Mole Concept-1 (a)

Mole concept-1

Introduction :

There are a large number of objects around us which we can see and feel.

Anything that occupies space and has mass is called matter.

Ancient Indian and Greek Philospher’s beleived that the wide variety of object around us are made from

combination of five basic elements : Earth, Fire, Water, Air and Sky.

Ancient Greek Philosphers also believed that all matter was composed of tiny building blocks which were hard and indivisible.The Greek philosphere Democritus named these building blocks as atoms, meaning indivisible.All these people have their philosphical view about matter, they were never put to experimental tests, nor ever explain any scientific truth.It was John Dalton who firstly developed a theory on the structure of matter, later on which is known as Dalton’s atomic theory.

I DALTON’S ATOMIC THEORY :

1. Matter is made up of very small indivisible particle called atoms.

2. All the atoms of a given element is idenctical in all respect i.e. mass, shape, size, etc.

3. Atoms cannot be created or destroyed by any chemical process or physical process.

4. Atoms of different elements are different in nature.

Classification of matter

on the basis of physical behaviour on basis the of chemical behaviour

Solids Liquids Gases Pure substances Mixtures

Element Compound

III Some Definitions

I .

RELATIVE ATOMIC MASS :

It is the ratio of the mass of 1 atom of a substance and 1/12 of mass of 1 atom of C12 _{isotope. For atoms this}

is done by expressing mass of one atom with respect to a fixed standard. Dalton used hydrogen as the standard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the reference.



C-12 ISOTOPE OF CARBON IS LATEST CHOSEN STANDARD SINCE 1961

Therefore relative atomic mass is given as

Relative atomic mass (R.A.M) =

atom C one of mass 12 1 element the of atom one of mass 12  = nucleon 1 of mass 12 12 1 nucleon 1 of mass nucleons of number total   

On Hydrogen scale :

Relative atomic mass (R.A.M) =

atom H one of mass element the of atom one of mass 2 Oxygen scale :

Relative atomic mass (R.A.M) =

atom 16 O one of mass 16 1 element the of atom one of mass  

I I .

ATOMIC MASS UNIT (OR AMU)

The atomic mass unit (amu) is equal to one twelfth       12 1

of the mass of one atom of carbon-12 isotope.

 1 amu =

12 1

× mass of one C-12 atom

~ mass of one neucleon in C-12 atom. = 1.66 × 10–24 _{gm or 1.66}

× 10–27 _kg



one amu is also called one Dalton (Da).TODAY , AMU HAS BEEN REPLACED BY ‘u’ WHICH IS KNOWN AS UNIFIED MASS

I I I .

ATOMIC MASS

It is the mass of 1 atom of a substance it is expressed in AMU. Atomic mass = R.A.M × 1 amu

Note : Relative atomic mass is nothing but the number of nucleons present in the atom. Example :

Find the relative atomic mass of ‘O’ atom and its atomic mass. Sol. The number of neucleons present in ‘O’ atom is 16.

 relative atomic mass of ‘O’ atom = 16.

Atomic mass = R.A.M × 1 amu = 16 × 1 amu = 16 amu

Q. Find the relative atomic mass, atomic mass of the following elements. (i) Na (ii) F (iii) H (iv) Ca (v) Ag

Ans. (i) 23, 23 amu (ii) 19, 19 amu (iii) 1, 1.008 amu

Q. How many neucleons are present in 5 atoms of an element which has atomic mass 14 amu

Ans. = 70

Note : _{NCERT Reading NCERT Exercise DPP No.}

Text. 1.6,1.7,1.7.1,1.7.2

Page 13 – –

Sheet

Exercise–1 Exercise–2 Exercise–3

Part–I– Part–I– Part–I–

I V.

MOLE :THE MASS–NUMBER RELATIONSHIP

Mole is a chemical counting S unit and defined as follows :

A mole is the amount of a substance that contains as many entities (atoms, molecules or other particles) as there are atoms in exactly 0.012 kg (or 12 gm) of the carbon-12 isotope.

From mass spectrometer we found that there are 6.023 × 1023 atoms are present in 12 gm of C-12 isotope.

The number of entities in 1 mol is so important that it is given a separate name and symbol known as Avogadro constant denoted by N_A.

i.e. on the whole we can say that 1 mole is the collection of 6.02 × 1023 entities. Here entities may represent

atoms, ions, molecules or even pens, chair, paper etc also include in this but as this number (N_A) is very large therefore it is used only for very small things.

1 mole x mass of 1 atom of C12 _{isotope = 12g}  

 Alternatively value of N_A can be

1 mole x 12 x mass of one nucleon = 12 g found in this fashion

 1 mole = 24 10 x 66 . 1 1  = 6.023 x 1023



Note : In modern practice gram-atom and gram-molecule termed as mole.

V.

GRAM ATOMIC MASS :

The atomic mass of an element expressed in gram is called gram atomic mass of the element.

For example for oxygen atom :

Atomic mass of ‘O’ atom = mass of one ‘O’ atom = 16 amu

gram atomic mass = mass of 6.02 × 1023‘O’ atoms

= 16 amu × 6.02 × 1023

= 16 × 1.66 × 10–24 g × 6.02 ×1023 = 16 g

( 1.66 × 10–24× 6.02 × 1023 ~ 1 )

Q. How many atoms of oxygen are their in 16 g oxygen.

24 10 x 66 . 1 x   = 16 g x = 24 10 x 66 . 1 1  = N_A or It is also defined as mass of 6.02 × 1023 atoms.

or

It is also defined as the mass of one mole atoms.



Now see the table given below and understand the definition given before.

Element R.A.M.

(Relative Atomic Mass)

Atomic mass

(mass of one atom) Gram Atomic mass/weight

N 14 14 amu 14 gm

He 4 4 amu 4 gm

C 12 12 amu 12 gm

Example :

What is the weight of 3-g atoms of sulphur R.A.M. of s = 32.

Ans. 96 g Example :

How many g atoms are present in 144 g of sulphur

Example :

The ratio of mass of a silver atom to the mass of a carbon atom is 9 : 1. Find the mass of 1 mole of C atom if molar mass of Ag is 108.

Ans. 12 Example :

Calculate mass of sodium which contains same number of atoms as are present in 4g of calcium. Atomic masses of sodium and calcium are 23 and 40 respectively.

Ans. 2.3 g

VI.

MOLECULES :

It is the smallest particle of matter which has free existence. Molecules can be further divided into its constituents atoms by physical & chemical process.

Number of atoms presents in molecule is called its atomicity. Element :

H

₂, O₂, O₃ etc.

Compound : KCl, H₂SO₄, KClO₄ etc.

Molecule Atomicity

KCl - 2

H₂SO₄ - 7

O₃ - 3

H₂ - 2

VII.

MOLECULAR MASS :

It is the mass of one molecule

Ex. Molecule Molecular mass

H₂ 2 amu

KCl (39 + 35.5) = 74.50 amo

H₂SO₄ (2 + 32 + 64) = 98 amu.

VI II . GRAM MOLECULAR MASS :

The molecular mass of a substance expressed in gram is called the gram-molecular mass of the substance.

or

It is also defined as mass of 6.02 × 1023 molecules or

It is also defined as the mass of one mole molecules. (molar mass)

For example for ‘O₂’ molecule :

Molecular mass of ‘O₂’ molecule = mass of one ‘O₂’ molecule

= 2 × mass of one ‘O’ atom

= 2 × 16 amu

= 32 amu

gram molecular mass = mass of 6.02 × 1023‘O₂’ molecules = 32 amu × 6.02 × 1023

= 32 × 1.66 × 10–24 _gm

× 6.02 × 1023 = 32 gm

Sol. Mass = mol × At. wt. = 3 × 65 gm = 195 gm Example :

How many atoms of copper are present in 0.5 mol of pure copper metal?

