Mole Concept-1 (a)

8/10/2019 Mole Concept-1 (a)

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Page # 1

LECTURE # 1

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 Philosphers 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 Daltonwho firstly developed a theory on the structure

of matter, later on which is known as Daltons atomic theory.

I DALTONS 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 C12isotope. For atoms thisis 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) =atomConeofmass

12

1

elementtheofatomoneofmass

12!

=nucleon1ofmass12

12

1

nucleon1ofmassnucleonsofnumbertotal

!!

! = Total Number of nucleons


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Page # 2

On Hydrogen scale :

Relative atomic mass (R.A.M) =atomHoneofmass


2

Oxygen scale :

Relative atomic mass (R.A.M) =atom16Ooneofmass

16

1


"!

I I . ATOMIC MASS UNIT (OR AMU)

The atomic mass unit (amu) is equal to one twelfth #$

%&'

(12

1of 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 1024gm or 1.66 1027kg

! one amu is also called one Dalton (Da).TODAY , AMU HAS BEEN REPLACED BY uWHICH IS

KNOWN AS UNIFIED MASS

II 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 Oatom and its atomic mass.Sol. The number of neucleons present in Oatom is 16.

) relative atomic mass of Oatom = 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) AgAns. (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 amuAns. = 70

Note : NCERT Reading NCERT Exercise DPP No.

Text. 1.6,1.7,1.7.1,1.7.2

Page 13

Sheet

Exercise1 Exercise2 Exercise3

Part

I

Part

I

Part

I

PartII PartII PartII


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Page # 3

LECTURE # 2

IV. MOLE :THE MASSNUMBER 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 otherparticles) 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 1023atoms 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 NA

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

atoms, ions, molecules or even pens, chair, paper etc also include in this but as this number (NA) is very

large therefore it is used only for very small things.

1 mole x mass of 1 atom of C12isotope = 12g

+,- Alternatively value of N

Acan be

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

.1 mole = 2410x66.1

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 Oatom = mass of one Oatom = 16 amu

gram atomic mass = mass of 6.02 1023Oatoms

= 16 amu 6.02 1023

= 16 1.66 1024g 6.02 1023= 16 g

(! 1.66 1024 6.02 1023~ 1 )

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

2410x66.1x "! = 16 g

x = 2410x66.11

" = NA

orIt is also defined as mass of 6.02 1023atoms.

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

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

ElementR.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

Ans. 4.5 g atoms


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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 : H2, O

2, O

3etc.

Compound : KCl, H2SO

4, KClO

4etc.

Molecule Atomicity

KCl - 2

H2SO

4- 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

H2SO

4(2 + 32 + 64) = 98 amu.

VIII. GRAM MOLECULAR MASS :

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

orIt is also defined as mass of 6.02 1023molecules

or

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

For example for O2molecule :

Molecular mass of O2molecule = mass of one O

2molecule

= 2 mass of one Oatom= 2 16 amu

= 32 amugram molecular mass = mass of 6.02 1023O

2molecules = 32 amu 6.02 1023

= 32 1.66 1024gm 6.02 1023 = 32 gm

Use the Y-map in the following example.


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Example:

Find the mass in grams of 3 mol of zinc. (GMM = 65)

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 1023= 3.01 1023

Example :

The molecular mass of H2SO

4is 98 amu. Calculate the number of moles of each element in 294 g of H

2SO

4.

SolutionGram molecular mass of H

2SO

4= 98 gm

moles of H2SO

4=

98

294= 3 moles

H2SO

4H S O

One molecule 2 atom one atom 4 atom

1 NA

2 NAatoms 1 N

Aatoms 4 N

Aatoms

)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 107molecules of methane. How many C2H6moleculesdoes the sample contain ?

Ans. n = 5.34 106

Example :

How many molecules of water are present in 252 mg of (H2C

2O

4.2H

2O)

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.AlsoCalculate 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

Page 14Q.No.-1.1,1.10,1.28,1.30,1.33 1

Sheet


PartI1,2,3,4,5,6,7. PartI PartI

PartII6,7,8,9,10,11,12 PartII PartII


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Page # 6

LECTURE # 3

GAY-LUSSACS 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) + Cl

2(g) /0/ 2HCl

1 vol 1 vol 2 vol

AVOGADROS HYPOTHESIS :

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

mathematically, for ideal gases, V 1n (CONSTANT T & P)

S.T.P. (Standard Temperature and Pressure):

At S.T.P. / N.T.P. condition :temperature = 0C 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 2atomic weight = gm2

gm20

= 10 mol

volume of hydrogen gas at STP = 10 22.4 lt.

