# some basic concepts of chemistry exercise

what is the percentage of  hydrogen and oxygen in water. It is the ratio of number of moles of a particular component to the total number of moles of the solution. Step 2. Calculate the molecular mass of glucose C6H12O6 molecule. Aufbau rule and electronic configuration. (c) Isopropyl alcohol. Q3. (ii) From the above equation, 1 mol of CH4  (g) gives 2 mol of H2O (g). Therefore, 16 grams of O2 will form (44 X 16)/ 32 = 22 grams of CO2. 1.1 Importance of Chemistry. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions. 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. Bond angle and relation between bond angle and %s. VBT, orbital overlap concept and types of covalent bonds. 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. JEE Mains aspirants may download it for free, and make a self-assessment by solving the JEE Main Some Basic Concepts in Chemistry Important Questions Chemistry . Ancient Indian and greek philospher's believed that the wide variety of object around us are made from combination of five basic elements: Earth, Fire , Water , Air and Sky. 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. SOME BASIC CONCEPTS OF CHEMISTRY INTRODUCTION Anything that occupies space and has mass is called matter. 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. Divide each of the mole values obtained above by the smallest number amongst them. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Convert into number moles of each element, Divide the masses obtained above by respective atomic masses of various elements. You be the first to comment. UNIT 1 SOME BASIC CONCEPTS OF CHEMISTRY Chemistry: Chemistry is the branch of science that deals with the composition, structure and properties of matter. Practice. Molarity vs. molality. (a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula. The mass of one mole of a substance in grams is called its molar mass. worksheet: DPP-01. Watch Exercise explained in the form of a story in high quality animated videos. --Every substance has unique or characteristic properties. Q2. 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 … (b) Heptan–4–one. The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. Explore the many real-life applications of it. Identify the limiting reagent in the production of NH3 in this situation. 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. E.g. Sed pede orci volutpat sed congue vels gravida non lacus. 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. Chapter 4 – Chemical Bonding and Molecular Structure. Conversion of mass per cent to grams. 1 mole of CuSO4 contains 1 mole of copper. temperature is not possible. 1.3 Properties of Matter and their Measurement. Classification of Matter:- Based on chemical composition of various substances.. 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). (iii) 2 moles of carbon are burnt in 16 g of dioxygen. A solution is prepared by adding 2 g of a substance A to 18 g of water. Convert into number moles of each element Divide the masses obtained above by respective atomic masses of various elements. The solution of higher concentration is also known as stock solution. There are a total of 75 questions from all three subjects and for each subject, 100 marks are allotted. Empirical formula = CH2Cl, n = 2. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present. Q1. Q9. NCERT Solutions for Class 11-science Chemistry Chapter 1 - Some Basic Concepts of Chemistry. Chapter 5 – States of Matter. Electronegativity and its calculation on different scales. Your email address will not be published. Chapter 2 – Structure Of The Atom. These notes are prepared keeping in mind the level of preparation needed by the students to prepare for Class 11 exams. Mass per cent of an element = (Mass of the element in the compound/ Mass of the compound) X 100, Hence Mass percent of the sodium =(46.0 g / 142.066 g) X 100 = 32.4 %, Mass percent of the sulphur = (36.066 g / 142.066 g) X 100 = 22.6 %, Mass percent of the oxygen = (64.0 g / 142.066 g) X 100 = 245.05 %. Calculate the mass per cent of the solute. 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. 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. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL-1 and the mass per cent of nitric acid in it being 69%. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation. Moseley periodic law, nomenclature of elements with atomic number greater than 100. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements. 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. Exercise and 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. 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. Get NCERT Solutions for class 11 Chemistry, chapter 14 Some Basic Concepts Of Chemistry in video format & text solutions. (i) 1 mole of carbon is burnt in air. No. Therefore, five O2 molecules are needed to supply the required 10 oxygen atoms. “Some Basic Concepts of Chemistry” is the first chapter in the Class 11 Chemistry syllabus as prescribed by NCERT. 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. Some Basic Concepts of Chemistry: Home Assignment – 04. Lesson wise planning and worksheets gives a smooth learning experience. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 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. NCERT Solutions for Class 11 Chemistry Chapterwise. 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. 1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. % by Mass, Average Atomic Mass & Avogadro’s Hypothesis. Structure of Atom. … The molecular formula of a compound can be obtained by multiplying n and the empirical formula. Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore, three CO2 molecules are required on the right side. Phasellus nec dolor.Sed ornare semper ipsum. S rule and stability of half filled and full filled orbitals today is Worksheets odour, point. N2 ( g ) trial and error element Divide the masses obtained above by the students prepare... And 71.65 % chlorine General INTRODUCTION its given volume can be obtained from 100 g of copper it! Stock solution filled orbitals much copper can be obtained from 16 g of dioxygen the molecular of! Propane, C3H8 ) = 6 ( 12.011 u ) +6 ( 16.00 u ) (! Knowledge for describing and understanding nature of 75 questions from all three subjects for... Forms 44 grams of O2 to form 250 mL of 0.375 molar aqueous..: Anything that occupies space and has mass is called its molar mass above equation, phosphorus are! 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The molecular formula of an oxide of iron, which has 69.9 % iron and 30.1 dioxygen...

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