01. Basic Concept of Chemistry [1-24]

Basic Concepts of Chemistry Chapter Objectives 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

Chemistry /Basic Concepts of Chemistry

... 1

Concept Notes Branches of Chemistry Chemistry may be defined as the branch of science dealing with the composition, structure and properties of matter. Chemistry has mainly three branches: 1. Physical chemistry 2. Organic chemistry 3. 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 chemical methods. A compound is a substance which can be decomposed into two or more dissimilar substances by chemical reactions. For 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. For 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. For 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 is equal to the total mass of the reactants. Or in other words, during any physical or chemical change, matter is neither created nor destroyed.

AgNO3 Silver nitrate

+ NaCl Sodium chloride

¾¾®

AgCl

+ NaNO3

Silver chloride

Sodium nitrate

Mass of reactants (AgNO3 + NaCl) = Mass of product (AgCl + NaNO3) Chemistry /Basic Concepts of Chemistry

... 2

Law of constant composition or law of definite proportions Louis Proust states that a pure chemical compound always contains same elements combined together in 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 of the 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 two compounds 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 combines directely with eachother.

S SO

H S

H

O

H O

H2S

→ 2 : 32, 1 : 16

SO2

→

H2S : SO2 →

32 : 32, 1 : 1

1 1 : = 1: 16 16 1

[H2S : SO2 ] : H2O →

1 8 1 × = 16 1 2

 1 This simple ratio   explain the law of reciprocal proportion. 2 Gay Lussacs law of combining gaseous volume Gay Lussac states that when gases combine to form gaseous products, a simple ratio exists between the volumes of the reactants and the products at constant temperature and pressure.

H2 (g) 1volume

+ Cl2 (g) → 2HCl(g) 1 volume

2 volume

The ratio is 1 : 1 : 2.


... 3

Daltons atomic theory Although the origin of idea that matter is composed of small indivisible particles called a-tomio (meaning 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. 2. 3. 4.

Matter consists of indivisible atoms. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass. Compounds are formed when atoms of different elements combine in a fixed ratio. 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 Mass

Since 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

12 6 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 of Atomic mass =

1 th mass of a 12

12 6 C

atom.

Mass of 1 atom of the element 1 × Mass of 1 atom of 12 6 C 12

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

1 th mass of a 12

12 6 C

atom.

1 1 g ; 1⋅ 67 × 10 24 g × Mass of one 12 6 C atom = 12 6 ⋅ 023 × 1023 Atoms of the same element may have different masses. They are known as isotopes. In such cases, the atomic mass of an element is taken as the average of the atomic masses of the various isotopes. 1 amu =

35 37 For example, chlorine has two isotopes 17 Cl and 17 Cl present in relative abundance ratio 3 : 1. Therefore,

the average atomic mass of chlorine atom =


35 × 3 + 37 × 1 = 35.5 4

... 4

1llustrative Example 22 and 10 Ne is in the ratio 90.51%, 0.28% and 9.21%, respectively. Calculate the average atomic mass of the element.

Example 1: The natural occurrence of the isotopes

Average atomic mass of Neon atom (Ne) =

20 21 10 Ne, 10Ne

90.51× 20 + 0.28 × 21 + 9.21× 22 = 20.187 amu (90.51 + 0.28 + 9.21)

Gram atomic mass

Atomic mass expressed in grams is called gram atomic mass. This amount of the element is also called one gram atom. 1 gram atom of oxygen = Gram atomic mass of oxygen = 16 g 1 gram atom of nitrogen = 14 g Atomic mass of oxygen = 16 Molecular Mass

( )

12 Relative mass of a molecule may be obtained in terms of the same standard 6 C used for defining

atomic mass. The molecular mass is a number which expresses how many times the mass of a given molecule is that of

1 th mass of a 12

12 6 C

atom.

