01. Basic Concept of Chemistry [1-24]

Chapter Objectives

Basic Concepts

of Chemistry

1. Introduction to Chemistry 2. Laws of chemical combination 3. Atomic mass, molecular mass 4. Avogadro’s 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

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 + N aCl _¾¾® AgCl + NaNO3

Silver

nitrate Sodium chloride Silver chloride Sodium nitrate Mass of reactants (AgNO₃ + NaCl) = Mass of product (AgCl + NaNO₃)

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 H O H S SO H O H₂S _→ 2 : 32, 1 : 16 SO_{2 →} 32 : 32, 1 : 1 2 2 1 1 H S : SO : 1: 16 16 1 → =

[

H S : SO : H O2 2

]

2 →_{16 1 2}1 8 1× =

This simple ratio   _{ }1₂ explain the law of reciprocal proportion. Gay Lussac’s 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.

2 2

2 volume 1volumeH (g) +1 volumeCl (g) → 2HCl(g)

Dalton’s 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.

Dalton’s 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 ₁₂1 th mass of a 12

6 C atom.

Atomic mass = ₁₂

6 Mass of 1 atom of the element

1 _{Mass of 1 atom of 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 ₁₂1 th mass of a 12

6 C atom.

12 –24

6 ₂₃

1 1

1 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, the atomic mass of an element is taken as the average of the atomic masses of the various isotopes. For example, chlorine has two isotopes 35

17Cl and 1737Cl present in relative abundance ratio 3 : 1. Therefore,

1llustrative Example

Example 1: The natural occurrence of the isotopes 20 21 22

10Ne, 10Ne and10 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 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

Relative mass of a molecule may be obtained in terms of the same standard

( )

126 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

12 mass of a 126 C atom.

Molecular mass =

12 6

Mass of a molecule of the substance 1 th mass of one 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 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

Avogadro’s Law

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

H₂ + I₂ → 2HI 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

Avogadro’s 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 Avogadro’s number, it is denoted by N_A = 6.023 × 1023_{. Since atomic mass and molecular mass are defined with reference to} 12

6 C atom.

Avogadro’s number is also defined relative to 12

6 C atom.

Avogadro’s number is defined as the number of atoms present in exact 12 g of 12₆ C . Mass of 6.023 × 1023 _{atoms of} 12 6 C = 12 g ∴Mass of 1 atom of 126 C 12 ₂₃ g 6.023 10 = × = 1.992 × 10–24 _g

On the amu scale, the mass of a 12

6 C atom = 12 amu

= 12 × 1.66 × 10–24

= 1.992 × 10–23 _g

(1) If 6.02 × 1023 _{hydrogen atoms were laid side by side, the total length would encircle the earth about}

a 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 _{. (Avogadro’s number)}

A mole is defined as the amount of any substance containing one Avogadro’s 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.

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: 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 = Avogadro’s number = 6.023 × 1023

∴ Mass of one molecule of H₂O = Gram molecular mass_{Avogadro's number}

= 18 g ₂₃ 6.023 10×

= 2.99 × 10–23 _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 22.7

× _{= 2.6 × 10}22 _[approx.]

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

23 17 6023 6023 molecules g 6.023 10 × ∴ = × 20 17 10− g = ×

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

2 mole of acetic acid [CH3COOH]

Solution: (a) 1 g= _{4 31×}1 moles of P4 = _{4 31×}1 ×6.023 10× 23 molecules

Since one molecule of P4 = 4 atoms

∴Number of atoms = 4 6.023 1023 1.94 10 (approx)22 4 31

× × _{= ×} ×

(b) 5.675 litre = 5.675_22.7 mole of H S2 =5.675_22.7 ×6.023 10 molecules× 23

Since one molecule of H2S = 3 atoms

∴No. of atoms = 3 5.675 6.023 10×_22.7 × × 23

23 23

3 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 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?

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

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 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 2 Mass of 30 ml of vapour V.D.= Mass of 30 ml of H at NTP ∴ 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: The vapour density of a compound having empirical formula CH is 39. Find its molecular formula.

Solution: 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

n

Empirical formula mass

∴ = ₌ 78

13 = 6 Hence, the molecular formula is C₆H₆.

Calculation of empirical formula from percentage composition

Example 3: An organic compound contains 57.8% carbon, 3.6% hydrogen.

Find the empirical formula. Solution:

Element Percentage Atomic ratio =

mass Atomic

Percentage

Simplest ratio Simplest whole number ratio C 57.8 4.82 12 8 . 57 = 2 41 . 2 82 . 4 _{= 2 × 2 = 4} H 3.6 3.6 1 6 . 3 = 1.49 41 . 2 6 . 3 _{= 2 × 1.49 » 3} O 100 – (57.8 + 3.6) = 38.6 16 2.41 6 . 38 = 1 41 . 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 density is 58. Calculate the empirical formula and molecular formula.