Mole

22

.4lt

222.

4lt

Volume at STP

2NA

NA

Number

mol. wt.

At. wt.2

2At. 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 1024molecules of SO

2are 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 CO2gas are mixed together. Find the volume of gaseous mixture at STP.

Ans. 22.4 Ltr.

Ex. 672 ml of ozonized oxygen (mix of O2and O

3) at N.T.P. were found to weight one gram. Calculate the volume

of ozone in the ozonized oxygen.

Ans. 56 ml

Note :


2

Sheet


PartI1 to 8 PartI PartI

PartII1 to 12 PartII PartII


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Page # 7

LECTURE # 4

II THE LAWS OF CHEMICAL COMBINATION

Atoine Lavoisier, John Daltonand 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 O

2(g) /0/ H

2O (l)

Before reaction 1 mole2

1mole

After the reaction 0 0 1 mole

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

2

1mole O

2(g)

= 2 + 16 = 18 gmmass 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 theirsource 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 steamgave 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

) 5.1g2.1

g80.1

oxygenof.wt

metalof.wt33

In the second sample of the oxide,

Wt. of metal = 1.50 g,

Wt. of oxygen = (2.50 1.50) g = 1 g.

) 5.1g1

g50.1

oxygenof.wt

metalof.wt33

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 thissimple 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 CO

2, 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 simpleratio of 16 : 32 or 1 : 2

Note : See oxidation number of carbon also have same ratio 1 : 2 in both the oxide.


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CONCEPTS RELATED TO DENSITY :

It is of two type.1. Absolute density2. Relative density

For liquid and solids

Absolute density =volume

mass

specific gravity =C4atwaterofdensity

cetansubstheofdensity

4For gases :

Absolute density (mass/volume) =gastheofvolumeMolar

gastheofmassMolar

* 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 DENSITY :

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

Ex. What is the V.D. of SO2with respect to CH

4

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 N

2O(g).

VAPOUR DENSITY :

Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature

and pressure.

Vapour density =2H

gas

d

d= volumemolar/HofmassMolar

volumemolar/gasofmassMolar

2

V.D. =2H

gasM

M=

2Mgas ; 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 gas7.5 litre = 16 gram

moles =M

16

4.22

5.73

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


PartI12,13,14. PartI2. PartI

PartII17,18,19,20,21. PartII33,35. PartII


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LECTURE # 5

PERCENTAGE COMPOSITION :

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

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

Ex. Every molecule of ammonia always has formula NH3irrespective 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 NH3always contains

14 gm of N and 3 gm of H. Now find out % of each element in the compound.

Mass % of N in NH3 = 100NHofmol1ofMass

NHmol1inNofMass

3

3 ! =17

14 100 = 82.35 %

Mass % of H in NH3= 100NHofemol1ofMass

NHmol1inHofMass

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 composition of the elements initially, we can

calculate the relative number of atoms of each element in the molecules of the compound. This gives us theempirical formula of the compound. Further if the molecular mass is known then the molecular formula caneasily be determined.

Thus, the empirical formula of a compound is a chemical formula showing the relative number of atoms in thesimplest 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 = massformulaempirical

massformulamolecular

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-1The empirical formula of the compound is CH

) Empirical formula mass= (1 12) + 1 = 13.

Molecular mass = 26

Step-2To calculate the value of n

n =massformulaEmpirical

massMolecular=

13

26= 2

Step-3To calculate the molecular formula of the compound.

Molecular formula = n (Empirical formula of the compound)= 2 CH = C

2H

2

Thus the molecular formula is C2H

2

Similarly for benzene

To calculate the value of n

n = massformulaEmpirical

massMolecular=

13

78= 6

thus the molecular formula is 6 CH = C6H

6


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Page # 10

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

Percentageof element

40.687

5.085

54.228

At. massof element

12

1

16

Relative no.Percentage

At. massof atoms =

40.687

12 = 3.390

5.085

1= 5.085

54.228

16= 3.389

Simplestatomic ratio

3.390

3.389 =1

5.085

3.389=1.5

3.389

3.389=1

Simplest wholeno. atomic ratio

2

3

2

) Empirical Formula is C2H

3O

2

Step-2

To calculate the empirical formula mass.

The empirical formula of the compound is C2H

3O

2.

) Empirical formula mass= (2 12) + (3 1) + (2 16) = 59.

Step-3

To calculate the value of n

n = massformulaEmpirical

massMolecular=

59

118= 2

Step-4

To calculate the molecular formula of the salt.

