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1PHYSICS_IJSO_PAGE # 1

FORCE AND NEWTON�S LAWS OF MOTION

EFFECTS OF FORCE

To define force first of all one has to see the effects offorce. By �effects of force� we mean what force can do

or what changes a force can bring about.

Effects of Force :

A force can produce the following effects :

(i) A force can move a stationary body.

(ii) A force can stop a moving body.

(iii) A force can change the speed of a moving body.

(iv) A force can change the direction of a moving body.

(v) A force can change the shape (and size) of a body.

Based on the effects of force, it may be defined as :

Force is a pull or push, which changes or tends tochange the state of rest or of uniform motion of a bodyor changes its direction or shape.

(a) Mathematical Representation of Force :

Mathematically, force F is equal to the product of mass,m of a body and acceleration a, produced in the bodydue to that force.

i.e. F = maWhere a = final velocity � initial velocity/time

(b) Units of Force :

(i) In C.G.S. system :

F = ma gram × cm/s2 = dyne

If m = 1 gram, a = 1 cm/s2, then F = 1 dyne

When a force is applied on a 1 gram body and theacceleration produced in the body is 1 cm/s2 then theforce acting on the body will be one dyne.

(ii) In S.I. system :

F = ma kg × m/s2 = Newton

If m = 1 kg and a = 1 m/s2 then by F = ma,

F = 1 × 1 = 1 kg × m/s2 = 1 Newton.

If a force is applied on a body of mass 1 kg andacceleration produced in the body is 1 m/s2 then theforce acting on the body will be one Newton.

Relationship between the newton and dyne 1 N = 1 kg × 1 m s�2

= 1000 g × 100 cm s�2

= 100000 g cm s�2

= 105 dyneThus 1 N = 105 dyne

ILLUSTRATIONS

1. A force produces an acceleration of 5.0 cm/s2 in a body

of mass 20g. Then find out the force acting on the body

in Newton.

Sol. Acceleration of the body, a = 5 cm/s = 0.05 m/s

Mass of the body, m = 20 g = 0.02 kg

F = ma F = 0.05 × 0.02 = 10�3 N

2. A force of 15 N acts on a body of mass 5 kg for 2s. What

is the change in velocity of body ?

Sol. Given : F = 15 N , t = 2s , m = 5 kg

F = ma a = mF

= 5

15 = 3 m/s2

a = t

uv v � u = at = 3 × 2 = 6m/s

RESULTANT FORCE

Many forces may be simultaneously applied on a body,

for example- several persons may jointly make an effort

to move a heavy body, each person pushes it i.e. each

person applies a force on it. t is also possible that a

stronger man pushes that body hard enough and

produces same acceleration in it. f a single force

acting on a body produces the same acceleration as

produced by a number of forces, then that single force is

called the resultant force of these individual forces .

FUNDAMENTAL FORCES

All forces observed in nature such as muscular force,

tension, reaction, friction, weight, electric, magnetic,

nuclear, etc., can be explained in terms of only following

four basic interactions.

(a) Gravitational Force :

The force of interaction which exists between two

particles of masses m1 and m

2, due to their masses is

called gravitational force. The gravitational force acts

over long distances and does not need, any intervening

medium. Gravitational force is the weakest force of

nature.

(b) Electromagnetic Force :

Force exerted by one particle on the other because of

the electric charge on the particles is called

electromagnetic force. Following are the main

characteristics of electromagnetic force

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(i) These can be attractive or repulsive.

(ii) These are long range forces.

(iii) These depend on the nature of medium between

the charged particles.

(iv) All macroscopic forces (except gravitational) which

we experience as push or pull or by contact are

electromagnetic, i.e., tension in a rope, the force

of friction, normal reaction, muscular force, and

force experienced by a deformed spring are

electromagnetic forces. These are manifestations

of the electromagnetic attractions and repulsions

between atoms/molecules.

