01. basic concept of chemistry [1-24]

7/28/2019 01. Basic Concept of Chemistry [1-24]

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... 1Chemistry /Basic Concepts of Chemistry

Chapter Objectives

Basic Conceptsof Chemistry

1. Introduction to Chemistry

2. Laws of chemical combination

3. Atomic mass, molecular mass

4. Avogadros law

5. Mole concept

6. Empirical and Molecular formula

7. Chemical equation and Limiting reagent

8. Concentration terms

9. Concept of equivalent mass

10. Normality


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Branches of Chemistry

Chemistry may be defined as the branch of science dealing with the composition, structure and propertiesof matter.

Chemistry has mainly three branches:1. Physical chemistry2. Organic chemistry3. Inorganic chemistry

Matter

Matter may be defined as anything which occupies space and has mass. Based on the physical state of

matter, it can be classified into solids, liquids and gases.

An element is a substance which cannot be decomposed into simpler substances by ordinary chemicalmethods.

A compound is a substance which can be decomposed into two or more dissimilar substances bychemical reactions.

or example, water (H2O).

Mixture

A mixture contains two or more components in varying amounts. Mixtures are of two types:a. Homogenous mixtures: Mixtures having the same or uniform composition throughout the sample.

or example, air is a mixture of gases like oxygen, nitrogen, carbon dioxide and water vapours.b. Heterogenous mixtures: Mixtures having different composition in different phases. or example, a

mixture of iron fillings and sulphur is a heterogeneous mixture.

Laws of chemical combination

Law of conservation of mass

Antoine Lavoisier states that during any physical or chemical change, the total mass of the products isequal to the total mass of the reactants.Or in other words, during any physical or chemical change, matter is neither created nor destroyed.

AgNO3 + NaCl

AgCl + NaNO3

Silvernitrate

Sodiumchloride

Silverchloride

Sodiumnitrate

Mass of reactants (AgNO3 + NaCl) = Mass of product (AgCl + NaNO3)

Concept Notes


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Law of constant composition or law of definite proportions

Louis Proust states that a pure chemical compound always contains same elements combined togetherin the same proportion by weight.

Pure sample of water, whatever may be its source, contains H and O in ratio of 1 : 8.

Law of multiple proportions

John Dalton states that when two elements combine to form two or more compounds, the mass of one ofthe elements, which combines with a fixed mass of the other, bear a simple whole number ratio.

The ratio between the different masses of copper combining with the same mass of oxygen in the twocompounds Cu2O and CuO is 8 : 4 = 2 : 1 (which is a simple whole number ratio).

Law of reciprocal proportions

The ratio of the weights of two elements A and B which combine separately with a fixed weight of the third

element C is either the same or some simple multiple of the ratio of weights in which A and B combinesdirectely with eachother.

S

H O

H S

SO

H O

H2S 2 : 32, 1 : 16

SO2 32 : 32, 1 : 1

2 21 1

H S : SO : 1: 1616 1

=

[ ]2 2 21 8 1

H S : SO : H O16 1 2

=

This simple ratio1

2

explain the law of reciprocal proportion.

Gay Lussacs law of combining gaseous volume

Gay Lussac states that when gases combine to form gaseous products, a simple ratio exists betweenthe volumes of the reactants and the products at constant temperature and pressure.

2 22 volume1volume 1 v olume

H (g) Cl (g) 2HCl(g)+

The ratio is 1 : 1 : 2.


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Daltons atomic theory

Although the origin of idea that matter is composed of small indivisible particles called a-tomio (mean-ing indivisible), dates back to the time of Democritus, a Greek Philosopher (460 - 370 BC), it result of

several experimental studies which led to the Laws mentioned above.

In 1808, Dalton published A New System of Chemical Philosophy in which he proposed the following :

1. Matter consists of indivisible atoms.2. All the atoms of a given element have identical properties including identical mass. Atoms of

different elements differ in mass.3. Compounds are formed when atoms of different elements combine in a fixed ratio.4. Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in

a chemical reaction.Daltons theory could explain the laws of chemical combination.

Atomic MassSince we cannot see atom, it becomes impossible to weigh it. Therefore, the relative mass of atoms are

used instead of their actual mass. International Union of Chemists selected 126 C isotope as the standard.

Based on this, atomic mass of an element is defined as a number, which expresses how many times the

mass of one atom of the element, is that of1

th12

mass of a 126 C atom.

