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Lecture-1

1.1 Chemical Arithmetic

Chemistry is the branch of physical science which deals with the study of matter, their physical and chemical properties, and their chemical composition.

The main branches of chemis try are :(i) Organic Chemistry

(ii) Inorganic Chemistry

(iii) Physical Chemistry

(iv) Analytical Chemistry

(i) Organic Chemis try: It is concerned with the study of

com pounds of carbon except carbonate, bicarbonate, cyanides,isocyanides, carbides and oxides of carbon.

(ii) Inorganic Chemis try: It deals with the study of all known elements and their com pounds except organic com pounds.

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(iii) Physical Chemistry: It is concerned with the physical properties, the laws of the chemical combination and theories which are governing the reaction.

(iv) Analytical Chemis try: It deals with various methods of analysis of chemical substances both quali ta tive andquan ti ta tive.

Some of the specialised branches are:

(i) Bio-Chemistry

(ii) Medicinal Chemistry

(iii) Soil and Agriculture

(iv) Industrial Chemistry

(v) Nuclear Chemistry

(vi) Polymer Chemistry

1.2 Matter and Energy

Mat ter is any thing that has mass and occu pies space. All the

bodies in the uni verse con form to this defini tion. Mass is thequan tity of mat ter in a par ticular sam ple of mat ter whether the term weight should not be used in place of mass.

Energy is de fined as the ca pacity of doing work. Any thing which has the ca pacity to push mat ter from one place to an other posses energy. There are various forms of energy such as heat,light, etc. Energy is nei ther be created nor be destroyed.

Classification of Matter

(i) Physical Classification: Matter can exist in any one of three forms Solid, liquid and Gas. In the solid state, substances arerigid. They have definite shape and fixed volume.In liquid4

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state, substances have no definite shape but possess fixed volume. In a gaseous state, substances have no definite shapeand volume. Depending on temperature and pressure, asubstance can exist in any one of the three forms of matter.

(ii) Chemical Classi fica tion: Matter exists in nature in the form of chemical substances. A pure substance is defined as a variety of matter, which have same compositions and properties,i.e., a material containing only one substance.

Metals, Non-Metals and Metalloids

All the elements may be classified into two groups i.e., metalsand non-metals. The division is based on both physical andchemical properties.

Metals are regarded as those elements which possess the following properties:

(i) They are generally solids at ordinary conditions,mercury is an exception which is in liquid state.

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MATTER

ChemicalClassification

PhysicalClassification

Solids Liquids Gases

Heterogeneous

Homogeneous

PureSubstances Mixtures

HeterogeneousHomogeneousCompoundsElements

(All Homogeneous)(Solutions)

OrganicCompounds

InorganicCompounds

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decomposed nor built from simpler substances by ordinary physical or chemical methods.

Com pounds are classified into two types:

(i) Organic Com pounds: The com pounds ob tained from livingsources are termed organic com pounds. The term organic isnow ap plied to hydrocarbons.

(ii) Inorganic Com pounds: The com pounds ob tained fromnon-living sources such as rocks and minerals are termedinorganic com pounds.

3. Mixture: A mixture may be defined as: A combination of two or more elements or compounds in any proportion so that the components do not lose their identity.

Mixtures are further classified into following types:

(i) Homogeneous mix tures: These have the same com posi tions throughout the sam ple. The com ponents of a mix ture can beseen even under a power ful microscope. Homogeneous mix turesare also called solu tion.

(ii) Hetrogeneous mix tures: They consist of two or more parts(called phases) which have dif fer ent com posi tions.

Dif ference be tween a com pound and a mix ture:

Compound Mixture

1. The constituents of a

compound are always present

in a fixed ratio by mass.

1. The constituents of a mixture

may be present in any ratio

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2. Compound is always

homogeneous in nature.

2. Mixture may or may not be

homogeneous in nature.

3. The properties of a compound

are different from those of itsconstituent elements.

3. The properties of a mixture are

midway between those of itsconstituents.

4. The constituents of a

compound cannot be easily

separated by simple

mechanical means. Energy

in the form of heat or light is

often required.

4. The constituents of a mixture

can be easily separated by

simple mechanical means.

5. Compound are formed as a

result of a chemical change.

5. Mixtures are formed as a

result of a physical change.

6. Formation of a compound is

always accompanied by

absorption or evolution of heat,

light or electrical energy.

6. When a mixture is formed, no

heat, light or electrical energy

is absorbed or evolved.

7. Chemical Compounds posses

sharp melting and boiling

points.

7. The melting and boiling

points of mixtures are

usually no sharp.

1.3 Measurement in Chemistry

‘A unit is defined as the standard of reference chosen in order to measure a definite physical quantity in chemistry.

Fundamental and Derived Units:The units of physical quantities depends on three basic units,i.e., units of mass, length and time. Since these are8

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(ii) Distance between earth and sun, i.e., 93 million miles.

(iii) Average between earth and sun, i.e., 5 feet 6 inches.

Solution: (i) The SI units for speed are m s − 1

1 mile = 1.60 km = 1.60 × 103 m

∴ Conversion factor = × 1 60 10 1

3. mmile

1 hr = 60 × 60 s = 3.6 × 103

∴ Conversion factor = ×3 6 10

1

3. s

hr

Now, speed= 120 miles

hr

= 120 miles

hr 1.60 10 m

1mile

3×

×

××

1hr

3.6 10 s3

= 53 3. m s− 1.

(ii) The Si units for distance is metre (m)

1 mile = 1.60 km = 1.60 × 103 m

Conversion factor = × 1.60 10 m

1mile

3

Distance = × 93 106 miles

= 93 10 miles 1.60 10 m

1miles

6 3× × ×

= 1.49 10 m. 11×10

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Sol.

This indicates that the compound contains oxygen alsoand its percentage is (100 – 92) = 8.

Ex-3:

Gun powder is a mixture of sulphur, charcoal and

potassium nitrate (KNO3). How would you separate itinto its constituents?

Sol:

Sulphur is soluble in CS 2 and is insoluble in water.KNO3 is soluble in water. Charcoal is insoluble in both water as well as CS 2.

Ex-4:

Vanadium metal is added to steel to impart strength.The density of Vanadium is 5.96 g/cm3. Express in

S.I. unit (kg / m3).Sol: d = 5 96 5 . /g cm m of vanadium = 5 96 10 3. × −

5 96. g cm−3 = ×

=−

−− 5 96 10

10 5960

3

6 33. kg

mm .

Vol of vanadium metal = − 10 6 3m .

And we know, d m v

=

Significant figures

Number of significant figures in a physical quantity dependsupon the least count of the instrument used for its

measurement.1. Common rules for counting significant figures:

Following are some of the common rules for countingsignificant figures in a given expression:

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Rule 1. All non zero digits are significant.Example: x = 1234 has four significant figures.

Again, x = 189 has only three significant figures.

Rule 2. All non zero digits occurring be tween two non zero digits are significant.

Example: x = 1007 has four signif icant figures.

Again, x = 1.0809 has five significant figures.

Rule 3. In a number, less than one, all zero's to the rightof decimal point and to the left of a non zero digit are not

significant.

Example: x = 0.0084 has only two significant fig ures.

Again, x = 1.0084 has five significant figures.

Rule 4. All zeroes on the right of the last non zero digit in the decimal part are significant.

Example: x = 0.00800 has three significant figures 8,0,0. The zeros be fore 8 are not significant.

Rule 5. All zeroes on the right of the non zero digit mayor may not be significant.

Example: We can write 20,000 in scien tific no ta tion as -

2 × 104 Having 1 significant figure

2 .0 × 104 Having 2 significant figure

2.00 × 104 Having 3 significant figure

Rule 6. All zeroes on the right of the last non zero digitbecome significant, they come from a measurablequan tity.

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Example: Sup pose dis tance be tween two sta tions ismeasured to be 3050m. It has four significant fig ures. the samedis tance can be ex pressed as 3.050 km or 3.050 ×10 5 cm.

2. Rounding off: While round ing off measurements, weuse the fol lowing rules by con ven tion :

Rule 1. If the digit to be dropped is less than 5, then the preceeding digit is left unchanged

Example: x = 7.82 is rounded off to 7.8.

Again, x = 3.94 is rounded off to 3.9.

Rule 2. If the digit to be dropped is more than 5, then the preceeding digit is raised by one.



Rule 3. If the digit to be dropped is 5 followed by digitsother than zero, then the preceeding digit is raised by one.

Example: x = 16.351 is rounded off to 16.4. Again, x = 6.758 is rounded off to 6.8.

Rule 4. If the digit to be dropped is 5 followed by zeroes, then the preceeding digit is left unchanged, if it is even.