Sol. No. of atoms = no. of moles × NA = 0.5 × 6.02 × 10

23 _{= 3.01} × 1023

Example :

The molecular mass of H₂SO₄ is 98 amu. Calculate the number of moles of each element in 294 g of H₂SO₄.

Solution

Gram molecular mass of H₂SO₄ = 98 gm moles of H₂SO₄ =

98 294

= 3 moles

H₂SO₄ H S O

One molecule 2 atom one atom 4 atom

1 × NA 2 × NA atoms 1 × NA atoms 4 × NA atoms

 one mole 2 mole one mole 4 mole

 3 mole 6 mole 3 mole 12 mole

Example :

A sample of (C₂H₆) ethane has the same mass as 107 _{molecules of methane. How many C}

2H6 molecules

does the sample contain ?

Ans. n = 5.34 × 106

Example :

How many molecules of water are present in 252 mg of (H₂C₂O₄.2H₂O)

Ans. 2.4 × 1021

Example :

From 48 g of the He sample ,6.023 x 1023 _{atoms of He are removed. Find out the moles of He left.Also}

Calculate the mass of carbon which contains same number of atoms as left over in this sample.

Ans. 11 mole, 132 g of C. Note :

NCERT Reading

NCERT Exercise

DPP No.

Text. Formula Mass

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Q.No.-1.1,1.10,1.28,1.30,1.33

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Exercise

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LECTURE # 3

GAY-LUSSAC’S LAW OF COMBINING VOLUME :

Gases combine in a simple ratio of their volumes provided all measurements should be done at the same temperature and pressure

H₂ (g) + Cl₂ (g)  2HCl

1 vol 1 vol 2 vol

AVOGADRO’S HYPOTHESIS :

Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressure condition.

mathematically, for ideal gases, V  n (CONSTANT T & P)

S.T.P. (Standard Temperature and Pressure): At S.T.P. / N.T.P. condition :

temperature = 0°C or 273 K

pressure = 1 atm = 760 mm of Hg

volume of one mole of an ideal gas = 22.4 litres (experimentally determined)

NOTE FOR FACULTY : The gas equation PV = nRT should never be used in this chapter. Ex. Calculate the volume in litres of 20 g hydrogen gas at STP.

Sol. No. of moles of hydrogen gas = mass  atomic weight = _2gm

gm 20

= 10 mol volume of hydrogen gas at STP = 10 × 22.4 lt.

Mole × 22 .4 lt  22 .4 lt Volume at STP  _N A × _N A Number × mol. wt. × At. wt.   At. wt. mol. wt. Mass

Ex. Calculate the volume in litres of 142 g chlorine gas at STP.

Ans. 44.8 lt.

Ex. Find the volume at STP occupied by 16 g of ozone at STP.

Ans. 3 4 . 22 = 7.5 

Ex. From 160 g of SO₂ (g) sample, 1.2046 x 1024 _{molecules of SO}

2 are removed then find out the volume of left

over SO₂ (g) at STP.

Ans. 11.2 Ltr.

Ex. 14 g of Nitrogen gas and 22 g of CO₂ gas are mixed together. Find the volume of gaseous mixture at STP.

Ans. 22.4 Ltr.

Ex. 672 ml of ozonized oxygen (mix of O₂ and O₃) at N.T.P. were found to weight one gram. Calculate the volume of ozone in the ozonized oxygen.

Ans. 56 ml

Note :

NCERT Reading NCERT Exercise DPP No.

– – 2

Sheet

Exercise–1 Exercise–2 Exercise–3

Part–I– 1 to 8 Part–I– Part–I–

II THE LAWS OF CHEMICAL COMBINATION

Atoine Lavoisier, John Dalton and other scientists formulate certain law concerning the composition of

matter and chemical reactions.These laws are known as the laws of chemical combination.

1 .

THE LAW OF CONSERVATION OF MASS :( ANTOINE LAVOISIER)

It states that matter can neither be created nor destroyed in chemical reaction i.e . In a chemical change , total mass remains conserved.i.e.

mass of all reactants = mass of products after reaction. ( In a closed system )

Example : H₂ (g) + 2 1

O₂ (g)  H₂O (l) Before reaction 1 mole

2 1

mole

After the reaction 0 0 1 mole

mass before reaction = mass of 1 mole H₂ (g) + 2 1

mole O₂ (g) = 2 + 16 = 18 gm

mass after reaction = mass of 1 mole water = 18 gm

2 .

LAW OF CONSTANT OR DEFINITE PROPORTION :{JOSEPH PROUST}

A given compound always contains exactly the same proportion of elements by weight irrespective of their source or method of preparation .

Ex. In water (H₂O), Hydrogen and Oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is tap water, river water or sea water or produced by any chemical reaction.

Ex. 1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion.

Sol. In the first sample of the oxide, Wt. of metal = 1.80 g, Wt. of oxygen = (3.0 – 1.80) g = 1.2 g  _1.2g 1.5 g 80 . 1 oxygen of . wt metal of . wt  

In the second sample of the oxide, Wt. of metal = 1.50 g, Wt. of oxygen = (2.50 – 1.50) g = 1 g.  _1g 1.5 g 50 . 1 oxygen of . wt metal of . wt  

Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed. Hence, the results follow the law of constant proportion.

3 .

THE LAW OF MULTIPLE PROPORTION : {JOHN DALTON}

When one element combines with the other element to form two or more different compounds, the mass of one elements, which combines with a constant mass of the other, bear a simple ratio to one another.

Note : Simple ratio here means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3, later on this

simple ratio becomes the valency and then oxidation state of the element.

Ex. Carbon and oxygen when combine, can form two oxides viz CO (carbonmonoxide), CO₂ (Carbondioxides) In CO, 12 gm carbon combined with 16 gm of oxygen.

In CO₂, 12 gm carbon combined with 32 gm of oxygen.

Thus, we can see the mass of oxygen which combine with a constant mass of carbon (12 gm) bear simple ratio of 16 : 32 or 1 : 2

CONCEPTS RELATED TO DENSIT Y :

It is of two type.

1. Absolute density

2. Relative density

For liquid and solids

Absolute density =

volume mass

specific gravity = density_densityof_ofthe_watersubs_attan₄ce_C

 For gases :

Absolute density (mass/volume) = _MolarMolar_volumemassof_ofthe_thegas_gas

* For simplification, we can conclude that the density and specific gravity of any substance is numerically same, but density has a definite unit, but specific gravity has no unit. (dimension less)

Ex. Specific gravity of a solution is 1.8 then find the mass of 100 ml of solution.

Ans. 180 gm.

RELATIVE DENSIT Y :

It is the density of a substance with respect to any other substance.

Ex. What is the V.D. of SO₂ with respect to CH₄ V.D. = 4 2 CH . W . M SO . W . M V.D = 16 64 = 4

Ex. Find the density of CO₂(g) with respect to N₂O(g).

