = ×× will be completely consumed in the reaction, producing 4/5 mol 2 2 3 of NO and 6/5 mol of H2O. 3 1 1 Weight of O2 = ×××=32 8 g 103. (c) : Milli mole of H SO = Milli mole of H SO 2 2 3 2 4 2 4 (Conc.) (Dil.) 18x 94. (d) : ×=100 55.9 10 × 18 = M × 1000 ⇒ M = 0.18 142 +18x 1 104. (a) : OH+→HO x = 10 2 22()g ()g2 95. (b) : On calcination, marble stone decomposes to give Initially V mL 4 mL quick lime (CaO). After reaction V( – 2) 0

CaCO3 → CaO + CO2 uncombined O2 ⇒ (V – 2) = 8 mL 1 mole of CaO ≡ 1 mole of CaCO3 V = 10 mL 56 g of CaO ≡ 100 g of CaCO3 105. (c) : 2.05 moles present in 1000 mL of solution 14 20. 5 The weight of calcium carbonate required ×100 = 25 g M = ×=1000 22.8 mol kg−1 56 ()1020 −120 Percentage purity is only 80% (Acetic acid exists as dimer) Some Basic Concepts of Chemistry 29

106. (b) : Reaction of mineral acid with zinc is given as 115. (c) : 2KClO3 → 2KCl + 3O2 ↑ Zn + 2HCl → ZnCl2 + H2 245 g KClO3 on heating shows a mass loss = 96 g (of O2) The net reaction is \ 100 g KClO3 on heating shows a mass loss 2H+ + 2e– → H 96 ×100 2 = g.= 39 18% 1 mole of H2 ≡ 2 moles of electrons 245 23 10 rupees ≡ (2 × 6.022) × 10 electrons 39 g 116. (d) : 39 g of benzene ==05. mol Number of electrons worth of one rupee −1 1 78 g mol = ××26..022 ×=1023 1 204 ×1023 10 2 mol of benzene requires 15 mol of oxygen → 05. ×15 107. (d) : CO2 + C 2CO 0.5 mol of benzene requires mol = 3.75 mol of oxygen (Red hot) 2 1 vol. 2 vol. The volume of 1 mol of oxygen at STP = 22.4 L, so the volume 26 cc 52 cc of 3.75 mol of oxygen at STP = 3.75 mol × 22.4 L/mol = 84 L. → 108. (c) : XXHO==08.,50 HSO .15 117. (d) : CxHy(g) + O2(excess) CO2(g) + H2O(g) 2 24 500 mL = 0.5 L 2.5 L 3L XHSO moles of H SO 01. 5 ratio of vol. 1 5 6 \ 2 4 ==2 4 X moles of H O 08. 5 C H + O → 5CO + 6H O HO2 2 x y(g) 2(excess) 2 2 Comparing both sides mole of HSO m = 24 x = 5 and y = 12 mass of H2O (in kg) Hence hydrocarbon would be C5H12. mole of HSO 1000 01. 5 ×1000 118. (d) : Suppose, x g of each gas is present. or m =×24 = = 98. mole of HO2 18 08. 51× 8 x Number of moles of H2 = 109. (c) : (500 × 1.5) + (500 × 1.2) = M × 1000 2 x \ M = 1.35 Number of moles of C2H6 = 30 3 8x 110. (a) : KClO32→+KClO Total number of moles = 2 15 3 15 mole or 33.6 litres of O2 is produced from 1 mole KClO3 Mole fraction of H2 = 2 16 1 1 Mole fraction of C2H6 = 11.2 litres of O2 is produced from mole of KClO3 16 3 Pressure exerted is directly proportional to mole fraction. 111. (c) : Normality = Molarity × valency factor 15 \ 1 M H PO = 3 N H PO 3 4 3 4 \ Fraction of total pressure exerted by H2 = 16 112. (b) : The stoichiometric equation is given as 119. (d) : Number of gram equivalents of HCl HCl + NaOH → NaCl + H O Normality × V 01. ×100 2 = = = 00. 1 1 mol of HCl ≡ 1 mol of NaOH 1000 1000 36.5 g of HCl = 40 g of NaOH Number of gram equivalents of metal carbonate NaOH is the limiting reagent. Therefore, the weight of NaCl = number of gram equivalents of HCl obtained is limited by NaOH. w 2 =⇒00..1 =⇒001 E = 200 1 mol of NaOH = 1 mol of NaCl EE 40 g of NaOH = 58.5 g of NaCl 2+ 120. (c) : Cu + H2S → Cu2S + 2H ↑ 4 The weight of sodium chloride obtained = 58..5 ×=585 g 63.5 g 40 = 1 mol

