Lec. Notes Mole Concept

LECTURE # 1Mole 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 fromcombination 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 werehard and indivisible.The Greek philosphere Democritus named these building blocks as atoms, meaningindivisible.All these people have their philosphical view about matter, they were never put to experimentaltests, nor ever explain any scientific truth.It was John Dalton who firstly developed a theory on the structureof 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 DefinitionsI . 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 thisis done by expressing mass of one atom with respect to a fixed standard. Dalton used hydrogen as thestandard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the reference.

C-12 ISOTOPE OF CARBON IS LATEST CHOSEN STANDARD SINCE 1961Therefore relative atomic mass is given as

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

121

elementtheofatomoneofmass12

= nucleon1ofmass12

121

nucleon1ofmassnucleonsofnumbertotal

= Total Number of nucleons

On Hydrogen scale :

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

elementtheofatomoneofmass

2

Oxygen scale :

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

161

elementtheofatomoneofmass

II . ATOMIC MASS UNIT (OR AMU)

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

121

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

1 amu = 121

mass of one C-12 atom

~ mass of one neucleon in C-12 atom.= 1.66 1024 gm or 1.66 1027 kg

one amu is also called one Dalton (Da).TODAY , AMU HAS BEEN REPLACED BY u WHICH ISKNOWN AS UNIFIED MASS

III . ATOMIC MASSIt 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 amuAns. = 70

Note : NCERT Reading NCERT Exercise DPP No.

Text. 1.6,1.7,1.7.1,1.7.2Page 13

SheetExercise1 Exercise2 Exercise3

PartI PartI PartI PartII PartII PartII

LECTURE # 2IV. 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 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 asAvogadro constant denoted by NA.i.e. on the whole we can say that 1 mole is the collection of 6.02 1023 entities. Here entities may representatoms, ions, molecules or even pens, chair, paper etc also include in this but as this number (NA) is verylarge therefore it is used only for very small things.

1 mole x mass of 1 atom of C12 isotope = 12g Alternatively value of NA can be

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

1 mole = 2410x66.11

= 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 amugram atomic mass = mass of 6.02 1023 O atoms

= 16 amu 6.02 1023 = 16 1.66 1024 g 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 1023 atoms.

orIt 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 sulphurAns. 4.5 g atoms

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 atomif 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. Atomicmasses 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 itsconstituents atoms by physical & chemical process.Number of atoms presents in molecule is called its atomicity.Element : H2, O2, O3 etc.Compound : KCl, H2SO4, KClO4 etc.

Molecule AtomicityKCl - 2H2SO4 - 7O3 - 3H2 - 2

VII. MOLECULAR MASS :It is the mass of one moleculeEx. Molecule Molecular mass

H2 2 amuKCl (39 + 35.5) = 74.50 amoH2SO4 (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 1023 molecules

orIt 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 1024 gm 6.02 1023 = 32 gm

Use the Y-map in the following example.

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

Sol. Mass = mol At. wt. = 3 65 gm = 157.3 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 H2SO4 is 98 amu. Calculate the number of moles of each element in 294 g of H2SO4.

SolutionGram molecular mass of H2SO4 = 98 gm

moles of H2SO4 = 98294

= 3 moles

H2SO4 H S OOne molecule 2 atom one atom 4 atom1 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 10

7 molecules of methane. How many C2H6 moleculesdoes 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.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 14 Q.No.-1.1,1.10,1.28,1.30,1.33 1


PartI 1,2,3,4,5,6,7. PartI PartI PartII 6,7,8,9,10,11,12 PartII PartII

LECTURE # 3GAY-LUSSACS LAW OF COMBINING VOLUME :

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

H2 (g) + Cl2 (g) 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 n (CONSTANT T & P)S.T.P. (Standard Temperature and Pressure):

At S.T.P. / N.T.P. condition :temperature = 0C or 273 Kpressure = 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.

Example :Calculate the volume in litres of 20 g hydrogen gas at STP.

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

= gm2gm20

= 10 mol

volume of hydrogen gas at STP = 10 22.4 lt.

Mole

22.4

lt

22.4 lt

Volume at STP

NA

NA

Number

mol. wt. At. wt.

At. wt. mol. wt.

