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    JEE Main Study Notes for Basic Concepts in Chemistry: Sample Questions, Last Minute Tips and Study Plan

    Anam Shams Anam Shams
    Content Curator

    Chemistry is the part of science that deals with substances, their properties, structures, and their transformation. Since all the objects in the universe are made of matter, so understanding their properties is also important. JEE Main also focuses on posing questions from basic concepts and each year 2 questions to a least are put forward. 

    Some of the basic concepts of chemistry include the ones which we have studied in class 11 and 12th. Thus NCERT is considered more than enough to prepare the chapter. Check JEE Main Chemistry Syllabus

    Some of the important topics that are included in basic chemistry include:

    • Matter and Its Nature

    • Significant Figures and Scientific Notation Introduction

    • Laws of Chemical Combination

    • Atoms and Molecules

    • Chemical Formulas

    • Mole Concept

    Chemistry is the part of science that deals with substances, their properties, structures, and their transformation. Since all the objects in the universe are made of matter, so understanding their properties is also important. JEE Main also focuses on posing questions from basic concepts and each year 2 questions to a least are put forward. 

    Also Check

    Matter: Nature and Properties 

    Matter: Nature and Properties 

    Anything that exhibits inertia is being referred to as matters. The quantity of matter is its mass, and it is fixed based on the various chemical compositions of substances.

    Everything in the world that has some mass and holds some space is being referred to as matter. Matter exists in three different physical forms including solid, liquid and gas.

    PropertySolidLiquidGas
    TightnessThey are tightly packedTightly packedThey are loosely packed
    Intermolecular spaceLowMediumHigh
    Force of attractionMaximumMediumHigh
    Kinetic EnergyMinimumMediumHigh
    DensityHighMediumMinimum
    VolumeSameFixedVariable
    ShapeFixedVariableVariable
    Compressibility factorMinimumMediumMaximum

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    Elements

    • Elements are the simplest types of matter.
    • The smallest unit of an element is being referred to as atom.
    • The total number of the known elements is 118, out of which 98 elements happen naturally and 20 are formed by artificial transmutation.
    • Some of the examples of elements include Na, K, Mg. Al, Si, P, C, F, Br, etc.

    Compound

    • Compounds are the non-elemental pure compound.
    • They are created with the chemical combination of two or more atoms of different elements in a fixed ratio.
    • Some of the examples of compounds include H2O, CO2, C6H12O6 etc.

    Mixture

    • Mixtures are formed by the physical combination of two or more pure substances in any ratio.
    • The chemical identity of the pure components will be maintained in mixtures.
    • Homogeneous mixtures are the ones whose composition for each part is the same.
    • Some of the examples include Aqueous and gaseous solutions.
    • Heterogeneous mixtures are those whose composition may vary for each part. Example, Soil, and concrete mixtures.
    Matters
    MixturePure Substances
    Homogenous MixtureElements
    Heterogeneous MixtureCompounds

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    Physical Quantities and Their Measurement

    Physical Quantities and Their Measurement

    Fundamental Units

    Fundamental units are not derived from one another; also they are not resolved into any further other units. There are a total of seven fundamental units of the S.I. system including:

    Physical quantityUnit nameUnit symbol
    TimeSecondS
    MassKilogramKg
    LengthMeterM
    TemperatureKelvinK
    Electric currentAmpereA
    Luminous intensityCandelaCd
    Amount of substanceMoleMol

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    Derived Units

    These units are the function of more than one fundamental unit:

    Quantity with SymbolUnit (S.I.)Symbol
    Velocity (v)Metre per secms-1
    Area (A)Square metrem2
    Volume (V)Cubic metrem3
    Density (r)Kilogram m-3Kg m-3
    Energy (E)Joule (J)Kg m2 s-2
    Force (F)Newton (N)Kg ms-2
    Frequency (n)HertzCycle per sec
    Pressure (P)Pascal (Pa)Nm-2
    Electrical chargeCoulomb (C)A-s (ampere – second)

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    Dalton’s Atomic Theory

    Dalton’s Atomic Theory

    • According to this theory, each matter consists of indivisible atoms.
    • Atoms can neither be created nor can they be destroyed.
    • Atoms of a given element are the same in their properties.
    • Atoms of different elements have different properties.
    • The atoms with different elements combine in a fixed ratio to produce molecule of a compound.