Example: x = 3.250 is Rounded off to 3.2

Again x = 12.650 is rounded off to 12 .6

Rule 5. If the digit to be dropped is 5 or 5 followed by zero, then the preceeding digit is raised by one.



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Not to be Copied by Others

Y our S ecrets M y S e c r e t s

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QuestionsQ.1: Express the following numbers upto three significant figures.

(i) 306.35 (ii) 0.0038816

(iii) 1.78975×104 (iv) 0.25400

(v) 2.65986×103

Ans: (i) 306.35 = 306

(ii) 0.0038816 = 3.88×10 −3

(iii) 1.78975×104 = 1.79×104[rounded off according to rule vi (b)]

(iv) 0.25400 = 2.54×10 −1

(v) 2.65986×103 = 2.66×103

[rounded off according to rule vi (b)].

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Q.2: The mass of a sample of iron metal is 5.932 g. If the density of iron is 7.8

g/cm3 , what is its volume?

Ans: Volume = = − Mass

Density g cm

593278 3

..

The smallest number of significant figures in this calculation is two. Hence theresult has been rounded off to two significant figures.

Q.3: "The star of India" sapphire weighs 563 carats. If one carat is equal to

200 mg, what is the weight of the gemstone in grams?

Ans: Weight one carat = 200 mg

∴ Weight of 563 carats = × =2001000

563 112 6. . g

Q.4 Express the following in scientific notation:

(i) 0.0048 (ii) 234,000 (iii) 8008

(iv) 500.0 (v) 6.0012

Ans. (i) 4.8×10−3 (ii) 2.34×105 (iii) 8.008×103

(iv) 5.000×102 (v) 6.0012×100

Q.5 How many significant figures are present in the following ?

(i) 0.0025 (ii) 208 (iii) 5005(iv) 126,000 (v) 500.0 (vi) 2.0034

Ans. (i) 2 (ii) 3 (iii) 4

(iv) 3 (v) 4 (vi) 5

Q.6 Round up the following upto three significant figures :

(i) 34.216 (ii) 10.4107

(iii) 0.04597 (iv) 2808

Ans. (i) 34.2 (ii) 10.4

(iii) 0.0460 (iv) 2810

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Q.7 How many significant figures should be present in the answer of the

following calculation ?

(i)0 02856 29815 0 112

0 5785

. . .

.

× ×(ii) 5×5.364

(iii) 0.0125 + 0.7864 + 0.0215

Ans.(i) The least precise term has three significant figures (i.e., in 0.112). Hence,

the answer should have three significant figures.

(ii) Leaving the exact number (5), the second term has four significantfigures. Hence, the answer should have four significant figures.

(iii) In the given addition, the least number of decimal places in the term is4. Hence, the answer should have four significant.

Q.8 Why are the atomic masses of most of the elements fractional ?

Ans. Atomic masses of most of the elements are fractional because most of elementsoccur in nature as a constant mixture of isotopes. The atomic masses of theisotropes actually the average relative masses of the isotopes depending on theirabundance.

Q.9 Express the following in the scientific notation :

(i) 0.0048 (ii) 234,000 (iii) 8008

(iv) 500.0 (v) 6.0012

Ans. (i) 4.8 × 10 −3 (ii) 2.34 × 105 (iii) 8.008 × 103

(iv) 5.00 × 102 (v) 6.0012 × 100

Q.10 How many significant figures are present in the following ?

(i) 0.0025 (ii) 208 (iii) 5005

(iv) 126,000 (v) 500.0 (vi) 2.0034

Ans. (i) 2 (ii) 3 (iii) 4

(iv) 6 (v) 4 (vi) 5

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Q.11 Round up the following upto three significant figures :

(i) 34.216 (ii) 10.4107

(iii) 0.04597 (iv) 2808

Ans. (i) 34.2 (ii) 10.4(iii) 0.04597 (iv) 281

Q.12 How many significant figures should be present in the answer of the

following calculations

(i)0 02856 29815 0 112

0 5785

. . .

.

× ×

(ii) 5×5.364

(iii) 0.0125 + 0.7864 + 0.0215

Ans.(i) The least precise figure (0.0112) has 3 significant figures. Therefore, the

answer should havethree significant figures.

(ii) The second figure (5.364) has 4 significant figures. Therefore, the answershould be reported upto four significant figures. The exact figure (5) isnot considered in this case.

(iii) In this case, the least precise figures (0.0125 and 0.0215) have 3significant figures.

Therefore, the answer should be reported upto three significant figures.

Q.13 When do zeros present in a number become insignificant ?

Ans. The zeros written to the left of the first non-zero digit in a number areinsignificant. For example, in the number 0.014, both the zeros areinsignificant.

Q.14 Is velocity a basic or derived quantity according to SI system ?

Ans. Velocity is a derived quantity and it depends upon two basic quantities i.e.,distance and time. The SI units of velocity are :

Velocity = Distance / Time = ms–1

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Q.15 Why is air not always regarded as homogeneous mixture ?

Ans. Air which contains certain suspended particles such as dust particles is ahetrogeneous mixture and not a homogeneous mixture.

Q.16 Statements given below pertain either to an element or mixture or

compound. Identify the statement which corresponds to

(a) an element

(b) mixture

(c) compound.

(i) The properties of the reactants are entirely different from the

properties of the products of their chemical reactions.

(ii) The constituents retain their individual chemical identity.

(iii) The pure substance which cannot be subdivided into two or more

substances by any chemical means.

Ans. (i) The statement corresponds to a compound.

(ii) The statement corresponds to a mixture.

(iii) The statement corresponds to an element light.

Q.17 What do you understand by limnochemistry ?

Ans. Chemistry of water reservoirs like rivers, lakes, etc. is called Limnochemistry. Inthis branch, we study the variation of pH with seasons and temperature.Moreover, we also study reactions occurring in aqueous medium of waterreservoir.

Q.18 What do you understand by Phytochemistry ?

Ans. Photochemistry is the branch of chemistry which involves the study of chemicalcomposition of various parts of plants.

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Lecture-2

1.4 Laws of Chemical Combinations

1. Law of Conservation of Mass

This law was stated by the french chemist ‘‘antoine Laurentlavoisier (1774)". This law states that:

During any physical or chemical changes, the total mass of the products remains equal to the total mass of the reactants.

Lavoisier showed that when mercuric oxide was heated it

produced free mercury and oxygen. The sum of masses of mercury and oxygen was found to be equal to the massmercuric oxide

Mercuric oxide Mercury Oxygen g

Heat

g g 100 92 6 7 4

→ +. .

Law of conservation of mass is also known as law of indestructibility of matter.

Example: Is law of conser va tion of mass al ways valid ?

Solution No, it is not valid for nuclear reac tions. In thesereac tions, a cer tain amount of mass gets con verted into energyknown as nuclear energy. There fore, mass is not conservedand it does not remain constant.

2. Law of Constant composition or Definite Proportions

This law deals with the composition of chemical compounds. It was discovered by the french chemist, Joseph Proust (1799).This law states that:

A chemical com pound al ways con tains same elementscombined together in same pro por tion by mass.

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4. Law of Reciprocal Proportions

The ratio of the masses of two elements A and B which combineseparately with a fixed mass of the third element C is either

same or some simple multiple of the ratio of the masses in which A and B combine directly with each other.

The elements H and O combine separately with the thirdelement S to form H 2 S and SO 2 and they combine directly with each other to form H 2O as shown in figure.

The masses of H and O which combine with the fixed mass of S, viz, 32 parts are 2 and 32, i.e., they are in the ratio 2 : 32 or 1 :

16.When H and O combine directly to form H 2O, the ratio of their combining masses is 2 : 16 or 1 : 8.

The two ratios are related to each other as 1

16 18

1 2: := i.e., they

are simple multiple of each other.

Example: Ammonia con tains 82.35% of ni trogen and 17.65%

of hydrogen. Wa ter con tains 88.90% of oxygen and 11.10% of

hydrogen.

Show that these data illustrate the law of reciprocal proportions.

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Solution:

In NH 3 , 17.65 g of H combine with N=82.35 g

Therefore, 1 g of H combine with

N g g= =

82 35 17 65 4 67

.

. .

In H 2O, 11.10 g of H combine with O=88.90 g

Therefore, 1 g of H combine with

O g g= =88 90 11 10

8 01..

.

Therefore, ratio of the masses of N and O which combine with fixed mass (=1 g) of H = 4.67 : 8.01 = 1 : 1.72.

In N 2O3 ratio of masses of N and O which combine with each

other = 36.85 : 63.15 = 1 : 1.71.