113. (d) : 3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl So, 1 mol H2S i.e, 34 g H2S is required. ↓ ↓ ↓ 121. (a) : 100 mL solution = 100 × 1.12 = 112 g 3 mol 2 mol 1 mol 112 g solution consists of 40 g NaCl Given 0.4 mol 0.2 mol 40 106 g solution consists of =×1065=×35. 710 ppm So, 0.3 mol of BaCl2 will react with 0.2 mol of Na3PO4 and gives 112 0.1 mol of Ba (PO ) . 3 4 2 122. (a) :

114. (b) : 20 g KCl present in (100-20) = 80 g of H2O Mass of HNO3 absorbing one mole of electron i.e., 20 Molecular weight Wt. of KCl in 60 g water = ×=40 10 g Equivalent mass = = 63/3 = 21 80 Change in oxidation number 30 Chemistry = 10 × 6.023 × 1023 atoms = 6.023 × 1024 atoms 123. (c) : 1 mole of C = 12 g Change in oxidation number = 5 1 23 Atomic weight M 60 g of C = ×=60 55 moles =×6.023 ×10 atoms Equivalent weight = Change in oxidation number = 5 12 1 mole of He = 4g 124. (c) : It means that water added = 9 g in 100 g oleum. 1 8 g of He = × 8 moles = 2 moles = 2 × 6.023 × 1023 atoms SO3 + H2O → H2SO4 4 80 g 18 g 98 g 9. (b) : 0.1 mol O2 = 3.2 g i.e., 9 g H2O can dissolve 40 g SO3 to form 49 g H2SO4 0.1 mol SO2 = 6.4 g 22 \ Mass of SO3 in 100 g oleum = 40 g 6.02 × 10 molecules of SO2 = 0.1 mol SO2 = 6.4 g 23 and % by mass of SO3 = 40% 1.024 × 10 molecules of O2 and % by mass of H2SO4 = 60% = 0.2 mol O2 = 6.4 g 125. (b) : Molality is the number of moles of the solute present Mass 10. (c) : Number of molecules = × NA in one kg of the solvent whereas molarity is the number of moles Molarmass of the solute present in one litre of the solution. Thus molality 16 N Number of molecules in 16 g oxygen =×N = A involves only masses which do not change with temperature. 32 A 2 Hence, molality is preferred over molarity. 16 N 16 g of CO = ×=N A 28 A 17. 5 LEVEL 2 28 28 g of N2 = ×=NN 28 AA 1. (d) : BaCl2 + H2SO4 = BaSO4 + 2HCl According to law of conservation of mass, 14 NA 14 g of N2 = ×=NA Total mass of reactants = Total mass of products 28 2 10 Suppose, the mass of BaCl2 decomposed be x g. 10 g of H2 = ×=NNAA5 So, x + 9.8 = 12.0 + 2.75 = 14.75 ⇒ x = 4.95 g 2 11. (d) 2. (b) : 100 cc of 0.5 M ethyl alcohol ≡ 100 × 0.5 × 10–3 mole = 5 × 10–2 mole 12. (d) : Weight of ethyl alcohol required = 5 × 10–2 × 46 g = 2.3 g Ca(OH)2 + 2NH4Cl → CaCl2 + 2NH3 + H2O [Q Molecular weight of ethyl alcohol = 46] 40 + 2 × (16 + 1) 2 × (14 + 4 + 35.5) 40 + 71 2 × 17 18 g = 34 g 23..230 = 74 g = 107 g = 111 g Mass 11. 5 2 ∴=d ⇒=⇒=V = 107 g NH4Cl reacts with slaked lime = 74 g Volume V 11. 5 \ 10 g NH4Cl will react with slaked lime \ Volume required = 2 cc 74 =×10 = 69.g 3. (a) 107 4. (d) : Mole ratio of iodine : oxygen \ Slaked lime left unreacted 254 112 = 10 – 6.9 = 3.1 g ==::27 Thus, NH Cl is the limiting reactant. 127 16 4 CaCl formed from 10 g NH Cl The oxide is I O . 2 4 2 7 111 When dissolved in water it can produce HIO4 or H5IO6 =×10 = 10.g4 107 I2O7 + H2O 2HIO4 NH formed from 10 g NH Cl I2O7 + 5H2O 2H5IO6 3 4 34 5. (d) : Marble (CaCO3) is a compound while wax (mixture =×10 = 32. g of hydrocarbon), brass (Cu + Zn) and iodized salt (NaCl + KI) 107 are mixtures. 13. (b) : The temperature is going down, so the reaction is 6. (b) endothermic (i.e., absorbing energy). 7. (d) : 35000 can be written as 3.5 × 104 or 3.50 × 104 14. (b) : S and O combine to form two given compounds, SO2 4 4 or 3.500 × 10 or 3.5000 × 10 . So it can have 2, 3, 4 or 5 and SO3. 1 significant figures. 15. (d) : CO+→OCO 2 22 8. (d) : 1 mole of Fe = 55.85 g 1 558.5 g Fe ≡ 10 moles 1 L of CO reacts with L of O2 to produce 1 L of CO2 2 Some Basic Concepts of Chemistry 31