VAPOUR DENSIT Y :

Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature and pressure. Vapour density = 2 H gas d d

= Molar_Molarmass_massof_ofgas_{H /}/_molarmolar_volumevolume

2 V.D. = 2 H gas M M = 2 Mgas ; M_gas = 2 V.D.

Ex. What is V.D. of oxygen gas

V.D = 2 32

= 16 unitss

Ex. 7.5 litre of the particular gas as S.T.P. as weight 16 gram. What is the V.D. of gas 7.5 litre = 16 gram moles = M 16 4 . 22 5 . 7  M = 48 gram V.D. 2 48 = 24.

Note : V.D. can be expressed with respect to some other gas other than hydrogen.

Note : _{NCERT Reading NCERT Exercise DPP No.}

Text. 1.5.1 to 1.5.5

Page 11 to 30. Q.No.1.21 3

Sheet

Exercise–1 Exercise–2 Exercise–3

Part–I–12,13,14. Part–I–2. Part–I–

% PERCENTAGE COMPOSITION :

Here we are going to find out the percentage of each element in the compound by knowing the molecular formula of compound.

We know that according to law of definite proportions any sample of a pure compound always possess constant ratio with their combining elements.

Ex. Every molecule of ammonia always has formula NH₃ irrespective of method of preparation or sources. i.e. 1 mole of ammonia always contains 1 mol of N and 3 mole of H. In other wards 17 gm of NH₃ always contains 14 gm of N and 3 gm of H. Now find out % of each element in the compound.

Mass % of N in NH₃ = 100 NH of mol 1 of Mass NH mol 1 in N of Mass 3 3  ₌ 17 14 × 100 = 82.35 %

Mass % of H in NH₃ = Mass_Massof_of1H_molin1_emol_ofNHNH 100

3 3  ₌ 17 3 × 100 = 17.65 %

Ex. What is the percentage of calcium and oxygen in calcium carbonate (CaCO₃) ?

Ans. 40%, 48%.

Ex. A compound of sodium contains 11.5% sodium then find the minimum molar mass of the compound.

Ans. 200 gm/mole.

EMPIRICAL AND MOLECULAR FORMULA :

We have just seen that knowing the molecular formula of the compound we can calculate percentage com-position of the elements. Conversely if we know the percentage comcom-position of the elements initially, we can calculate the relative number of atoms of each element in the molecules of the compound. This gives us the empirical formula of the compound. Further if the molecular mass is known then the molecular formula can easily be determined.

Thus, the empirical formula of a compound is a chemical formula showing the relative number of atoms in the simplest ratio, the molecular formula gives the actual number of atoms of each element in a molecule. The molecular formula is generally an integral multiple of the empirical formula.

i.e. molecular formula = empirical formula × n

where n = molecular_{empiricalformula}formula_massmass

Ex. Acetylene and benzene both have the empirical formula CH. The molecular masses of acetylene and ben-zene are 26 and 78 respectively. Deduce their molecular formulae.

Sol.  Empirical Formula is CH

Step-1

The empirical formula of the compound is CH

 Empirical formula mass

= (1 × 12) + 1 = 13.

Molecular mass = 26

Step-2

To calculate the value of ‘n’

n = _EmpiricalMolecular_formulamass_mass = 13 26

= 2

Step-3

To calculate the molecular formula of the compound.

Molecular formula = n × (Empirical formula of the compound)

= 2 × CH = C2 H2

Thus the molecular formula is C₂ H₂

Similarly for benzene To calculate the value of ‘n’

n = _EmpiricalMolecular_formulamass_mass = 13 78

= 6 thus the molecular formula is 6 × CH = C6H6

Ex. An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition. C = 40.684% ; H = 5.085% and O = 54.228%

The molecular weight of the compound is 118. Calculate the molecular formula of the compound.

Sol. Step-1

To calculate the empirical formula of the compound.

Element _Symbol Carbon C Hydrogen H Oxygen O Percentage of element 40.687 5.085 54.228 At. mass of element 12 1 16 Relative no. Percentage At. mass of atoms = 40.687 12 = 3.390 5.085 1 = 5.085 54.228 16 = 3.389 Simplest atomic ratio 3.390 3.389 =1 5.085 3.389 =1.5 3.389 3.389 =1 Simplest whole no. atomic ratio

2 3

2

 Empirical Formula is C₂ H₃ O₂ Step-2

To calculate the empirical formula mass. The empirical formula of the compound is C₂ H₃ O₂ .

 Empirical formula mass

= (2 × 12) + (3 × 1) + (2 × 16) = 59. Step-3

To calculate the value of ‘n’

n = _EmpiricalMolecular_formulamass_mass = 59 118

= 2

Step-4

To calculate the molecular formula of the salt.

Molecular formula = n × (Empirical formula) = 2 × C₂ H₃ O₂ = C₄ H₆ O₄

Thus the molecular formula is C₄ H₆ O_4.

Ex. An oxide of nitrogen gave the following precentage composition : N = 25.94

and O = 74.06

Calculate the empirical formula of the compound.

Ans. N₂O₅

Ex. Hydroquinone, used as a photographic developer, is 65.4%C, 5.5% H, and 29.1%O, by mass. What is the empirical formula of hydroquinone ?

Ans. C₃H₃O Note :

NCERT Reading

NCERT Exercise

DPP No.

Text. 1.9,1.9.1

Page 15

Q.No.1.2,1.3,1.8.

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CHEMICAL REACTION :

It is the process in which two or more than two substances interact with each other where old bonds are broken and new bonds are formed.

VI CHEMICAL EQUATION :

All chemical reaction are represented by chemical equations by using chemical formule of reactants and products. Qualitatively a chemical equation simply describes what the reactants and products are. However, a balanced chemical equation gives us a lot of quantitative information mainly the molar ratio in which reactants combine and the molar ratio in which products are formed.

Example :

When potassium chlorate (KClO₃) is heated it gives potassium chloride (KCl) and oxygen (O₂). KClO₃  KCl + O2 (unbalanced chemical equation )

2KClO₃  2 KCl + 3 O2 (balanced chemical equation) Attributes of a balanced chemical equation: (From NCERT PAGE - 17)

(a) It contains an equal number of atoms of each element on both sides of equation.(POAC) (b) It should follow law of charge conservation on either side.

(c) Physical states of all the reagents should be included in brackets.

(d) All reagents should be written in their standard molecular forms (not as atoms ) (e) The coefficients give the relative molar ratios of each reagent.

Balancing a chemical equation

According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides

4 Fe(s) + 3O₂(g)  2Fe₂O₃(S) (a) balanced equation

2 Mg(s) + O₂(g)  2MgO(S) (b) balanced equation

P₄(s) + O₂(g)  P₄O₁₀(S) (c) unbalanced equation

Equations (a) and (b) are balanced since there are same number of metal and oxygen atoms on each side of equations. However equation (c) is not balanced. In this equation. phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation.

P₄(S) + 5O₂(g)  P₄O₁₀(S) balanced equation

Now let us take combustion of propane, C₃H₈, This equation can be balanced in steps.

Step 1. Write down the correct formulas of reactants and products. Here propane and oxygen are reactants, and

carbon dioxide and water are products.

C₃H₈(g) + O₂(g)  CO₂ (g) + H₂O (l) unbalanced equation

Step 2. Balance the number of C atoms : Since 3 carbon atoms are in the reactant, therefore, three CO₂ mol-ecules are required on the right side.