Atomic mass =12

6

Mass of 1 atom of the element1

Mass of 1 atom of C12

Atomic mass unit

Atomic mass can also be expressed in a unit called atomic mass unit (amu). Atomic mass unit is

defined as exactly1

th12


12 246 23

1 11 amu Mass of one C atom g 1 67 10 g

12 6 023 10= =

;

Atoms of the same element may have different masses. They are known as isotopes. In such cases, theatomic mass of an element is taken as the average of the atomic masses of the various isotopes.

or example, chlorine has two isotopes 3517 Cl and3717 Cl present in relative abundance ratio 3 : 1. Therefore,

the average atomic mass of chlorine atom =35 3 37 1

4

+ = 35.5


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1llustrative Example

Example 1: The natural occurrence of the isotopes 20 21 2210 10 10Ne, Ne and Ne is in the ratio 90.51%,

0.28% and 9.21%, respectively. Calculate the average atomic mass of the element.

Average atomic mass of Neon atom (Ne) =90.51 20 0.28 21 9.21 22

(90.51 0.28 9.21)

+ + + + = 20.187 amu

Gram atomic massAtomic mass expressed in grams is called gram atomic mass. This amount of the element is also calledone gram atom.1 gram atom of oxygen = Gram atomic mass of oxygen = 16 g1 gram atom of nitrogen = 14 gAtomic mass of oxygen = 16

Molecular Mass

Relative mass of a molecule may be obtained in terms of the same standard ( )126 C used for definingatomic mass. The molecular mass is a number which expresses how many times the mass of a given

molecule is that of1

th12


Molecular mass =126

Mass of a molecule of the subs tance1

th mass of one C atom12

Relative molecular mass of a molecule is expressed in grams and the actual mass of a molecule isexpressed in amu scale. Relative molecular mass is obtained by adding the atomic masses of all theatoms present in the given molecule.or example, a molecule of water consists of two atoms of hydrogen and one atom of oxygen. Therefore,molecular mass of water = (2 1.008 + 16 1)

= 18.016

Gram molecular mass

Molecular mass expressed in grams is called gram molecular mass. Thus, 18.016 g of water is 1 g

molecule of water.Molecular mass of O2 = 32Gram molecular mass of O2 = 32 g


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Avogadros Law

Equal volume of all gases, under the same conditions of temperature and pressure, contains the samenumber of molecules.

H2 + I2 2HI1 volume 1 volume 2 volume

1 volume of H2 = 1 mole of H2 = 6.023 1023 molecule

1 volume of I2 = 1 mole of I2 = 6.023 1023 molecule

Avogadros number

The number of atoms present in 1 g atom of an element as well as the number of molecules present in1 g mole of any substance is a constant. This is known as Avogadros number, it is denoted by

NA = 6.023 10

23

. Since atomic mass and molecular mass are defined with reference to

12

6 C atom.Avogadros number is also defined relative to 126 C atom.

Avogadros number is defined as the number of atoms present in exact 12 g of126 C .

Mass of 6.023 1023 atoms of 126 C = 12 g

Mass of 1 atom of126 23

12C g

6.023 10=

= 1.992 1024 g

On the amu scale, the mass of a126 C atom = 12 amu

= 12 1.66 1024

= 1.992 1023 g

(1) If 6.02 1023 hydrogen atoms were laid side by side, the total length would encircle the earth abouta million times.

(2) The mass of 6.02 1023 olympic shotput balls is almost equal to the mass of the earth.(3) The volume of 6.02 1023 cricket balls is about half the volume of the earth.

Mole Concept

Mole, like dozen or century is a number equal to 6.023 1023

. (Avogadros number)A mole is defined as the amount of any substance containing one Avogadros number of particles(atom, ions or molecules).1 mol atoms or molecules = 6.023 1023 atoms or molecules respectively.Mole of a substance can be related to (a) mass and (b) the volume of the substance in gaseous formexpressed in grams.


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Molar mass

The mass of one mole is the molar mass.

or example, molar mass of Na = 23 g

Molar mass of Cl2 = 71 g, molar mass of [K4e(CN)6] = 3 39 + 56 + 6 (12 +14) = 329 g

Molar volume

At STP, one mole of a gas (6.023 1023 particles) is found to occupy a volume of 22.7 L. or example,1 mole of oxygen (32 g), occupies 22.7 L at STP. Similarly 28 g of nitrogen, 44 g of carbon dioxide or 1 gmole of any gaseous element or compound occupy a volume of 22.7 L at STP.