Molecular formula = n (Empirical formula)= 2 C2H

3O

2= C

4H

6O

4

Thus the molecular formula is C4H

6O

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. N2O

5

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

Ans. C3H

3O

Note :


Text. 1.9,1.9.1

Page 15Q.No.1.2,1.3,1.8. 4

Sheet


PartI9,10,11. PartI3,11,13. PartI

PartII13,14,15,16. PartII3,4,7,8,9,14,28,29. PartII


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Page # 11

LECTURE # 6

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 whichreactants combine and the molar ratio in which products are formed.

Example :

When potassium chlorate (KClO3) is heated it gives potassium chloride (KCl) and oxygen (O

2).

KClO3 /0/5 KCl + O2 (unbalanced chemical equation )

2KClO3 /0/5 2 KCl + 3 O

2(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 trialand error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides

4 Fe(s) + 3O2(g) 0 2Fe

2O

3(S) (a) balanced equation

2 Mg(s) + O2(g) 0 2MgO(S) (b) balanced equation

P4(s) + O

2(g) 0P

4O

10(S) (c) unbalanced equation

Equations (a) and (b) are balanced since there are same number of metal and oxygen atoms on eachside 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 leftside of the equation to balance the oxygen atoms appearing on the right side of the equation.

P4(S) + 5O

2(g) 0 P

4O

10(S) balanced equation

Now let us take combustion of propane, C3H

8, This equation can be balanced in steps.

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

C3H

8(g) + O

2(g) 0 CO

2(g) + H

2O (l) unbalanced equation

Step 2. Balance the number of C atoms : Since 3 carbon atoms are in the reactant, therefore, three CO2mol-

ecules are required on the right side.C

3H

8(g) + O

2(g) 0 3CO

2(g) + H

2O (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 hydrogenatoms on the right side.

C3H

8(g) + O

2(g) 0 3CO

2(g) + 4H

2O (l)

Step 4. Balance the number of O atoms : There are ten oxygen on the right side (3 2 = 6 in CO2and 4 1 = 4

in water). Therefore, five O2molecules are needed to supply to supply the required ten oxygen atoms.

C3H

8(g) + 5O

2(g) 0 3CO

2(g) + 4H

2O (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 anequation.


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Page # 12

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 toclearly understand this analysis.Now consider again the decomposition of KClO

3.

2KClO3 /0/ 2KCl + 3O

2

In very first step of mole-mole analysis you should read the balanced chemical equation like2 moles KClO

3on decomposition gives you 2 moles KCl and 3 moles O

2.and from the stoichiometry of

reaction we can write

2

KClOofMoles 3=

2

KClofMoles=

3

OofMoles 2

Now for any general balance chemical equation like

a A + b B /0/ c C + d D

you can write.

a

reactedAofMole=

b

reactedBofmoles=

c

reactedCofmoles=

d

reactedDofmoles

Ex. 3 moles (367.5 gm) of KClO3when heated how many moles KCl and O

2is produced.

Sol. The reaction is

2KClO3 /0/ 2KCl + 3O2

2

KClOofMoles 3=

2

KClofMoles. Moles of KCl produced = 3

Now,2

KClOofMoles 3=

3

OofMoles 2

mole of O2produced =

2

33 != 4.5 moles

Ex. CH4+ 2O

20CO

2+ 2H

2O (from NCERT Page - 18)

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

7One mole of CH4(g) reacts with two moles of O

2(g) to give one mole of CO

2(g) and two moles of H

2O (g)

7One molecule of CH4(g) reacts with 2 molecues of O2(g) to give one molecule of CO2(g) and 2 moleculesof H

2O (g)

722.4 L of CH4(g) reacts with 44.8 L of O

2(g) to give 22.4L of CO

2(g) and 44.8 L of H

2O (g)

716 g CH4(g) reacts with 232 g of O

2(g) to give 44 g of CO

2(g) and 2 18 g of H

2O (g).

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

Mass 2 At. wt. / Mol. Wt.

MoleMole-molerelationshipof equation

Mole

22.4 lt

Volume at STP

m

ol.wt./A

t.wt.

Mass

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

(a) How many grams O2is produced

(b) How many litre of O2is produced at STP

Sol. Now consider the balanced chemical equation

2KClO3 /0/ 2KCl + 3O

2

Now go with the above chart


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Page # 13

367.5 gmKClO

3

2 122.5 gm 3 moleKClO

3

Mole of KClO3

2=

Mole of O2

3

(Mole-mole relationship of equation)

9/2 mole of O2

(a) 144 gm32gm(mol.wt.)