\ 10 L O2 will react with 20 L of CO to produce 20 L of CO2 44 \ 1 g carbon gives = 36.6 g carbon dioxide. \ CO left = 10 L and CO2 produced = 20 L 2 16. (d) 25. (d) : All of these have two significant figures. 17. (c) : Gay Lussac’s law is true only for gaseous substances. 26. (d) 18. (d) : 8 g of O2 = 0.25 mol, 11 g of CO2 = 0.25 mol, 27. (b) : 1 L of 0.1 M H2SO4 contains = 0.1 mol of H2SO4 22 g of CO2 = 0.50 mol, 7 g of CO = 0.25 mol, 1 L of 0.1 M NaOH contains = 0.1 mol of NaOH (Same number of moles will contain the same number of According to the given equation, 1 mol of H2SO4 reacts with molecules). 2 mol of NaOH. Hence, 0.1 mol of NaOH will react with 19. (a) : FeSO4.7H2O(A) and ZnSO4.7H2O(C) are isomorphous. 0.05 mol of H2SO4 (and 0.05 mol of H2SO4 will be left unreacted), 4 i.e., NaOH will produce 0.05 mol of Na2SO4 20. (d) : 1 mol of O2 will react with mol of NH3 to produce 5 = 0.05 × (46 + 32 + 64) g = 0.05 × 142 g = 7.10 g 4 6 Volume of solution after mixing = 2 L mol of NO and mol of H2O. Hence, all the oxygen is 5 5 H2SO4 left unreacted in the solution = 0.05 mole consumed. \ Molarity of the solution = 0.05/2 = 0.025 mol L–1. 21. (c) : N1V1 + N2V2 = N3V3 1 28. (c) : HH22+→ΟΟ2 or, 0.1 × 100 + 0.2 × 100 = N3 × 200 2 03. ×100 2 g 16 g 18 g or, N3 = = 01. 5 200 10 g H2 requires O2 = 80 g. Normality of a solution is equal to the number of gram- 64 g O2 requires 8 g of H2 and H2 left = 2 g equivalents of solute present in one litre solution. Thus, 2O is the limiting reactant and H2 is the excess reactant. Hence, H2O formed from 64 g of O2 22. (c) : 2CO + O2 → 2CO2. The residual gas is CO. 18 72 Volume of CO oxidised = 2 × 30 = 60 mL; =×64 ==72 gmolm= 4 ol. Volume of CO = 60 + 10 = 70 mL 16 18