C₃H₈(g) + O₂ (g)  3CO₂ (g) + H₂O (l)

Step 3. Balance the number of H atoms : on the left there are 8 hydrogen atoms in the reactants however, each

molecule of water has two hydrogen atoms , so four molecules of water will be required for eight hydrogen atoms on the right side.

C₃H₈ (g) + O₂ (g)  3CO₂ (g) + 4H₂O (l)

Step 4. Balance the number of O atoms : There are ten oxygen on the right side (3 × 2 = 6 in CO2 and 4 × 1 = 4

in water). Therefore, five O₂ molecules are needed to supply to supply the required ten oxygen atoms. C₃H₈ (g) + 5O₂ (g)  3CO₂ (g) + 4H₂O (l)

Step 5. Verify that the number of atoms of each element is balanced in the final equation.

Always remember that subscripts in formula of reactants and products cannot be changed to balance an equation.

INTERPRETATION OF BALANCED CHEMICAL EQUATIONS :

(STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS,

NCERT, PAGE - 17

)

3 . Mole-mole analysis :

This analysis is very much important for quantitative analysis point of view. Students are advised to clearly understand this analysis.

Now consider again the decomposition of KClO₃ .

2KClO

₃  2KCl + 3O₂

In very first step of mole-mole analysis you should read the balanced chemical equation like

2 moles KClO₃ on decomposition gives you 2 moles KCl and 3 moles O_2. and from the stoichiometry of reaction we can write

2 KClO of Moles ₃ = 2 KCl of Moles = 3 O of Moles 2

Now for any general balance chemical equation like a A + b B  c C + d D

you can write. a reacted A of Mole = b reacted B of moles = c reacted C of moles = d reacted D of moles

Ex. 3 moles (367.5 gm) of KClO₃ when heated how many moles KCl and O₂ is produced.

Sol. The reaction is

2KClO₃  2KCl + 3O₂ 2 KClO of Moles ₃ = 2 KCl of Moles  Moles of KCl produced = 3 Now, 2 KClO of Moles ₃ = 3 O of Moles 2 mole of O₂ produced = 2 3 3 = 4.5 moles

Ex. CH₄ + 2O₂ CO2 + 2H2O (from NCERT Page - 18)

following conclusions can be drawn from above reaction by observing its stoichiometry

 One mole of CH₄ (g) reacts with two moles of O₂ (g) to give one mole of CO₂ (g) and two moles of H₂O (g) One molecule of CH₄ (g) reacts with 2 molecues of O₂ (g) to give one molecule of CO₂ (g) and 2 molecules

of H₂O (g)

 22.4 L of CH₄ (g) reacts with 44.8 L of O₂ (g) to give 22.4L of CO₂ (g) and 44.8 L of H₂O (g)  16 g CH₄ (g) reacts with 2×32 g of O₂ (g) to give 44 g of CO₂ (g) and 2 × 18 g of H₂O (g).

Note : In fact mass-mass and mass-vol analysis are also interpreted in terms of mole-mole analysis you can use

following chart also.

Mass At. wt. / Mol. Wt. Mole

Mole-mole relationship of equation Mole × 22.4 lt Volume at STP × m ol. w t./At. w t. Mass

Ex. 367.5 gm KClO₃ (M = 122.5) when heated

(a) How many grams O₂ is produced (b) How many litre of O₂ is produced at STP

Sol. Now consider the balanced chemical equation

2KClO₃  2KCl + 3O₂ Now go with the above chart

(Mole-mole relationship of equation) 9/2 mole of O2 (a) 144 gm × _{32 gm} (mol. w_t.) (b) 100.8 lt × 22.4 lt (volume a t STP)

Ex. Iron in the form of fine wire burns in oxygen to form iron (III) oxide 4Fe(s) + 3O₂(g)  2Fe₂O₃(s)

How many moles of O₂ are needed to produce 5 mol Fe₂O₃ ?

Ans. 7.5 mol O₂

Ex. Nitric acid, HNO₃, is manufactured by the Ostwald process, in which nitrogen dioxide, NO₂, reacts with water.

3NO₂(g) + H₂O(

l

)  2HNO₃(aq) + NO(g)

How many grams of nitrogen dioxide are required in this reaction to produced 6.3 g HNO₃ ?

Ans. 6.9g NO₂

Ex. How many grams of Fe₂ O₃ is formed by heating 18 gm FeO with Oxygen. 4FeO + O₂  2Fe₂ O₃

Ans. 20. gm

Ex. How many litre O₂ at N.T.P. is required for complete combustion of 1 mole C₅ H₁₀.

Ans. 168 lt.

Ex. Calculate the weight of residue obtained when CaCO₃ is strongly heated and 5.6 litre CO₂ is produced at N.T.P.

Ans. 14 gm

Ex. When sodium bicarbonate is heated 1.806 x 1024 _{molecules of water is obtained. Then find the volume of}

CO₂(g) obtained at STP and amount of NaHCO₃ needed for this reaction.

Sol. 2NaHCO₃  Na₂CO₃ + H₂O + CO₂ So volume of CO₂ = 3 × 22.4 = 67.2 Lt.

Mass of NaHCO₃ needed = 6 × 84 = 504 gm. Problem 1.3 (NCERT, Page 18)

Calculate the amount of water (g) produced by the combustion of 16 g of methane.

Solution : The balanced equation for combustion of methane is : CH₄ (g) + 20₂ (g)  CO2 (g) + 2H2O (g)

(i) 16 g of CH₄ corresponds to one mole. (ii) From the above equation, 1 mol of

CH₄ (g) gives 2 mol of H₂O (g)

Problem 1.4 (NCERT, Page 18)

How many moles of methane are required to produce 22 g CO₂ (g) after combustion?

Solution: According to the chemical equation, CH₄ (g) + 20₂ (g)  CO₂ (g) + 2H₂O (g)

44g CO₂ (g) is obtained from 16 g CH₄ (g).

[  ] 1mol CO2 (g) is obtained from 1 mol of CH4 (g)

mole of CO₂ (g) = 22g CO₂ (g) × _{44gCO (g)} ) g ( molCO 1 2 2 = 0.5 mol CO₂ (g)

Hence, 0.5 mol CO₂ (g) would be obtained from 0.5 mol CH₄ (g) or 0.5 mol of CH₄ (g) would be required to produce 22 g CO₂ (g).

Note :

NCERT Reading

NCERT Exercise

DPP No.

Text. 1.10

Page 17

Q.No. 1.3,1.4.

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PRINCIPLE OF ATOM CONSERVATION (POAC) :

POAC is based on law of mass conservation if atoms are conserved, moles of atoms shall also be conserved hence mass of atoms is also conserved.

This principle is fruitful for the students when they don’t get the idea of balanced chemical equation in the

problem. This principle can be under stand by the following example.

Consider the decomposition of KClO₃ (s)  KCl (s) + O2 (g) (unbalanced chemical reaction)

Apply the principle of atom conservation (POAC) for K atoms. Moles of K atoms in reactant = moles of K atoms in products or moles of K atoms in KClO₃ = moles of K atoms in KCl. Now, since 1 molecule of KClO₃ contains 1 atom of K

or 1 mole of KClO₃ contains 1 mole of K, similarly,1 mole of KCl contains 1 mole of K. Thus, moles of K atoms in KClO₃ = 1 × moles of KClO₃

and moles of K atoms in KCl = 1 × moles of KCl.

 moles of KClO3 = moles of KCl

or 3 3 KClO of wt. mol. g in KClO of wt. = _mol.wt.of_wt.KCl_ofin_KClg



The above equation gives the mass-mass relationship between KClO₃ and KCl which is important in stoichio-metric calculations.