    Precision and Accuracy

    Precision: Closeness of outcomes of different measurements taken for the same quantity.

    Accuracy: Accuracy means the agreement of experimental value to the real value


    Significant figures- Rules

    • All non-zero digits are important.
    • Zeroes preceding the first non-zero digit are not important.
    • Zeroes between two non-zero digits are important.
    • Zeroes at the end of a number are significant when they are on the right side of the decimal point.
    • The counting numbers of objects include infinite significant figures.

    Scientific Notation

    • The numbers are represented in N × 10n form.
    • In this case, N means Digit term while n means exponent having positive or negative value.
    • Some of the examples of scientific notation include 12540000 = 1.254 × 107
    • 0.00456 = 4.56 ×10-3

    Mathematical Operations of Scientific Notation

    Mathematical Operations of Scientific Notation

    • Multiplication and Division
    • Follow the same rules which are for exponential number.
    • Example: (7.0 ×103 ) × (8.0×10-7 ) = ( 7.0×8.0) × ( 10[3 + (-7)] ) = 56.0 × 10-4
    • The result don’t have more digits to the rite of decimal point than any of the original numbers
    • (7.0 ×103 ) / (8.0×10-7 ) = ( 7.0/8.0) × ( 10[3 - (-7)] ) = 0.875 ×1010 = 0.9 ×1010

    Addition and Subtraction

    • Numbers are written in a way that it results in the same exponent and after that coefficients are either added or removed.
    • (5 ×103 ) + (8×105 ) = (5 ×103 ) + (800×103 ) = (5+800) ×103 = 805×103
    • The results need to be reported with no more significant figures as there in the original number with few significant figures.

    What are the rules for limiting the result of mathematical operations?

    • In case the rightmost digit is removed by more than 5, the preceding number is increased by one.
    • If the rightmost digit is removed by less than 5, the preceding number will remain the same.
    • If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but is increased by one if it is an odd number.

    Atomic and Molecular Masses

    Atomic and Molecular Masses

    Atomic Mass

    • Mass of an atom.
    • Atomic mass is reported in the atomic mass unit “amu” or in unified mass “U”
    • One atomic mass unit, amu, is the mass which is equal to one-twelfth the mass of one carbon-12 atom.

    Molecular Mass

    • This mass is equal to the total sum of atomic masses of all the elements that are present inside the molecule.

    Formula Unit Mass

    • Mass of a molecule of an ionic compound
    • Formula Unit Mass will remain equal to the sum of atomic masses of all the elements that are present in the molecule

    Laws of Chemical Combination

    Laws of Chemical Combination

    Law of conservation of mass- For any chemical change, the total mass of active reactants will always remain equal to the mass of the product formed.

    Law of constant proportions- This law states that the chemical compound will always contains the same elements in definite proportion by mass and they will not depend on the source of the compound.

    Law of multiple proportions- When two elements are combined together to form two or more than two different compounds the different masses of one element B combines with fixed mass of the other element.

    Law of reciprocal proportion- If two elements B and C react with the same mass of a third element (A), then, the ratio with which they have reacted will remain the same.

    Practice with JEE Main Question Paper


    Previous Year Solved Questions

    Previous Year Solved Questions from Basic Concepts of Chemistry

    Question 1- Ammonia contains 82.65 % N2 and 17.65% H2. In case the law of constant proportions is right, then the mass of zinc required to give 10 g Ammonia will be:

    • 8.265 g
    • 0.826 g
    • 82.65 g
    • 826.5 g

    Solution- The mass of zinc required to give 10 g ammonia will be 0.826 g

    Question 2- No matter what type of source it is, pure sample of water will results in 88.89% mass of oxygen and 11.11% mass of hydrogen. This fact is explained by the law of:

    • conservation of mass
    • constant composition
    • multiple proportion
    • constant volume

    Solution- The correct answer is B

    Question 3- Copper oxide was prepared by the beneath ways:

    a. In the first case, 1.75 g of the metal is dissolved in nitric acid and igniting the residual copper nitrate producing 2.19 g of copper oxide.

    b. In the second situation, 1.14 g of metal dissolved in nitric acid was precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying, and heating results in 1.43 g of copper oxide.

    c. In the third case, 1.45 g of copper on heating strongly in a current of air results 1.83 g of copper oxide. Explain that the given data illustrates the law of constant composition.

    Solution:

    Step 1: In the first experiment.

    • 2.19 g of copper oxide include 1.75 g of Cu.
    • 100 g of copper oxide contained

    Step 2: In the second experiment.

    • 1.43 g of copper oxide include 1.14 g of copper.
    • 100 g of copper oxide contained

    Step 3: In the third experiment.

    • 1.83 g of copper oxide include 1.46 g of copper
    • 100 g of copper oxide contained

    The percentage of copper in copper oxide obtained in all of them is almost the same. Hence, the above data explains the law of constant composition.

    Question 1: Which of the following laws explains that “For any chemical change total mass of active reactants are always equal to the mass of the product formed?”

    • Law of Conservation of Mass
    • Law of Constant Composition
    • Law of Multiple Proportions
    • Gay Lussac’s law

    Solution- Correct option is Law of Conservation of Mass

    Question 2- What will be the mass of water formed by reaction of two gram hydrogen with 16 gram oxygen?

    • 16 g

    • 32 g

    • 18 g

    • 20 g

    Solution- Correct option is 18g

    Question 3: The statement “Chemical compound has the same elements in definite proportion by mass and it is not dependent on the source of compound” is explained in

    • Law of Conservation of Mass

    • Law of Constant Composition

    • Law of Multiple Proportions

    • Gay Lussac’s law

    Solution- Correct option is Law of Constant Composition

    Question 4- Hydrogen is combined with chlorine to produce HCl. Also, hydrogen is combined with sodium to produce NaH. If sodium and chlorine are combined with each other, they will do the same in the ratio

    • 23 : 35.5

    • 35.5 : 23

    • 1 : 1

    • 23 : 1

    Correct option is 23 : 35.5


    JEE Main 2020 Chemistry Preparation Tips

    • Chemistry is one of the most scoring subjects for most of the JEE Main aspirants as it includes the basics with the difficulty curve remaining the same almost every year. The questions asked in Chemistry are more theoretical than calculative which makes it easy to solve them.
    • JEE Main Chemistry Syllabus is less about conceptual applications and more about systemic planning. If you want to get a good score in this section, you must memorize all the formulas, named reactions, chemical equations, and periodic table trends as all these concepts constitute a major portion of IIT JEE Chemistry syllabus. A lot of direct and indirect questions from these concepts are asked in the exam.
    • The best way to solve chemistry problems includes developing a broad understanding of the fundamental concepts. Students with a solid conceptual foundation can easily solve the IIT JEE Chemistry problems. All you need is to memorize different theories and formulas that will help the JEE aspirants to develop their problem-solving skills.
    • The JEE Chemistry study material is offered to the students with an aims to develop their understanding and evaluation of different concepts included in Organic, Inorganic, and Physical Chemistry. With the help of JEE Chemistry important topics, the aspirants can get a detailed idea of the marks allocation and weightage of different concepts covered in the IIT JEE chemistry syllabus.
    • While preparing for the JEE entrance exam, the candidates must practice more previous years’ question papers and mock tests that play an important role in preparations. More practicing will help the student in realising their stronger and weaker points and how they should approach a question.
    • It is important to have a better and detailed understanding of the type of questions that can come in the exam so that the students get mentally prepared for the same and can know how to solve these questions. Students can practice more last year papers, which is extremely important step while preparing for JEE Main Chemistry.
    • The last section of the chemistry includes general topics like polymers, environmental chemistry, and chemistry in everyday life. Such chapters are very easy for candidates who are preparing for JEE. Memorising some certain like the Periodic Table, Atomic Weights, Atomic Symbols, Atomic Numbers, etc. are important and must be included while preparing for the Chemistry section