Thus, the two ratios are the same. Hence, it illustrates the law of

reciprocal proportions.Example: Carbondioxide con tains 27.27% of car bon,

carbondisulphide con tains 15.79% of carbon and

sulphurdioxide con tains 50% of sul phur. Show that the data is

in agree ment with law of re ciprocal pro por tions.

Solution:

In CS 2

C : S mass ratio is 15.79 : 84.21

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15.79 parts of carbon combine with sulphur = 84.21

∴ 27.27 parts will combine with S

= × =84 21

15 79 27 27 145 434

.

.. .

hence, ratio of S : O is 145.434 : 72.73 i.e., 2 : 1

In SO 2 the ratio of S : O is 1 : 1

Since, the ratio of S : O is a simple whole number ratio, therefore law of reciprocal proportions is proved.

5. Gay Lussac’s Law of Gaseous Volumes

When gases react together, they always bear a simple ratio to

one another and to the volumes of the products, if these product

are also gases, provided all measurement of volumes are done

under similar conditions of temperature and pressure.

Consider, for illustration, the following example:

Combination between hydrogen and chlorine. One volume of hydrogen and one volume of chlorine always combine to form

two volumes of hydrochloric acid as:

25

C 27.27 15.79

84.21 72.73 50 50CS 2

CS 2 CO 2

S O

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Ex.

Which law co-relates the mass and volume of a gas ?

Sol.

It is the Avogardo’s Law and states that equal volumes of all gases under similar conditions of temperature and

pressure contain equal number of moles (or molecules).

Add to your Knowledge

Avogadro’s law is also helpful in developing the relationshipbetween

(a) Molecular mass and vapour density

(b) Mass and volume of gas

(a) Rela tionship be tween molecular mass and vapour density

By definition

Vapour density (V.D.)

= Density of gas Density of hydrogen

=Mass of some volume of gas at S.T.P.

Mass of same volume of H 2 at S.T.P.

=Mass of N molecules of gas

Mass of N molecules of H 2

=Mass of 1 molecule of gas

Mass of 1 H molecule 2

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=×

Mass of 1 molecule of gas

2 Mass of H atom

= × 1

2 Molecular mass of gas

Thus, Mol.mass=2 × V.D.

(b) Relationship be tween mass and volume of gas

Mol. mass = × 2 V.D.

= × 2 mass of 1 L of gas at S.T.P.(1atm)

Mass of 1 L of H at S.T.P. 2

= × 2 Mass of 1 L of gas at S.T.P.

0.089 g

= × 22 4. MAss of 1 L of gas at S.T.P.

= Mass of 22.4 L of gas at S.T.P.

Thus, Mass of 22.4 L of gas at S.T.P.

= Molar mass (in g mol− 1)

1.5 Dalton’s Atomic Theory

The main points of this theory are as follows:

1. Matter is made up of extremely small invisible particlescalled atoms.

2. Atoms of the same element are identical in all respects,i.e., size, shape, and mass.

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3. Atoms of different element have different masses, sizesand also possess different chemical properties

4. Atoms of the same or different elements combine together

to form compound atoms (now called as molecules). 5. When atoms combine with one another to form compound

atoms (molecules), they are in simple whole number ratios, such as 1 : 1, 2 : 1, 2 : 3 and so on.

6. Atoms of two elements may combine in different ratios to form more than one compound. For example, sulphur combines with oxygen to form sulphur dioxide and

sulphur trioxide, the combining ratios being 1 : 2 and 1 :3 respectively.

7. An atom is the smallest particle that takes part in achemical reaction. In other words, whole atoms, rather than fractions of atoms take part in a chemical reaction.

Ex plana tion of the Laws of Chemical Combina tion by Dal ton’s

Atomic Theory: 1. Law of Conservation of Mass. Mat ter is made up of atoms

(pos tulate 1) which can nei ther be created nor destroyed(pos tulate 8). Hence, mat ter can nei ther be created nor destroyed.

2. Law of Constant composition. It follows directly from postulate 5.

3. Law of Multiple proportions. As follows directly from postulate 6.

4. Law of Reciprocal proportions. As atoms combine with eachother in simple multiple of each other.

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Limitations of Dalton’s Atomic Theory

The main drawbacks of Dalton’s Atomic Theory are :(i) It could explain the laws of chemical combination by mass

but failed to explain the law of gaseous volumes.(ii) It could not explain why atoms of different elements havedifferent masses, sizes, valencies etc.

(iii) It could not explain, why do atoms of the same or differentelements combine to form molecules?

(iv) It could not explain, what is the nature of binding forcebetween atoms and molecules which accounts for theexistence of matter in three states, i.e., solids, liquids and

gases.(v) It makes no distinction between the ultimate particles of

an element or a compound.

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QuestionsQ.1 Pressure is determined as force per unit area of the surface. The SI unit

of pressure, pascal is as shown below : 1 Pa = 1 Nm−2

If the mass of air at sea level is 1034 g, calculate the pressure in pascal.

Ans. Pressure = = ×Force

AreaMass Accleration of gravity

AreaMass of air = 1034 g = 1.034 kg

Acceleration of gravity = 9.806 ms−2

Area = 1 cm2 = 10−4 m2

Q.2 The following data is obtained when dinitrogen and dioxygen react to

gether to form different compounds :

Mass of dinitrogen 14 g 14 g 28 g 28 g

Mass of dioxygen 16 g 32 g 32 g 80 g

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Ans. By keeping 14 g as the fixed mass of dinitrogen (N2), the ratios by mass of dioxygen (O2) combining with 14 g of dinitrogen are : 16 : 32 : 16 : 40 or 2 : 4: 2 : 5. Since this ratio is simple whole number, the data obeys the Law of Multiple proportions.

Q.3 If the speed of light is 3.0×108

ms−1

, calculate the distance covered by light in 2.00 ns.

Ans. Distance travelled by light in 1 s = 3.0 × 108 m

Distance travelled by light in 2.0 × 10 −9s

= (3.0 × 108 m)×(2.0 × 10 −9s) / (1s)

= 0.600 m.

Q.4 Convert the following into basic units

(i) 28.7 pm (ii) 15.15 µs (iii) 25365 mg.

Ans.(i) 1 pm = 10−12 m

28.7 pm = (10 −12 m)× (28.7 pm) / (1.0 pm)

= 2.87×10−11 m

(ii) 1µ s = 10−6 s

15.15 µ s = (15.15)× (10−6 s) / (1.0 µs)

= 1.515×10−5 s

(iii) 1 mg = 10−6 kg

25365 mg = (25365 mg)× (10−6 kg) / (1mg)

= 2.5365×10−2 kg

Q.5 We breathe in fresh air in the morning walk ? Is it pure as well ?

Ans. Fresh air may be regarded as pure by an ordinary person but not by a chemist.

Actually fresh air is the mixture of number of gases like oxygen, nitrogen, noblegases, carbon dioxide, water vapours etc. It is therefore, not pure from the angleof a chemist.

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Lecture-3

1.6 Atomic and Molecular Mass

The atomic mass of an element can be defined as the number whichindicates how many times the mass of one atom of the element isheavier in comparison to the mass of one atom of hydrogen.

Atomic mass of an element also can be defined as the number which indicates how many times the mass of one atom of theelement is heavier in comparison to 1/12 th part of the mass of one atom of carbon-12( 12C).

A = Atomic mass of an element

=Massofoneatomof theelement

thpart of themassof onea 1

12 tomof carbon − 12

AMassof oneatomof theelements

Massof oneatomof carbon =

− 12 12×

Ex

Calculate the molar mass of glucose (C H O6 12 6) and thenumber of atoms of each kind in it.

Sol:

Molecular mass of glucose (C H O6 12 6)

= × + × + ×6 12 011 12 1 008 6 16 00( . ) ( . ) ( . )u u u

= 180 162. u

Calculation of number of atoms of each kind 1 mole of glucose (C H O6 12 6)

= 6 moles of carbon + 12 moles of hydrogen +6 molesof oxygen

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Hence, Atoms of carbon = × ×6 6 02 10 23.

= ×36.12 10 23

Atoms of hydrogen = × × 12 6 02 10 23.

= × 72 24 10 23. Atoms of oxygen oxygen = × ×6 6 02 10 23.

= ×36 12 10 23.

Atomic Mass Unit:The quantity of 1/12th mass of an atom of carbon-12 ( 12C) is

known as the atomic mass unit and is abbreviated as amu. Theactual mass of one atom of carbon-12 is 1.9924×10 − 23 g or 1.9924×10− 26 kg.

i) Atom: The smallest particle of an element that can take part in chemical change but generally cannot exist freelyas such.

ii) Molecule: The Smallest particle of a substance (element or

compound) which has free or independent existence and possess all characteristics properties of the substance.

iii) Molecular Mass: Molecular mass of a substance may bedefined as:

The average relative mass of its molecule as compared to the mass of an atom of carbon(C 12) taken as 12.