(b) 100.8 lt

22.4lt

(volumeatSTP)

Ex. Iron in the form of fine wire burns in oxygen to form iron (III) oxide

4Fe(s) + 3O2(g) /0/ 2Fe

2O

3(s)

How many moles of O2are needed to produce 5 mol Fe

2O

3?

Ans. 7.5 mol O2

Ex. Nitric acid, HNO3, is manufactured by the Ostwald process, in which nitrogen dioxide, NO

2, reacts with

water.

3NO2(g) + H2O(l) 0 2HNO3(aq) + NO(g)How many grams of nitrogen dioxide are required in this reaction to produced 6.3 g HNO

3?

Ans. 6.9g NO2

Ex. How many grams of Fe2O

3is formed by heating 18 gm FeO with Oxygen.

4FeO + O2 /0/ 2Fe

2O

3

Ans. 20. gm

Ex. How many litre O2at N.T.P. is required for complete combustion of 1 mole C

5H

10.

Ans. 168 lt.

Ex. Calculate the weight of residue obtained when CaCO3is strongly heated and 5.6 litre CO

2is produced at

N.T.P.

Ans. 14 gm

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

CO2(g) obtained at STP and amount of NaHCO

3needed for this reaction.

Sol. 2NaHCO3 /0/ Na

2CO

3+ H

2O + CO

2

So volume of CO2 = 3 22.4 = 67.2 Lt.

Mass of NaHCO3needed = 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) + 20

2(g) 0 CO

2(g) + 2H

2O (g)

(i) 16 g of CH4corresponds to one mole.

(ii) From the above equation, 1 mol of

CH4(g) gives 2 mol of H

2O (g)

2 mol of water (H2O) = 2 (2 + 16) = 2 18 = 36 g


14/31

Page # 14

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) + 20

2(g) 0 CO

2(g) + 2H

2O (g)

44g CO2(g) is obtained from 16 g CH

4(g).

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

4(g)

mole of CO2(g) = 22g CO

2(g) )g(gCO44

)g(molCO1

2

2

= 0.5 mol CO2(g)

Hence, 0.5 mol CO2(g) would be obtained from 0.5 mol CH

4(g) or 0.5 mol of CH

4(g) would be required to

produce 22 g CO2(g).

Note :


Text. 1.10

Page 17Q.No. 1.3,1.4. 5

Sheet


PartI15,16,17. PartI PartIPartII22,23,24. PartII16,17,18,26,27. PartII


15/31

Page # 15

LECTURE # 7

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 dont 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) + O

2(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 KClO3contains 1 atom of K

or 1 mole of KClO3contains 1 mole of K, similarly,1 mole of KCl contains 1 mole of K.

Thus, moles of K atoms in KClO3= 1 moles of KClO

3

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

) moles of KClO3= moles of KCl

or3

3

KClOofwt.mol.

ginKClOofwt.=

KClofwt.mol.

ginKClofwt.

! 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 KClO

3

moles of O in O2= 2 moles of O

2

) 3 moles of KClO3= 2 moles of O

2

or 3 3

3

KClOof.wt.mol

KClOof.wt= 2

.)lt4.22(.volmolardardtans

NTPatOof.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 + P40P4O10+ N2(ii) P

4+ HNO

30H

3PO

4+ NO

2+ H

2O

Example :

27.6 g K2CO

3was treated by a series of reagents so as to convert all of its carbon to K

2Zn

3[Fe(CN)

6]2.

Calculate the weight of the product.

[mol. wt. of K2CO

3= 138 and mol. wt. of K

2Zn

3[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

K2CO

3

StepsSeveral//// 0/ K

2Zn

3[Fe(CN)

6]2

Since C atoms are conserved, applying POAC for C atoms,

moles of C in K2CO

3= moles of C in K

2Zn

3[Fe(CN)

6]2

1 moles of K2CO

3= 12 moles of K

2Zn

3[Fe(CN)

6]2

(! 1 mole of K2CO

3contains 1 moles of C)

32

32

COKof.wt.mol

COKof.wt= 12

productof.wt.mol

producttheof.wt

wt. of K2Zn

3[Fe(CN)

6]

2 =

138

6.27

12

698= 11.6 g


16/31

Page # 16

Q.1 0.32 mole of LiAlH4in ether solution was placed in a flask and 74 g (1 moles) of t-butyl alcohol was added.