MassExample :

Calculate the volume in litres of 142 g chlorine gas at STP.Ans. 44.8 lt.Example :

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

Ans. 34.22

= 7.5

Example.From 160 g of SO2 (g) sample, 1.2046 x 10

24 molecules of SO2 are removed then find out the volume of leftover SO2 (g) at STP.Ans. 11.2 Ltr.

Example.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 volumeof ozone in the ozonized oxygen.

Ans. 56 ml

Note :

NCERT Reading NCERT Exercise DPP No.

2


PartI 1 to 8 PartI PartI PartII 1 to 12 PartII PartII

LECTURE # 4II THE LAWS OF CHEMICAL COMBINATION

Atoine Lavoisier, John Dalton and other scientists formulate certain law concerning the composition ofmatter and chemical reactions.Theses 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) + 21

O2 (g) H2O (l)

Before reaction 1 mole 21

mole

After the reaction 0 0 1 mole

mass before reaction = mass of 1 mole H2 (g) + 21

mole O2 (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 .

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

Example: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

5.1g2.1g80.1

oxygenof.wtmetalof.wt

In the second sample of the oxide,Wt. of metal = 1.50 g,Wt. of oxygen = (2.50 1.50) g = 1 g.

5.1g1g50.1

oxygenof.wtmetalof.wt

Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen a fixed. Hence, theresults 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 ofone 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.

Example : 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 simpleratio of 16 : 32 or 1 : 2

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

CONCEPTS RELATED TO DENSITY :It is of two type.1. Absolute density2. Relative densityFor liquid and solids

Absolute density = volumemass

specific gravity = C4atwaterofdensitycetansubstheofdensity

For gases :

Absolute density (mass/volume) = gastheofvolumeMolargastheofmassMolar

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

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

RELATIVE DENSITY :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.MSO.W.M

V.D = 1664

= 4

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

VAPOUR DENSITY :Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperatureand pressure.

Vapour density = 2H

gas

dd

= RT/PMRT/PM

2H

gas

V.D. = 2H

gas

MM

= 2

Mgas

Mgas = 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 = M16

4.225.7

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


PartI 12,13,14. PartI 2. PartI PartII 17,18,19,20,21. PartII 33,35. PartII

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.

Example :Every molecule of ammonia always has formula NH3 irrespective of method of preparation or sources. i.e. 1mole of ammonia always contains 1 mol of N and 3 mole of H. In other wards 17 gm of NH3 always contains14 gm of N and 3 gm of H. Now find out % of each element in the compound.

Mass % of N in NH3 = 100NHofmol1ofMassNHmol1inNofMass

3

3 = 17

gm14 100 = 82.35 %

Mass % of H in NH3 = 100NHofemol1ofMassNHmol1isHofMass

3

3 = 173

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 cancalculate the relative number of atoms of each element in the molecules of the compound. This gives as theempirical formula of the compound. Further if the molecular mass is known then the molecular formula caneasily de 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. Themolecular formula is generally an intergal multiple of the empirical formula.

i.e. molecular formula = empirical formula n

where n = massformualempiricalmassformulamolecular

Example:Acetylene and benzene both have the empirical formula CH. The molecular masses of acetylene and benzeneare 26 and 78 respectively. Deduce their molecular formulae.

Sol. Empirical Formula is CHStep-1The empirical formula of the compound is CH Empirical formula mass

= (1 12) + 1 = 13.Molecular mass = 26Step-2

To calculate the value of n

n = massformulaEmpiricalmassMolecular

= 1326

= 2

Step-3To calculate the molecular formula of the compound.Molecular formula = n (Empirical formula of the compound)

= 2 CH = C2 H2Thus the molecular formula is C2 H2Similarly for benzeneTo calculate the value of n


= 1378

= 6

thus the molecular formula is 6 CH = C6H6

Example :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-1To 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. massof element

12

1

16

Relative no.Percentage

At. massof atoms =

40.687 12

= 3.390

5.0851

= 5.085

54.22816

= 3.389

Simplestatomic ratio

3.3903.389

=1

5.0853.389

=1.5

3.3893.389

=1

Simplest wholeno. atomic ratio

2

3

2

Empirical Formula is C2 H3 O2Step-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-3To calculate the value of n


= 59118

= 2

Step-4To calculate the molecular formula of the salt.