Mass of a molecule of the subs tance 1 th mass of one 12 6 C atom 12 Relative molecular mass of a molecule is expressed in grams and the actual mass of a molecule is expressed in amu scale. Relative molecular mass is obtained by adding the atomic masses of all the atoms present in the given molecule. For 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

Molecular mass =

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 = 32 Gram molecular mass of O2 = 32 g


... 5

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

→ 2HI + I2 H2 1 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 in 1 g mole of any substance is a constant. This is known as Avogadros number, it is denoted by NA = 6.023 × 1023. Since atomic mass and molecular mass are defined with reference to Avogadros number is also defined relative to

12 6 C

12 = 6 C

12 ∴ Mass of 1 atom of 6 C =

atom.

atom.

Avogadros number is defined as the number of atoms present in exact 12 g of Mass of 6.023 × 1023 atoms of

12 6 C

12 . 6 C

12 g

12

g 6.023 × 1023 = 1.992 × 1024 g

On the amu scale, the mass of a

(1) (2) (3)

12 atom 6 C

= 12 amu = 12 × 1.66 × 1024 = 1.992 × 1023 g

If 6.02 × 1023 hydrogen atoms were laid side by side, the total length would encircle the earth about a million times. The mass of 6.02 × 1023 olympic shotput balls is almost equal to the mass of the earth. 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 form expressed in grams.


... 6

Molar mass The mass of one mole is the molar mass. For example, molar mass of Na = 23 g Molar mass of Cl2 = 71 g, molar mass of [K4Fe (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. For 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 g mole of any gaseous element or compound occupy a volume of 22.7 L at STP.

1llustrative Examples Example 1: Solution:

Calculate the mass of one molecule of water. 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

18 g

=

6.023 × 1023 = 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 =

6.023 × 1023 = 2.6 × 1022 [approx.] 22.7

Realize that the answer is independent of nature of the gas i.e.whether we have NH3 or 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 NH3 = 17 g 1 mole = 1 molar mass (17 g) = 6.023 × 1023 molecules

∴ 6023 molecules =

17 × 6023 6.023 × 10 23

g

= 17 × 10 − 20 g


... 7

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

given number 6.023 × 10

[See Ex. 3]

!

given mass of substance

Number of moles

Molar mass given volume of gas at NTP in litre 22.7

Example 4:

Find number of atoms in (a) 1 g yellow phosphorous [P4] (b) 5.675 litres of H2S gas at S.T.P. (c)

Solution:

[See Ex. 2]

(a)

1 mole of acetic acid [CH3COOH] 2 1 1 moles of P4 = × 6.023 × 10 23 molecules 4 × 31 4 × 31 Since one molecule of P4 = 4 atoms 1g =

∴ Number of atoms = (b)

5.675 5.675 mole of H2S = × 6.023 × 1023 molecules 22.7 22.7 Since one molecule of H2S = 3 atoms

5.675 litre =

∴ No. of atoms =

(c)

4 × 6.023 × 1023 = 1.94 × 1022 (approx) 4 × 31

3 × 5.675 × 6.023 ×1023 22.7

3 × 5.675 × 6.023 × 1023 = 4.5 × 1023 (approx.) 22.7 Do it yourself. Ans = 24.1 × 1023 (approx.)

You must have used the information that atomic mass of sulphur = 32 or molar mass of SO2 = 64. Can you say that mass of 1 atom of sulphur is 32 g or that mass of 1 molecule of SO2 is 64 g?


... 8

Empirical Formula and Molecular Formula Empirical formula is the simplest ratio of all types of atoms present in compound whereas molecular formula gives the exact number of atoms of each type present in the compound. For example, empirical formula of benzene is CH and molecular formula of benzene is C6H6. Molecular formula = n × Empirical formula Molecular formula is either identical with the empirical formula or a simple multiple of it. For some molecules like C2H2, C6H6, the empirical formula is the same, i.e. CH. But the molecular formula 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 gas =

mass of a given volume of vapour of the substance mass of same volume of Hydrogen under similar condition

1llustrative Examples Example 1: Solution:

0.1g of a gaseous substance occupies 30 ml at S.T.P. What is its V.D. and molecular mass. Mass of 22700 ml of H2 at S.T.P. = 2 g [molecular mass] 2 = 0.0009 g 22700 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 Mass of 30 ml of vapour ∴ V.D.= Mass of 30 ml of H2 at NTP