Volume of CO2 intially present in the mixture 29. (b) : (i) 1 mol of Fe2O3 = 2 × 56 + 3 × 16 g = 160 g = 100 – 70 = 30 mL \ 0.1 mol of Fe2O3 = 16 g 23 Volume of CO2 formed = 60 mL; (ii) 1 mole of SO2 = 6.023 × 10 molecules = 64 g Volume of CO2 absorbed by KOH 23 64 23 = 30 + 60 = 90 mL \ 3.01 × 10 molecules = ××30.1103= 2 g 6.023 ×1023 23. (d) : 1 mole of CO = 44 g 2 (iii) 1 mol of O2 = 32 g 1 a.m.u. = 1.66 × 10–24 g \ 0.5 mol of O2 = 16 g 1 25 (iv) 1 mole of O at STP = 22400 mL = 48 g 44 g of CO2 = ×=44 26. 51× 0 u 3 16. 61× 0−24 48 \ 11200 mL of O3 at STP =×11200 g = 24 g 24. (b) : 2H2 + C → CH4 22400 Here hydrogen is the limiting reagent. 30. (b) : K2Cr2O7 and KMnO4 both are good oxidizing agents 4g hydrogen gives 16 g methane. in acidic medium. The reaction is as follows : 2– + – 3+ 16 Cr2O7 + 14H + 6e → 2Cr + 7H2O Q 1 g hydrogen gives ==4 g CH4 2– + – 2+ 4 MnO4 + 8H + 5e → Mn + 4H2O 2– 2H2 + O2 → 2H2O Normality of Cr2O7 = Molarity × no. of electrons involved Here oxygen is the limiting reagent. = 0.04 × 6 = 0.24 N – 32 g oxygen gives 36 g water. Normality of MnO4 = Molarity × no. of electrons involved 36 = 0.03 × 5 = 0.15 N Q 1 g oxygen gives = 1.g125 water 32 N1V1 = N2V2 (K Cr O ) (KMnO ) N2 + 3H2 → 2NH3. 2 2 7 4 02. 45× 0 3 Here nitrogen is the limiting reagent. ⇒ 0.24 × 50 = 0.15 × V2 ⇒ V2 = = 80 cm 28 g nitrogen gives 34 g ammonia. 01. 5 31. (d) : Statement (c) should be one mole of hydrogen atom 34 Q 1 g nitrogen gives = 12. 1 g ammonia contains Avogadro’s number of atoms. 28 C + O2 → CO2 32. (d) : Volatile component of CH3CH2OH = 46 g Here carbon is the limiting reagent. 46 = 1 mol 12 g carbon gives 44 g carbon dioxide. 46 32 Chemistry

1 mole = NA molecules = 9NA atoms Moles of O2 = 0.044 – 0.002 = 0.042 Thus, (c) is correct. 38. (b): Weight of O3 = 0.002 × 48 = 0.096 g Non-volatile component is H2O = 54 g 39. (a) : Weight of O = 0.042 × 32 = 1.344 g 54 2 ==3moles 0.096 18 % of O = ×=100 67% Thus, (a) is correct. 3 (.1 344 + 0.)096 Non-volatile component of H O = 3N molecule 2 A 40. (d) : As we know, = 3 × 3NA atoms 23 22400 cc of N2O contain 6.02 × 10 molecules = 9NA atoms Thus, (b) is correct. 60. 21× 023 \ 1 cc of N2O contain molecules 22400 33. (a) : The amount of CO2 in 3.38 g CO2 33. 8 g Since in N2O molecule there are 3 atoms ==0.m077 ol 23 44.m0 g ol−1 36××.02 10 \ 1 cc N2O = atoms 1 mol CO2 contains 1 mol C atoms 22400 So, the amount of C atoms in 3.38 g CO2 = 0.077 mol 18. ×1022 0.690 g = atoms The amount of 2H O in 0.690 g water ==0.m0383 ol 224 18 g mol −1 No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22 As 1 mol H2O contains 2 mol H atoms, the amount of H atoms 60. 21× 023 in 0.690 g water = 2 × 0.0383 mol = 0.077 mol. Hence, no. of electrons = × 22 electrons So, the atomic ratio C : H in the welding gas 22400 = 0.077 mol : 0.077 mL = 1 : 1 13. 2 23 \ The empirical formula of the welding gas is CH. = ×10 electrons 224 34. (d) : The mass of 22.4 L of the welding gas at STP 31.4 11.g6 41. (b) : Mole of CO2 = = 07. 1 =×22.L42= 60.g 44 10.L0 12.9 Hence, the molar mass of the welding gas = 26.0 g Mole of H2O = = 07. 1 18 35. (a) : If the molecular formula of the gas is (CH)n, then its \ Ratio = 0.71 : 0.71 = 1 : 1 molar mass = n × (12.0 + 1.0) = 26.0 g 42. (c) : It will consist of CO2, H2O and hydrocarbon. \ n = 2, and the molecular formula of the gas is (CH)2, or Since, the hydrocarbon is C2H4 (the atomic ratio C : H is 1 : 2 C2H2. and so the hydrocarbon is C2H4.) 36. (d) : Z2O3 + 3H2 → 2Z + 3H2O Valency of metal in Z2O3 = 3 0.1596 g of Z2O3 react with 6 mg of H2. –3 1 mg = 0.001 g = 10 g Since, the mole of O2 required is three times the mole of 0.1596 hydrocarbon so in a mixture containing equal number \ 1 g of H2 react with = = 26.6 g of Z2O3 0.006 of moles of hydrocarbon and oxygen, hydrocarbon will \ Eq. wt. of Z2O3 = 26.6 be in excess and some of it will remain unreacted while Now, Eq. wt. of Z + Eq. wt. of O = Eq. wt. of Z + 8 = 26.6 whole of O2 will be consumed. Thus, the mixture in vessel ⇒ Eq. wt. of Z = 26.6 – 8 = 18.6 after the completion of reaction will consist of products  Atomic wt.  (i.e., CO2 and H2O) and excess of hydrocarbon that has remained \ At. wt. of Z = 18.6 × 3 = 55.8 Eq. wt =   Valency of metal  unreacted. 25..22++525250. 37. (b) : O3 + 2KI + H2O 2KOH + I2 + O2 43. (c) : Average titre value = = 25.15 I2 + 2Na2S2O3 Na2S4O6 + 2NaI 3 Applying dilution law, According to rounding off rule of significant figures, if the right Millimoles of O3 ≡ millimole of I2 most digit to be removed is 5, then the preceding number is not 1 changed, if it is even; but increased by one, if it is odd. ≡ millimole of Na22SO3 2 \ 25.15 can be rounded off to 25.2 and has three significant 1 1 ××40 ≡≡20millimoles .002 mol of O figures. 2 10 3 44. (a) : The weight % of available Cl from the given sample PV = nRT (ideal gas equation) 2 1 × 1 = n × 0.0821 × 273 of bleaching powder on reaction with dil. acids or CO2 is called n = 0.044 mol (moles of O2 and O3 in mixture) available chlorine. Some Basic Concepts of Chemistry 33