Molecular formula = n (Empirical formula) = 2 C2 H3 O2 = C4 H6 O4Thus the molecular formula is C4 H6 O4.

Example :An oxide of nitrogen gave the following precentage composition :

N = 25.94and O = 74.06

Calculate the empirical formula of the compound.Ans. N2O5

Example :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. C3H3O

Note :

NCERT Reading NCERT Exercise DPP No.Text. 1.9,1.9.1

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


PartI 9,10,11. PartI 3,11,13. PartI PartII 13,14,15,16. PartII 3,4,7,8,9,14,28,29. PartII

LECTURE # 6CHEMICAL REACTION :

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

VI CHEMICAL EQUATION :All chemical reaction are represented by chemical equations by using chemical formule of reactants andproducts. Qualitatively a chemical equation simply describes what the reactants and products are. However,a balanced chemical equation gives us a lot of quantative information mainly the molar ratio in which reac-tants 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 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 equationAccording to the law of conservation of mass, a balanced chemical equation has the same number ofatoms 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) 2Fe2O3(S) (a) balanced equation2 Mg(s) + O2(g) 2MgO(S) (b) balanced equationP4(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 eachside of equations. However equation (c) is not balanced. In this equation. phosphorus atoms are balancedbut 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) + 5O2(g) P4O10(S) balanced equationNow 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, andcarbon dioxide and water are products.

C3H8(g) + O2(g) 3CO2 (g) + H2O (l) unbalanced equationStep 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, eachmolecule of water has two hydrogen atoms , so four molecules of water will be required for eight hydrogenatoms 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 formuls of reactants and products cannot be changed to balance anequation.

INTERPRETATION OF BALANCED CHEMICAL EQUATIONS :(STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS, NCERT, PAGE - 17)

3 . Mole-mole analysis :This analysis is very much important for quantative analysis point of view. Students are advised to

clearly understand this analysis.Now consider again the decomposition of KClO3 .

2KClO3 2KCl + 3O2In very first step of mole-mole analysis you should read the balanced chemical equation like2 moles KClO3 on decomposition gives you 2 moles KCl and 3 moles O2. and from the stoichiometry ofreaction we can write

2KClOofMoles 3 =

2KClofMoles

= 3OofMoles 2

Now for any general balance chemical equation likea A + b B c C + d D

you can write.

areactedAofMole

= breactedBofmoles

= creactedCofmoles

= dreactedDofmoles

Example : 3 moles (367.5 gm) of KClO3 when heated how many moles KCl and O2 is produced.Sol. The reaction is

2KClO3 2KCl + 3O2

2KClOofMoles 3 =

2KClofMoles

Moles of KCl produced = 3

Now,2

KClOofMoles 3 = 3OofMoles 2

mole of O2 produced = 233

= 4.5 moles

Example : 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 moleculesof 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 232 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 usefollowing chart also.

Mass At. wt. / Mol. Wt. MoleMole-mole relationship of equation

Mole

22.4 lt

Volume at STP

mol.

wt./At.

wt.

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 equation2KClO3 2KCl + 3O2Now go with the above chart

367.5 gm KClO3

122.5 gm 3 mole KClO3

Mole of KClO32

=Mole of O2

3

(Mole-mole relationship of equation)

9/2 mole of O2

(a) 144 gm 32 gm(mol. wt.)

(b) 100.8 lt 22.4

lt

(volume at ST

P)

Ex. Iron in the form of fine wire burns in oxygen to form iron (III) oxide4Fe(s) + 3O2(g) 2Fe2O3(s)

How many moles of O2 are needed to produce 5 mol Fe2O3 ?Ans. 7.5 mol O2Ex. Find the number of moles of oxygen obtained from the decomposition of 3 moles of KClO3 :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 NO2Ex. How many grams of Fe2 O3 is formed by heating 18 gm FeO with Oxygen.

4FeO + O2 2Fe2 O3Ans. 20. gmEx. How many litre O2 at N.T.P. required to 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 atN.T.P.

Ans. 14 gm

Ex. When sodium bicarbonate is heated 1.806 x 1024 molecules of water is obtained. Then find the volume ofCO2(g) obtained at STP and amount of NaHCO3 needed for this reaction.