Solution:

Element Percentage Atomic ratio =

mass Atomic Percentage Simplest ratio C 41.4 41.4 3.45 12 = 3.45 1.13.1 = »



H 3.1 3.1 3.1 1 = 3.1 13.1= O 100 – (41.4 + 3.1) = 55.5 55.5 3.4616 = 3.46 1.13.1 = »





∴ 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 n

Empirical formula mass

∴ = ₌ 116

29 = 4

Dulong and Petit’s law

Atomic heat = Atomic mass × specific heat Atomic heat of most elements ≈ 6.4 (approx.) Approximate atomic mass = _{Specific heat}6.4 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 3

2SO +O → 2SO

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_{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, during reaction.

2 2 2

2H +O →2H O

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.

1llustrative Example

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 moles 1 mole}2H2 + O2 →_{2 moles}2H O2

Number of moles of H₂ present = _{Molar mass =} Mass 4 2 mol 2 =

Number of moles of O₂ present = _{Molar mass =} Mass 40 1.25 mol 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 100 w% Mass of solution× =W Percentage by volume = Volume of solute 100 v%

Volume of solution× = V Percentage mass by volume = Mass of solute 100

1llustrative Example

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 = ₁₂₀20 100 16.67%× =

Strength of Solution

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

Strength =_{Volume of solution in litre}Mass of solute (ingram)

₍

₎

1llustrative Examples

Example 1: 4 gm of NaOH was dissovled in water and volume was made upto 500 ml. Calculate strength of NaOH solution.

Solution: Strength of solution = _{Voluem of solution}Mass of solute = 4 1000 8gm/L×₅₀₀ =

Example 2: 0.02 moles of H2SO4 is 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/L×₈₀₀× =

Molarity

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

M

Volume of solution (in litres)

=

(

)

Mass of solute 1000 M

Molar mass of solute Volume in millilitres

× =

×

Molar mass= _{M V (in millilitres)×} W 1000× Also moles of solute = M × V (in litres) Moles of solute = _{molar mass}Mass

Milli moles of solute = M × V (in millilitres) In terms of molarity,

1llustrative Examples

Example 1: Calculate molarity of a solution containing 98 g of H₂SO₄ present in 500 ml of solution. Solution: Molarity = _{molar mass Volume in millilitres}Mass of solute 1000_× ×

₍

₎

= 98 1000 2 M_{98 500}× =

Example 2: Calculate the strength of 0.01 M NaOH solution.

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).

m = _{Mass of solvent in kg}Moles of solute

₍

₎

m = _{Molar mass of solute Mass of solvent in grams}Mass of solute 1000_× ×

₍

₎

1llustrative Examples

Example 1: Find the molality of H₂SO₄ solution in which 98 g of H₂SO₄ is dissolved in 500 g of solvent. Solution: Molality 98 1000 2 98 500 × = = × m

Example 2: Find 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 1000 m 1000 × = _{= 1 molal}

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

Solution: 100 ml solution contains 95 g H₂SO₄. Moles of H₂SO₄ = 95₉₈

Mass of solution = 100 × 1.98 = 198 g Mass of water = 198 – 95 = 103 g Molality = 95 1000_{98 103}×_× = 9.412 m

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

Let’s consider a binary solution in which n moles of solute and N moles of solvent are present, then,

Mole fraction of solute (X_A) =_{moles of solute moles of solvent n N}moles of solute₊ = ₊n

Mole fraction of solvent (X_B) 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. X_A + X_B = 1

1llustrative Example

Example 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 2 4

2 4 massof H SO molar massof H SO = 98 1 98 = =

number of moles of H₂O in solution massof water 900 50 molar massof water 18

= = =

mole fraction of H₂O 50 50 50 1 51

= = +

mole fraction of H₂SO₄=_{50 1 51}1₊ = 1 (also mole fraction of H₂SO₄ = 1–50 1 51 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 fraction of solute (M_A) = _{mass of solute mass of solvent}mass of solute₊ = _{w W+}w

Mass fraction of solvent (M_B) = _{mass of solute mass of solvent}mass of solvent₊ = _{w W}W₊ The sum of mass fractions of all the components is unity ie. M_A + M_B = 1

1llustrative Example

Example 1: 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. Solution: Mass Fraction of H₂SO₄ =

2 4 2 4

mass of H SO 0.1 98 9.8 _0.001

mass of H SO 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 10_{number of grams of solution}× 6

1llustrative Example

Example 1: A 104 _{kg sample of hard water contains 96 grams of SO}

42– and 183 grams of HCO3– .

Calculate concentration of SO₄2– _{and HCO}

3– in ppm.

Solution: Concentration in ppm =

6

number of gramsof solute 10 number of gramof solution

× Concentration of SO₄2– ₌ 96 106 _{9 6 ppm} 10000 1000 × _{= ⋅} × Concentration of HCO₃– 6 7 183 10 _{18 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 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.

1llustrative Examples

Example 1: 1 g magnesium when heated in excess of oxygen for long time produces 5 g

3 of magnesium oxide. What is the equivalent mass of Mg?

Solution: Weight of oxygen that reacted with Mg 5 ₁ 2 _g

3 3

= − =

Now, 2 g₃ oxygen reacts with 1 g Mg

∴ 8 g oxygen would react with _{2 / 3}1 × =8 12 g

Example 2: 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.?