1llustrative Examples

Example 1: Calculate the mass of one molecule of water.Solution: Gram molecular mass of water (H2O) = 18 g

Number of molecule in 1 g molecular mass = Avogadros number = 6.023 1023

Mass of one molecule of H2O =Gram molecular mass

Avogadro 's number

= 2318 g

6.023 10= 2.99 1023 g

Example 2: How many molecules of CO2

would be present in one litre of CO2

at S.T.P.?

Solution: Since 22.7 litres of CO2

= 1 mole = 6.023 1023 molecules

In 1 litre number of molecules =236.023 10

22.7

= 2.6 1022 [approx.]

Realize that the answer is independent of nature of the gas i.e.whether we have NH3or

CO2

or CH4, number of molecules in any gas at S.T.P. = 2.6 1022 (approx)

Example 3: What is the mass of 6023 molecules of NH3

gas?Solution: Molar mass of NH

3= 17 g

1 mole = 1 molar mass (17 g) = 6.023 1023 molecules

23

17 60236023 molecules g

6.023 10

=

2017 10 g=


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To find number of moles in (a) a certain number of molecules (b) certain mass or (c)certain volume of gas at S.T.P. we can use the following relationship

Number of moles

given number

6.023 10!

given mass of substance

Molar mass

given volume of gas at NTP in litre

22.7

[See Ex. 3]

[See Ex. 2]

Example 4: ind number of atoms in(a) 1 g yellow phosphorous [P

4]

(b) 5.675 litres of H2S gas at S.T.P.

(c)1

2mole of acetic acid [CH

3COOH]

Solution: (a)23

41 1

1 g moles of P 6.023 10 molecules4 31 4 31

= =

Since one molecule of P4

= 4 atoms

Number of atoms =23

224 6.023 10 1.94 10 (approx)4 31

=

(b) 5.675 litre = 2325.675 5.675mole of H S 6.023 10 molecules22.7 22.7=

Since one molecule of H2S = 3 atoms

No. of atoms =233 5.675 6.023 10

22.7

23 233 5.675 6.023 10 4.5 10 (approx.)22.7

=

(c) Do it yourself. Ans = 24.1 1023 (approx.)

You must have used the information that atomic mass of sulphur = 32 or molar mass ofSO2

= 64. Can you say that mass of 1 atom of sulphur is 32 g or that mass of 1molecule of SO

2is 64 g?


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Empirical ormula and Molecular ormula

Empirical formula is the simplest ratio of all types of atoms present in compound whereas molecularformula gives the exact number of atoms of each type present in the compound. or example, empirical

formula of benzene is CH and molecular formula of benzene is C6H6.Molecular formula = n Empirical formulaMolecular formula is either identical with the empirical formula or a simple multiple of it.or some molecules like C2H2, C6H6, the empirical formula is the same, i.e. CH. But the molecularformula is different.Also, for molecules like CH4, CO, etc., both the empirical and molecular formula are the same.The molecular mass is calculated from vapour density.Molecular mass = 2 Vapour density

Vapour density of a gasmass of a given volume of vapour of the substance

mass of same volume of Hydrogen under similar condition=


Example 1: 0.1g of a gaseous substance occupies 30 ml at S.T.P. What is its V.D. and molecularmass.

Solution: Mass of 22700 ml of H2

at S.T.P. = 2 g [molecular mass]

Mass of 1 ml of H2 at S.T.P. =2

= 0.0009 g22700

This is a very useful information. The student must memorize it.

Mass of 30 ml of H2 at S.T.P. = 30 .00009 g = 0.0027 g

2

Mass of 30 ml of vapourV.D.=

Mass of 30 ml of H at NTP

0.137 [approx.]

0.0027= =

Molecular mass = 2 37 = 74 g.Realize that (1) V.D. has no units and (2) does not depend on temperatureor pressure etc.

Example 2: The vapour density of a compound having empirical formula CH is 39. ind its molecular

formula.Solution: Empirical formula = CHEmpirical formula mass = 12 1 + 1 1 = 13Molecular mass = 2 Vapour density= 2 39 = 78Molecular formula mass = n Empirical formula mass

Molecular formula massn

Empirical formula mass = =

78

13= 6

Hence, the molecular formula is C6H6.


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Calculation of empirical formula from percentage composition

Example 3: An organic compound contains 57.8% carbon, 3.6% hydrogen.ind the empirical formula.