CaOCl2 + H2SO4 → CaSO4 + H2O + Cl2 48. (b) : CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) 35..51××00125 1 mol 2 mol % of available Cl = = 5% 2 35. 5 40 + 12 + 3 × 16 2 × (1 + 35.5) × 25 = 100 g = 73 g 100 According to above equation,

45. (c) : 23 g Na = 1 mol 100 g of CaCO3 required 73 g of HCl \ 46 g Na = 2 mol 73 23 20 g of CaCO3 requires =×20 = 14.g6 22.4 L of O2 contains 6.023 × 10 molecules 100 23 \ 5.6 L of O2 contains 1.505 × 10 molecules But amount of HCl actually present = 20.0 g 23 = 3.01 × 10 atoms Therefore, CaCO3 is limiting reactant and HCl is excess reactant. 44 g CO2 = 1 mol Now, let us calculate the amount of CO2 produced when entire \ 22 g CO2 = 0.5 mol quantity of limiting reactant reacts. 23 100 g CaCO3 = 1 mol = 6.023 × 10 molecules CaCO3(s) + 2HCl(g) → CaCl2(aq) + H2O(l) + CO2(g) 4 g He contains 6.023 × 1023 atoms 1 mol 1 mol 100 g 44 g \ 52 g He contains 7.83 × 1024 atoms. 100 g of CaCO3 produces CO2 = 44 g 46. (d) : MClx + xAgNO3 → xAgCl + M(NO3)x 44 20 g of CaCO will produce CO =×20 =≈88.09 gg Mole of MCl Mole of AgNO 3 2 100 x = 3 1 x 49. (c) 1 04. 50. (d) : W = 3.2 g 01.(=×05..08) ; or x ==4 x 01. T = 450 + 273 = 723 K 723 47. (c) : (A) : Zn + 2HCl → ZnCl2 + H2 P ==09.a5 tm 760 1 mole of Zn produces 2 g of H 2 R = 0.082 L-atm K–1 mol–1 0.5 mole of Zn will produce 1 g of H 2 V = 780 mL = 0.78 L (B) : C H 70 22 WRT 32..××0 082 723 Molar mass = 862 Now M = = = 256 g/ mol PV 07..80× 95 Mass of atoms = 862/6.023 × 1023 = 1.43 × 10–21 g So, suppose the formula is S (C) : 70 g of Cl = 6.023 × 1023 molecules x 2 x × 32 = 256 35.5 g of Cl = 3.01 × 1023 molecules 2 256 == (D) : Molar mass of SO2 = 64 = 1 mole x 8 23 32 64 g of SO2 = 6.023 × 10 molecules Hence, formula = S8