Problem 1.3 (NCERT, Page 18)Calculate the amount of water (g) produced by the combustion of 16 g of methane.

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

Problem 1.4 (NCERT, Page 18)How many moles of methane are required to produce 22 g CO2 (g) after combustion?

SolutionAccording 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) )g(gCO44)g(molCO1

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 toproduce 22 g CO2 (g).

Note :

NCERT Reading NCERT Exercise DPP No.Text. 1.10Page 17 Q.No. 1.3,1.4. 5


PartI 15,16,17. PartI PartI PartII 22,23,24. PartII 16,17,18,26,27. PartII

LECTURE # 7PRINCIPLE OF ATOM CONSERVATION (POAC) :

POAC is based on law of mass conservation if atoms are conserved, moles of atoms shall also beconserved 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 theproblem. 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 productsor moles of K atoms in KClO3 = moles of K atoms in KCl.Now, since 1 molecule of KClO3 contains 1 atom of Kor 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 KClO3and moles of K atoms in KCl = 1 moles of KCl. moles of KClO3 = moles of KCl

or3

3KClOofwt.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 KClO3moles of O in O2 = 2 moles of O2 3 moles of KClO3 = 2 moles of O2

or 3 3

3

KClOof.wt.molKClOof.wt

= 2 .)lt4.22(.volmolardardtansNTPatOof.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 reactionsbut we know about initial reactant and final product accordingly

K2CO3 StepsSeveral 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]21 moles of K2CO3 = 12 moles of K2Zn3 [Fe(CN)6]2( 1 mole of K2CO3 contains 1 moles of C)

32

32

COKof.wt.molCOKof.wt

= 12 productof.wt.molproducttheof.wt

wt. of K2Zn3 [Fe(CN)6]2 = 1386.27

12698

= 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 gNote :


Q.No. 1.7,1.34,1.36. 6


PartI 26,27. PartI PartI PartII 31 to 34. PartII 1,2,5,6,11,13,15,20,22 PartII

LECTURE # 8LIMITING REAGENT :

The reactant which is consumed first and limits the amount of product formed into the reaction, and istherefore 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 balance chemical equation then if number of moles of reactants are not in theratio of stoichiometric coefficient of balalnced chemical equation, then there should be one reactant whichshould 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 + H2Ogiven moles 3 mol 6 molgiven mole ratio 1 : 2Stoichiometric 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

1CONaofMoles 32 =

1oducedPrCOofMole 2

Moles of CO2 produced = 3volume 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. Thereaction is

Na2 CO3 + 2HCl 2 NaCl + CO2 + H2O

Sol. From the reaction : Na2 CO3 + 2HCl 2 NaCl + CO2 + H2Ogiven mole of reactant 6 : 4give molar ratio 3 : 2Stoichiometric coefficient ratio 1 : 2

See here given number of moles of reactants are not in stoichiometric coefficient ratio. Therefore there shouldbe 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. Accordingto 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 limitingreagent for you.

Step :Now once you find limiting reagent then your focus should be on limiting reagentFrom Step & Na2 CO3 HCl

16

= 624

= 2 (division is minimum)

HCl is limiting reagentFrom Step

From 2

HClofMole =

1producedCOofMoles 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 Y4Above 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 reactedX + Y X3 Y4

Ans. B, C, DSol. 3X + 4Y X3 Y4

6 mole 6 mole6 4.5 0 1.5 mole1.5 moleleft 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*) 21 mole of AA3B2C2 is formed

(C*) 1 mole of A3B2 is formed (D*) 21 mole of AA3B2 is left finallyAns. B, C, DSol. 3A + 2B AA3 B2

3 mole 3 mole 1 mole formedA3B2 + 2C AA3 B2 C21 mole 1 mole0.5 mole 0 0.5 mole

Ex. CS2 and Cl2 in the weight ratio 1 : 2 are allowed to react according to equation find the freaction of excessreagent left behind.

CS2 + 3Cl2 CCl4 + S2Cl2

mole 76w

71w2

76w

371w2

L.R. = Cl2 .

Moles of CS2 = 371w2

remaining = 76w

213w2

fraction of Cl2 left = 100x

76w

213w2

76w

= 28.6%.