Again, applying the principle of atom conservation for O atoms, moles of O in KClO₃ = 3 × moles of KClO3

moles of O in O₂ = 2 × moles of O₂

 3 × moles of KClO3 = 2 × moles of O2

or 3 × 3 3 KClO of . wt . mol KClO of . wt

= 2 × _{standardmolarvol.(22.4lt.)}

NTP at O of . vol 2



The above equations thus gives the mass-volume relationship of reactants and products.

Q. Write POAC equation for all the atoms in the following reaction. (i) N₂O + P₄ P₄O₁₀ + N₂

(ii) P₄ + HNO₃ H₃PO₄ + NO₂ + H₂O Example :

27.6 g K₂CO₃ was treated by a series of reagents so as to convert all of its carbon to K₂ Zn₃ [Fe(CN)₆]₂. Calculate the weight of the product.

[mol. wt. of K₂CO₃ = 138 and mol. wt. of K₂Zn₃ [Fe(CN)₆]₂ = 698]

Sol. Here we have not knowledge about series of chemical reactions but we know about initial reactant and final product accordingly K₂CO₃ Steps Several       K2Zn3 [Fe(CN)6]2

Since C atoms are conserved, applying POAC for C atoms, moles of C in K₂CO₃ = moles of C in K₂Zn₃ [Fe(CN)₆]₂ 1 × moles of K₂CO₃ = 12 × moles of K₂Zn₃ [Fe(CN)₆]₂

( 1 mole of K2CO3 contains 1 moles of C)

3 2 3 2 CO K of . wt . mol CO K of . wt = 12 × _{mol.wt.of product} product the of . wt wt. of K₂Zn₃ [Fe(CN)₆]₂ = 138 6 . 27 × 12 698 = 11.6 g

Q.1 0.32 mole of LiAlH₄ in ether solution was placed in a flask and 74 g (1 moles) of t-butyl alcohol was added. The product is LiAlHC₁₂H₂₇O₃ . Find the weight of the product if lithium atoms are conserved.

[Li = 7, Al = 27, H = 1, C = 12, O = 16]

Ans. 81.28 g

Note :

NCERT Reading

NCERT Exercise

DPP No.

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Q.No. 1.7,1.34,1.36.

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LIMITING REAGENT :

The reactant which is consumed first and limits the amount of product formed into the reaction, and is therefore called limiting reagent.

Limiting reagent is present in least stoichiometric amount and therefore controls amount of product. The remaining or leftout reactant is called the excess reagent.

When you are dealing with balanced chemical equation then if number of moles of reactants are not in the ratio of stoichiometric coefficient of balanced chemical equation, then there should be one reactant which should be limiting reactant.

Example :

Three mole of Na₂ CO₃ is reacted with 6 moles of HCl solution. Find the volume of CO₂ gas produced at STP. The reaction is

Na₂ CO₃ + 2HCl  2 NaCl + CO₂ + H₂O

Sol. From the reaction : Na₂ CO₃ + 2HCl  2 NaCl + CO₂ + H₂O

given moles 3 mol 6 mol

given mole ratio 1 : 2

Stoichiometric coefficient ratio 1 : 2



See here given moles of reactant are in stoichiometric coefficient ratio therefore none reactant left over. Now use Mole-mole analysis to calculate volume of CO₂ prdouced at STP

1 CO Na of Moles _{2 3} = 1 oduced Pr CO of Mole ₂ Moles of CO₂ produced = 3 volume of CO₂ produced at STP = 3 × 22.4 L = 67.2 L Example :

6 moles of Na₂ CO₃ is reacted with 4 moles of HCl solution. Find the volume of CO₂ gas produced at STP. The reaction is

Na₂ CO₃ + 2HCl  2 NaCl + CO₂ + H₂O

Sol. From the reaction : Na₂ CO₃ + 2HCl  2 NaCl + CO₂ + H₂O

given mole of reactant 6 : 4

give molar ratio 3 : 2

Stoichiometric coefficient ratio 1 : 2



See here given number of moles of reactants are not in stoichiometric coefficient ratio. Therefore there should be one reactant which consumed first and becomes limiting reagent.

But the question is how to find which reactant is limiting, it is not very difficult you can easily find it according to the following method.

HOW TO FIND LIMITING REAGENT :

Step : 

Divide the given moles of reactant by the respective stoichiometric coefficient of that reactant.

Step : 

See for which reactant this division come out to be minimum. The reactant having minimum value is limiting reagent for you.

Step : 

Now once you find limiting reagent then your focus should be on limiting reagent

From Step  &  Na2 CO3 HCl

1 6 = 6 2 4 = 2 (division is minimum)  HCl is limiting reagent From Step  From 2 HCl of Mole = 1 produced CO of Moles ₂

 mole of CO2 produced = 2 moles

 volume of CO2 produced at S.T.P. = 2 × 22.4 = 44.8 lt. Examples on limiting reagent :

Ex. The reaction 2C + O₂  2CO is carried out by taking 24g of carbon and 96g O₂, find out :

(a) Which reactant is left in excess ?

(b) How much of it is left ?

(c) How many mole of CO are formed ?

Ex. For the reaction 2P + Q  R, 8 mol of P and 5 mol of Q will produce

(A) 8 mol of R (B) 5 mol of R (C*) 4 mol of R (D) 13 mol of R

Ex. X + Y  X3 Y4

Above reaction is carried out by taking 6 moles each of X and Y respectively then

(A) X is the limiting reagent (B) 1.5 moles of X₃ Y₄ is formed

(C) 1.5 moles of excess reagent is left behind (D) 75% of excess reagent reacted X + Y  X3 Y4 Ans. B, C, D Sol. 3X + 4Y  X₃ Y₄ 6 mole 6 mole 6 – 4.5 0 1.5 mole 1.5 mole left formed Ex. A + B A3B2 (unbalanced) A₃B₂ + C A3B2C2 (unbalanced)

Above two reactions are carried out by taking 3 moles each of A and B and one mole of C. Then (A) 1 mole of A₃B₂C₂ is formed (B*) 12 mole of AA₃B₂C₂ is formed

(C*) 1 mole of A₃B₂ is formed (D*) 12 mole of AA₃B₂ is left finally

Ans. B, C, D

Sol. 3A + 2B  AA₃ B₂

3 mole 3 mole 1 mole formed

A₃B₂ + 2C  AA₃ B₂ C₂ 1 mole 1 mole

0.5 mole 0 0.5 mole

Ex. CS₂ and Cl₂ in the weight ratio 1 : 2 are allowed to react according to equation, find the fraction of excess reagent left behind.

CS₂ + 3Cl₂ CCl₄ + S₂Cl₂ mole 76 w 71 w 2 ; 76 w 3 71 w 2  L.R. = Cl₂ . Moles of CS₂ = 3 71 w 2   remaining = 76 w – 213 w 2 fraction of Cl₂ left = x100 76 w 213 w 2 76 w  = 28.6%.

Ans. Vol of H₂ = 22.4 litre Al reacted = 66.66%

Ex. Equal weights of carbon and oxygen are heated in a closed vessel producing CO and CO₂ in a 1 : 1 mol ratio. Find which component is limiting and which component left and what is its percentage out of total weight taken.