∴ Mass of 1 ml of H2 at S.T.P. =

0.1 = 37 [approx.] 0.0027 ∴ Molecular mass = 2 × 37 = 74 g. Realize that (1) V.D. has no units and (2) does not depend on temperature or pressure etc. =

Example 2: Solution:

The vapour density of a compound having empirical formula CH is 39. Find its molecular formula. Empirical formula = CH Empirical formula mass = 12 × 1 + 1 × 1 = 13 Molecular mass = 2 × Vapour density = 2 × 39 = 78 Molecular formula mass = n × Empirical formula mass

Molecular formula mass 78 Empirical formula mass = 13 = 6 Hence, the molecular formula is C6H6. ∴n =


... 9

Calculation of empirical formula from percentage composition Example 3: Solution:

An organic compound contains 57.8% carbon, 3.6% hydrogen. Find the empirical formula. Atomic ratio =

Percentage Atomic mass

Simplest whole

Element

Percentage

C

57.8

57.8 = 4.82 12

4.82 =2 2.41

2×2=4

H

3.6

3 .6 = 3.6 1

3 .6 = 1.49 2.41

2 × 1.49 » 3

38.6 = 2.41 16

2.41 =1 2.41

2×1=2

O

100 (57.8 + 3.6) = 38.6

Simplest ratio

number ratio

∴ The empirical formula is C4H3O2.


An organic compound on analysis gave C = 41.4% and H = 3.1%. Its vapour density is 58. Calculate the empirical formula and molecular formula.

Atomic ratio =

Percentage Atomic mass

Simplest ratio

Element

Percentage

C

41.4

41.4 = 3.45 12

3.45 = 1.1 » 3.1

H

3.1

3.1 = 3.1 1

3.1 =1 3.1

55.5 = 3.46 16

3.46 = 1.1 » 3.1

O

100 (41.4 + 3.1) = 55.5

∴ Empirical formula is CHO. ∴ Empirical formula mass = 12 × 1 + 1 × 1+ 16 × 1 = 29 Molecular mass = 2 × vapour density = 2 × 58 = 116 Molecular formula mass = n × Empirical formula mass Molecular formula mass 116 Empirical formula mass = 29 = 4 ∴ Molecular formula is C4H4O4 ∴n =


... 10

Dulong and Petits law Atomic heat = Atomic mass × specific heat Atomic heat of most elements ≈ 6.4 (approx.)

6.4 Approximate atomic mass = 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. 2SO2 + O2 → 2SO3

1llustrative Example Example 1:

Calculate the mass of calcium oxide and carbon dioxide obtained by the decomposition of 20 g of CaCO3 .

Solution:

CaCO3 → CaO + CO2

Mol. Mass (40 + 12 + 3 × 16) (40 + 16) (12 + 2 × 16) = 100 = 56 = 44 The above balanced equation shows that 100 g (1 mole) of CaCO3 gives 56 g (1 mole) of CaO and 44 g (1 mole) of CO2.

∴ 20 g of CaCO3 gives =

56 g × 20 g of CaO 100 g

= 11.2 g of CaO Similarly, 20 g of CaCO3 gives =

44 g × 20 g of CO2 100 g

= 8.8 g of CO2

Limiting Reagent The reagent which determines the yield of a reaction is called limiting reagent. It is consumed first, during reaction. 2H2 + O2 → 2H2O Suppose the reaction mixture contains 2 moles of H2 and 2 moles of O2. From the above equation, it is evident that 2 moles of H2 require only 1 mole of O2 to complete the reaction. Therefore, 1 mole of O2 will be in excess. In this case, H2 is said to be the limiting reagent.


... 11


Solution:

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. 2H2 + O2 2 moles 1 mole

→ 2H2O

2 moles

Mass 4 Number of moles of H2 present = Molar mass = = 2 mol 2 Mass 40 = 1.25 mol Number of moles of O2 present = Molar mass = 32 (i) Identification of limiting reactant According to the equation, 2 moles of H2 require 1 mol of O2. Number of moles of O2 actually present = 1.25 mol, i.e. O2 is in excess and thus, H2 is 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. O2) 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 in smaller 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 =

w Mass of solute × 100 = % V Volume of solution


... 12


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. Mass of solute (in gram) Strength = Volume of solution (in litre )

1llustrative Examples Example 1:

Solution: Example 2: Solution:

4 gm of NaOH was dissovled in water and volume was made upto 500 ml. Calculate strength of NaOH solution.