Ex. 27 gm Al is heated with 48.5 ml of H2SO4 (sp. gr = 2) produces H2 gas. Calculate the volume ofH2 gas at N.T.P. and % of Al reacted with H2SO4Ans. 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 a1 : 1 mol ratio. Find which component is limiting and which component left and what is itspercentage out of total weight taken.Ans. 50%

Ex. Three moles of Phosphorus is reacted with 2 moles of iodine to form P3 according to the reactionP + 2 P3

P3 formed in the above reaction is further reacted with 27 g of water. According to the reaction.P3 + 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 + 2Ans. 3/8 NA.

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. Identifythe limiting reagent in the production of NH3 in this situation. Take reaction N2 + 3H2 2NH3.

SolutionA 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 22

2 Nkg1Ng1000N

2

2

Ng0.28Hmol1

= 17.86 102 mol

moles of H2 = 10.00 kg H2 22

Hkg1Hg1000

2

2

Hg016.2Hmol1

= 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 N2, the moles of H2 (g) required would be

17.86 102 mol N2 )g(molN1)g(Hmol3

2

2= 5.36 103 mol H2

But we have only 4.96103 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 molSince 3 mol H2 (g) gives 2 mol NH3 (g)

4.96 103 mol H2 (g) )g(Hmol3)g(HNmol2

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) )g(molNH1)g(gNH0.17

3

3. = 3.30 103 17 g NH3 (g)

= 56.1 103 g NH3 = 56.1 kg NH3Note :

NCERT Reading NCERT Exercise DPP No.Text. 1.10.1.

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


PartI 18 to 25. PartI 7. PartI PartII 25 to 30. PartII 16,23,25,26. PartII

LECTURE # 9SOLUTIONS :

A mixture of two or more substances can be a solution. We can also say that a solution is a homogeneousmixture 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 expressed the concentration of a solution. These are :1. strength of solution

2. Molarity (M)3. Molality (m)4. Mole fraction (x)5. % calculation6. Normality (N)7. ppm

Remember that all of these concentration terms are related to one another. By knowing one concentrationterm 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 meanings65% 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% H2SO4, 65% HNO3 etc, if anything is not specified than percentage bymass/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 = 1931000193

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 themolarity of the solution.

i.e., Molarity of solution = litre in solution of volumemoles 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 = Mw

V ml water have Mw

mole of solute

1000 ml water have inmlVM

1000w

Molarity (M) = inmlV)solute of wt .Mol(

1000w

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 250mL of the solution.

SolutionSince molarity (M)

= eslitr in solution of Volumesolute of moles of No.

= L 0.250NaOH of mass NaOH/Molar of Mass

= L250.0g 40 / g 4

L250.0mol1.0

= 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)solute of .wt .Mol(soluteofmass

= (Molarity of solution Vinml)

Molarity is an unit that depends upon temperature .it varies inversely with temperature .mathematically : molarity decreases as temperature increases.

Molarity etemperatur1

volume1

Molarity of solution may also given as :

ml in solution of volume Totalsolute of millimole of Number

(i) If a particulars solution having volume V1 and molarity = M1 is diluted to V2 mL theM1V1 = M2V2M2 : Resultant molarity

(ii) If a solution having volume V1 and molarity M1 is mixed with another solution of same solute havingvolume V2 mL & molarity M2then M1V1 + M2V2 = MR (V1 + V2)MR = Resultant molarity

= 21

2211

VVVMVM

Example149 gm of potassium chloride (KCl) is dissolved in 10 Lt of an aqueous solution. Determine the molarity of thesolution (K = 39, Cl = 35.5)

SolutionMolecular mass of KCl = 39 + 35.5 = 74.5 gm

Moles of KCl = gm 5.74gm 149

= 2

Molarity of the solution = 102

= 0.2 M

Question117 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 H2SO4Sol. For HCl

M1 = 1 MV1 = 200 mL

For H2SO4M2 = 1V2 = 200 mL

nHCl = MHCl + VHClHCl H+ + Cl

Hn = 0.2 (from HCl)

424242 SOHSOHSOH VMn = 1 0.2 = 0.2 mole

H2SO4 2H+ + SO42

Hn = 0.2 (from H2SO4)