Solution:

Element Percentage Atomic ratio =massAtomic

Percentage Simplest ratio

Simplest whole

number ratio

C 57.8 82.412

8.57= 2

41.2

82.4= 2 2 = 4

H 3.6 6.31

6.3= 49.1

41.2

6.3= 2 1.49 3

O100 (57.8 +

3.6) = 38.6 41.2166.38= 141.2 41.2

= 2 1 = 2

The empirical formula is C4H3O2.

Example 4: An organic compound on analysis gave C = 41.4% and H = 3.1%. Its vapour densityis 58. Calculate the empirical formula and molecular formula.

Solution:

Element Percentage Atomic ratio =massAtomic

Percentage Simplest ratio

C 41.441.4

3.4512

= 3.45

1.13.1

=

H 3.13.1

3.11

= 3.1

13.1

=

O100 (41.4 +

3.1) = 55.5

55.53.46

16=

3.461.1

3.1=

Empirical formula is CHO. Empirical formula mass = 12 1 + 1 1+ 16 1 = 29

Molecular mass = 2 vapour density = 2 58 = 116Molecular formula mass = n Empirical formula mass

Molecular formula massn

Empirical formula mass = =

116

29= 4

Molecular formula is C4H4O4


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Dulong and Petits law

Atomic heat = Atomic mass specific heatAtomic heat of most elements 6.4 (approx.)

Approximate atomic mass =6.4

Specific heat

This is an empirical relation.

Chemical equation

A chemical equation is a representation of a chemical reaction by using symbols and molecular formulae.

2 2 32SO O 2SO+


Example 1: Calculate the mass of calcium oxide and carbon dioxide obtained by the decompositionof 20 g of CaCO3 .

Solution: 3 2CaCO CaO CO +Mol. Mass (40 + 12 + 3 16) (40 + 16) (12 + 2 16)

= 100 = 56 = 44The above balanced equation shows that 100 g (1 mole) of CaCO3 gives 56 g (1 mole) ofCaO and 44 g (1 mole) of CO2.

20 g of CaCO3 gives =56 g 20 g

100 g

of CaO

= 11.2 g of CaO

Similarly, 20 g of CaCO3 gives =44 g 20 g

100 g

of CO2

= 8.8 g of CO2

Limiting Reagent

The reagent which determines the yield of a reaction is called limiting reagent. It is consumed first, duringreaction.

2 2 22H O 2H O+ Suppose the reaction mixture contains 2 moles of H2 and 2 moles of O2. rom the above equation, it isevident that 2 moles of H2 require only 1 mole of O2 to complete the reaction. Therefore, 1 mole of O2 willbe in excess. In this case, H2 is said to be the limiting reagent.


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Example 1: 4 g H2 reacts with 40 g O2 to yield H2O.(i) Which is the limiting reagent?

(ii) Calculate the maximum amount of H2O that can be formed.(iii) Calculate the amount of one of the reactants, which remains unreacted.

Solution: 2 2 22 moles 1 mole 2 moles2H O 2H O+

Number of moles of H2 present =Mass

Molar mass =4

2 mol2

=

Number of moles of O2 present =Mass

Molar mass =40

1.25 mol32

=

(i) Identification of limiting reactantAccording to the equation, 2 moles of H

2require 1 mol of O

2.

Number of moles of O2

actually present = 1.25 mol, i.e. O2

is in excess and thus,H

2is the limiting reactant.

(ii) Calculation of maximum amount of H2O formed

2 moles of H2

form 2 moles of H2O. Limiting reagent decides the amount of the

product.(iii) Calculation of amount of one of the reactants (i.e. O

2) which remain unreacted.

Number of moles of O2

actually present = 1.25 moles Number of moles of O2 reacted = 1 mol Number of moles of O2 unreacted = 1.25 1.0 = 0.25 moles

Concentration terms

If a solution consists of only two components, it is called a binary solution. The component present insmaller amount is called the solute while the other present in larger amount is called the solvent.

The concentration of a solution can be expressed in a number of ways as follows:-

Concentration in terms of percentage

Percentage by mass =Mass of solute w

100 %Mass of solution W =

Percentage by volume =Volume of solute v

100 %Volume of solution V

=

Percentage mass by volume =Mass of solute

100Volume of solution

=w

%V


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Example 1: 20g NaOH is dissolved in 100g of water. Calculate % by mass of NaCl in the solution.