(c) Nuclear Force :

It is the strongest force. It keeps nucleons (neutronsand protons) together inside the nucleus inspite oflarge electric repulsion between protons. Radioactivity,fission, and fusion, etc. results because of unbalancingof nuclear forces. It acts within the nucleus that tooupto a very small distance. It does not depends oncharge and acts equally between a proton and proton,a neutron and neutron, and proton and neutron,electrons does not experience this force. It acts for veryshort distance order of 10�15 m.

(d) Weak Force :

It acts between any two elementary particles. Under itsaction a neutron can change into a proton emitting anelectron and a particle called antineutrino. The rangeof weak force is very small, in fact much smaller thansize of a proton or a neutron.It has been found that for two protons at a distance of 1fermi :

FN:F

EM:F

W:F

G::1:10�2:10�7:10�38

On the basis of contact forces are classified into two

categories

(i) Contact forces

(ii) Non contact or field forces

(a) Contact force :

Forces which are transmitted between bodies by short

range atomic molecular interactions are called contact

forces. When two objects come in contact they exert

contact forces on each other. e.g. Normal, Tension etc.

(b) Field force :

Force which acts on an object at a distance by the

interaction of the object with the field produced by other

object is called field force. e.g. Gravitational force,

Electro magnetic force etc.

DETAILED ANALYSIS OF CONTACT FORCE

(a) Normal force (N) :

It is the component of contact force perpendicular to

the surface. It measures how strongly the surfaces in

contact are pressed against each other. It is the

electromagnetic force.

e.g.1 A table is placed on Earth as shown in figure

Here table presses the earth so normal force exerted

by four legs of table on earth are as shown in figure.

e.g.2 A boy pushes a block kept on a frictionlesssurface.

Here, force exerted by boy on block is electromagneticinteraction which arises due to similar chargesappearing on finger and contact surface of block, it isnormal force.

A block is kept on inclined surface. Component of itsweight presses the surface perpendicularly due towhich contact force acts between surface and block.

Normal force exerted by block on the surface of inclinedplane is shown in figure. Here normal force is acomponent of weight of the body perpendicular to theinclined surface i.e. N = mgsin

Force acts perpendicular to the surface


3. Two blocks are kept in contact on a smooth surface asshown in figure. Draw normal force exerted byA on B.

Sol. In above problem, block A does not push block B, sothere is no molecular interaction between A and B.Hence normal force exerted by A on B is zero.

Note :

� Normal is a dependent force it comes in role whenone surface presses the other.

(b) Tension :

Tension is the magnitude of pulling force exerted by astring, cable, chain, rope etc. When a string isconnected to a body and pulled out, the string said tobe under tension. It pulls the body with a force T, whosedirection is away from the body and along the length ofthe string. Usually strings are regarded to be masslessand unstretchable, known as ideal string.

Note : (i) Tension in a string is an electromagneticforce and it arises only when string is pulled. If amassless string is not pulled, tension in it is zero.(ii) String can not push a body in direct contact.

(c) Force Exerted by spring :

A spring is made of a coiled metallic wire having adefinite length. When it is neither pushed nor pulledthen its length is called natural length.At natural length the spring does not exert any force onthe objects attached to its ends.f the spring is pulledat the ends, its length becomes larger than its naturallength, it is known as stretched or extended spring.Extended spring pulls objects attached to its ends.

A B

A B

A B

Normal spring

Stretched spring

Compressed spring

Spring force on A Spring force on B

Spring force on A Spring force on B

If the spring is pushed at the ends, its length becomesless than natural length. It is known as compressedspring. A compressed spring pushes the objectsattached to its ends.

F

x

FF

x

F

F = 0 spring in naturallength does not exertsany force on its ends

F = � kx

x = compression in spring

F = � kx ;k = spring

constant or stiffness constant (unit = N/m)x = extension in spring

Fext

Fext

Note : Spring force is also electromagnetic in nature :

(d) Friction force :

When a body is moving on a rough surface resistance

to the motion occurs because of the interaction

between the body and its surroundings. We call such

resistance as force of friction. Friction is also

considered as component of contact force which acts

parallel to the surfaces in contact.