Mass of solute 4 × 1000 Strength of solution = Voluem of solution = 500 = 8 gm / L 0.02 moles of H2SO4 is dissolved in water and volume of solution was made upto 800 ml. Calculate strength of acid solution. Mass of solute = number of moles × molar mass Strength of acid solution =

0.02 × 98 × 1000 = 2.45 gm / L 800

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

M=

Moles of solute Volume of solution (in litres)

M=

Mass of solute × 1000 Molar mass of solute × Volume (in millilitres )

W × 1000 Molar mass= M × V (in millilitres) Also moles of solute = M × V (in litres) Mass Moles of solute = molar mass Milli moles of solute = M × V (in millilitres) In terms of molarity, Strength (g/L) = Molarity ×Molar mass Chemistry /Basic Concepts of Chemistry

... 13


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

Solution:

Mass of solute × 1000 98 1000 × =2M Molarity = molar mass × Volume (in millilitres ) = 98 500


Calculate the strength of 0.01 M NaOH solution. Strength = Molarity × Molar mass = 0.01 × 40 = 0.40 g/L

Molality

It is defined as number of moles of solute present in 1 kg (or 1000 g) of solvent. It is represented by m (small letter). Moles of solute m = Mass of solvent (in kg ) Mass of solute × 1000 m = Molar mass of solute × Mass of solvent (in grams )

1llustrative Examples Example 1: Solution: Example 2: Solution:

Find the molality of H2SO4 solution in which 98 g of H2SO4 is dissolved in 500 g of solvent. 98 × 1000 =2 m Molality = 98 × 500 Find the molality of one molar solution of NaOH. Density of solution is 1.04 g/ml. 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 m=


1× 1000 = 1 molal 1000

Find the molality of H2SO4 solution whose specific gravity(density) is 1.98 g/ml and contains 95% mass by volume H2SO4. 100 ml solution contains 95 g H2SO4. 95 Moles of H2SO4 = 98 Mass of solution = 100 × 1.98 = 198 g Mass of water = 198 95 = 103 g Molality =

95 × 1000 = 9.412 m 98 × 103


... 14

Mole Fraction The ratio of number of moles of one component to the total number of moles of all the component present in 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


98 gram of H2SO4 is dissolved in 900 ml of water. Calculate mole fraction of H2SO4 and water in given solution.

Solution:

number of moles of H2SO4 in solution = number of moles of H2O in solution = mole fraction of H2O =

98 massof H2SO4 = =1 molar massof H2SO4 98

massof water 900 = = 50 molar massof water 18

50 50 = 50 + 1 51

mole fraction of H2SO4 =

1 1 50 1 = = (also mole fraction of H2SO4 = 1 ) 51 51 50 + 1 51

Mass Fraction The ratio of mass of one component to the total mass of all components present in solution is called as mass fraction of that component: Lets consider a binary solution containing w g of sloute in W g of solvent, then.

mass of solute w Mass fraction of solute (MA) = mass of solute + mass of solvent = w + W mass of solvent W Mass fraction of solvent (MB) = mass of solute + mass of solvent = w + W The sum of mass fractions of all the components is unity ie. MA + MB = 1 Chemistry /Basic Concepts of Chemistry

... 15


Solution:

0.1 mole of H2SO4 is dissolved in 970.2 ml of water. calculate mass fraction of each component. Given density of water is 1 g/ml. Mass Fraction of H2SO4 =

mass of H2SO4 0.1× 98 9.8 = = = 0.001 mass of H2SO4 + mass of water 9.8 + 970.2 980 Mass fraction of water = 1 0.001 = 0.999

Concentration in ppm

The 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 =

number of gramsof solute × 106 number of grams of solution


Solution:

A 104 kg sample of hard water contains 96 grams of SO42 and 183 grams of HCO3 . Calculate concentration of SO42 and HCO3 in ppm.