Solution: Percentage by mass of NaCl in the solution = 20 100 16.67%120 =

Strength of Solution

Amount of solute present in one litre of solution is called strength of solution.

Strength = ( )

Mass of solute (ingram)

Volume of solution in litre


Example 1: 4 gmof NaOH was dissovled in water and volume was made upto 500 ml. Calculatestrength of NaOH solution.

Solution: Strength of solutionMass of solute 4 1000

8gm/LVoluem of solution 500

= = =

Example 2: 0.02 moles of H2SO

4is dissolved in water and volume of solution was made upto 800 ml.

Calculate strength of acid solution.Solution: Mass of solute = number of moles molar mass

Strength of acid solution =

0.02 98 1000

2.45gm/L800

=

Molarity

It is defined as the number of moles of solute present in one litre of solution.

Moles of soluteM

Volume of solution (in litres)=

( )

Mass of solute 1000M

Molar mass of solute Volume in millilitres

=

Molar mass=W 1000

M V (in millilitres)

Also moles of solute = M V (in litres)

Moles of solute =Mass

molarmassMilli moles of solute = M V (in millilitres)

In terms of molarity,

Strength (g/L) = Molarity Molar mass


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Example 1: Calculate molarity of a solution containing 98 g of H2SO4 present in 500 ml of solution.

Solution: Molarity = ( )Mass of solute 1000

molar mass Volume in millilitres

=98 1000 2 M98 500

=

Example 2: Calculate the strength of 0.01 M NaOH solution.Solution: Strength = Molarity Molar mass = 0.01 40 = 0.40 g/L

MolalityIt is defined as number of moles of solute present in 1 kg (or 1000 g) of solvent. It is represented by m(small letter).

m = ( )

Moles of solute

Mass of solvent in kg

m = ( )

Mass of solute 1000

Molar mass of solute Mass of solvent in grams


Example 1: ind the molality of H2SO4 solution in which 98 g of H2SO4 is dissolved in 500 g ofsolvent.

Solution:98 1000

Molality 2

98 500

= =

m

Example 2: ind the molality of one molar solution of NaOH. Density of solution is 1.04 g/ml.

Solution: One molar solution means one mole of solute present in one litre of solution

Mass of one litre solution = 1000 1.04 = 1040 g

Mass of solute = 1 40 = 40 g

Mass of solvent = 1040 40 = 1000 g

1 1000m

1000

= = 1 molal

Example 3: ind the molality of H2SO4 solution whose specific gravity(density) is 1.98 g/ml andcontains 95% mass by volume H2SO4.Solution: 100 ml solution contains 95 g H2SO4.

Moles of H2SO4 =95

98Mass of solution = 100 1.98 = 198 gMass of water = 198 95 = 103 g

Molality =95 1000

98 103

= 9.412 m


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Mole raction

The ratio of number of moles of one component to the total number of moles of all the component presentin solution is called as mole fraction of that component

Lets consider a binary solution in which n moles of solute and N moles of solvent are present,

then,

Mole fraction of solute (XA)moles of solute n

moles of solute mole s of solvent n N= =

+ +

Mole fraction of solvent (XB)moles of solvent N

moles of solute moles of solvent n N= =

+ +

The sum of mole fractions of all components of a given solution is unity i.e.

XA + XB = 1


Example 1: 98 gram of H2SO4 is dissolved in 900 ml of water. Calculate mole fraction of H2SO4 andwater in given solution.

Solution: number of moles of H2SO4 in solution2 4

2 4

massofH SO

molar massof H SO=

981

98= =

number of moles of H2O in solutionmassof water 900

50molar massof water 18

= = =

mole fraction of H2O50 50

50 1 51= =

+

mole fraction of H2SO41 1

50 1 51= =

+(also mole fraction of H2SO4 =

50 11

51 51= )

Mass raction

The ratio of mass of one component to the total mass of all components present in solution is called asmass fraction of that component:

Lets consider a binary solution containing w g of sloute in W g of solvent, then.

Mass fraction of solute (MA) =mass of solute w

mass of solute mass of solvent w W=

+ +

Mass fraction of solvent (MB) =mass of solvent W

mass of solute mass of solvent w W=

+ +

The sum of mass fractions of all the components is unity ie. MA + MB = 1


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Example 1: 0.1 mole of H2SO4 is dissolved in 970.2 ml of water. calculate mass fraction of eachcomponent.

