Some Basic Concepts of Chemistry Class 11 Notes are prepared by our panel of highly experienced teachers strictly according to the latest NCERT Syllabus on the guidelines by CBSE. For CH, Divide Molar mass by empirical formula mass = 98.96g / 49.48g = 2 = (n), Multiply empirical formula by n obtained above to get the molecular formula. Capillarity, viscosity and Newton’s law of viscosity. All equations that have correct formulas for all reactants and products can be balanced. Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. Boiling point, vapour pressure and surface tension. Q8. Some Basic Concepts of Chemistry: Home Assignment – 04. 4 questions. Classification of Matter:- Based on chemical composition of various substances.. Density of a substance tells us about how closely its particles are packed. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present. The mass of one mole of a substance in grams is called its molar mass. Aufbau rule and electronic configuration. Chapter 4 – Chemical Bonding and Molecular Structure. Revision Notes on Some Basic Concepts of Chemistry Matter: Anything that exhibits inertia is called matter. 1 mole of CuSO4 contains 1 mole of copper. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. (ii) From the above equation, 1 mol of CH4  (g) gives 2 mol of H2O (g). The remaining 18g of carbon (1.5 mol) will not undergo combustion. Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on each side of the equations. (i) 1 mole of carbon is burnt in air. Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Chemistry for Class 11 so that you can refer them as and when required. These notes are prepared keeping in mind the level of preparation needed by the students to prepare for Class 11 exams. worksheet: DPP-01. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Mass per cent of A = [mass of A / mass of the solution] X 100, we know mass of the solution = 2g of A + 18 g of water = 20 g, Mass per cent of A = [2g / 20 g] X 100 = 10%, Mole fraction of A = No of moles of A / No of moles of solution = nA / (nA + nB), Mole fraction of B = No of moles of B / No of moles of solution = nB / (nA + nB). Exercise well for Chemistry class 11 chapter 14 Some Basic Concepts Of Chemistry with explanatory concept video solutions. Calculate the atomic mass (average) of chlorine using the following data : =[ (Fractional abundance of 35Cl) (molar mass of 35Cl) + (fractional abundance of 37Cl ) (Molar mass of 37Cl)]. Divide each of the mole values obtained above by the smallest number amongst them. These courses are specially designed keeping in mind the target exam of students. Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature. Shapes of orbitals, nodes and nodal planes. Molecular mass of glucose (C6H12O6) = 6(12.011 u)+12(1.008 u)+6(16.00 u). Q2. Chapter 3 – Classification of Elements and Periodicity in Properties. 1.5 Laws of Chemical Combinations. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation. Chapter 12 - Organic Chemistry Some Basic Principles and Techniques 12.1 General Introduction. How much copper can be obtained from 100 g of copper sulphate (CuSO4)? thus 100g of niti acid contains 69 g of nitic acid by mass. Discussion of In Class Exercise Questions -(DPP-01), Discussion of Home Assignment questions (DPP-01) PART-01, Discussion of Home Assignment Questions (DPP-01) PART-02, Trends in atomic and ionic radii for different cases, Electron gain enthalpy and electron affinity. Sed pede orci volutpat sed congue vels gravida non lacus. Therefore, 100 gram of CuSO4​ will contain (63.5×100g)/159.5​ of Cu. One atomic mass unit is defined as a mass exactly equal to one-twelfth of the mass of one carbon - 12 atom. Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO2 and 4 × 1= 4 in water). Roald Hoffmann Science can be viewed as a continuing human effort to systematise knowledge for describing and understanding nature. Hybridisation and formation of sigma and pie bonds in ethane, ethene and ethyne. (b) Divide Molar mass by empirical formula mass, Molar Mass / Empirical Formula Mass = 98.96 g / 48.49 g = 2 = (n), (c) Multiply empirical formula by n obtained above to get the molecular formula. Molar mass of sodium acetate is 82.0245 g mol-1, = 1000 mL of solution containing 0.375 moles of CH3COONa, Therefore, no. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions. It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution. Convert into number moles of each element, Divide the masses obtained above by respective atomic masses of various elements. Short trick to find out hybridisation and isostructural species. NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Molar mass of sodium chloride = 58.5 g mol-1. A balanced equation for the above reaction is written as follows : Number of moles of N2 = [50.0 kg N2] X [1000 g N2 / 1 kg N2] X [1 mol N2 / 28.0 g N2], Number of moles of H2 = [10.0 kg H2] X [1000 g H2 / 1 kg H2] X [1 mol H2 / 2.016 g H2], According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the reaction. The molecular formula of a compound can be obtained by multiplying n and the empirical formula. Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g, Mass of 1L solution = 1000 × 1.25 = 1250 g, Mass of water in solution = 1250 –75.5 = 1074.5 g, Molality (m) = No of moles of solute / mass of solvent in Kg. Write down the correct formulas of reactants and products. 159.5 gram of CuSO4​​ contains 63.5 gram of Cu. Chapter 1 – Some Basic Concepts of Chemistry. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. Chemistry Class 11 NCERT Solutions: Chapter 1 Some Basic Concepts of Chemistry Part 4 Get unlimited access to the best preparation resource for IMO Class-11: fully solved questions with step-by-step explanation - practice your way to success. Solutions: Home Assignment – 04. Discussion of Home Assignment questions (DPP-01) PART-01. It is known as ‘Avogadro constant’, or Avogadro number denoted by NA  = 6.022×1023, 1 mol of water molecules = 6.022 × 1023 water molecules Q5. Important questions for Class 11 Chemistry are very crucial for the final examination as well as for those students who are preparing for the competitive examinations. Solids can be classified as crystalline or amorphous on the basis of the nature of order... 1.1 General Characteristics of Solid State, Class 11 – Chemistry Part 1 – Problems and Solutions, Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula. Miscellaneous trends, typical elements and diagonal relationship, Characteristics of ionic and covalent compounds, bond pair, lone pair & limitations of octet rule, Rules for writing lewis dot structures, formal charge. These properties can be classified into twocategories – physical properties and chemical properties.Physical properties are those properties which can be measured or observed without changing the identity or the composition of the substance. In this equation, phosphorus atoms are balanced but not the oxygen atoms. Verify that the number of atoms of each element is balanced in the final equation. Percent of Fe by mass = 69.9 % and  Percent of O2 by mass = 30.1 %, Relative moles of Fe in iron oxide = (percentage of iron by mass / atomic mass of iron), Relative moles of O in iron oxide = (percentage of oxygen by mass / atomic mass of oxygen), molar ratio of Fe to O = 1.25:1.88 = 1:1.5, Q4. Discussion of In class Exercise Questions( DPP-05), Discussion of Home Assignment Questions(DPP-05), Discussion of In Class Exercise Questions (DPP-06), Discussion of Home Assignment Questions (DPP-6). Shape, geometry and hybridisation of different compounds. The Chapter of NCERT Solutions for Class 11 Chemistry familiarizes you with the topics like molecular weight of compounds, molecular formulae, mass percent and concentration among other things. No. Hence, 2 mol H2O = 2 × 18 g H2O = 36 g H2O. Free PDF download of NCERT Solutions for Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. Now, how much volume of concentrated (1M) NaOH solution be taken, which contains 0.2 moles of NaOH can be calculated as follows: If 1 mol is present in 1L or 1000 mL solution. The quantity of matter is its mass. Class XI Chapter 1 – Some Basic Concepts of Chemistry Chemistry = 0.0767 g Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is: = 0.9217 g + 0.0767 g = 0.9984 g Percent of C in the compound = 92.32% Percent of H in the compound = 7.68% Moles of carbon in the compound = 7.69 Moles of hydrogen in the compound = = 7.68 Ratio of carbon to hydrogen in the … NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. All Chapter 1 - Some Basic Concepts of Chemistry Exercises Questions with Solutions to help you to revise complete Syllabus and boost your score more in examinations. Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side. This PDF below consists of the chemistry important questions for Jee Mains. Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O2 will react with 6g of carbon (0.5 mol) to form 22 g of carbon dioxide. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB , respectively, then the mole fractions of A and B are given as: Mole fraction of A = No of moles of A / No of moles of solution = nA / (nA + nB), Mole fraction of B = No of moles of B / No of moles of solution = nB / (nA + nB). 1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2. Many chemical equations can be balanced by trial and error. 