number of gramsof solute × 106 Concentration in ppm = number of gramof solution Concentration of SO42 = Concentration of HCO3 =

96 × 106 = 9 ⋅ 6 ppm 10000 × 1000

183 × 106 107

= 18 ⋅ 3ppm

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 or products 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 on This 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 an element reacts with 1 g H or 35.5 g chlorine has to be its respective equivalent mass. Equivalent mass of oxygen is 8 when it forms oxide but 16 when it forms a peroxide. Chemistry /Basic Concepts of Chemistry

... 16


1 g magnesium when heated in excess of oxygen for long time produces

Solution:

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

5 2 −1= g 3 3

Now,

2 g oxygen reacts with 1 g Mg 3

∴ 8 g oxygen would react with Example 2: Solution:

5 g of magnesium 3

1 × 8 = 12 g 2/3

Equivalent mass of calcium is 20. What volume of hydrogen at STP would be liberated by 40 g calcium if it reacts with excess of dil. HCl.? Since equivalent mass of hydrogen = 1 20 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 = And volume of 1 mole of H2 at STP = 22.4 litres

2 =1 2

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 molecule of acid


Calculate the equivalent mass of HCl, H2SO4 and H3PO4.

Solution:

HCl → H+ + Cl−

Equivalent mass of HCl =

1 + 35.5 = 36.5 1

H2SO4 → 2H+ + SO42− Equivalent mass of H2SO4 =


2 × 1 + 32 + 16 × 4 = 49 2

... 17

H3PO 4 → 3H+ + PO 34 −

3 × 1 + 30 + 4 × 16 97 = = 32.33 3 3

Equivalent mass of H3PO4 = 2.

Equivalent mass of base

Molar mass of base Equivalent mass of base = Number of replacable OH− Acidity ( )


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 = = 40 40 1

Ca (OH)2 → Ca2+ + 2OH− Equivalent mass of Ca(OH)2 =

40 + 2 × 16 + 2 × 1 74 = = 37 2 2

Al (OH )3 → Al3 + + 3OH− Equivalent mass of Al(OH)3 = 3.

27 + 3 × 16 + 3 × 1 78 = = 26 3 3

Equivalent mass of salt

Molar mass of salt Equivalent mass of salt = Total positive charge on cations or total negative charge on anions

Illustrative Examples Example 1:

Calculate the equivalent mass of NaCl, MgCl2 and Na2SO4.

Solution:

MgCl2 → Mg++ + 2Cl−

Equivalent mass of MgCl2 =

24 + 2 × 35.5 = 95 = 47.5 2 2

Na 2 SO 4 → 2Na + + SO 24−

Equivalent mass of Na2SO4 = Chemistry /Basic Concepts of Chemistry

2 × 23 + 32 + 4 × 16 142 = = 71 2 2 ... 18

NaCl → Na+ + Cl Equivalent mass of NaCl = Example 2: Solution:

23 + 17 = 40 1

Calculate the equivalent mass of FeSO4.(NH4 )2 SO4.6H2O. (Mohrs salt) Equivalent mass of FeSO4.( NH4)2SO4.6H 2O

=

392 = 98 gram 4

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

Number of gram equivalent of solute Volume of solution (in litre )

N=

Mass of solute Equivalent mass of solute × Volume (in litre )

N=

W × 1000 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


If 20 ml of 0.1 N BaCl2 is mixed with 30 ml of 0.2 N Al2(SO4)3, how many gram of BaSO4 is formed?

Solution:

BaCl2

+

Al2 (SO4 )3

Equivalents before reaction = 2 × 103 = 6 × 103 Equivalents after reaction

→

AlCl3

+

BaSO4

20 30 × 0.1 × 0.2 1000 1000 0 0 2 × 103 0 4 × 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 × Chemistry /Basic Concepts of Chemistry

233 = 0.233 g 2 ... 19

Example 2:

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

Solution:

N=

Equivalent × 1000 Volume (in ml )

N=

0.050 × 1000 = 0.25 N 200

Example 3:

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

Solution:

N=

=

Mass of solute × 1000 Equivalent mass × Volume (in ml )

2 × 1000 = 0.0625 N 40 × 800

Relation between normality and molarity Normality =

Mass of solute × 1000 Molecular mass of solute × Volume (in ml) n factor

Normality = Molarity × n factor N=M×n For monovalent compound, n = 1 For monovalent compound normality and molarity is same.