The product is LiAlHC12

H27

O3. 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 :


Q.No. 1.7,1.34,1.36. 6

Sheet


PartI26,27. PartI PartI

PartII31 to 34. PartII1,2,5,6,11,13,15,20,22 PartII


17/31


18/31

Page # 18

Step :# $$$Now once you find limiting reagent then your focus should be on limiting reagentFrom Step *& ** Na

2CO

3HCl

1

6= 6

2

4= 2 (division is minimum)

HCl is limiting reagentFrom Step ***

From2

HClofMole=

1

producedCOofMoles 2

) mole of CO2produced = 2 moles

) volume of CO2produced at S.T.P. = 2 22.4 = 44.8 lt.

Examples on limiting reagent :

Ex. The reaction 2C + O2 0 2CO is carried out by taking 24g of carbon and 96g O

2, find out :

(a) Which reactant is lef t in excess ?(b) How much of it is left ?(c) How many mole of CO are formed ?

Ex. For the reaction 2P + Q 0R, 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 0 X3Y

4

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 X

3Y

4is formed

(C) 1.5 moles of excess reagent is left behind (D) 75% of excess reagent reactedX + Y 0 X

3Y

4

Ans. B, C, D

Sol. 3X + 4Y /0/ X3Y

4

6 mole 6 mole6 4.5 0 1.5 mole1.5 moleleft formed

Ex. A + B 0 A3B

2(unbalanced)

A3B

2+ C 0 A

3B

2C

2(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 A3B2C2is formed (B*) 21 mole of AA3B2C2is formed(C*) 1 mole of A

3B

2is formed (D*) 21 mole of AA

3B

2is left finally

Ans. B, C, D

Sol. 3A + 2B /0/ AA3B

2

3 mole 3 mole 1 mole formed

A3B

2+ 2C /0/ AA

3B

2C

2

1 mole 1 mole0.5 mole 0 0.5 mole

Ex. CS2and Cl

2in the weight ratio 1 : 2 are allowed to react according to equation, find the fraction of excess

reagent left behind.

CS2+ 3Cl

2 /0/ CCl

4+ S

2Cl

2

mole76

w

71

w2;

76

w

371

w2

!

L.R. = Cl2.

Moles of CS2=

371

w2

! . remaining =

76

w

213

w2

fraction of Cl2left = 100x

76

w213

w2

76

w"

= 28.6%.


19/31

Page # 19

Ex. 27 gm Al is heated with 49 ml of H2SO

4 (sp. gr = 2) produces H

2 gas. Calculate the volume of

H2gas at N.T.P. and % of Al reacted with H

2SO

4

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*3according to the reaction

P + *20P*3P*

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

P*3+ H

2O 0H

3PO

3+ H*

HI formed in the above reaction is collected in the gaseous form. At higher temperature HI dissociated 50%

then find the molecules of H2gas liberated

H*0H2+ *

2

Ans. 3/8 NA.

Problem 1.5 (NCERT Page - 19)

50.0 kg of N2(g) and 10.0 kg of H

2(g) are mixed to produce NH

3(g) Calculate the NH

3(g) formed. Identify

the limiting reagent in the production of NH3in this situation. Take reaction N

2+ 3H

202NH

3.

Sol. A balanced equation for the above reaction is written as follows :

Calculation of moles :N

2(g) + 3H

2(g)02NH

3(g)

moles of N2

= 50.0 kg2

22

Nkg1

Ng1000N !

2

2

Ng0.28

Hmol1= 17.86 102mol

moles of H2

= 10.00 kg H2

2

2

Hkg1

Hg1000

2

2

Hg016.2

Hmol1= 4.96 103mol

According to the above equation, 1 mol N2(g) requires 3 mol H

2(g), for the reaction, Hence, for 17.86

102mol of N2, the moles of H

2(g) required would be

17.86 102mol N2

)g(molN1

)g(Hmol3

2

2= 5.36 103mol H

2

But we have only 4.96103mol H2. Hence, dihydrogen is the limiting reagent in this case. So NH

3(g)

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

Since 3 mol H2(g) gives 2 mol NH

3(g)

) 4.96 103mol H2(g) )g(Hmol3

)g(HNmol2

2

3

= 3.30 103mol NH3(g) is obtained.

If they are to be converted to grams, it is done as follows :

3.30 103mol NH3(g) )g(molNH1

)g(gNH0.17

3

3. = 3.30 103 17 g NH

3(g)

= 56.1 103g NH3

= 56.1 kg NH3

Note :


Text. 1.10.1.