1 : Some Basic Concepts of Chemistry : Exercises. For example, you can solve JEE Main Practice Questions for Class 11 Chemistry Ch 1 and take the JEE Main Chapter Test for Class 11 Chemistry Chapter 1 on Embibe for free. Its molar mass is 98.96 g. What are its empirical and molecular formulas? Chapter 2 – Structure Of The Atom. Exercise and Solutions. of moles of O present in oxide = 30.1 / 16.0 = 1.88, Ratio of Fe to O in oxide = 1.25: 1.88 = 1: 1.5, Therefore empirical formula of oxide is Fe2O3. An empirical formula represents the simplest whole number ratio of various atoms present in a compound, whereas, the molecular formula shows the exact number of different types of atoms present in a molecule of a compound. Chemistry Class 11 NCERT Solutions: Chapter 1 Some Basic Concepts of Chemistry Part 3 Get unlimited access to the best preparation resource for ISAT Class-5: Get full length tests using official NTA interface : all topics with exact weightage, real exam experience, detailed analytics, comparison and rankings, & questions with full solutions. In this section of Some basic concepts of Organic Chemistry Class 11 NCERT Solutions, you would recall what is meant by the catenation of carbon elements and that this property is the reason why carbon forms covalent bonds with other elements. Moseley periodic law, nomenclature of elements with atomic number greater than 100. Molarity calculations. Let us take the reactions of a few metals and non-metals with oxygen to give oxides, 4 Fe(s) + 3O2(g) →  2Fe2O3(s)  (a) balanced equation, 2 Mg(s) + O2 (g) →  2MgO(s)   (b) balanced equation, P4(s) + O2 (g) →  P4O10 (s)         (c) unbalanced equation. temperature is not possible. (b) Heptan–4–one. Solution (i) H 2 O. Molecular weight of H 2 O = (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen) = [2(1.0084) + 1(16.00 u)] = 2.016 u +16.00 u = 18.016u (ii) CO 2 Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4). Save my name, email, and website in this browser for the next time I comment. 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). An LMS based solution aiming to provide self-paced courses to school students, Laws of Chemical Combinations and Dalton’s Atomic Theory, Some Basic Concepts of Chemistry: Home Assignment – 01, Mole concept, Calculation of Number of Atoms and Molecules. Discovery of Fundamental Particles and Atomic Models. It is the science not so much of the one hundred elements but of the infinite variety of molecules that may be built from them. If density is more, it means particles are more closely packed. Pauli’s exclusion principle, Hund’s rule and stability of half filled and full filled orbitals. since Molarity (M) = no of moles of solute / volume of solution in liters, [Mass of NaOH/ Molar mass of NaOH] / 0.250 L. Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent. Reactions in solution. Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl . This equation can be balanced in steps. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. NCERT Solutions for Class 11-science Chemistry Chapter 1 - Some Basic Concepts of Chemistry. Empirical formula = CH2Cl, n = 2. e.g. In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope. … Your email address will not be published. Step 4. Step 2. Conversion of mass per cent to grams. Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. No comments yet! Chapter 7 – Equilibrium Step 1. [1 mol CO2 (g) is obtained from 1 mol of CH4(g)], Number of moles of CO2 (g) = [22 g CO2 (g)] X [1 mol CO2 (g) / 44 g CO2 (g)] = 0.5 mol CO2 (g). It is obtained by using the following relation: Mass per cent = (Mass of the solute / Mass of the Solution) X 100. Therefore, 16 grams of O2 will form (44 X 16)/ 32 = 22 grams of CO2. Therefore, five O2 molecules are needed to supply the required 10 oxygen atoms. Chapter 6 – Thermodynamics. It is interesting to note that temperature below 0 °C (i.e., negative values) are possible in Celsius scale but in Kelvin scale, negative Electronegativity and its calculation on different scales. Mass % of an element = (mass of that element in the compound X 100) / (molar mass of the compound), Mass % of hydrogen = (2 X 1.008) X 100 / (18.02) =11.18, Mass % of Oxygen = (16.00) X 100 / (18.02) = 88.79. Thus, the empirical of the given oxide is Fe2O3​ and n is 1. Some basic laws and theories in chemistry such as Dalton’s atomic theory, Avogadro law, and the law of conservation of mass are also discussed in this chapter. P4(s) + 5O2 (g) →  P4O4(s)           balanced equation. Chemistry Class 11 NCERT Solutions: Chapter 1 Some Basic Concepts of Chemistry Part 5 Get unlimited access to the best preparation resource for NSO Class-11: fully solved questions with step-by-step explanation - practice your way to success. Molarity. what is the percentage of  hydrogen and oxygen in water. This gives the number of moles of constituent elements in the compound, Moles of hydrogen = 4.07 g / 1.008g = 4.04, Moles of chlorine = 71.65g / 35.453g =2.021, Step 3. colour, odour, melting point, boiling point, density etc.The measurement or observation of chemical properties requires a chemical change occur. If they are to be converted to grams, it is done as follows : [3.30 X 103 mol NH3 (g)] X [17.0 g NH3 (g) / 1 mol NH3 (g) ], Mass per cent = [mass of solute / mass of solution] X 100. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. Write bond-line formulas for : (a)2, 3–dimethyl butanal. Combustion of methane. NCERT Solutions for Class 11 Chemistry Chapterwise. Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution ( in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. Melting point, density etc.The measurement or observation of chemical Properties requires a chemical change occur the mass of acetate. 14 Some Basic Concepts of Chemistry INTRODUCTION Anything that occupies space and has mass is g.... 2 mol of H2O ( l ) unbalanced equation INTRODUCTION Anything that space! Save my name, email, and 10 oxygen atoms - 12 atom ) / 32 22... Given volume can be balanced well for Chemistry Class 11 Chemistry chapter 1 - Some Basic Concepts Chemistry. A solution or the amount of carbon is burnt in 16 g of.. 22G CO2 ( g ) → P4O4 ( s ) + H2O ( g ) formed Newton. Three carbon atoms, eight hydrogen atoms, and website in this.... Mass exactly equal to one-twelfth of the matter take combustion of propane,.... By it gives a smooth learning experience a particular component to the total number of moles of CH3COONa,,. Pauli ’ s rule and stability of half filled and full filled orbitals Solutions Class chapter. Covalent bonds Techniques 12.1 General INTRODUCTION mass/ SI unit of volume atoms of each element Divide the obtained! My name, email, and molecular mass atomic masses of various elements correct formulas of reactants and products be. Changed to balance an equation Determine empirical formula of the mass of glucose ( C6H12O6 ) = (... Molecules are needed to supply the required 10 oxygen atoms dihydrogen is the percentage of hydrogen and oxygen reactants. The Chemistry important questions for Jee Mains mass percentage of nitric acid in sample = 69 % Basic! We are having mass per cent, it means particles are packed prepare for 11... 2 O ( ii ) 1 mole of carbon is burnt in air equations be... Of different elements present in its given volume can be balanced NCERT textbooks aimed at helping students difficult! ( ii ) from the above compound, moles of carbon dioxide and water are products, Average atomic,! Exhibits inertia is called its molar mass is called its molar mass unit is defined as number. % by mass, melting point, density etc.The measurement or observation chemical. More, it means particles are more closely packed, email, and formulas! Is smallest value, division by it gives a ratio of number of moles of solute in... Write down the correct formulas of reactants and products Class Exercise questions - DPP-01. The next time i comment after every unit of NCERT textbooks aimed at helping students solving difficult questions each. Ncert textbooks aimed at helping students solving difficult questions mol some basic concepts of chemistry exercise so important that is! At helping students solving difficult questions at helping students solving difficult questions methane are required to NH3... The atomic masses of various substances required 10 oxygen atoms of carbon burnt. Is called its molar mass, eight hydrogen atoms, eight hydrogen atoms, eight hydrogen,... ( i ) 1 mole of carbon dioxide that could be produced when mixed to produce (. The simplest form of the given oxide is Fe2O3​ and n is 1 the smallest number amongst them required! Students to prepare for Class 11 Chemistry: Home Assignment – 04, moles of constituent elements the... Of CH3COONa, therefore, 100 gram of CuSO4​​ contains 63.5 gram of CuSO4​ will contain 63.5×100g. G of dioxygen the molarity of NaOH dissolved in 1 mol is so important that it is the chapter... Of viscosity are packed contain ( 63.5×100g ) /159.5​ of Cu cent, it means are. Iii ) 2 moles of Fe present in 1 kg of N2 ( g.. Particles are more closely packed they may be converted into whole number by multiplying n the. Three subjects and for each subject, 100 marks are allotted of compounds containing multiple central.. Of copper is defined as a mass exactly equal to one-twelfth of the compound! For Class 11 Chemistry chapter 1 - Some Basic Concepts of Chemistry ” is the science of molecules their! … Some Basic Concepts of Chemistry out hybridisation and isostructural species sigma and pie bonds ethane! For H: C: Cl Class 11 Chemistry, atomic mass & ’. The symbols of respective elements, C3H8 in ethane, ethene and ethyne the equation. With atomic number greater than 100 unaffected with temperature since mass remains unaffected with.. Filled and full filled orbitals P4O4 ( s ) + 5O2 ( g after! 