Ans. Limiting reagent is O₂, carbon will left behind, 50%

Ex. Three moles of Phosphorus is reacted with 2 moles of iodine to form P₃ according to the reaction

P + ₂ P₃

P3 formed in the above reaction is further reacted with 27 g of water. According to the reaction.

P3 + H2O  H3PO3 + H

HI formed in the above reaction is collected in the gaseous form. At higher temperature HI dissociated 50% then find the molecules of H₂ gas liberated

H H2 + 2 Ans. 3/8 N_A.

Problem 1.5 (NCERT Page - 19)

50.0 kg of N₂ (g) and 10.0 kg of H₂ (g) are mixed to produce NH₃ (g) Calculate the NH₃ (g) formed. Identify the limiting reagent in the production of NH₃ in this situation. Take reaction N₂ + 3H₂ 2NH3.

Sol. A balanced equation for the above reaction is written as follows : Calculation of moles : N₂ (g) + 3H₂ (g)  2NH3 (g) moles of N₂ = 50.0 kg 2 2 2 N kg 1 N g 1000 N  _× 2 2 N g 0 . 28 H mol 1 = 17.86 × 102 mol moles of H₂ = 10.00 kg H₂× 2 2 H kg 1 H g 1000 × 2 2 H g 016 . 2 H mol 1 = 4.96 × 103 mol

According to the above equation, 1 mol N₂ (g) requires 3 mol H₂ (g), for the reaction, Hence, for 17.86 ×

102 _{mol of N}

2, the moles of H2 (g) required would be

17.86 × 102 mol N2× _{1molN (g)} ) g ( H mol 3 2 2 = 5.36 × 103 mol H2

But we have only 4.96×103 mol H2. Hence, dihydrogen is the limiting reagent in this case. So NH3 (g)

would be formed only from that amount of available digydrogen i.e., 4.96 × 103 mol

Since 3 mol H₂ (g) gives 2 mol NH₃ (g)

 4.96 × 103 mol H₂ (g) × _{3molH (g)} ) g ( H N mol 2 2 3 = 3.30 × 103 mol NH3 (g) is obtained.

If they are to be converted to grams, it is done as follows : 3.30 × 103 mol NH3 (g) × _{1molNH (g)} ) g ( gNH 0 . 17 3 3 . = 3.30 × 103× 17 g NH3 (g) = 56.1 × 103 g NH₃ = 56.1 kg NH₃ Note :

NCERT Reading

NCERT Exercise

DPP No.

Text. 1.10.1.

Page 18

Q.No. 1.4,1.23,1.24,1.26.

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LECTURE # 9

SOLUTIONS :

A mixture of two or more substances can be a solution. We can also say that a solution is a homogeneous mixture of two or more substances ‘Homogeneous’ means ‘uniform throughout’. Thus a homogeneous

mix-ture, i.e., a solution, will have uniform composition throughout.

CONCENTRATION TERMS :

The following concentration terms are used to express the concentration of a solution. These are :

1. strength of solution 2. Molarity (M) 3. Molality (m) 4. Mole fraction (x) 5. % calculation 6. Normality (N) 7. ppm



Remember that all of these concentration terms are related to one another. By knowing one concentration term you can also find the other concentration terms. Let us discuss all of them one by one.

1.

STRENGTH OF SOLUTION :

The concentration of solution in gram/litre is said to be strength of solution.

(a) A 65% solution has the following meanings

65% by weight i.e. 100 gm solution contain 65 gm solute 65% by volume i.e. 100 ml of solution contain 65 ml solute 65% by strength i.e. 100 ml of solution contain 65 gm solute If, anything is not specified, 65% generally mean 65% by mass

(b) For concentrated acids, like 98% H₂SO₄, 65% HNO₃ etc, if anything is not specified than percentage by mass/volume is usually considered.

(c) For the calculation of strength (% w/w, %w/v etc) the solute must be completely dissolved into the solution, otherwise, the given terminologies will be invalid. For example, the specific gravity of gold = 19.3 gm/cm3_{, if}

we add 193 gm gol powder in 1 litre of water, its % w/w =

193 1000

193

 x 100 = 16.17 is appears to be correct,

but gold is not dissolvable in water, its % w/w in water cannot be calculated.

2 .

MOLARIT Y (M) :

The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity of the solution.

i.e., Molarity of solution =

litre in solution of volume moles of number

Let a solution is prepared by dissolving w gm of solute of mol.wt. M in V ml water.

 Number of moles of solute dissolved =

M w  V ml water have M w mole of solute  1000 ml water have ml V M 1000 w    Molarity (M) = ml V ) solute of wt . Mol ( 1000 w  

mL of the solution. Solution Since molarity (M) = es litr in solution of Volume solute of moles of No. = L 0.250 NaOH of mass NaOH/Molar of Mass = L 250 . 0 g 40 / g 4 L 250 . 0 mol 1 . 0  = 0.4 mol L–1 _{= 0.4 M}

Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent.

Some other relations may also useful.

Number of millimoles = 1000 ) solute of . wt . Mol ( solute of mass

 = (Molarity of solution × Vinml)

Molarity is an unit that depends upon temperature .it varies inversely with temperature .

mathematically : molarity decreases as temperature increases.

Molarity  _temperature 1



volume 1



Molarity of solution may also given as :

ml in solution of volume Total solute of millimole of Number

(i) If a particulars solution having volume V₁ and molarity = M₁ is diluted to V₂ mL then M₁V₁ = M₂V₂

M₂ : Resultant molarity

(ii) If a solution having volume V₁ and molarity M₁ is mixed with another solution of same solute having volume V₂ mL & molarity M₂

then M₁V₁ + M₂V₂ = M_R (V₁ + V₂) M_R = Resultant molarity = 2 1 2 2 1 1 V V V M V M  

Ex. 149 gm of potassium chloride (KCl) is dissolved in 10 Lt of an aqueous solution. Determine the molarity of the solution (K = 39, Cl = 35.5)

Sol. Molecular mass of KCl = 39 + 35.5 = 74.5 gm

 Moles of KCl = _74.5gm

gm 149

= 2

 Molarity of the solution =

10 2

= 0.2 M

Q. 117 gm NaCl is dissolved in 500 ml aqueous solution. Find the molarity of the solution.

Ans. 0.4 M

Ex. Calculate molarity of the following :

(a) 0.74 g of Ca(OH)₂ in 5 mL of solution [ 2M ]

(b) 3.65 g of HCl in 200 ml of solution [0.5M ]

(c) 1/10 mole of H₂SO₄ in 500 mL of solution [0.2 M ]

Ex. Calculate the resultant molarity of following : (a) 200 ml 1M HCl + 300 ml water

(b) 1500 ml 1M HCl + 18.25 g HCl (c) 200 ml 1M HCl + 100 ml 0.5 M H₂SO₄ (d) 200 ml 1M HCl + 100 ml 0.5 M HCl

Ex. Calculate the molarity of H+ _{ion in the resulting solution when 200 ml 1M HCl is mixed with 200 ml 1M H} 2SO4 Sol. For HCl M₁ = 1 M V₁ = 200 mL For H₂SO₄ M₂ = 1 V₂ = 200 mL n_HCl = M_HCl× VHCl = 1 × 0.2 HCl H+ + Cl–  H n = 0.2 (from HCl) 4 2 4 2 4 2SO HSO HSO H M V n   = 1 × 0.2 = 0.2 mole H₂SO₄ 2H+ + SO₄–2  H n = 0.2 × 2 (from H2SO4) Total H+ ₌  H n (from HCl) + n_H(from H₂SO₄) = 0.2 + 0.4 = 0.6 Total volume = 200 + 200 = 400 mL = 0.4 L M_R = Resultant molarity = 4 . 0 6 . 0 V n solution H   = 1.5 Ans.