Page 18Q.No. 1.4,1.23,1.24,1.26. 7

Sheet


PartI18 to 25. PartI7. PartI

PartII25 to 30. PartII16,23,25,26. PartII


20/31

Page # 20

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 Homogeneousmeans 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 solute65% by volume i.e. 100 ml of solution contain 65 ml solute65% by strength i.e. 100 ml of solution contain 65 gm soluteIf, anything is not specified, 65% generally mean 65% by mass

(b) For concentrated acids, like 98% H2SO

4, 65% HNO

3etc, 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 =1931000

1938

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. MOLARITY (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 =litreinsolutionofvolume

molesofnumber

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 haveM

wmole of solute

) 1000 ml water havemlVM

1000w

!

!

) Molarity (M) =mlV)soluteofwt.Mol(

1000w

!

!


21/31

Page # 21

Problem 1.7 (NCERT Page - 20)Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250

mL of the solution.Solution

Since molarity (M)

=eslitrinsolutionofVolume

soluteofmolesofNo.=

L0.250

NaOHofmassNaOH/MolarofMass

=L250.0

g40/g4

L250.0

mol1.03 = 0.4 mol L1 = 0.4 M

Note that molarity of a solution depends upon temperature because volume of a solution is temperaturedependent.Some other relations may also useful.

Number of millimoles = 1000)soluteof.wt.Mol(

soluteofmass! = (Molarity of solution V

inml)

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

mathematically : molarity decreases as temperature increases.

Molarity 1etemperatur

11

volume

1

! Molarity of solution may also given as :

mlinsolutionofvolumeTotal

soluteofmillimoleofNumber

(i) If a particulars solution having volume V1and molarity = M

1is diluted to V

2mL then

M1V

1= M

2V

2

M2: Resultant molarity

(ii) If a solution having volume V1and molarity M

1is mixed with another solution of same solute having

volume V2mL & molarity M

2

then M1V

1 + M

2V

2= M

R(V

1+ V

2)

MR= Resultant molarity

= 21

2211

VV

VMVM

8

8

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

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

) Moles of KCl = gm5.74

gm149= 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)2in 5 mL of solution [ 2M ](b) 3.65 g of HCl in 200 ml of solution [0.5M ](c) 1/10 mole of H2SO4in 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


22/31

Page # 22

Ex. Calculate the molarity of H+ion in the resulting solution when 200 ml 1M HCl is mixed with 200 ml 1M H2SO4Sol. For HCl

M1= 1 MV1= 200 mL

For H2SO4M2= 1V2= 200 mL

nHCl= MHCl VHCl= 1 0.2

HCl/0/

H

+

+ Cl

8Hn = 0.2 (from HCl)

424242 SOHSOHSOHVMn !3

= 1 0.2 = 0.2 mole

H2SO4 /0/ 2H++ SO4

2

8Hn = 0.2 2 (from H2SO4)

Total H+= 8Hn (from HCl) + 8Hn (from H2SO4)

= 0.2 + 0.4 = 0.6

Total volume = 200 + 200 = 400 mL = 0.4 LMR= Resultant molarity

=4.0

6.0

V

n

solution

H 38

= 1.5 Ans.

Ex. What are the final concentration of all the ion when following are mixed50 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

226.0100312.050 !!8!!

= 250702505218 38 = 0.28

[Cl] = 0.12 M[Mg++] = 0.104 M

[Fe3+] = 0.064 M

Ex. Calculate the molarity of water

H2O /0/ 18 gm

= 1 moleVolume 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 NaOHsolution.

Ex. CaCO3

reacts with aq. HCl to give CaCl2and CO

2according to reaction

CaCO3(s) + 2HCl(aq) /0/ CaCl2+ CO2+ H2O

How much mass of CaCO3is required to react completly with 100 ml of 0.5 m HCl


23/31

Page # 23

Sol. millimole of HCl100 0.5 = 50

2 mole of HCl reacts /0/ 1 moles CaCO3

1 mole of HCl reacts /0/ 2

1

50 mmole of HCl reacts /0/ 2

1 50 = 25 mmole of CaCO

3

mole of CaCO3=

1000

25

mass of CaCO3=

1000

25 100 = 2.5 gm.

Ex. Na2CO

3+ 2HCl /0/ 2NaCl + CO2+ H2O

(a) moles of NaCl formed when 10.6 gm of Na2CO

3is 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 CO

2liberated at S.T.P.