Calculate normality of 0.3 M AlCl3 solution.

Solution:

For AlCl3 → Al+++ + 3Cl− n factor = 3 Hence, normality = 0.3 × 3 = 0.9 N

d × 10 × x Equivalent mass

Also

Normality =

And

d × 10 × x Molarity = Molecular mass where, d = Density of solution x = Percentage by mass of solute


... 20

1llust rative Examples Example 1: Solution:

A mixture is obtained by mixing 500 ml 0.1M H2SO4 and 200 ml 0.2M HCl at 25°C. Find the normality of the mixture. N2 V2 + N1V1 We know, N = V1 + V2 For the mixture, N =

500 × 0.1× 2 + 200 × 0.2 × 1 = 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 the resulting solution?

Solution:

HCl + NaOH → NaCl + H2O

Equivalents of HCl = 500 × 0.2 × 10−3 Equivalents of NaOH = 250 × 0.2 × 10−3 Equivalent of excess HCl = (500 × 0.2 × 10−3 − 250 × 0.2 × 10−3 egv) = 50 × 10−3

500 × 10−3 × 103 = 0.067 N 750 Strength of HCl = .067 × 36.5 g/litre = 2.44 g/litre

Normality of HCl (excess) =

Alternate Method: Normality of HCl (excess), N =

N1V1 − N2 V2 V1 + V2

0.2 × 1 × 500 - 0.2 × 1 × 250 500 + 250 N = 2.44 N Strength of HCl = .067 × 36.5 grams/litre = 2.44 grams/litre N=

Relationship between normality and strength (gram/litre of solution) In terms of normality, Strength = Normality × Equivalent mass


... 21

Take-Off 1.

A plant virus is found to consist of uniform cylinderical particles of 150Å in diameter and 5000Å long. 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 carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number of moles in a gram of tetra-hydrocannabinol is 0.00318. Determine its molecular formula.

3.

Assuming fully decomposed, the volume of CO2 released in litres at STP on heating 9.85 g of (Atomic mass Ba = 137) BaCO3 will be (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 residue weighing (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 CH4 are 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.

7.

Determine the volume of dil. HNO3 (d = 1.11 g/ml, 19% HNO3) that can be prepared by diluting with water. 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 solution is 1.84g/mL. (1990)

9.

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

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)

11.

(2003)

(1983) the molalities of Na+ and S2O32− ions. A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12 H22 O11) Calculate (i) molal concentration 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 density 1077.2 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) ... 22

14.

The equivalent weight of MnSO4 is half its molecular weight when it is converted to: (a) Mn2O3

(b) MnO2

(c) MnO−4

(1988)

(d) MnO24−

15.

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

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 H2SO4, 12 ml of which neutralises 15 ml of

*18.

N NaOH solution? 10 The formula mass of an acid is 82.0. 100 ml of a solution of this acid containing 39.0 g of the acid per litre 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.

Find out the equivalent mass of H3PO4 in the following reaction.

Ca (OH)2 + H3PO4 → CaHPO4 + 2H2O 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 complete conversion of all the CO2 into Na2CO3. How much NaOH would it require for conversion into Na2CO3 if the mixture (one mole) is completely oxidised to CO2? (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 to precipitate 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 (b) FeCl2 (a) Fe2Cl3 (c) FeCl3 (d) FeCl

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 1 5 (b) (a) 6 8 1 (c) (d) None of these 3 0.722 g of bromide of an element A having atomic weight 91.2 in a series of reaction gives 1.32 g of AgBr. Find the valency of element A. (a) 2 (b) 4 (c) 3 (d) 5

*24.


... 23

Answer Keys Key Basic Concepts of Chemistry 1. 70.96 × 106 g/mol 3. d 4. a

2. C21H30O2 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


... 24

01. Basic Concept of Chemistry [1-24]

Recommend Documents