Ex-1: Atomic mass of Ca is 40 gm/mole, means 1 mole of Ca & having mass = 40 g.

Sol: In 1 mole number of Ca = ×6 023 10 23. .

So, 6 023 10 23. × atoms of Ca having mass = 40 g.

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1 atom of Ca having mass =×

40

6 023 10 23g

.

=×

40 1

6 023 10 23.g

= × × −40 1 67 10 24. g

= 40 amu 1

6 023 10 23. × is constant for all,

means 1 67 10 24. × − gm

40 AMU means, mass of one atom of Ca

So, 1 AMU = × − 1 67 10 24. gm= × − 1 67 10 27 . kg

Ex-2:

Find out which is having minimum mass.

(a) 10 mole H 2(b) 10 g H 2(c) 10 amu H 2

(d) All sameSol: (c) i.e., 10 1 67 10 24× × −. gm

(iv) Vapour density (v.d): It is the density of a gas or vapour w.r.t. H 2 at constant pressure and temperature and given as

v d

mol wt of gas

mol wt of H

. .

. .

. .

= 2

2 2

let, mol. wt. of gas = Mmol. wt. of H 2 = 2

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v d

M

. .= 2 2 2

v d M. .= 2

it is unit less

Ex-3:

Find out v.d. of O3 , O 2 and CO 2.

Sol: v dMol wt

. .. .

= 2

v d. . of O3 = =48 2 24

v d of O. . 232 2

16= =

v.d. of CO 2 = =44 2

22

Ex-4:

V.d. of gas is 40. Find out molecular weight.

Sol: V.d. = Mol wt. . 2

Mol. wt. = × 2 v d. .

= × 2 40

= 80

Ex-5:

v.d. of sulphur vapour is 64. Find out formula of

sulphur vapour.Sol:

We know that sulphur vapour is represented as-

S S S S S , , , , 2 4 6 8.First of all find out molecular wt. of sulphur vapour.

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v dMol wt

. .. .

= 2

64 2

=mol wt. .

mol. wt. = × = 2 64 128mol. wt. = 128Mol. wt. = No. of sulphur × wt. of one SLet, formula is S x 128 32= × xor, x = = 128 32 4 / S4 = Formula of vapour

Ex-6:

If v.d. of gas is 10 4 2( )O = . Find out molecular weightof the gas.

Sol:

v.d. of gas is 10. When we assume that v.d. of oxygen is 4.

But, Real v.d. of O 2 is 32/2 = 16mean, real v.d. = 16given v.d. is = 4

We have to find out the factor between actual and given v.d.

give v.d. ×F = Real v.d.F

al v d

given v d=

Re . .

.

F = = 16

44

means, for the given gas,Real v.d. = ×F given v.d.

Real v.d. = × =4 10 40 So, mol. wt. = × 2 v d. .

= × 2 40= 80

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1.7 Empirical Formula and Molecular Formula

Em pirical formula: The formula which gives the sim plest wholenumber ra tio of the atoms of various elements present in one

molecule of the com pound is called em pirical for mula.Example : Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass (At. masFe=56, O=16).

Ans. Fe O 2 3

Element Symbol %age At.

mass

Rel. no.

of atoms

Simplest

ratio

Whole no.

ratioIron Fe 69.9 56 699

56

=1.248

1 248 1 248

.

.

=1

1×2

=2

Oxygen O 30.1 16 30 1 16

.

=1.881

1881 1224

=1.50

1.5×2

=3

Molecular formula: The formula which gives the ac tualnumber of various elements present in one molecule of thecom pound is called molecular formula.

Percentage Com posi tion: Mass %age of an element

= ×Massof that element in thecompound

Molar massof thec

100

ompound

Mass %age of H in H 2O

= × ×

= 2 1 008 100

18 02 11 18

..

. %

Mass %age of O = 100 − = 11 18 88 79. . %38

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1.8 Stoichiometry of Chemical ReactionWhen chemical equations is written in the balanced form, itgives quantitative relationships between the various reactantsand products in terms of moles masses, molecules and volumes. This is called stoichiometry

(Greek word meaning ‘ to measure an element’). The coefficientsof the balanced chemical equa tion are called stoichiometriccoef ficients. For exam ple, a balanced chemical equa tion along with the quan ti ta tive in forma tion con veyed by it is given below:

CaCO3 + 2HCl → CaCl 2 + H 2O + CO 2

1mole 1moles 1moles 1mole 1mole40 12 3 16× + × 2 1 35 5 ( . )+ 40 2 35 5 + × . 12 16 32+ ×

= 100gm = 72 g = 111g = 18g

The problems involving these calculations may be classifiedinto the following different types:

1. Volume: Volume Rela tionships, i.e.,volume of one of the

re ac tants or products is given and the volume of the other is tobe calculated.The general method of calcula tions for all the problems of the above types consists of the following steps:

(1) Balance the reaction

(2) Find out cofficient

(3) Calculate according to cofficient

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Note: Molal solu tion means m = 1

Deci. molal solution means m = 1

10

Centi. molal solution means m = 1

100

2. Atom: It is the smallest part of the com pound which isneutral and having following proper ties.

(1) It is having three fundamental particle.(a) Protons (b) Neutrons (c) Electrons

(2) Generally represented as z A X .

Where Z = No. of protons (P) or atomic No. A = P + N = Atomic mass

3. Molecule: It is the combina tion of atoms and having neu tralin na ture.

Example: O 2 , N 2 , Cl 2 , CO 2

4. Atomic Weight: It is the weight of 6.023 × 10

23

atoms of the given.Example: If weight of 3.0115 × 10 20 atoms is 20mg then

find out atomic weight of the atom.

Solution: Since 3 0115 10 20. × atoms having wt. = 20 mg

1 atom is having weight = ×

×

− 20 103 0115 10

3

20gm

.

Hence, 6 023 10

23

. ×

atoms are having weight=

××

× ×− 20 10

3 0115 106 023 10

3

20 23

..

= × × ×− 20 10 2 103 3

= 40 gm/mole

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Not to be Copied by OthersY our S ecrets M y S e c r e t s

5. Molecular Weight: It is the weight of 6 023 10 23. × moleculeof the given.

Example: If Weight of 0 06023 10 18. × molecules of a gas is 1 µg

, then find out the molecular weight of the given gas.Solution: Since, 0 06023 10 18. × molecule of gas is having

Weight = 1µg

1 molecule is having weight = ×

×

− 1 100 06023 10

6

18gm

.

Hence, weight of 6 023 10 23. × molecule of gas

= ××

× ×− 1 100 06023 10

6 023 106 18 23

..

= × × =− 1 10 10 106 7 gm /mole

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

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QuestionsQ.1 What will be the mass of one 12C atom in g ?

Ans. 1 mole of 112C atoms = 6.022×1023 atoms = 12g

Thus, 6.022×1023 atoms of 12C have mass = 12g1 atom of 12C will have mass = 12 / 6.022×1023 g = 1.9927×10 −23 g.

Q.2 Which one of the following will have largest number of atoms ?

(i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g Cl2 (g)

(Atomic masses : Au = 197, Na = 23, Li = 7, Cl = 35.5 amu)

Ans. (i) 1gAu 1197

=mol= 1197

6.02 1023× × atoms

(iii) 1gNa =

1

23 mol=

1

23 6.02 1023

× × atoms(iii) 1gLi=

17

mol=17

6.02 1023× × atoms

(iv) 1gCl =171

mol=171

6.02 10 molecules=271

6.02 10223 23× × × × atoms

Q.3 How many moles of methane are required to produce 22g CO2 (g) after

combustion ?

Ans. The balanced chemical equation for the combustion of methane is;

CH4 + 2O2 → CO2 + 2H2O1 mole 1 mole

12 + 2 ×16 = 44g

Thus, to produce 44 g of CO2 , CH4 required = 1 mole

Therefore, to produce 22g of CO2 , CH4 required = (1/44)×22 mole = 0.5 mole.

Q.4 How much copper can be obtained from 100g of copper sulphate (CuSO )4 ?

Ans. Molecular mass of CuSO4

= Atomic mass of Cu + Atomic mass of S + 4× Atomic mass of O= 63.5 + 32 + 4 ×16 = 159.5 u.

Gram molecular mass of CuSO4 = 159.5 g

Now, 159.5 g of CuSO4 have Cu = 63.5 g

∴ 100 g of CuSO4 have Cu = (63.5 g) × 100 g / 159.5 g = 39.81 g

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Q.8 Which of the following has largest number of atoms ?