Given density of water is 1 g/ml.

Solution: Mass raction of H2SO4 =

2 4

2 4

mass of H SO 0.1 98 9.80.001

mass of H SO mass of water 9.8 970.2 980

= = =

+ +

Mass fraction of water = 1 0.001 = 0.999

Concentration in ppmThe number of grams of solute present in 106 grams (one million grams) of solution is called as concentration

of solute in ppm.

Concentration in ppm =6number of gramsof solute 10

number of grams of solution


Example 1: A 104 kg sample of hard water contains 96 grams of SO42 and 183 grams of HCO3

.Calculate concentration of SO4

2 and HCO3 in ppm.

Solution: Concentration in ppm =

6numberof gramsof solute 10

numberof gramof solution

Concentration of SO42 =

696 109 6 ppm

10000 1000

=

Concentration of HCO3

6

7

183 1018 3ppm

10

= =

Equivalent Mass

Equivalent mass of an atom is defined as that mass which either reacts with or displaces 1 g hydrogen or

35.5 g chlorine, from respective molecules.

Now the main concept behind this definition is in genral whatever be the molar ratio of reactants orproducts in a balanced chemical equation, 1 equivalent of any substance A has to react with 1 equivalent

of B to produce 1 equivalent of C and so forth or so onThis is a very important and useful law of chemical reactions.

Coming back to the definition, since equivalent mass of H is 1 or of chlorine is 35.5 so whatever mass of anelement reacts with 1 g H or 35.5 g chlorine has to be its respective equivalent mass. Equivalent mass ofoxygen is 8 when it forms oxide but 16 when it forms a peroxide.


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Example 1: 1 g magnesium when heated in excess of oxygen for long time produces5

g3

of magnesium

oxide. What is the equivalent mass of Mg?Solution: Weight of oxygen that reacted with Mg

5 21 g

3 3= =

Now,2

g3

oxygen reacts with 1 g Mg

8 g oxygen would react with1

8 12 g2 / 3

=

Example 2: Equivalent mass of calcium is 20. What volume of hydrogen at STP would be liberated by40 g calcium if it reacts with excess of dil. HCl.?

Solution: Since equivalent mass of hydrogen = 120 g calcium will liberate 1 g hydrogen from HCl 40 g calcium will liberate 2 g hydrogen.

As per mole concept, number of moles in 2 gm hydrogen =2

12

=

And volume of 1 mole of H2 at STP = 22.4 litres

Equivalent mass of compound is calculated in different ways.

1. Equivalent mass of acid

Equivalent mass of acid =( )

Molar mass of acid

Number of replacable H Basicity per moleculeof acid+


Example 1: Calculate the equivalent mass of HCl, H2SO4 and H3PO4.

Solution: HCl H Cl+ +

Equivalent mass of HCl =1 35.5

36.51

+=

+ + 22 4 4H SO 2H SO

Equivalent mass of H2SO4 =2 1 32 16 4

492

+ + =


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+ + 33 4 4H PO 3H PO

Equivalent mass of H3PO4 =3 1 30 4 16 97

32.333 3

+ + = =

2. Equivalent mass of base

Equivalent mass of base = ( )

Molar mass of base

Number of replacable OH Acidity


Example 1: Calculate the equivalent mass of NaOH, Ca(OH)2 and Al(OH)3

NaOH Na OH+ +

Solution: Equivalent mass of NaOH =23 16 1 40

4040 1

+ += =

( ) + +22Ca OH Ca 2OH

Equivalent mass of Ca(OH)2 =40 2 16 2 1 74

372 2

+ + = =

( ) + +33Al OH Al 3OH

Equivalent mass of Al(OH)3 =27 3 16 3 1 78

263 3

+ + = =

3. Equivalent mass of salt

Equivalent mass of salt =Molar mass of salt

Total positive charge on cations or total negative charge on anions

IIIIIllustrative Examples

Example 1: Calculate the equivalent mass of NaCl, MgCl2 and Na2SO4.Solution: 2MgCl Mg 2Cl

++ +

Equivalent mass of MgCl2 =24 2 35.5

2

+ 95 47.52

= =

+ + 22 4 4Na SO 2Na SO

Equivalent mass of Na2SO4 =2 23 32 4 16 142

712 2

+ + = =


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NaCl Na+ + Cl

Equivalent mass of NaCl =23 17

401

+=

Example 2: Calculate the equivalent mass of eSO4.(NH4 )2 SO4.6H2O.Solution: Equivalent mass of eSO4.( NH4)2SO4.6H 2O (Mohrs salt)

392

98gram4

= =

Normality

It is defined as the number of gram-equivalents of a solute present in one litre of solution.