(i) Origin of friction : The frictional force arises due to

molecular interactions between the surfaces at the

points of actual contact. When two bodies are placed

one over other, the actual area of contact is much

smaller then the total surface areas of bodies. The

molecular forces starts operating at the actual points

of contact of the surfaces. Molecular bonds are formed

at these contact points. When one body is pulled over

the other, these bonds are broken, and the material

get deformed and new bonds are formed. The local

deformation sends vibrations into the bodies. These

Vibrations ultimately dumps out and energy of

vibrations appears as heat. Hence to start or carry on

the motion, there is a need of force.

Body 1

Body 2

Actual area of contact

(ii) Statics and Kinetic Frictions :

� Experiment :

(A) Consider a block placed on a table, and a small

force F1 is acted on it. The block does not move. It

indicates that the frictional force fs starts acting in

opposite direction of applied force and its magnitude

is equal of F1(figure b). That is for the equilibrium of


the block, we have

F1 � f

s = 0 or F

1 = f

s

The force of friction when body is in state of rest over

the surface is called static friction (fs).

(B) As the applied force increases the frictional forcealso increases. When the applied force is increasedup to a certain limit (F

2) such that the block is on the

verge of motion. The value of frictional force at thisstage is called limiting friction f

lim (figure c).

(C) Once the motion started, the smaller force is nownecessary to continue the motion (F

3) and thus

frictional force decreases. The force of frictionwhen body is in state of motion over the surface iscalled kinetic or dynamic friction f

k (figure d).

(iii) More about frictional force :

(A) About static friction

1. The limiting friction depends on the materialsof the surfaces in contact and their state ofpolish.

2. The magnitude of static friction isindependent of the apparent area of contactso long as the normal reaction remains thesame.

3. The limiting friction is directly proportional tothe magnitude of the normal reactionbetween the two surfaces i.e. f

lim=

SN. Here

s is coefficient of static friction.

We can write, s =

Nflim

(B) About kinetic friction :

1. The kinetic friction depends on the materialsof the surface in contact.

2. It is also independent of apparent area ofcontact as long as the magnitude of normalreaction remains the same.

3. Kinetic friction is almost independent of thevelocity, provided the velocity is not too largenot too small.

4. The kinetic friction is directly proportional to

the magnitude of the normal reaction

between the surfaces.

fk =

k N. Here

k is coefficient of kinetic friction.

We can write, k =

Nfk

� There are two types of kinetic frictions:

(i) Sliding friction : The force of friction when one

body slides over the surface of the another body is

called sliding friction.

(ii) Rolling friction : When a wheel rolls without

slipping over a horizontal surface, there is no

relative motion of the point of contact of the wheel

with respect to the plane. Theoretically for a rolling

wheel the frictional force is zero. This can only

possible when bodies in contact are perfectly rigid

and contact of wheel with the surface is made

only at a point. But in practice no material body is

perfectly rigid and therefore bodies get deformed

when they pressed each other. The actual area of

their contact no longer remains a point, and thus

a small amount of friction starts acting between

the body and the surface. Here frictional force is

called rolling friction. It is clear from above

discussion that rolling friction is very much smaller

than sliding friction.

flim > fkinetic > frolling.

Note : s and

k are dimensionless quantities and

independent of shape and area of contact . It is a

property of two contact surfaces. s

will always be

greater than k .Theoretical value of can be o to but

practical value is 0 < 1.6


(a) Conservative Force :

A force is said to be conservative if the amount of work

done in moving an object against that force is

independent on the path. One important example of

conservative force is the gravitational force. It means

that amount of work done in moving a body against

gravity from location A to location B is the same

whichever path we may follow in going from A to B. This

is illustrated in figure.

A force is conservative if the total work done by the

force on an object in one complete round is zero,

i.e. when the object moves around any closed path

(returning to its initial position).