(i) 1 g of Au (ii) 1 g of Na (iii) 1 g of Li (iv) 1g of Cl2

Ans.

(i) 197 g of Au have atoms = 6.022×1023 (Gram atomic mass of Au = 197 g)

∴ 1 g Au has atoms = × × ×6 022 10 123. ( g)(197g) = 3.06 1021 atoms

(ii) 23 g of Na have atoms = 6.022×1023 (Gram atomic mass of Na = 23 g)

∴ 1 g Na has atoms = × × = ×6 022 101

2 62 1023 22.( )

.g

(23g) atoms

(iii) 7 g of Li have atoms = 6.022×1023 (Gram atomic mass of K = 39 g)

∴ 1 g Li has atoms = × × = ×6 022 101

860 1023 22.( )

.g

(7g) atoms

(iv) 71 g of Cl2 have molecules = 6.022×1023 (Gram molecules mass of Cl2

= 71 g)71 g of Cl2 have atoms = 2×6.022 ×10

23

∴ 1 g of Cl2 has atoms = × × × = ×2 6 022 101

167 1023 22.( )

.g

(71g) atoms

Thus, 1 g of lithium (Li) has the largest number of atoms.

Q.9 What will be mass of one 12C in g ?

Ans. 6.022×1023 atoms of carbon have mass = 12 g

1 atoms of carbon have mass = ×

× = × −(12g) (1atom)

atom g( . ) .6 022 10 1993 1023 23

Q.10 Use the data given in the following table to calculate the molar mass of

naturally occurring argon,

Isotope Isotopic molar mass Abundance

36 Ar 35.96755 g mol −1 0.337 %

38 Ar 37.96272 g mol −1 0.063 %

40 Ar 39.9624 g mol −1 99.600 %

Ans. Molar mass of argon in the average molar mass and may be calculated as:

= × + × ×( . ) ( . ) ( . ) ( . ) ( . )

( .0 337 35 96755 0 063 99 6 39 9624

00 337 0 063 99 600+ +. . )

= × +

=12121 23916 398026

10039947

. . .. g mol –1

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Q.11 A welding fuel gas contains carbon and hydrogen only, Burning a small

sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g water and no

other products. A volume of 10.0 L (measured at NTP) of this welding

gas is found to weight 110.6 g. Calculate

(i) empirical formula (ii) molar mass (iii) molecular formula of the gas.

Ans. Step-1. Calculation of mass percent of carbon and hydrogen.

CO2 ≡ C

44g 12g

3 38 3 38. .g =12g44g

g = 0.9218g of C×

H2O ≡ 2H

18g 2g

0 690 0 69. .g = 2g18g g =0.0766 gof H×

Total mass of fuel gas = 0.9218 + 0.0766 = 0.9984 g

Percentage of carbon = ×Mass of carbonMass of fuel gas

100

= × =0921809984

100 92 33..

. %gg

Percentage of hydrogen = ×Mass of hydrogenMass of fuel gas

100

= × =0076609984

100 767..

. %gg

Step-2. Determination of empirical formula of fuel gas.

Element % Atomic

mass

Gramn atoms

(moles)

Atomic ratio

(molar ratio)

Simplest

whole

no. ratio

C

92.33 129233

12769

..=

769769

1..

= 1

Empirical formula of the fuel gas = CH.

Step-3. Calculation of molecular mass of the fuel gas.

10.0 L of the fuel gas at NTP weigh = 11.6 g

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10.0 L of the fuel gas at NTP weigh = × =11610 0

22 4 25 98..

. . g

∴ Molecular mass of the fuel gas = 2598. g = 26.0 g = 26 amu

Step-4. Calculation of molecular formula of the gas.

Empirical formula mass = 12 + 1 = 13 amuMolecular formula = 26 amu

nMolecular mass

Empirical formula mass= = =

2613

2

∴ Molecular formula = n × Empirical formula =2 × CH = C2H2

Q.12 What is the difference between the mass of a molecule and molecular

mass ?

Ans. Mass of a molecule is that of a single molecule also known as its actual mass. But

molecular mass is the mass of Avogardo’s number (6.022×1023) of molecules.

Q.13 The average atomic mass of chlorine is 35.5 amu. Do we come across a

sample of the element with atomic mass 35.0 amu ?

Ans. No, it is not possible. The fractional atomic mass of an element is its averagemass and not the actual mass. In this case, the element chlorine exists as twoisotopes with atomic mass 35 amu and 37 amu respectively in the ratio of 3 : 1.The average comes out to be fractional i.e., 35.5 amu.

Q.14 Are the atomic masses of the elements their actual masses ?

Ans. No, atomic masses of the elements are not actual masses.

These are only relative masses because the actual masses are very small. Fordetail, consult text part.

Q.15 Give an example of a molecule in which the ratio of the molecular

formula is six times the empirical formula.

Ans. The compound is glucose. Its molecular formula is C6H12O6 while empiricalformula is CH2O.

Q.16 How many oxygen atoms are present in 96 amu of ozone ?

Ans. 96 amu of ozone (O3) contain96

6amu

16 amu = atoms of oxygen.

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Lecture-4

1.9 Mole Concept

It is denoted by ‘n’ 1 mole of a gas at NTP is having volume 22.4 L 1 mole of a compound is having weight equal to it’s

molecular weight. So on the basis of different concept molehaving following types.

Example:

20 gm of He sample is having moles.

Solution:

n

weight of sample

atomic weight of He=

So, n = = 20

4 5

Mole of molecule sample

Mole(n) = weight of sample

Molecular weight

Example:

880 gm sample of CO 2 gas is there. Find out molesof the gas.

Solution:

Mole(n)= weight of sample

Molecular weight

n = =880

44

20

If number of particles, ions atom or molecules are given then:

Mole n Number of Particles

No( )

.( . )=

×6 023 10 23

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So, n V

= = 24 22 4.

V= × 24 22 4. or V = 537 6. Litre

Example: If 448 litres of CO 2 gas sample is there at NTP then, find out mass of CO 2 sample.

Solution:

Moles of the CO 2 at NTP

n = =448

22 4 20

.or n = 20

mass of the CO 2 gas

n weight of COMolecular weight

= 2 or n x=44

x = 880 gm

2) Prob lems related with num ber of atoms or molecules

Example:

If 300 gm CaCO3 is there then find out number of atoms in CaCO3

Solution: To find number of atoms in CaCO 3 first of all.Find out No. of atoms in one CaCO3 then number of CaCO3

Toal atom = No. of CaCO3 × No. of atoms in oneCaCO3

Moles of CaCO3 (n) =Weight of CaCO

Molecular weight of CaCO

3

3

n =300 100

or n = 3

Number of CaCO3 =?

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n Number of CaCO

N=

×3

0 236 023 10( . )

3 3

0

=Number of CaCO

N Number of CaCO3= 3 0NNumber of atoms in one CaCO3= 5 CaCO3= Ca + C + 3.0

1 1 3 5 =

So, total number of atoms in CaCO3

= No. of CaCO3 × No. of atoms in one CaCO 3= × =3 5 15 0 0N N

= × × 15 6 023 10 23. or 9 0345 10 24. ×

3) Problems related with charge on the ions and num ber of

electrons and pro tons

Example: If 540 gm of Al+++

is there then find out number of electrons and charge on Al+++ sample.

Solution: One Al+++ is having charge = −3e (+ ve charge) in Al number of electrons = 13

So, in Al+++ number of electrons = 10

First of all find out number of Al+++

Moles(n)= =Weight 27

540 27

Moles n ( ) = 20

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Moles( ).

n No of Al

N=

+++

0

No. of Al+++= 20 0N

Total charge = No. of Al+++ × charge of 1 Al+++

= × 20 30N electron

= −60 0N e

= × × × × −60 6 023 10 1 6 10 23 19. .

= × 578 2 104. coulomb

= × 5 78 106. coulomb

No. of electron in one Al+++= 10

So, total electron = No. of Al+++× electron in 1 Al+++

20 10 2000 0N N× = .

Example: If a gas sample of CO3−− at NTP is having 120 gm

weight, then find out following:

(a) Volume of gas sample at NTP.

(b) Number of atoms in the sample.

(c) Number of electrons.

(d) Number of protons.

(e) Number of charge.

