# Current Affairs JEE Main & Advanced

#### Atomic, Molecular and Equivalent masses

Atomic, Molecular and Equivalent masses          (1) Atomic mass : It is the average relative mass of atom of element  as compared with an atom of carbon –12 isotope taken as 12.            $\text{Atomic mass }=\frac{\text{Average mass of an atom}}{\text{1/12}\times \text{Mass of an atom of }{{C}^{12}}\,}$          Average atomic mass : If an elements exists in two isotopes having atomic masses ‘a’ and ‘b’ in the ratio   m : n, then average atomic mass =$\frac{(m\times a)+(n\times b)}{m+n}.$ Since the atomic mass is a ratio, it has no units and is expressed in amu, 1 amu = $1.66\times {{10}^{-24}}g$. One atomic mass unit (amu) is equal to $\frac{1}{12}th$ of the mass of an atom of carbon-12 isotope.          Gram atomic mass (GAM) : Atomic mass of an element expressed in grams is called Gram atomic mass or gram atom or mole atom.            (i) Number of gram atoms =$\frac{\text{Mass of an element}}{\text{GAM}}$            (ii) Mass of an element in gm. = No. of gm. atom $\times$ GAM            (iii) Number of atoms in 1 GAM = 6.02 $\times$${{10}^{23}}$            $\therefore \,\,\,\,$Number of atoms in a given substance            = No. of  GAM $\times$6.02 $\times$${{10}^{23}}$ = $\frac{\text{Mass }}{\text{GAM}}\,$$\times$6.02 $\times$${{10}^{23}}$          (iv) Number of atoms in 1gm of element  =$\frac{6.02\,\times \,{{10}^{23}}}{\text{Atomic mass}}$            (v) Mass of one atom of the element (in gm.) $=\frac{\text{GAM}}{\text{6}\text{.02}\,\,\times \,\text{1}{{\text{0}}^{\text{23}}}}$            Methods of determination of atomic mass          (i) Dulong and Pettit's method : According to Dulong and Pettit's law            Atomic mass $\times$Specific heat = 6.4 (approx.)            Atomic mass (approx.)  = $\frac{6.4}{\text{Specific heat (in cals}\text{.)}}$          This law is applicable to solid elements only except Be, B, C and Si because their specific heat is variable with temperature.            Atomic mass = Equivalent mass $\times$Valency                   $\text{Valency }=\frac{\text{Approximate atomic mass}}{\text{Equivalent mass}}$            (ii) Vapour density method : It is suitable for elements whose chlorides are volatile.            Valency of the element =$\frac{\text{Molecular mass of chloride}}{\text{Equivalent mass of chloride}}$                                                      =$\frac{2\,\times \,\text{Vapour density of chloride}}{\text{Equivalent mass of metal }+\text{ 35}\text{.5}}$            Atomic mass = Equivalent mass of metal $\times$Valency            (iii) Specific heat method : It is suitable only for gases. The two types of specific heats of gases are CP (at constant pressure) and ${{C}_{v}}$ (at constant volume). Their ratio is known as g whose value is constant (1.66 for monoatomic, 1.40 for diatomic and 1.33 for triatomic gases).            $\text{Atomic mass of a gaseous element }=\frac{\text{Molecular mass}}{\text{Atomicity}}$          (iv) Isomorphism method : It is based on law of isomorphism which states that compounds having identical crystal structure have similar constitution and chemical formulae.            Example : ${{K}_{2}}S{{O}_{4}},\,\,{{K}_{2}}CrO{}_{4}$ and ${{K}_{2}}Se{{O}_{4}}$ (valency of S, Cr, Se = 6),                $ZnS{{O}_{4}}.\,7\,{{H}_{2}}O,\,\,MgS{{O}_{4}}.7\,{{H}_{2}}O,\,\,FeS{{O}_{4}}.\,7\,{{H}_{2}}O$ (valency of Zn, Mg, Fe = 2).            (2) Molecular mass : Molecular mass of a molecule, of an element or a compound may be defined as a number which indicates how many times heavier is a molecule of that element or compound as compared more...