Total H+ = Hn (from HCl) + Hn (from H2SO4)= 0.2 + 0.4 = 0.6

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

= 4.06.0

Vn

solution

H

= 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

= 25070

2505218

= 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 LitreMass = 1000 gm

mole = 181000

Molarity of water = 181000

= 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 CaCl2 and CO2 according to reactionCaCO3(s) + 2HCl(aq) CaCl2 + CO2 + H2O

How much mass of CaCO3 is required to react completly with 100 ml of 0.5 m HCl

Sol. millimole of HCl100 0.5 = 502 mole of HCl reacts 1 moles CaCO3

1 mole of HCl reacts 21

50 mmole of HCl reacts 21

50 = 25 mmole of CaCO3

mole of CaCO3 = 100025

mass of CaCO3 = 100025

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 = 1066.10

1000 = 100 m mole

HCl = 100 0.5 = 50 milli mole

limiting reagent HCl 2 HCl reacts = 2 NaCl50 mmole HCl reacts = 50 mmole of Na2CO3remaining Na2CO3 = 100 25 = 75 mmole

(a) moles of NaCl = 100050

= 0.05

(c) volume of CO2

= 4.22100025

= 0.556 L

[Na+] = 100200

10015050

= 2 M

[Cl] = 10050

= 0.5 M

(CO32] = 100

75= 0.75 M

Note :

NCERT Reading NCERT Exercise DPP No.Text. 1.10.2.

Page 19 8


PartI 28 to 41 PartI PartI PartII 35 to 41. PartII 19. PartII

LECTURE # 103. MOLALITY (M) :

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

1000gram in solvent of masssolute 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 moleof the solute are dissolved in x gm of the solvent. Hence

Molality = 1000xm

Y

Example

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)

Solution.Mass of urea = 5 gm Molecular mass of urea = 60

Number of moles of urea = 605

= 0.083

Mass of solvent = (255 5) = 250 gm

Molality of the solution = 1000gram in solvent of Masssolute of moles of Number

= 1000250083.0

= 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.Solution

M = 3 mol L1Mass of NaCl in 1 L solution = 3 58.5 = 175.5 gMass 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 = kg in solvent of Masssolute of moles of No.

= kg 1.0745mol 3

= 2.79 m

Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution ofknown 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 numberof moles present in the solution is known as the mole fraction of substances concerned.

Let number of moles of solute in solution = nNumber of moles of solvent in solution = N

Mole fraction of solution (x1) = Nnn

Mole fraction of solvent (x2) = NnN

also x1 + x2 = 1Note : mole fraction is a pure number its also independent of temperature changes.

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 = 100gm in solution of massgm in solute of mass

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

i.e., % w/v = 100ml in solution of massgm 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 = 100solution of Volumeml in solute of Volume

Example0.5 g of a substance is dissolved in 25 g of a solvent. Calculate the percentage amount of the substance inthe solution.

Solution.Mass of substance = 0.5 gMass of solvent = 25 g

percentage of the substance (w/w) = 100255.05.0

= 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 thesolute.

Solution

Mass per cent of A = solutionofMassAofMass

100 = gofwater18gofA2g2

100

100g20

g2 = 10 %

Example20 cm3 of an alcohol is dissolved in80 cm3 of water. Calculate the percentage of alcohol in solution.

SolutionVolume of alcohol = 20 cm3Volume of water = 80 cm3

percentage of alcohol = 100802020

= 20.

Interconversion of Concentration units :Representing solvent and solute in a binary solution by subscripts 1 and 2 respectively, the variousconversion expressions are as follows :

DERIVE THE FOLLOWING CONVERSION :

1. Mole fraction of solute into molarity of solution M = 2211

2

xMMx1000x

Sol. Mole fraction into molarity Mmole fraction of solvent and solute are X1 and X2so X1 + X2 = 1supposs total mole of solution is = 1mole of solute and solute and solvent is X2, X1weight of solute = X2M2 , weight of solvent = X1M1total wt of solution = X1M1 + X2M2

volume of solution = 2211 MXMX ml

volume in L = 1000NXNX 2211

molarity (M) = 2211

2

NXNX1000X

2. Molarity into mole fraction x2 = 2

1

MM10001000MM

Sol. Molarity into mole fractionmolarity (M) = moles solute in 1000 ml of solutionso moles of solution = Mmass of solution = x 1000wt. of solute = MM2wt. of solvent = 1000 MM2