( )Number of gram equivalent of soluteN

Volume of solution in litre=

( )

Mass of soluteN

Equivalent mass of solute Volume in litre=

( )

W 1000N

E V in ml

=

Also number of equivalents = N V (in litre)Milli equivalent = N V (in ml)

Mass of solute = Number of equivalent of solute Equivalent mass of solute


Example 1: If 20 ml of 0.1 N BaCl2 is mixed with 30 ml of 0.2 N Al2(SO4)3, how many gram of BaSO4is formed?

Solution: ( )2 2 4 3 43BaCl Al SO AlCl BaSO+ +

Equivalents before reaction20

0.11000

30

0.21000

= 2 103

= 6 103

0 0Equivalents after reaction 0 4 103 2 103 2 103

Since BaCl2 is limiting reagent

Hence, mass of BaSO4 = 2 103 Equivalent mass of BaSO4

Since BaSO4 is a divalent

n factor of BaSO4 = 2

Mass of BaSO4 = 2 103

233

2= 0.233 g


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Example 2: Calculate the normality of H2SO4 solution having 0.05 equivalents in 200 ml.

Solution: ( )

Equivalent 1000N Volume in ml

=

N =0.050 1000

0.25 N200

=

Example 3: Calculate the normality of NaOH solution when 2 g of it are present in 800 ml solution.

Solution:( )

Mass of soluteN 1000

Equivalent mass Volume in ml=

=2 1000

0.0625 N40 800 =

Relation between normality and molarity

Normality =Mass of solute 1000

Molecular mass of soluteVolume (in ml)

n factor

Normality = Molarity n factor

N = M n

or monovalent compound, n = 1or monovalent compound normality and molarity is same.


Example 1: Calculate normality of 0.3 M AlCl3 solution.

Solution: or 3AlCl Al 3Cl+++ +

n factor = 3

Hence, normality = 0.3 3 = 0.9 N

Also Normality =d 10 x

Equivalent mass

And Molarity =d 10 x

Molecular mass

where,d = Density of solutionx = Percentage by mass of solute


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1llust rative Examples

Example 1: A mixture is obtained by mixing 500 ml 0.1M H2SO4 and 200 ml 0.2M HCl at 25C. indthe normality of the mixture.

Solution: We know, N =2 2 1 1

1 2

N V N V

V V

++

or the mixture,

500 0.1 2 200 0.2 1N 0.2

700

+ = =

Example 2: 500 ml 0.2 N HCl is neutralized with 250 ml 0.2 N NaOH. What is the strength of theresulting solution?

Solution:2HCl NaOH NaCl H O+ +

Equivalents of HCl = 3500 0.2 10

Equivalents of NaOH = 3250 0.2 10

Equivalent of excess HCl = 3 3(500 0.2 10 250 0.2 10 egv) 350 10=

Normality of HCl (excess)3 3500 10 10

750

= = 0.067 N

Strength of HCl = .067 36.5 g/litre= 2.44 g/litre

Alternate Method:

Normality of HCl (excess), N =1 1 2 2

1 2

N V N V

V V

+

0.2 1 500 - 0.2 1 250N =

500 + 250

N = 2.44 NStrength of HCl = .067 36.5 grams/litre = 2.44 grams/litre

Relationship between normality and strength (gram/litre of solution)

In terms of normality,Strength = Normality Equivalent mass


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1. A plant virus is found to consist of uniform cylinderical particles of 150 in diameter and 5000long. The specific volume of the virus is 0.75 cm3/g. If the virus is considered to be a single particle,find its molar mass. (1999)

2. A drug marijuana owes its activity to tetra-hydrocannabinol, which contains 70% as many carbonatoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number ofmoles in a gram of tetra-hydrocannabinol is 0.00318. Determine its molecular formula.