#### Important hypothesis

Important hypothesis (1) Atomic hypothesis : Keeping in view various law of chemical combinations, a theoretical proof for the validity of different laws was given by John Dalton in the form of hypothesis called Dalton's atomic hypothesis. Postulates of Dalton's hypothesis is as followes, (i) Each element is composed of extremely small particles called atoms which can take part in chemical combination. (ii) All atoms of a given element are identical i.e., atoms of a particular element are all alike but differ from atoms of other element. (iii) Atoms of different elements possess different properties (including different masses). (iv) Atoms are indestructible i.e., atoms are neither created nor destroyed in chemical reactions. (v) Atoms of elements take part to form molecules i.e., compounds are formed when atoms of more than one element combine.  (vi) In a given compound, the relative number and kinds of atoms are constant.   (2) Modern atomic hypothesis : The main modifications made in Dalton’s hypothesis as a result of new discoveries about atoms are, (i) Atom is no longer considered to be indivisible. (ii) Atoms of the same element may have different atomic weights. e.g., isotopes of oxygen ${{O}^{16}},\,\,{{O}^{17}},\,\,\text{and }{{O}^{18}}$. (iii) Atoms of different element may have same atomic weights. e.g., isobars $C{{a}^{40}}\,\text{and }A{{r}^{40}}$. (iv) Atom is no longer indestructible. In many nuclear reactions, a certain mass of the nucleus is converted into energy  in the form of a, b and g rays. (v) Atoms may not always combine in simple whole number ratios. e.g., in sucrose $({{C}_{12}}{{H}_{22}}{{O}_{11}})$, the elements carbon, hydrogen and oxygen are present in the ratio of 12 : 22 : 11 and the ratio is not a simple whole number ratio.   (3) Berzelius hypothesis : “Equal volumes of all gases contain equal number of atoms under same conditions of temperature and pressure”.  When applied to law of combining volumes, this hypothesis predicts that atoms are divisible and hence it is contrary to Dalton's hypothesis.            (4) Avogadro’s hypothesis : “Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.” Avogadro hypothesis has been found to explain as follows, (i) Provides a method to determine the atomic weight of gaseous elements. (ii) Provides a relationship between vapour density (V.D.) and molecular masses of substances. $\text{Molecular mass }=\text{ }2\,\,\times \,\,\text{vapour density}$ (iii) It helps in the determination of mass of fixed volume of a particular gas. Mass of $1\ ml$ gas = V.D. ´ 0.0000897 gm. (iv) It also helps in the determination of molar volume at N.T.P. $\because$ V.D. ´ 0.0000897 gm. gas has volume = $1\ ml$ $\therefore$ 2 ´ V.D.(i.e., molecular mass) gm. has $\text{volume}=\frac{1\ \times \ 2\ \times \ V.D.}{V.D.\ \times \ 0.0000897}\ ml=22400\,ml$ $\therefore$ Molar mass of a gas or its 1 mole occupies 22.4 L volume at S.T.P. (v) It helps in determination of molecular formulae of gases and is very useful in gas analysis. By knowing the molecular volumes more...

#### Laws of chemical combination

Laws of chemical combination   Various chemical reactions take place according to the certain laws, known as the Laws of chemical combination. (1) Law of conservation of mass : It was proposed by Lavoisier and verified by Landolt. According to this law, Matter is neither created nor destroyed in the course of chemical reaction though it may change from one form to other.  The total mass of materials after a chemical reaction is same as the total mass before reaction.            (2) Law of constant or definite proportion : It was proposed by Proust. According to this law, A pure chemical compound always contains the same elements combined together in the fixed ratio of their weights whatever its methods of preparation may be. (3) Law of multiple proportion : It was proposed by Dalton and verified by Berzelius. According to this law, When two elements A and B combine to form more than one chemical compounds then different weights of A, which combine with a fixed weight of B, are in proportion of simple whole numbers.                (4) Law of equivalent proportion or law of reciprocal proportion : It was proposed by Ritcher. According to this law, The weights of the two or more elements which separately react with same weight of a third element are also the weights of these elements which react with each other or in simple multiple of them. (5) Gay-Lussac’s law : It was proposed by Gay–Lussac and is applicable only for gases. According to this law, When gases combine, they do so in volumes, which bear a simple ratio to each other and also to the product formed provided all gases are measured under similar conditions. The Gay-Lussac’s law, was based on experimental observation.

#### Units for measurement

Units for measurement              The chosen standard of measurement of a quantity which has essentially the same nature as that of the quantity is called the unit of the quantity. Following are the important types of system for unit,            (1) C.G.S. System : Length (centimetre), Mass (gram), Time (second)            (2) M.K.S. System : Length (metre), Mass (kilogram), Time (second)            (3) F.P.S. System  : Length (foot), Mass (pound),      Time (second)            (4) S.I. System : The 11th general conference of weights and measures (October 1960) adopted International system of units, popularly known as the SI units. The SI has seven basic units from which all other units are derived called derived units. The standard prefixes which helps to reduce the basic units are now widely used. Dimensional analysis : The seven basic quantities lead to a number of derived quantities such as pressure, volume, force, density, speed etc. The units for such quantities can be obtained by defining the derived quantity in terms of the base quantities using the base units. For example, speed (velocity) is expressed in distance/time. So the unit is $m/s$ or $m{{s}^{-1}}$. The unit of force $\text{(mass}\times \text{ acceleration)}$ is $kg\ m{{s}^{-2}}$ and the unit for acceleration is $m{{s}^{-2}}$. Seven basic S.I. units
Length Mass Time Temperature Electric Current Luminous Intensity Amount of substance
metre (m) Kilogram (kg) Second more...