    Best Books for JEE Main Chemistry

    As the Chemistry preparation for JEE Main is done majorly through books and comprises theoretical parts, studying from good books has an important role. The candidates must not juggle between several books as that will only lead to confusion. They should refer to certain books that are reputed and acclaimed to be the best publications for the subject. Some of the best books of Chemistry for the preparation of JEE Main are given below:

    BookAuthor
    Modern Approach to Chemical CalculationsR.C. Mukherjee
    Organic ChemistryO. P. Tandon
    Concept of Physical ChemistryP. Bahadur
    Physical ChemistryP. W. Atkins
    Concise Inorganic ChemistryJ. D. Lee
    Organic ChemistryMorrison and Boyd

    Last Minute Preparation Tips

    Last Minute Preparation Tips for JEE Main Chemistry

    We are discussing some of the approaches that will help to ace the exam.

    • Revision & Practice- Revision is the most essential thing to remember all the concepts and formulas. So, pay more emphasis on all the important topics, especially when JEE Mains is just a month away.
    • Don’t study any new topics especially in the last few days. Just make sure that you are revising and paying a detailed focus on improving the topics that you know very well.
    • Solving more last year’s papers and the mocks test will help you to know the difficulty level and type of questions. You must solve more sample papers and take online mock test as per your time frame.
    • Take Short Break- Don’t make your preparation more stressful, rather divide the study time into equal parts with fixed intervals. Taking small breaks between your studies will help you to stay more focused. You can listen to your favorite tasks in between your studies, you can listen to your favourite music, or can go for a short walk, besides doing other activities as well.
    • Be Healthy- Stay healthy while doing preparation, do not skip your meal. Taking healthy food is important as it will help you to stay healthy and fit.
    • Stay Confident: Do feel demotivated while doing the preparations; stay confident that you will crack the JEE Mains with flying colors.

    JEE Main Chemistry 45 Day Study Plan

    JEE Main Chemistry 45 Day Study Plan

    Candidates can make a study plan for the last 45 days remaining for JEE Main 2020. During the last days, most of the candidates are almost done with their syllabus and are planning to start the revisions. Your revisions also need to be planned so that you can cover all the topics. So here we have prepared a 45 days study plan which is as follows :

    WeeksActivity
    July 5 to 15, 2020Physical Chemistry has the most weightage in JEE Main Chemistry. Study important chapters from this section such as Atomic Structure, Electrochemistry, and Equilibrium first. Then, move on to study the chapters with lesser weightage like Surface Chemistry and Solid State. Choose two topics everyday and go through them. Revise the basics well. Use NCERT books.
    July 16 to 26, 2020Study Organic Chemistry. Pay more attention to important chapters such as Aromatic Compounds, Alkyl Halides, Alcohol, and Ether. Then, move on to study less important chapters such as Biomolecules and Carbonyl Compounds. Use NCERT books.
    July 27 to August 8, 2020Study Inorganic Chemistry. Start by studying the most important chapters like Chemical Bonding, p Block, and s Block. Then study less important chapters like Metallurgy and Qualitative Analysis. Use NCERT books.
    August 9 to 20, 2020Revise all the formulas and important concepts. Solve as many previous year’s papers as you can. Pick up questions from the Chemistry section and attempt them as fast as you can (within 45 minutes-1 hour).
    August 21 to 31, 2020Attempt mock tests everyday. Do one final revision.  
    September 1 to 6, 2020Research the analysis of the ongoing papers.

    Chemistry is one of the scoring parts of the JEE Main exam; following these preparation tips and strategies will help JEE aspirants to prepare Chemistry in a more accurate way. In case you are also planning to get admission to the top-most engineering colleges or universities in India, you must follow a strong study plan to reach your goals.

    Study hard to fulfill your dreams. Best of luck!!

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