Sol. molar mass of Na2CO

3= 106 gm

Na2CO

3=

106

6.10 1000 = 100 m mole

HCl = 100 0.5 = 50 milli mole

limiting reagent /0/ HCl /0/

2 HCl reacts = 2 NaCl50 mmole HCl reacts = 50 mmole of Na

2CO

3

remaining Na2CO

3= 100 25 = 75 mmole

(a) moles of NaCl =1000

50= 0.05

(c) volume of CO2

= 4.221000

25! = 0.556 L

[Na+] =100

200

100

150503

8= 2 M

[Cl] =100

50= 0.5 M

(CO3

2] =100

75= 0.75 M

Note :


Text. 1.10.2.

Page 19

8Sheet


PartI28 to 41 PartI PartI

PartII35 to 41. PartII19. PartII


24/31

Page # 24

LECTURE # 10

3 . MOLALITY (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 =

1000graminsolventofmass

soluteofmolesofnumber!

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 moleof the solute are dissolved in x gm of the solvent. Hence

Molality = 1000xm

Y !!

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

Sol. Mass of urea = 5 gmMolecular mass of urea = 60

Number of moles of urea =60

5= 0.083

Mass of solvent = (255 5) = 250 gm

) Molality of the solution = 1000

graminsolventofMass

soluteofmolesofNumber! = 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 mL1. Calculate molality of the solution.

Sol. M = 3 mol L1

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 mL1)Mass of water in solution = 1250 175.5 = 1074.5 g

Molaity = kginsolventofMass

soluteofmolesofNo.= kg1.0745

mol3= 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. Notethat 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 thesolution.

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) =

Nn

n

8

) Mole fraction of solvent (x2) =

Nn

N

8! also x

1+ x

2= 1

Note : mole fraction is a pure number its also independent of temperature changes.


25/31

Page # 25

5 . CALCULATION :

The concentration of a solution may also expressed in terms of percentage in the following way.

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

i.e. % w/w = 100gminsolutionofmass

gminsoluteofmass!

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

i.e., % w/v = 100mlinsolutionofmass

gminsoluteofmass!

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

i.e., % V/V = 100solutionofVolume

mlinsoluteofVolume!

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) = 100255.0

5.0

!8 = 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 =solutionofMass

AofMass 100 =

gofwater18gofA2

g2

8 100

100g20

g2!3 = 10 %

Example

20 cm3of an alcohol is dissolved in80 cm3of water. Calculate the percentage of alcohol in solution.Solution

Volume of alcohol = 20 cm3

Volume of water = 80 cm3

) percentage of alcohol = 1008020

20!

8= 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 =2211

2

xMMx

1000x

8

!9

Sol. Mole fraction into molarity M

mole fraction of solvent and solute are X1and X

2

so X1+ X

2= 1

supposs total mole of solution is = 1

mole of solute and solute and solvent is X2, X

1

weight of solute = X2M

2, weight of solvent = X

1M

1

total wt of solution = X1M

1+ X

2M

2


26/31

Page # 26

volume of solution = 9

8 2211 MXMXml

volume in L =1000

NXNX 2211!9

8

molarity (M) =2211

2

NXNX

1000X

8

!9!

2. Molarity into mole fraction x2=2

1

MM1000

1000MM

"!9

!

Sol. Molarity into mole fraction

molarity (M) = moles solute in 1000 ml of solution

so moles of solution = M

mass of solution = 9 x 1000

wt. of solute = MM2

wt. of solvent = 1000 9 MM2

moles of solvent =1

2

M

MM1000 "9

3. Mole fraction into molality m =11

2

Mx1000x !

Sol. Mole fraction into molarity

mole fraction of solute X2and solvent X

1

mole is n2& n

1

molality =11

2

Mn

nx 1000 :

;

?3

1

2

1

2

x

x

n

n

=11

2

Mx

xx 1000

4. Molality into mole fraction x2=

1

1

mM1000

mM

8

Sol. Moalrity into mole fraction

molality = moles of solute in 1000 gm of solute = m

mole of solvent =1M

1000

mole fraction X2=

mM

1000

m

1

8=

1mM1000

mM

8

5. Molality into molarity M =2mM1000

1000m

8!9

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 =9

8 2mM1000


27/31

Page # 27

volume in (L) =1000

mM1000 2!9

8

molarity =2mM1000

1000m

8!9!

6. Molarity into Molality m =2MM1000

1000M

"9!

M1

and M2

are molar masses of solvent and solute. 9is density of solution (gm/mL)

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

2= 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 9

mass of solvent = 1000 9 MM2

molality = solventofwt

soluteofmolesx 1000

m =2MM1000

1000M "9!