#### The mole concept

The mole concept            One mole of any substance contains a fixed number $(6.022\times {{10}^{23}})$ of any type of particles (atoms or molecules or ions) and has a mass equal to the atomic or molecular weight, in grams. Thus it is correct to refer to a mole of helium, a mole of electrons, or a mole of $N{{a}^{+}}$, meaning respectively Avogadro?s number of atoms, electrons or ions.             $\therefore$ Number of moles $=\frac{\text{Weight (grams)}}{\text{Weight of one mole (g/mole)}}$                                   $=\frac{\text{Weight}}{\text{Atomic or molecular weight}}$

#### Chemical stoichiometry

Chemical stoichiometry            Stoichiometry (pronounced “stoy-key om-e-tree”) is the calculation of the quantities of reactants and products involved in a chemical reaction. That means quantitative calculations of chemical composition and reaction are referred to as stoichiometry.            Basically, this topic involves two types of calculations.            (a) Simple calculations (gravimetric analysis) and            (b) More complex calculations involving concentration and volume of solutions (volumetric analysis).            There is no borderline, which can distinguish the set of laws applicable to gravimetric and volumetric analysis. All the laws used in one are equally applicable to the other i.e., mole as well as equivalent concept. But in actual practise, the problems on gravimetric involves simpler reactions, thus mole concept is convenient to apply while volumetric reactions being complex and unknown (unknown simple means that it is not known to you, as it’s not possible for you to remember all possible reactions), equivalent concept is easier to apply as it does not require the knowledge of balanced equation.            (1) Gravimetric analysis : In gravimetric analysis we relate the weights of two substances or a weight of a substance with a volume of a gas or volumes of two or more gases.          Problems Involving Mass-Mass Relationship          Proceed for solving such problems according to the following instructions,            (i) Write down the balanced equation to represent the chemical change.            (ii) Write the number of moles below the formula of the reactants and products. Also write the relative weights of the reactants and products (calculated from the respective molecular formula), below the respective formula.            (iii) Apply the unitary method to calculate the unknown factor (s).          Problems Involving Mass-Volume Relationship            For solving problems involving mass-volume relationship, proceed according to the following instructions,            (i) Write down the relevant balanced chemical equations (s).            (ii) Write the weights of various solid reactants and products.            (iii) Gases are usually expressed in terms of volumes. In case the volume of the gas is measured at room temperature and pressure (or under conditions other than N.T.P.), convert it into N.T.P. by applying gas equation.            (iv) Volume of a gas at any temperature and pressure can be converted into its weight and vice-versa with the help of the relation, by $PV=\frac{g}{M}\times RT$ where $g$ is weight of gas, $M$ is mole. wt. of gas, $R$ is gas constant.            Calculate the unknown factor by unitary method.          Problems Based on Volume-Volume Relationship            Such problems can be solved according to chemical equation as,            (i) Write down the relevant balanced chemical equation.            (ii) Write down the volume of reactants and products below the formula to each reactant and product with the help of the fact that $1gm$ molecule of every gaseous substance occupies 22.4 litres at N.T.P.            (iii) In case volume of the gas is measured under particular (or room) temperature, convert it to volume at NTP by using ideal gas equation.            Take the help of Avogadro’s hypothesis “Equal volume of more...

#### Significant figures

Significant figures   In the measured value of a physical quantity, the digits about the correctness of which we are surplus the last digit which is doubtful, are called the significant figures. Number of significant figures in a physical quantity depends upon the least count of the instrument used for its measurement. (1) Common rules for counting significant figures Following are some of the common rules for counting significant figures in a given expression                   Rule 1. All non zero digits are significant.            Example : $x=1234$ has four significant figures. Again $x=189$ has only three significant figures.                   Rule 2. All zeros occurring between two non zero digits are significant.            Example : $x=1007$ has four significant figures. Again $x=1.0809$ has five significant figures.                       Rule 3. In a number less than one, all zeros to the right of decimal point and to the left of a non zero digit are not significant.            Example : $x=0.0084$ has only two significant digits. Again, $x=1.0084$ has five significant figures. This is on account of rule 2.                   Rule 4. All zeros on the right of the last non zero digit in the decimal part are significant.            Example : $x=0.00800$ has three significant figures 8, 0, 0. The zeros before 8 are not significant again 1.00 has three significant figures.                   Rule 5. All zeros on the right of the non zero digit are not significant.            Example : $x=1000$ has only one significant figure. Again $x=378000$ has three significant figures.                   Rule 6. All zeros on the right of the last non zero digit become significant, when they come from a measurement. Example : Suppose distance between two stations is measured to be 3050 m. It has four significant figures. The same distance can be expressed as 3.050 km or $3.050\times {{10}^{5\,}}\,cm$. In all these expressions, number of significant figures continues to be four. Thus we conclude that change in the units of measurement of a quantity does not change the number of significant figures. By changing the position of the decimal point, the number of significant digits in the results does not change. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.                       (2) Rounding off: While rounding off measurements, we use the following rules by convention                   Rule 1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged.            Example : $x=7.82$ is rounded off to 7.8, again $x=3.94$ is rounded off to 3.9.                   Rule 2. If the digit to be dropped is more than 5, then the preceding digit is raised by one.            Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8.                   Rule 3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit more...

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