A force is conservative if there is no change in kinetic

energy in one complete round. KE = 0

This definition illuminates an important aspect of a

conservative force viz. Work done by a conservative

force is recoverable. Thus in figure, we shall have to

do mgh amount of work in taking the body from A to B.

However, when body is released from B, we recover

mgh of work.

Other examples of conservative forces are spring force,

electrostatic force etc.

(b) Non-Conservative Force :

A force is non-conservative if the work done by that force

on a particle moving between two points depends on

the path taken between the points.

The force of friction is an example of non-conservative

force. Let us illustrate this with an instructive example.

Suppose we were to displace a book between two points

on a rough horizontal surface (such as a table). If the

book is displaced in a straight line between the two

points, the work done by friction is simply FS where :

F = force of friction ;

S = distance between the points.

However, if the book is moved along any other path

between the two points (such as a semicircular path),

the work done by friction would be greater than FS.

Finally, if the book is moved through any closed path,

the work done by friction is never zero, it is always

negative. Thus the work done by a non-conservative

force is not recoverable, as it is for a conservative force.

GALILEO�S EXPERIMENTS

Experiment 1 :

It was observed by Galileo that when a ball is rolled

down on an inclined frictionless plane its speed

increases, whereas if it is rolled up an inclined

frictionless plane its speed decreases .If it is rolled on

a horizontal frictionless plane the result must be

between the cases describe above i.e. the speed

should remain constant. It can be explain as :

v�v

v� = v

moving down : speed increases moving up : speeddecreases moving horizontal : speed remains constant

Experiments 2 :

When a ball is released on the inner surface of asmooth hemisphere, it will move to the other side andreach the same height before coming to restmomentarily. f the hemisphere is replaced by a surfaceshown in figure(b) in order to reach the same heightthe ball will have to move a larger distance.

(a)

h

(b)

h

(c)

v v

If the other side is made horizontal, the ball will neverstop because it will never be able to reach the sameheight, it means its speed will not decrease. It willhave uniform velocity on the horizontal surface. Thus, ifunbalanced forces do not act on a body, the body willeither remain at rest or will move with a uniform velocity.It will remain unaccelerated.

Newton concluded the idea suggested by Galileo andwas formulated in the laws by Newton.


NEWTON�S FIRST LAW OF MOTION

Every body remain in its state of rest or uniform motionin a straight line unless it is compelled by some externalforce.

It means a body remain unaccelerated if and only if,the resultant force on it is zero.

In such a case the body is said to be in equilibrium.

INERTIA

(a) Definition of Inertia :

The tendency of the body to oppose the change itsstates of rest or uniform motion in a straight line iscalled inertia. Newton�s first law of motion is also called

law of inertia.(b) Description :

It follows from first law of motion that in absence of anyexternal force, a body continues to be in its state of restor in uniform motion along a straight line. In otherwords, the body cannot change by itself its position ofrest or of uniform motion.

(c) Inertia Depends upon Mass :

We know that it is difficult to move a heavier body thanthe lighter one. Similarly it is difficult to stop a movingheavier body than a lighter body moving with the samevelocity. Thus, we conclude that mass of the body isthe measure of inertia, more the mass, more theinertia.

TYPES OF INERTIA

There are three types of Inertia which are :

(a) Inertia of Rest :

The tendency of the body to oppose the change in itsstate of rest when some external unbalance force isapplied on it, is called the inertia of rest.Example based on Inertia of rest :

A person sitting in a bus falls backwards when the bussuddenly starts. The reason is that lower part of hisbody begins to move along with the bus but the upperpart of his body tends to remain at rest due to inertia ofrest.

(b) Inertia of Motion :

The tendency of the body to oppose its state of motionwhen some unbalance forces are applied on it, iscalled the inertia of motion.

Example based on Inertia of motion :

A man carelessly getting down a moving bus fallsforward, the reason being that his feet come to restsuddenly, whereas the upper part of his body retainsthe forward motion.