3. Assuming fully decomposed, the volume of CO2released in litres at STP on heating 9.85 g of

BaCO3will be (Atomic mass Ba = 137)

(a) 0.842 (b) 2.242(c) 4.062 (d) 1.122

4. 2.76 g of silver carbonate (atomic mass of Ag = 108) on being heated strongly yields a residueweighing(a) 2.16 g (b) 2.48 g(c) 2.32 g (d) 2.64 g

5. The molarity of a solution in which 2 moles of CH4are dissolved in 1,000 ml of solution will be

(a) 1 (b) 2(c) 3 (d) 1.5

6. Calculate the molarity of water if its density is 1000 kg/m3. (2003)

7. Determine the volume of dil. HNO3 (d = 1.11 g/ml, 19% HNO3) that can be prepared by diluting withwater. 50 ml of Conc. HNO3

(d = 1.42 g/ml, 69.8% HNO3).

8. Calculate the molality of 1 litre solution of 93% H2SO4 (Weight/volume). The density of the solutionis 1.84g/mL. (1990)

9. A solution 6.90 M of KOH in water contains 30% by mass of KOH. Calculate the density of thesolution.

10. The density of a 3 M sodium thiosulphate solution (Na2S2O3) is 1.25 gml1. Calculate (i) the

percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii)

the molalities of Na+ and 22 3S O ions. (1983)

11. A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12 H22 O11) Calculate (i) molalconcentration and (ii) mole-fraction of sugar in the syrup. (1988)

12. 8.0575 102 kg of Glaubers salt is dissolved in water to obtain 1 dm3 of a solution of density1077.2 kg m3. Calculate the molarity, molality and mole-fraction of Na2SO4 in the solution.

(1994)

13. How many ml of 0.5 MH2SO4 are needed to dissolve 0.5 g of copper (II) carbonate? (1999)

Take-Off


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14. The equivalent weight of 4MnSO is half its molecular weight when it is converted to: (1988)

(a) 2 3Mn O (b) 2MnO (c) 4MnO (d) 24MnO

15. 300 ml of 0.1 M HCl and 200 ml of 0.3 M H2SO4 are mixed. Calculate the normality of the resultingmixture.

16. How much of NaOH is required to neutralise 1,500 ml of 0.1 N HCl? (Na = 23)(a) 40 g (b) 4 g(c) 6 g (d) 60 g

17. What is the strength in grams per litre of a solution of H2SO

4, 12 ml of which neutralises 15 ml of

N

10NaOH solution?

*18. The formula mass of an acid is 82.0. 100 ml of a solution of this acid containing 39.0 g of the acid perlitre were completely neutralised by 95.0 ml of aqueous NaOH containing 40.0 g of NaOH per litre.What is the basicity of the acid?

19. ind out the equivalent mass of H3PO

4in the following reaction.

( ) 3 4 4 22Ca OH H PO CaHPO 2H O+ +

20. How much AgCl will be formed by adding 200 ml of 5 N HCl to a solution containing 1.7 g AgNO3?

21. One mole of a mixture of CO and CO2

requires exactly 20 g of NaOH in solution for completeconversion of all the CO

2

into Na2

CO3

. How much NaOH would it require for conversion into Na2

CO3if the mixture (one mole) is completely oxidised to CO

2?

(a) 60 g (b) 80 g(c) 40 g (d) 20 g

22. A compound of iron and chlorine is soluble in water. An excess of silver nitrate was added toprecipitate the chloride ion as silver chloride. If 134.8 mg of the compound gave 304.8 mg of AgCl,the formula of the compound is(a) e

2Cl

3(b) eCl

2

(c) eCl3

(d) eCl

23. One mol of N2

and 4 mol of H2

are allowed to react in a vessel and after reaction, H2O is added.

Aqueous solution required 1 mol of HCl. Mole fraction of H2 in the gaseous mixture after reaction is

(a)1

6(b)

5

8

(c)1

3(d) None of these

*24. 0.722 g of bromide of an element A having atomic weight 91.2 in a series of reaction gives 1.32 g ofAgBr. ind the valency of element A.(a) 2 (b) 4(c) 3 (d) 5


24/24

Answer KeysAnswer Key

Basic Concepts of Chemistry

1. 70.96 106 g/mol 2. C21

H30

O2

3. d4. a 5. b

6. 55.55 M

7. 235 mL 8. 10.42

9. 1.288 g/mL

10. (i) 37.92% (ii) 0.065 (iii) 7.732 m and 3.865 m

11. (i) 0.55 (ii) 0.0099

12. 0.25 M, 0.239 m, 0.0043

13. 8.097 L 14. b 15. 0.3N

16. c 17. 6.125 g/L

18. 2 19. 49

20. 1.435 g 21. b

22. b 23. b

24. b

01. basic concept of chemistry [1-24]

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