Note :


Q.No. 1.5,1.6,1.11,1.12,1.25,1.29,1.35. 9

Sheet


PartI42 to 50. PartI PartI

PartII42 to 45. PartII21,24,30,31,32. PartII


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Page # 28

LECTURE # 11

Mix Questions :

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

Calculate (1) Molality(2) mole fraction of CH

3COOH acid

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

Sol. Let volume & solution = 1 Litre

Molarity =solutionsolute

v

n= 2

nsolute

= 2 mole (CH3COOH)

soluteofmassmolar

soluteofmass= 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 (H

2O)

moles of solvent = 18

1080= 60 mole

(H2O)

(1) Molality = 08.1

2

w

n

)kg(solvent

solute 3 = 1.85 m Ans.

(2) COOHCH3X = 602

2

nn

n

OHCCOHCH

CCOHCH

23

3

83

8 = 0.0322

(3) % w/w =)gm(solutionof.wt

)gm(soluteof.wt 100 =

1200

120100 = 10%

(4) % w/v =)mL(solutionof.volume

)gm(soluteof.wt 100 =1000120 100 = 12%

2. Mole fraction for aqueous glucose soltuion is 0.1 specific gravity is 1.1Find (1) Molarity (2) Molality

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

Sol. XGlucose

= 0.1 = 10

1

nn

n

OHecosGlu

ecosGlu

2

38

Let 1 mole of glucose is present is solutionn

Glucose= 1

mGlucose

= 1 180 = 180 gm

OH2n + n

Glucose=

1.01

1.0n ecosGlu = 10

So OH2n = 9

OH

2

2M

OHofmass = 9.

OH2M = 9 10 = 162 gm = 0.162 kg

mass of solution = mass of solute + mass of solvent= mas of Glucose + mass of H

2O

= 180 + 162 = 342 gm


29/31

Page # 29

volume of solution = solutionofDensity

solutionofmass= mL/g1.1

g342

= 311 mL {Sp. gravity = 1.1Density = 1.1 g/mL}= 0.311 Litre.

(1) molarity = 311.0

1

v

n

solution

ecosGlu 3 = 3.215 M

(2) molality = 162.0

1

)kg(w

n

OH

ecosGlu

2

3 = 6.17 M

(3) % (w/w) = 342

180100

)gm(solutionof.wt

)gm(soluteof.wt3! 100

= 52.63%

(4) % (w/v) =311

180100

)mL(solutionof.volume

)gm(soluteof.wt3! 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 means10 g urea is present in 100 gm of solutionLet 100 gm of solution is taken

msolution

= 100 gm

vsolution

= mL/g1.1

gm100

Density

msolution 3 = 91 mL

= 0.91 LM

urea= 10 gm

nurea

=6

1

60

10

M

ureaofMass

urea

33

msolution= msolute + msolvent= 100

msolvent

=18

90

MM

OHn 2 3 = 5 mole

(1) % (w/v) =)mL(solutionof.volume

ureaof.wt 100 = 100

91

10! = 10.99%

(2) molarity = 91.0

6/1

v

n

solution

urea 3 = 1.83 M

(3) molarity = 9.0

6/1

w

n

solution

urea 3 = 1.85 M

(4) mole fraction of urea

Xurea

=OHurea

urea

2nn

n

8 = 6/316/1

56/1

6/13

8

= 1/31 Ans.

Q. Molality of HNO3soltuion is 2 (spf gr = 1.50) then find :

(a) Molarity of the solution

(b) Mole fraction of the solute

Ans. (a) 2.66 M. (b) 0.0347.


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Page # 30

4. AVERAGE/ MEAN ATOMIC MASS :

The weighted average of the isotopic masses of the elements naturally occuring isotopes.

Mathematically, average atomic mass of X (Ax) =

100

xa.....xaxa nn2211 888

Ex. Naturally occuring chlorine is 75% Cl35which has an atomic mass of 35 amu and 25% Cl37which 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

massatomicitsxisotopeIof%massatomsitsxisotopeof% *8*

=100

37x2535x75 8

= 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 =n21

nn2211

n....nn

Mn......MnMn

88

88

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

Gas At. wt. mole percentage compositionOxygen 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. Mavg

=

@

@3

3

3

3

nj

1j

j

nj

1j

jj

n

Mn

Here @3

3

nj

1j

jn = 100

)Mavg

=100

1x643x4428x8032x16 888=

100

641322240512 888=

100

2948= 29.48 Ans.

Note :


Q.No. 1.9,1.32. 10

Sheet


PartI4,5,10. PartI PartI

PartII34. PartII12. PartII


31/31

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Mole Concept-1 (a)