Solution:

First of all find out moles of the substance andsample.

n Weight of sample

Molar mass=

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n = 12060

n = 20

(a) Volume at NTP,

We know that,

moles( ).

n V

L =

22 4

2 22 4

=V:

V = × = 2 22 4 44 8. . Litre

(b) Moles( )n Number of CO

N=

−−3

0

Number of CO3−−= 2 0N ( . )N0

236 023 10= ×

In this number of atoms

= Total number of CO3−− is having atoms = 4

= × = 2 4 80 0N N

because one CO3−− is having atoms = 4

(c) In one CO3−−

Total number of electrons

In C = 6 electrons

In O = 8 electrons

So. total electrons in,

CO3 6 8 3 2−− = + × + (free negative charge)

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Total electrons = 32 in one CO3−−

Number of CO N3 0 2−− =

So, total number of electrons = × 2 320N = 64 0N

(d) Number of protons,

In one CO3−−

Total number of protons

In C = 6 protons

In O = 8 protons So. total electrons in,

CO3−−= + ×6 8 3

= +6 24 = 30

Total number of protons = total number of CO3−−

× protons in one CO3−−

= × = 2 30 600 0N N

(e) Charge on one CO3−−

q e= − 2 where e− = × − 1 6 10 19. coulomb

total charge = total number of CO3−−

Q N e= × − 2 2

= −4 0e N total charge.

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1.10 Limiting Reagent

Many a time, the reactions are carried out when the reactantsare not present in the amounts as required by a balanced

chemical reaction. In such situations one reactant is in excessover the other. The reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction Hence the reactant, which gets consumed,limits the amount of product formed and is therefore, called thelimiting reagent.

Example:

If a reaction is as,

A + 2B → C

Ratio is 1 2 1

So, if we use 40 moles of A then 80 moles of Bshould be there to produce 40 moles of C

Example: For, 2A + B → 2C, in the reaction if 40 moles of A reacts with 60 moles of B then find out limitingreactant.

Solution:

Given moles for the reaction

2A + B → 2C

40 60 0 initial.

If we use 1 mole of B then 2 moles of A should be.

Means if we use 60 moles of b then A should be 120.Which is 40, therefore, A is less in amount hence,

A is limiting Reagent So, reaction is according to it's amount.

In this reaction proper ratio for the production is, 55

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2A + B → 2C 2 1 2

So, 2X X 2X

40 20 40

B is used only 20 so remaining B is = 60 - 20 = 40.

Finding formula of hydrocarbon on the basis of percentage.

Hydrocarbons are of four types mainly.

1. Alkane 2. Alkene

3. Alkyne 4. Benzene ringAlkane:

General formula C H n n 2 2+

Where, n = 1, 2, 3,........

if n = 1 we get, CH 4 mol. wt. = 16

So, general mol. wt. of hydrocarbon,

C H n n 2 2+ atomic weight of C = 12, Atomic weight of H = 1.

= + + 12 1 2 2n n ( )

= + + 12 2 2n n

= + 14 2n

Alkene:

General formula C H n n 2

where, n = 1, 2, 3,........

So, general mol. wt. of hydrocarbon C H n n 2

= + 12 1 2n n ( )

= 14n

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Alkyne:

General formula C H n n 2 2−

where, n = 1, 2, 3,.......

So, general mol. wt. of hydrocarbon

C H n n 2 2−= + − 12 1 2 2n n ( )

= + − 12 2 2n n

= − 14 2n

Example:

If in a compound percentage of carbon is 80% andhydrogen is 20% then find out formula of

hydrocarbon. If molecular weight is 30.Solution:

C = 80% H = 20%

% of mass moles mole ratio

C

H

80

20

80/12

20/1

1

3

So, empirical formula is CH 3

Mol. wt. = 30 Empirical wt. = 15

f mol wt

Empericalwt=

. ..

f = =30

15

2

Molecular formula = Empirical formula × f

= × =CH C H 3 2 6 2

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Al terna tive Method

For a hydrocarbon molecular weight of the hydrocarbon may be

14 2n +

14n

14 2n —

So, mol. wt. should be according to these three.

14 2 30n + = 14 28n = or n = 2

14 30n =

14 2 30n — = or 14 32n =means, only first equation is satisfy the mol. wt. So, it isalkane having n = 2 i.e., C H n n 2 2+ or C H 2 6.

Mole concept used in the reac tion By using POAC

1.11 Principle of At oms Conservation

In this POAC, we assume that moles of the atoms in reactant as well as in product side are conserved.

Example: CaCO3 → CaO + CO 2

For the given reaction moles of Ca in CaCO3 are same as that ofCa in CaO.

First of all find out relation between moles of Ca and moles of CaCO3.

In CaCO3 → Ca + C + 3 O

1 1 1 3

X X X 3X

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Moles of Ca = moles of CaCO3

Moles of C = moles of CaCO3

Moles of O=

3 × moles of CaCO3In CaO → Ca + O

1 1 1

X X X

Moles of CaO = moles of Ca

Moles of caO=

moles of OIn CO 2 → C + 2 O

1 1 2

X X 2X

Moles of C = moles of CO 2

Moles of O = 2 × moles of CO 2

For CaCO3 → CaO + CO 2

If we conserve Ca then moles of Ca in CaCO3

= moles of Ca in CaO

n n Ca CaCO= 3 n n Ca CaO=

So, n n CaCO CaO3 =

Conserving C ⇒

Moles of C in CaCO3 = Moles of C in CO 2

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n n C CaCO= 3 | n n C CO= 2Means, n n CaCO CO3 2=

Conserving O⇒

Moles of O in CaCO3 = Moles of O in CO 2

+ moles of O in CaO

n n CaCO0 3 3= × | n n CO0 2 2= | n n CaO0 =

So, 3 2 13 2× = × + ×n nCO n CaCO CaO

Example: If 400 gm CaCO3 is heated then find out wt. of CaO and CO 2 formed.

Solution: CaCO3 → CaO + CO 2

For Ca n Ca in CaCO3 = n Ca in CaO

1 13

× = ×n n CaCO CaO

400 100 56= X

X g= 224

For C n C in CaCO n inCOC 3 2=

1 13 2

× = ×n n CaCO CO

400 100 44=

y

X g= 176

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Combustion Reaction

In the combustion reaction POAC is applied.

Example.In the combustion of methane, why is methane

regarded as the limiting reagent ?Sol.

Methane (CH 4) is regarded as the limiting reactantbecause air or oxygen is always present in excess. Theamounts of CO 2 and H 2O formed in the reaction dependupon the amount of methane only. Therefore, it isregarded as the limiting reagent.

Example: CH 4

is having combustion then find out therelation between moles of all.

CH4 + O 2 → CO 2 + H 2O

So, For, C If POAC is there.

1 14 2

× = ×n n CH CO

For, H If POAC is there.

4 24 2× = ×n n CH H O

For, O If POAC is there.

2 2 1 2 2 2

× = + ×n n n O CO H O

Example:

If 20 moles of a hydro carbon on combustion gives40 mols of CO 2 and 40 moles of H 2O, then findout formula of hydro carbon.

Solution:

Let the hydro carbon is C H x y.

If combustion reaction is as,

C xHy + O 2 → CO 2 + H 2O

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Not to be Copied by OthersY our S ecrets M y S e c r e t s

So, For, C If POAC is there

x n n C H CO x y× = × 1 2For, H y n n C H H O x y× = × 2 2

x and y are number of C and H used in the hydrocarbon.

For calculation

n C H x y = 20 n CO 2 40= n H O 2 30=

X × = × 20 1 40 or X = 2

y × = × 20 2 40 or y = 4

So, formula of hydrocarbon is C H 2 4

Note: If in place of moles we are using volume then. This volume is equivalent to moles.

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Lecture-5

1.12 Equivalent weight and Equivalents

Equivalent Weight (E): It is the amount of the given element ormetal which reacts with 1 gm of hydrogen.

Example:

Suppose we have to find equivalent weight of oxygen. Then combination of hydrogen andoxygen will give H O 2 .

Solution: From this formula,

2 gm of hydrogen reacts with 16 gm oxygen.

1 gm of hydrogen will reacts with 16 2

= 8 gm oxygen.

Means Equivalents weight of oxygen is = 8.

Example:

If we have to find equivalent weight of Al then,

Solution: Al reacts with hydrogen to give AlH 3.

From this, 3 gm H reacts with = 27 gm Al

1 gm H reacts with = = 27

3 9 gm Al

Means, Equivalent weight of Al = 9

Equivalent Weight New Con cept)

For metals equivalent weight =atomic weight

imumch emax argExample: Al E = =

27 3

9

Ca E = =40 2

20

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For elements equivalents weight =atomic weight

imumch emax arg

Example: O 2

E = = 16

28

or E = =324

8

For Acid Emolecular weight

Number of H replaced= +

Example: H SO 2 4 E

M=

2

HCl EM

= 1

EM

=3

H PO3 4

EM

=

2

H PO3 3

EM

= 1

H PO3 2

In case of a reaction,

NaOH + H SO 2 4 → NaHSO H O4 2+

Find out Equivalent weight of H SO 2 4

EMolecular weight

No of H replaced= +.