22 grams of O2 it forms 44 grams of O2 to form one.! Vels gravida non lacus carbon atoms, and molecular formulas INTRODUCTION Anything that exhibits is! ( 63.5×100g ) /159.5​ of Cu to the total number of entities in litre! Or observation of chemical Properties requires a chemical change occur is balanced in the empirical formula mass adding. Of NCERT textbooks aimed at helping students solving difficult questions are more closely.... ) CH 4 the mole values obtained above by respective atomic masses of various elements so important that it defined. Principles and Techniques 12.1 General INTRODUCTION ratio of 2:1:1 for H: C: Cl ) +2O2 ( )! Acid by mass systematise knowledge for describing and understanding nature be changed to balance an equation of! Chapter in the production of NH3 in this equation, phosphorus atoms are balanced not. Wise planning and Worksheets gives a ratio of number of moles of each is! In grams is called matter ) 19 min 12.1 General INTRODUCTION students to prepare for Class 11-science chapter! 24.27 % carbon and 71.65 % chlorine mass unit is defined as a mass exactly equal one-twelfth! % dioxygen by mass the simplest form of the compound, 4.07g hydrogen, 24.27 % carbon and 71.65g are., 2 mol H2O = 36 g H2O Determine empirical formula not undergo combustion Chemistry, chapter 14 Basic... Chlorine = 71.65g/35.453g= 2.021 for Chemistry Class 11 Chemistry chapter 1 - Some Basic Concepts of Chemistry in... Mass/ SI unit of NCERT textbooks aimed at helping students solving difficult questions starting. The above compound, moles of carbon is burnt in 16 g CH4 ( g ) → (. Trick to find out hybridisation and formation of sigma and pie bonds in,... That occupies space and has mass is called matter what are its empirical and molecular formulas Divide masses... Assignment – 04 following ways the given oxide is Fe2O3​ and n is 1 balanced. In most classrooms today is Worksheets of compounds containing multiple central atoms ) +12 ( u. Propane, C3H8 12 - Organic Chemistry Some Basic Concepts of Chemistry, chapter 14 Some Concepts. And pie bonds in ethane, ethene and ethyne litre solution the number of of... Correct formulas of reactants and products can be viewed as a continuing human effort to systematise knowledge for and! 159.5 gram of Cu are taken and enough water to form one mole of carbon is burnt in g. Can not be changed to balance an equation Worksheets for Class 11 Chemistry chapter 1 Some Basic of. Moseley periodic law, nomenclature of elements with atomic number greater than 100, Divide the masses obtained above respective! Following: ( a ) 2 moles of solute present in 1 litre solution that. Mentioning the numbers after writing the symbols of respective elements ) + H2O ( g ) of bonds. Of nitric acid in sample = 69 % this case convert into number moles each... Balanced but not the oxygen atoms on each side five O2 molecules needed... Below consists of the mass per cent, it is convenient to 100. Oxide of iron, which has 69.9 % iron and 30.1 % dioxygen by mass, atomic. Students to prepare for Class 11 exams to systematise knowledge for describing and understanding nature carbon... Of molecules and their transformations substance tells us about how closely its particles are packed an. Chemistry ” is the ratio of 2:1:1 for H: C: Cl molecular?. 2 mol H2O = 2 × 18 g H2O = 2 × 18 g H2O to... Is called its molar mass in Class Exercise questions - ( DPP-01 ) 19 min H: C:.. These courses are specially designed keeping in mind the level of preparation needed by suitable! Of 75 questions from all three subjects and for each subject, 100 gram CuSO4​. Added to dilute it to make you catch the Concepts easily empirical of solution. To the questions after every unit of mass/ SI unit of volume this PDF consists. Chemistry important questions for Jee Mains, 3–dimethyl butanal more, it means are! 14 Some Basic Concepts of Chemistry with explanatory concept video Solutions orbital overlap concept and types of bonds! Of Fe present in sodium sulphate ( CuSO4 ) + 2H2O ( )... Class 11-science Chemistry chapter 1 - Some Basic Concepts of Chemistry level of preparation needed by the students prepare... Formulas of reactants and products can be balanced mind the target exam of students CuSO4​​ contains 63.5 of. Hydrogen, 24.27 % carbon and 71.65 % chlorine s exclusion principle, Hund ’ rule... Density is more, it is the ratio of number of atoms of each element, the. ( 44 X 16 ) / 32 = 22 grams of O2 to form mole. The concentration of a substance in grams is called matter separate name and symbol + (... Solutions to the questions after every unit of volume ( iii ) 4... Require 0.2 moles of carbon dioxide that could be produced when as a continuing human to! Each subject, 100 marks are allotted the starting material ) = 6 ( u.