Solution: 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 = 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+

1llustrative Example

Example 1: Calculate the equivalent mass of HCl, H₂SO₄ and H₃PO₄. Solution: HCl→H+ +Cl− Equivalent mass of HCl = 1 35.5 36.5+₁ = + − → + 2 2 4 4 H SO 2H SO Equivalent mass of H₂SO₄ = 2 1 32 16 4 49 2 × + + _{× =}

+ −

→ + 3

3 4 4

H PO 3H PO

Equivalent mass of H₃PO₄ = 3 1 30 4 16 97 32.33× + ₃+ × = ₃ =

2. Equivalent mass of base

Equivalent mass of base = _{Number of replacable OH Acidity}Molar mass of base−

₍

₎

1llustrative Example

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 40+₄₀ + = =₁

( )

_→ 2+₊ −

2

Ca OH Ca 2OH

Equivalent mass of Ca(OH)₂ = 40 2 16 2 1 74 37+ ×₂ + × = =₂

( )

_→ 3+₊ −

3

Al OH Al 3OH

Equivalent mass of Al(OH)₃ = 27 3 16 3 1 78 26+ ×₃ + × = =₃ 3. Equivalent mass of salt

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

IIIII

llustrative Examples

Example 1: Calculate the equivalent mass of NaCl, MgCl₂ and Na₂SO₄. Solution: MgCl₂ →Mg++ +2Cl− Equivalent mass of MgCl₂ = 24 2 35.5 2 + × _{95 47.5} 2 = = + − → + 2 2 4 4 N a SO 2N a SO Equivalent mass of Na₂SO₄ = 2 23 32 4 16 142 71× + ₂ + × = ₂ =

NaCl →Na+ _{+ Cl}–

Equivalent mass of NaCl = 23 17₁+ =40

Example 2: Calculate the equivalent mass of FeSO₄.(NH₄ )₂ SO₄.6H₂O. Solution: Equivalent mass of FeSO₄.( NH₄)₂SO₄.6H ₂O (Mohr’s salt)

392 98gram 4

= =

Normality

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

(

)

Number of gram equivalent of solute N

Volume of solution in litre

=

(

)

Mass of solute N

Equivalent mass of solute Volume in litre

= ×

(

)

W 1000 N 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

1llustrative Examples

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 BaSO₄ is formed?

Solution: BaCl₂ + Al SO₂

(

₄

)

₃ → AlCl₃ + BaSO₄ Equivalents before reaction 20 0.1

1000×

30 _0.2 1000× = 2 × 10–3 _{= 6 × 10}–3 _{0 0}

Equivalents after reaction 0 4 × 10–3 _{2 × 10}–3 _{2 × 10}–3

Since BaCl₂ is limiting reagent

Hence, mass of BaSO₄ = 2 × 10–3 _{× Equivalent mass of BaSO} 4

Since BaSO₄ is a divalent n factor of BaSO₄ = 2

∴ Mass of BaSO₄ = 2 × 10–3 _× 233

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 N₂₀₀× =

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

Solution: N1000Mass of solute

₍

₎

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.

1llustrative Example

Example 1: Calculate normality of 0.3 M AlCl₃ solution. Solution: For AlCl₃ →Al+++ +3Cl−

n factor = 3

Hence, normality = 0.3 × 3 = 0.9 N

Also Normality = _{Equivalent mass}d 10 x× ×

And Molarity = _{Molecular mass}d 10 x× × where,

d = Density of solution

1llust rative Examples

Example 1: 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.

Solution: We know, N = 2 2 1 1 1 2 N V N V V V + +

For the mixture, N0.2500 0.1 2 200 0.2 1 700

× × + × ×

= =

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 H O+ ₂

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 egv)× ×} −3 ₃ 50 10− = × Normality of HCl (excess) 3 3 500 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 × 250 N = 500 + 250 N = 2.44 N

Strength of HCl = .067 × 36.5 grams/litre = 2.44 grams/litre Relationship between normality and strength (gram/litre of solution) In terms of normality,

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 BaCO3 will 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 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_{. (2003)} 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 gml–1_{. Calculate (i) the}

percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of Na+ _and 2

2 3

S O − ions. (1983)

11. 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 ×10–2 _{kg of Glauber’s salt is dissolved in water to obtain 1 dm}3 _{of a solution of density}

1077.2 kg m–3_{. Calculate the molarity, molality and mole-fraction of Na}

2SO4 in the solution.

(1994) 13. How many ml of 0.5 MH₂SO₄ are needed to dissolve 0.5 g of copper (II) carbonate? (1999)

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

(a) Mn O_{2 3} (b) MnO₂ (c) MnO₄− (d) 2

4

MnO −

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 N

10 NaOH 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 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.

( )

₂ 3 4 4 2

Ca 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 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

(a) Fe2Cl3 (b) FeCl2

(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

(a) 1₆ (b) 5₈

(c) ₃1 (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 of AgBr. Find the valency of element A.

(a) 2 (b) 4

Answer Keys

Answer Key

Basic Concepts of Chemistry

1. 70.96 × 106 g/mol 2. C21H30O2 3. d 4. 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