moles of solvent = 1

2

MMM1000

3. Mole fraction into molality m = 11

2

Mx1000x

Sol. Mole fraction into molaritymole fraction of solute X2 and solvent X1mole is n2 & n1

molality = 11

2

Mnn

x 1000

1

2

1

2

xx

nn

= 11

2

Mxx

x 1000

4. Molality into mole fraction x2 = 11

mM1000mM

Sol. Moalrity into mole fractionmolality = moles of solute in 1000 gm of solute = m

mole of solvent = 1M

1000

mole fraction X2 = m

M1000

m

1

= 1mM1000

mM

5. Molality into molarity M = 2mM1000

1000m

Sol. Molality in molaritymolality = moles of solute in 1000 gm of solventmole of solute = mwt. of solute = mM2wt. of solution = 1000 + mM2

volume of solution =

2mM1000

volume in (L) = 1000mM1000 2

molarity = 2mM1000

1000m

6. Molarity into Molality m = 2MM1000

1000M

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 solutionmoles of solute = Mwt. of solute = MM2wt. of solution = 1000mass of solvent = 1000 MM2

molality = solventofwtsoluteofmoles

x 1000

m = 2MM1000

1000M

Note :


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


PartI 42 to 50. PartI PartI PartII 42 to 45. PartII 21,24,30,31,32. PartII

LECTURE # 11Mix 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

vn

= 2

n solute = 2 mole (CH3COOH)

soluteofmassmolarsoluteofmass

= 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 = 181080

= 60 mole

(H2O)

(1) Molality = 08.12

wn

)kg(solvent

solute = 1.85 m Ans.

(2) COOHCH3X = 6022

nnn

OHCCOHCH

CCOHCH

23

3

= 0.0322

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

100 = 1200120

100 = 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 = 101

nnn

OHecosGlu

ecosGlu

2

Let 1 mole of glucose is present is solutionn Glucose = 1m Glucose = 1 180 = 180 gm

OH2n + nGlucose = 1.01

1.0n ecosGlu = 10

So OH2n = 9

OH

2

2M

OHofmass = 9.

OH2M = 9 10 = 162 gm = 0.162 kgmass of solution = mass of solute + mass of solvent

= mas of Glucose + mass of H2O

= 180 + 162 = 342 gm

volume of solution = solutionofDensitysolutionofmass

= mL/g1.1g342

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

(1) molarity = 311.01

vn

solution

ecosGlu = 3.215 M

(2) molality = 162.01

)kg(wn

OH

ecosGlu

2

= 6.17 M

(3) % (w/w) = 342180100

)gm(solutionof.wt)gm(soluteof.wt

100

= 52.63%

(4) % (w/v) = 311180100

)mL(solutionof.volume)gm(soluteof.wt

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/g1.1gm100

Densitymsolution = 91 mL

= 0.91 LMurea = 10 gm

nurea = 61

6010

MureaofMass

urea

msolution = msolute + msolvent= 100

msolvent = 1890

MMOHn 2 = 5 mole

(1) % (w/v) = )mL(solutionof.volumeureaof.wt

100 = 1009110

= 10.99%

(2) molarity = 91.06/1

vn

solution

urea = 1.83 M

(3) molarity = 9.06/1

wn

solution

urea = 1.85 M

(4) mole fraction of urea

Xurea = OHureaurea

2nn

n = 6/31

6/156/1

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

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

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) = 100xa.....xaxa nn2211

Ex. Naturally occuring chlorine is 75% Cl35 which has an atomic mass of 35 amu and 25% Cl37 which has a massof 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 = 100massatomicitsxisotopeIof%massatomsitsxisotopeof%

= 10037x2535x75

= 35.5 amuNote : (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....nnMn......MnMn

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 and32 respectively.

Sol. Mavg =

nj

1jj

nj

1jjj

n

Mn

Here

nj

1jjn = 100

Mavg = 1001x643x4428x8032x16

= 100641322240512

= 1002948

= 29.48 Ans.

Note :


Q.No. 1.9,1.32. 10


PartI 4,5,10. PartI PartI PartII 34. PartII 12. PartII

Documents

Lec. Notes Mole Concept