Ex. What are the final concentration of all the ion when following are mixed

50 ml of 0.12 M Fe(NO₃)₃ + 100 ml of 0.1 M FeCl₃ + 100 ml of 0.26 M Mg(NO₃)₂

[NO₃–_{] =} 250 2 26 . 0 100 3 12 . 0 50     = 250 70 250 52 18   = 0.28 [Cl–_{] = 0.12 M} [Mg++_{] = 0.104 M} [Fe3+_{] = 0.064 M} Ex. Calculate the molarity of water

H₂O 18 gm

= 1 mole Volume of water = 1 Litre

Mass = 1000 gm mole = 18 1000 Molarity of water = 18 1000 = 55.55 M

Ex. Find the minimum volume of 0.2 M HCl solution for the complete neutralisation of 0.4 M, 40 ml of NaOH solution.

Ex. CaCO₃ reacts with aq. HCl to give CaCl₂ and CO₂ according to reaction CaCO₃(s) + 2HCl(aq) _CaCl₂ + CO₂ + H₂O

2 mole of HCl reacts _1 moles CaCO₃ 1 mole of HCl reacts _ 2 1 50 mmole of HCl reacts _ 2 1 × 50 = 25 mmole of CaCO3 mole of CaCO₃ = 1000 25 mass of CaCO₃ = 1000 25 × 100 = 2.5 gm. Ex. Na₂CO₃ + 2HCl _2NaCl + CO₂ + H₂O

(a) moles of NaCl formed when 10.6 gm of Na₂CO₃ is mixed with 100 ml of 0.5 M HCl solution. (b) Calculate the concentration of each ion in the solution after the reaction

(c) Volume of CO₂ liberated at S.T.P.

Sol. molar mass of Na₂CO₃ = 106 gm Na₂CO₃ = 106 6 . 10 × 1000 = 100 m mole HCl = 100 × 0.5 = 50 milli mole limiting reagent HCl  2 HCl reacts = 2 NaCl

50 mmole HCl reacts = 50 mmole of Na₂CO₃ remaining Na₂CO₃ = 100 – 25 = 75 mmole

(a) moles of NaCl =

1000 50 = 0.05 (c) volume of CO₂ = 22.4 1000 25  = 0.556 L [Na+_{] =} 100 200 100 150 50   = 2 M [Cl–_{] =} 100 50 = 0.5 M (CO₃2–_{] =} 100 75 = 0.75 M Note :

NCERT Reading

NCERT Exercise

DPP No.

Text. 1.10.2.

Page 19

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LECTURE # 10

3 .

MOLALIT Y (M) :

The number of moles of solute dissolved in1000 gm (1 kg) of a solvent is known as the molality of the solution. i.e., molality = 1000 gram in solvent of mass solute of moles of number 

Let y gm of a solute is dissolved in x gm of a solvent. The molecular mass of the solute is m. Then Y/m mole of the solute are dissolved in x gm of the solvent. Hence

Molality = 1000 x m Y  

Ex. 225 gm of an aqueous solution contains 5 gm of urea. What is the concentration of the solution in terms of molality. (Mol. wt. of urea = 60)

Sol. Mass of urea = 5 gm

Molecular mass of urea = 60 Number of moles of urea =

60 5

= 0.083 Mass of solvent = (255 – 5) = 250 gm

 Molality of the solution = _{Mass of solvent in gram} 1000

solute of moles of Number  ₌ 1000 250 083 . 0  = 0.332

Note : molality is independent of temperature changes. Problem 1.8 (NCERT Page - 21)

The density of 3 M solution of NaCl is 1.25 g mL–1_{. Calculate molality of the solution.}

Sol. M = 3 mol L–1

Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g

Mass of 1L solution = 1000 × 1.25 = 1250 g (since density = 1.25 mL–1₎

Mass of water in solution = 1250 – 175.5 = 1074.5 g

Molaity = kg in solvent of Mass solute of moles of No. = kg 1.0745 mol 3 = 2.79 m

Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that molality of a solution does not change with temperature since mass remains unaffected with tempera-ture.

Q. 518 gm of an aqueous solution contains 18 gm of glucose (mol.wt. = 180). What is the molality of the solution.

Ans. 0.2 m

Q. Molality of an NaOH solution is 4. Find the wt. of solution if solvent is 500 g.

4 .

MOLE FRACTION (x) :

The ratio of number of moles of the solute or solvent present in the solution and the total number of moles present in the solution is known as the mole fraction of substances concerned.

Let number of moles of solute in solution = n Number of moles of solvent in solution = N

 Mole fraction of solution (x1) = _{n N}

n



 Mole fraction of solvent (x2) = _{n N}

N





also x₁ + x₂ = 1

(i) % weight by weight (w/w) : It is given as mass of solute present in per 100 gm of solution.

i.e. % w/w = _massmass _ofof _solutionsolute in _in gm_gm100

(ii) % weight by volume (w/v) : It is given as mass of solute present in per 100 ml of solution.

i.e., % w/v = 100 ml in solution of mass gm in solute of mass 

(iii) % volume by volume (V/V) : It is given as volume of solute present in per 100 ml solution.

i.e., % V/V = 100 solution of Volume ml in solute of Volume  Example

0.5 g of a substance is dissolved in 25 g of a solvent. Calculate the percentage amount of the substance in the solution.

Solution.

Mass of substance = 0.5 g Mass of solvent = 25 g

 percentage of the substance (w/w) = 100

25 5 . 0 5 . 0   = 1.96

Problem 1.6 (NCERT Page - 19)

A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.

Solution

Mass per cent of A =

solution of Mass A of Mass × 100 = _{2gofA 18gofwater} g 2  × 100 100 g 20 g 2   _{= 10 %} Example

20 cm3 _{of an alcohol is dissolved in80 cm}3 _{of water. Calculate the percentage of alcohol in solution.} Solution Volume of alcohol = 20 cm3 Volume of water = 80 cm3  percentage of alcohol = 100 80 20 20   = 20.

Interconversion of Concentration units :

Representing solvent and solute in a binary solution by subscripts 1 and 2 respectively, the various conversion expressions are as follows :

DERIVE THE FOLLOWING CONVERSION :

1. Mole fraction of solute into molarity of solution M =

2 2 1 1 2 x M M x 1000 x   

Sol. Mole fraction into molarity M

mole fraction of solvent and solute are X₁ and X₂ so X₁ + X₂ = 1

supposs total mole of solution is = 1

mole of solute and solute and solvent is X₂, X₁ weight of solute = X₂M₂ , weight of solvent = X₁M₁ total wt of solution = X₁M₁ + X₂M₂

volume of solution =   2 2 1 1M X M X ml volume in L = 1000 N X N X_{1 1 2 2}    molarity (M) = 2 2 1 1 2 N X N X 1000 X    

2. Molarity into mole fraction x₂ =

2 1 MM 1000 1000 MM    

Sol. Molarity into mole fraction

molarity (M) = moles solute in 1000 ml of solution so moles of solution = M mass of solution =  x 1000 wt. of solute = MM₂ wt. of solvent = 1000 – MM2 moles of solvent = 1 2 M MM 1000