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Here H SO 2 4 is having 2H + but number of H + replaced is

one.

So, E

Molecular weight

= 1

For, NaOH H PO Na HPO H O+ → +3 4 2 4 2

Find out Equivalent weight of H PO3 4.

So, EMolecular weight

Number of H replaced= +

In this number of H + replaced are 2H +

EMolecular weight

= 2

Equivalent Weight of the Compound or Salt (E)

For Compound Equivalent weight

EMolecular weight of compound

Charge on cation NUmber of =

× cation

Example:

Find out equivalent weight of

(a) CaCO3 (b) Na SO 2 4

(c) Al SO 2 4 3( )

Solution: (a) For CaCO3

EMolecular weight

F =

F X y= × 67

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X, Charge on cation (Ca++) = 2

y, Number of cations (Ca++) = 1

So, E = × = =

100

1 2

100

2 50

(b) For Na SO 2 4

EMolecular weight

F =

F x y= ×

x, Charge on cation (Na

+

)=

1y, Number of cations (Na+) = 2

So, EM

=× 1 2

or EM

= 2

(c) For Al 2(SO4)3

EMolecular weight

F

=

F x y= ×

x, Charge on cation (Al+++) = 3

y, Number of cations (Al+++) = 2

So, EM

=× 2 3

or EM

=6

Note: If compound is ionic then equivalent weight of compound (AB)= +E Ecation anion

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Example:

If Al O 2 3 is there, find out equivalent wt. of Al O 2 3.

Solution: E E E Al O Al 2 3 0= +

= +

9 8 = 17

or EMolecular weight

Charge on cation Number of catio Al O 2 3 =

× ns

Molecular weight of Al O 2 3 27 2 16 3= × + ×

= + = 54 48 102

X = Charge on cation i.e., + 3 (Al+++)

y = Number of cations i.e., 2 ( 2Al)

So, F = x y× or F = × =3 2 6

E Al O 2 3 102

6 17 = =

Finding equiv alent of the given :Equivalence of the given =

Weight of the given

E

Example:

Find out equivalence of 200g CaCO3

Solution: Equivalence. = weight of the given

E

E Molecular weightF

= Where, F x y= ×

x = Charge on cation = + 2

y = Number of cations = 1

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Equivalence of O 2 = wtEO

.

2

Equivalence of O 2 = x

8From the concept,

Equivalence of Ca = Equivalence of O 2400 20 8

= x

or x

8 20=

or x gm= 160

wt. of O 2 used is 160 gm.

Relation between Equivalence and Moles

If x gram of a given is there. Then, moles

n wt of subs ce

atomic wt molecular wt=

. tan

./ . ....(i)

Equivalence (eq.)=Wt of subs ce

Equivalent E

. tan

( )....(ii)

If,( )( )III

Eq n

atomic wt molecular wt

E. . / .

=

Eq n

Atomic wtE

molecular wtE

. . .= =

For, Acid,

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EMolecular wt

No of H replaced=

.

. *

For, Base,

EMolecular wt

No of OH replaced= −

..

For, Metal / Element,

Eatomic wt

Maximumch e=

.

arg

In case of compound,Example: Find out equivalent wt. of Al 2(SO4)3

Solution: EMolecular wt

Ch e on cation NO of cations=

×.

arg .

EMol wt Mol wt

=×

=. . . .

3 2 6

For, Al 2 (SO4)3

E Atomic wt molecular wt

F =

./ .

F No of H for acid= +.

= −No of OH for base.

= Ch e on metalarg

So, F Atomic wt molecular wt

E=

./ .

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So,Eq n

atomic wt

E

Molecular wt

E. . .

= =

Eq

n

F .

= or F Eq

n

=.

Example:

20 moles of Ca is having...........Equivalence.

Solution:

Eq n

F .

= or Eq. 20

2=

Where, F = No. of charge on Ca i.e., = 2

∴ Eq.= 40

Example: Find out No. of Eq. of 30 moles of H 3 PO4.

Solution:

Eq n

F .

= or Eq.30

3=

Where, F = 3

∴ Eq.= 90

Note: If for a reaction, moles are given then find out.equivalence first. Then calcula te,

Example:

20 moles of Ca reacts with O 2 then find weight ofO 2 used.

Solution: Ca + O 2 → CaO. Find out equivalence first.

Eq n

F . = for Ca F= 2

Eq n

.= 2 Eq. = × = 2 20 40

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Eq. of Ca = Eq. of O 2

40 2= Eq of O.

408

= = wt

E

x.

x gm= 320

Example:

If 20 moles of Al reacts with 20 moles of O 2 then find out.

(a) Moles Al 2O3 formed

(b) Weight and name of remainder.

Solution: Al + O 2 → Al 2O3

Eq n

F .

= for Al F = 3

Eq. 20

3= Eq. = × =3 20 60

Eq.= 60Eq n

F .

= for O 2 F = 4

Eq n

.= 4 Eq. = × =4 20 80

Eq.= 80

Means, Al + O 2 → Al 2O3Eq. used 60 80 0

Final 0 20 60

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(a) Eq. of Al 2O3 = 60Eq n

f .

= f = ?

E Mol WtNo of cations ch eon cations= ×

. .. arg

F = 6 For Al O 2 3

EMol wt M

=×

=. .

2 3 6

So,Eq

n

. ,= 6

606

n = n = 10

Remainder is O 2 and remaining Equivalence is 20.

For O 2

Eq wt

E.

.= = 20

20

8

= wt.

Wt. = 160gm.

1.13 Reactions in Solutions

A majority of reactions in the laboratories are carried out in solutions. The concentration of a solution or the amount of substance present in its given volume can be expressed in anyof the following ways.

1. Mass per cent or weight per cent (w/w %)

2. Mole fraction

3. Molarity

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4. Molality

Let us now study each one of them in detail

(i) Mass per cent : It is obtained by using the following

relation:

Mass per cent = ×mass of solute

massof solution 100

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

Solution: Mass per cent of A = ×

mass of A

mass of solution 100

=+

× 2

2 18 100

ggof A gof water

= × 2

20 100

gg

= 10%

(ii) Mole Fraction: It is the ratio of number of moles of a particular component to the total number of mole of thesolution. If a substance A dissolves in substance B and their number of moles are n A andn B respectively then the mole fraction of A and B are given as

Mole fraction of A =No.of moles of A

No. of moles of solution

=+

n n n

A

A B

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Mole fraction of B =No.of moles of B

No. of moles of solution

=+

n

n n

B

A B

(iii) Molarity: It is the most widely used unit and is denotedby M. It is defined as the number of oles of the solute in I litreof the solution. Thus,

Molarity (M) = No. of moles of solute

volume of solution in litres

Suppose we have 1 M solution of a substance say NaOH and we want to prepare a 0.2 M solution from it.

1 M NaOH means 1 mol of NaOH present in 1 litre of thesolution for 0.2 M solution we require 0.2 moles of NaOH in 1litre solution.

Hence, we have to take 0.2 moles of NaOH and make the

solution to 1 litre.Now how much volume of concentrated (1M) NaOH solution be taken which contains 0.2 moles of NaOH can be calculated as follows:

If 1 mole is present in 1 L or 1000 mL

then 0.2 mol is present in 1000

10 2

mL

molmol× .

= 200mL

Thus, 200 mL of 1 M NaOH are taken and enough water isadded to dilute it to make it 1 litre.

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In fact for such calculations, a general formula,

M V M V 1 1 2 2× = × where M and V are molarity and volumerespectively can be used in this case, M 1 is equal to 0.2;

V mL 1 1000= and M V 2 2 1 0= . ; is to be calculated substituting the value in the formula:

0 2 1000 1 0 2. .M mL M V× = ×

∴ VM mL

MmL 2

0 2 1000 1 0

200= ×

=.

.

Problem: Calculate the molarity of NaOH in the solu tion

pre pared by dissolving its 4g in enough wa ter to form 250 mL of the solu tion.

Solution: Since molarity (M)

= No. of moles of solute

volume of solution in litres

= Mass of NaOH / Molar mass of NaOH 0.250L

= =4 400 250

0 1

0 250g g

L

mol

L /

.

.

.

= 0 4 1. –mol L

= 0 4. MNote: that molarity of a solution depends upon temperaturebecause volume of a solution is temperature dependent.

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(iv) Molality: It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.

Thus, molality (m)=No. of moles of solute

Mass of solvent in kgNote: Molality is temperature independent

Problem: The density of 3 M solu tion of NaCl is 1.25 g mL –1.Calculate molality of the solu tion.