3. Mole fraction into molality m =

1 1 2 M x 1000 x 

Sol. Mole fraction into molarity

mole fraction of solute X₂ and solvent X₁ mole is n₂ & n₁ molality = 1 1 2 M n n x 1000        1 2 1 2 x x n n = 1 1 2 M x x x 1000

4. Molality into mole fraction x₂ =

1 1 mM 1000 mM  Sol. Moalrity into mole fraction

molality = moles of solute in 1000 gm of solute = m mole of solvent = 1 M 1000 mole fraction X₂ = m M 1000 m 1  = 1 mM 1000 mM 

5. Molality into molarity M =

2 mM 1000 1000 m   

Sol. Molality in molarity

molality = moles of solute in 1000 gm of solvent mole of solute = m wt. of solute = mM₂ wt. of solution = 1000 + mM₂ volume of solution =  mM2 1000

  molarity = 2 mM 1000 1000 m    

6. Molarity into Molality m =

2 MM 1000 1000 M   

M₁ and M₂ are molar masses of solvent and solute.  is density of solution (gm/mL)

M = Molarity (mole/lit.), m = Molality (mole/kg), x₁ = Mole fraction of solvent, x₂ = Mole fraction of solute

Sol. Molarity (M) into molality (m)

molarity = mole of solute in 1000 ml of solution moles of solute = M

wt. of solute = MM₂ wt. of solution = 1000

mass of solvent = 1000 – MM2

molality = moles_wtofof_solventsolute x 1000

m = 2 MM 1000 1000 M    Note :

NCERT Reading

NCERT Exercise

DPP No.

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Q.No. 1.5,1.6,1.11,1.12,1.25,1.29,1.35.

9

Sheet

Exercise

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Exercise

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Exercise

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Part

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– 42 to 50.

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Part

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LECTURE # 11

Mix Questions :

1. Density for 2 M CH₃COOH solution is (1.2 g/mL)

Calculate (1) Molality

(2) mole fraction of CH₃COOH acid (3) % (w/w) for the solution

(4) % (w/v) for the solution

Sol. Let volume & solution = 1 Litre Molarity = solution solute v n = 2 n _solute = 2 mole (CH₃COOH)

solute of mass molar solute of mass = 2 mass of solute = 2 × 60 = 120 g.

mass of solution = volume of solution × density of solution

= 1200 g

= mass of solute + mass of solvent

 mass of solvent = 1200 – 120 = 1080 g = 1.08 kg (H₂O) moles of solvent = 18 1080 = 60 mole (H₂O) (1) Molality = _1.08 2 w n ) kg ( solvent solute  _{= 1.85 m Ans.} (2) XCH3COOH= 2 60 2 n n n O H CCOH CH CCOH CH 2 3 3    = 0.0322 (3) % w/w = _wtwt_.of.of_solutionsolute(gm_(gm)₎ × 100 = 1200 120 ×100 = 10%

(4) % w/v = _volumewt.of_.ofsolute_solution(gm₍)_mL) × 100 =

1000 120

×100 = 12%

2. Mole fraction for aqueous glucose soltuion is 0.1 specific gravity is 1.1

Find (1) Molarity (2) Molality

(3) % (w/w) for the solution (4) % (w/v) for the solution

Sol. X_Glucose = 0.1 = ₁₀ 1 n n n O H e cos Glu e cos Glu 2  

Let 1 mole of glucose is present is solution n _Glucose = 1 m _Glucose = 1 × 180 = 180 gm O H2 n + n_Glucose = 1 . 0 1 1 . 0 nGlucose = 10 So nH2O= 9 O H 2 2 M O H of mass = 9. O H2 M _{= 9 × 10 = 162 gm= 0.162 kg}

mass of solution = mass of solute + mass of solvent = mas of Glucose + mass of H₂O = 180 + 162 = 342 gm

= 311 mL {Sp. gravity = 1.1Density = 1.1 g/mL} = 0.311 Litre. (1) molarity = n_{v 0.311}1 solution e cos Glu  _{= 3.215 M} (2) molality = _0.162 1 ) kg ( w n O H e cos Glu 2  _{= 6.17 M} (3) % (w/w) = _wtwt_.of.of_solutionsolute(₍gm_gm)₎100180_{342 × 100} = 52.63%

(4) % (w/v) = _volumewt.of_.ofsolute_solution(gm₍)_mL)100180_{311 × 100 = 57.87%}

3. 10% w/w urea solution is 1.2 g/mL calculate

(1) % w/v

(2) molarity

(3) molality

(4) mole fraction of urea

Sol. % w/w = 10%

It means

10 g urea is present in 100 gm of solution Let 100 gm of solution is taken

m_solution = 100 gm

v_solution = m_Densitysolution ₁100_.1g/gm_{mL = 91 mL}

= 0.91 L M_urea = 10 gm n_urea = 6 1 60 10 M urea of Mass urea  

m_solution = m_solute + m_solvent= 100 m_solvent = 18 90 MM O H n 2  = 5 mole

(1) % (w/v) = _volumewt_.of.of_solutionurea _(mL) × 100 = 100

91 10  = 10.99% (2) molarity = 91 . 0 6 / 1 v n solution urea  _{= 1.83 M} (3) molarity = _w 1₀/_.96 n solution urea  _{= 1.85 M}

(4) mole fraction of urea

X_urea = O H urea urea 2 n n n  = _31/6 6 / 1 5 6 / 1 6 / 1   = 1/31 Ans.

Q. Molality of HNO₃ soltuion is 2 (spf gr = 1.50) then find : (a) Molarity of the solution

(b) Mole fraction of the solute

4. AVERAGE/ MEAN ATOMIC MASS :

The weighted average of the isotopic masses of the element’s naturally occuring isotopes.

Mathematically, average atomic mass of X (A

_x) =

100 x a ... x a x a1 1 2 2  n n

Ex. Naturally occuring chlorine is 75% Cl35 _{which has an atomic mass of 35 amu and 25% Cl}37 _{which has a mass}

of 37 amu. Calculate the average atomic mass of chlorine

-(A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu

Sol. (A) Average atomic mass =

100 mass atomic its x isotope I of % mass atoms its x isotope of %    = 100 37 x 25 35 x 75  = 35.5 amu

Note : (a) In all calculations we use this mass. (b) In periodic table we report this mass only.

6.

MEAN MOLAR MASS OR MOLECULAR MASS:

The average molar mass of the different substance present in the container =

n 2 1 n n 2 2 1 1 n .... n n M n ... M n M n    

Ex. The molar composition of polluted air is as follows :

Gas At. wt. mole percentage composition

Oxygen 16 16%

Nitrogen 14 80%

Carbon dioxide - 03%

Sulphurdioxide - 01%

What is the average molecular weight of the given polluted air ? (Given, atomic weights of C and S are 12 and 32 respectively. Sol. M_avg =





    n j 1 j j n j 1 j j j n M n Here

_

  n j 1 j j n _{= 100}  Mavg = ₁₀₀ 1 x 64 3 x 44 28 x 80 32 x 16    = 100 64 132 2240 512   = 100 2948 = 29.48 Ans. Note :

NCERT Reading

NCERT Exercise

DPP No.

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Q.No. 1.9,1.32.

10

Sheet

Exercise

–1

Exercise

–2

Exercise

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Part

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– 4,5,10.

Part

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Part

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COURSE - VIKAAS (A)

(LECTURE No. 1 TO 11)