Solution: M=3 mol L –1

Mass of NaCl

in 1 L solution = 3×58.5 =175.5g mass of

1L solution = 1000× 1.25 =1250g

(since density = 1.25 g mL –1) mass of

water in solution = 1250–175.5 =1074.5g

Molality =No. of moles of soluteMass of solvent of kg

=3

1 074 5

mol

kg. .

= 2 79. m

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Relation between Moles fraction and Molality

Example:

If moles fraction of urea in water is 0.2 find outmolality of urea solution.

Solution: We know that molality

mn molesof solute

wt of solvent kg=

( )

. ( )

If moles of solute are ‘n’ and moles of solvent are ‘N’ then, mole fraction of urea

Xn

n N=

+n= moles of urea

N = moles of water

Now, given is, x = 0.2

0 2. =+n

n N

SO, n N n n + = =0 2

5 .

or n N n + = 5

or N n = 4 .....(i)

For molality,

mn

wt of solvent kg= . ( )

Let the wt. of solvent or water is y Then,

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moles Ny

mol wt=

. .

mol. wt. = 18 for H O 2

Ny=

18

or y N gm= 18

Therefore, molality,

mn

wt of solvent kg=

. ( )

mn

= 80 1000

We know that N n = 4

mn

n =× 18

4 1000

=×

= = 1000

18 4 55 55

4 14

.

Example:

If molality of urea in water is 2m. Find out mole fraction.

Solution:

mn molesof solute

wt of solvent kg=

( )

. ( )

Mole fraction xn

n N=

+

Let the weight of solvent is 1 kg. So,

mn

= 1

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or n m= = 2 ....(i)

Weight of solvent is 1kg, therefore,

Moles N wt

mol wt= =

.

. .

1000

18

Hence, x =+

= 2

1000 18

2

2 57 5 .

Example:

600 gm of urea is dissolved in 100 litre of solution.

Mn

V

=

Solution: n = =60060

10 V L = 100

Moles of urea,

Mol. wt. of urea = 60

M = 10 / 100 So, M = 0.1

Example: 20% urea solution is there having density 2g/ml.Find out molality, molarity of solution.

Solution:

20% urea solution means, 20 g of urea in 100g of solution.

Hence, 80g of solvent

Therefore, Molality mn molesof solute

Wt of solvent kg

=( )

. ( )

n = = 2060

13

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QuestionsQ.1 Calculate the mass percentage of different elements present in sodium

sulphate (Na2SO4 ).

Ans. Molecular mass of Na So =22 4 × Atomic mass of Na+ Atomic mass of S + 4 × Atomic mass of O

= 2×23 + 32 + 4×16 = 46 + 32+ 64 = 142 u.

The percentage of different elements present can be calculated as:

Percentage of sodium (Na)= ×Mass of Na

Molecular mass of Na So2 4100

= ×46

142100

= 3239. %

Percentage of sulphur(S) = ×Mass of S

Molecular mass of Na SO2 4100

= ×32142

100

= 2554. %

Percentage of oxygen(O) = ×Mass of OMolecular mass of Na SO2 4

100

= ×64

142100

= 4507. %

Q.2 Calculate the mass of sodium acetate (CH3COONa) required to make

500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate

is 82.0 g mol −1.

Ans. Molar mass of CH3COONa = 82 g mol−1 (given)

Molarity of solution = 0.375 M = 0.375 mol L−1

Volume of solution = 500 mL = 500/1000 = 0.5 L

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Molarity of solution(M) =Mass of CH COONa / Molar mass

Volume of Solution in litres3

( . )) ( . )

03750 5

molLW

(82gmol L –1

–1= ×

W =(0.375molL g mol L g –1) ( ) ( . ) .× × =−82 0 5 15 3751 .

Q.3 Calculate the concentration of nitric acid in moles per litre in a sample

which has a density 1.41 g mL −1 and the mass percent of nitric acid in it

being 69%.

Ans. Mass of solution = 100 g

Density of solution = 1.41 g mL −1

Volume of solution = = =−Mass of solution

Density of solution

g

1.41gML

100

71 0 92. mL

Concentration of HNO3 in moles per litre means molarity

Molarity(M) =

Mass of HNOMolar Mass of HNO

Volume of solution

3

3in litre

g

g mol

L=

−69

6370921000

1

.

= −1544 1. mol L =15.44 M

Q.4 What is the concentration of sugar (C H O12 22 11) in mol L –1 if 20 g of it

are dissolved in enough water to make final volume upto 2L ?

Ans. The concentration in mol L −1 means molarity (M).

From the available data, it can be calculate as :

Mass of sugar = 20 g

Molar mass of sugar (C H O12 22 11) = 12×12 + 22×1 + 16×11 = 342 g mol−1

Volume of solution in litre = 2 L

Molarity of solution =Mass of sugar Molar massVolume of solution inlitres

/

=20g

(342mol ) (2L)mol L =0.029M.−

−

× =1

10029.

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Q.5 If the density of methanol is 0.793 kg L –1, what is its volume needed for

making 2.5 L of its 0.25 M solution ?

Ans. Step-1. Calculation of mass of methanol (CH3OH)

Molar mass of methanol (CH3OH) = 12 + 4×1 + 16 = 32 g mol −1

Volume of solution = 2.5 L

Molarity of solution =Mass of methanol Molar mass

Volume of solution in litre /

0.25mol L =W

32 g mol 2.5L –1

− ×1

W =0.25 mol L 32 g mol 2.5 L =20 g –1 –1× ×

Step-2. Calculation of volume of methanol

Mass of methanol = 20 g = 0.002 kg

Density of methanol = 0.793 kg L −1

Volume of methanol = = =MassDensity

kg

kgLL –1

0002

079300025

.

..

Q.6 A sample of drinking water was found to be severely contaminated with

chloroform CHCl3 , supposed to be carcinogen.

The level of contamination was 15 ppm (by mass)

(i) Express this in percent by mass

(ii) Determine the molality of chloroform in the water sample.

Ans. (i) Calculate of percent by mass 15 ppm level of contamination means that15 parts or 15 g of chloroform (CHCl3) are present in 10

6 parts or 106

g of the sample i.e., water.

∴ Mass percent = × = × −15

100 15 10 3g

10 g

6 .

(ii) Calculate of molarity of the solution

Mass of chloroform = 1.5 × 10–3 g

Molar mass of chloroform (CHCl3) = 12 + 1 + (3×35.5) = 119.5 g mol−1

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Mass of sample i.e., water = 100 g

Molarity of solution (m) =

Mass of chloroformMolar mass of chloroform

Mass of solvent in kg

g

119.5g mol

kg

–1=

× −15 10

100

1000

3.

= × = ×− − −125 10 125 104 1 4. .mol kg m

Q.7 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different ?

Ans. 0.50 mol Na2CO3 represent concentration in moles.

0.50 mol Na2CO3 represent concentration in moles/litre (molarity).

Q.8 If 10 volumes of dihydrogen react with five volumes of dioxygen gas,

how many volumes of water will be produced ?

Ans. 2H2(g) + O2(g) → 2H2O(g)

2 vol 1 vol 2 vol

10 vol5 vol 10 vol

10 volumes of water vapours will be produced

Q.9 Calculate the molarity of a solution of ethanol in water in which mole

fraction of ethanol is 0.04.

Ans. Mole fraction of etahnol (X B) may be given as :

X =n

n +nBB

A B

Here, nB = no. of moles of ethanol ; n A = no. of moles of water

According to available data, X =0.04,n =1000g

(18gmol=55.55 molB A —1)

0.04=n

55.55+nB

B

2.222+(0.04)n =nB B

or n +0.04n =2.222 or 0.96n =2.222B B B

or n =2.2220.96

=2.31B

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2.31 moles of ethanol are dissolved in 1000 g (or 1000 mL) of water or 1000mL of the solution. In this case, the volume of solution is considered to be thesame as that of the solvent i.e., water. In other words, the solution is regardedas dilute solution,

Molarity of solution = 2.31 M.

Q.10 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2according to the reaction:

CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)

What mass of CaCO3 is required to react completely with 25 mL of 0.75

M HCl ?

Ans. Step-1. Calculation of the mass of HCl present

Molarity of HCl solution = 0.75 M = 0.75 mol L −1

Volume of HCl solution = 25 mL = 25/1000 = 0.025 L

Molarity of solution (M)=Mass of HCl Molar massVolume of solution in litres

/

(0.75mol LMass of HCl

g mol L –1)

( . ) ( . ) – =

×36 5 0 0251

Mass of HCl = (0.75 g mol L −1)×(36.5 g mol −1)×(0.025 L) = 0.684 g

Step-2. Calculation of the mass of CaCO3 reacted

some basic

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