(c) Inertia of Direction :

The tendency of a body to oppose any change in itsdirection of motion is known as inertia of direction.

Example based on Inertia of direction :

Tie a stone to one end of a string and holding otherend of the string in hand, rotate the stone in a horizontalcircle. If during rotation, the string breaks at certainstage, the stone is found to fly off tangentially at thatpoint of the circle.

StringBreaks

String breaks, stone goes awaytangentially

Definition of force from first law of motion :

According to first law of motion, if there is no force, thereis no change in state of rest or of uniform motion. Inother words, if a force is applied, it may change the stateof rest or of uniform motion. If the force is not sufficient,it may not produce a change but only try to do so. Henceforce is that which changes or tries to change the stateof rest or of uniform motion of a body in straight line.


MOMENTUM

Definition :

Momentum of a particle may be defined as the quantityof motion possessed by it and it is measured by theproduct of mass of the particle and its velocity.

Momentum is a vector quantity and it is represented

by p

vmp

Unit of momentum :

(In C.G.S. system) p = mv gram × cm/s = dyne × s

(In M.K.S. system) p = mv kg × m/s = Newton × s

4. A ball of mass 100 gm. is moving with a velocity of15 m/s. Calculate the momentum associated with theball.

Sol .

Mass of the ball = 100 gm. = 1000100

kg.

= 0.1 kg.Velocity of the ball = 15 m/sSo, momentum = mass of the ball × velocity of

the ball= 0.1 kg. × 15 m/s

= 1.5 kg. m/s

NEWTON�S SECOND LAW OF MOTION

The rate of change of momentum of a body is directly

proportional to the applied unbalanced forces i.e. Rate

of change of momentum Force applied

Let a body is moving with initial velocity u and after applying

a force F on it, its velocity becomes v in time t.Initial momentum of the body p

1 = mu

Final momentum of the body p2 = mv

Change in momentum in time t is mv � mu

So rate of change of momentum = t

mu�mv

But according to Newton�s second law, t

mu�mv F

or Ft

)u�v(mHere,

tu�v

= a (acceleration)

So Fma

or F = kma (Here k is proportionality constant.

If 1N force is applied on a body of mass 1 kg and the

acceleration produced in the body is 1 m/s2, then

1 = k × 1 × 1 or k = 1

Hence, F = ma

So the magnitude of the resultant force acting on a

body is equal to the product of mass of the body and

the acceleration produced. Direction of the force is

same as that of the acceleration.

UNITS OF FORCE

(a) In C.G.S. System :

F = ma gm × cm/s2 = Dyne

Definition of one dyne :

If m = 1 gm, a = 1 cm/s2, then F = 1 dyne.

When a force is applied on a body of mass 1 gram and

the acceleration produced in the body is 1 cm/s2, then

the force acting on the body will be one dyne.

(b) In S.I. System :

F = ma kg × m/s2 = Newton

Definition of one Newton :

If m = 1 kg and a = 1 m/s2 then by, F = ma

F = 1 × 1 = 1 kg × m/s2 = 1 N.

If a force is applied on a body of mass 1 kg and

acceleration produced in the body is 1 m/s2, then the

force acting on the body will be one Newton.

(c) Kilogram Force (kgf) :

Kilogram force (kgf) or Kilogram weight (kg. wt.) is forcewith which a mass of 1 kg is attracted by theearth towards its centre.

1kgwt = 1kgf = 9.8 N

(d) Gram Force (gf) :

Gram force or gram weight is the force with which amass of 1 gram is attracted by the earth towardsits centre.

1gwt = 1gf = 981 dyne

Abou both the units are called gravitational unit of force.

Relation between Newton and dyne.

We know :1 N = 1kg × 1ms-2

or 1 N = 1000 g × 100 cms-2

or 1 N = 105 g cms-2 = 105 dyne 1 N = 105 dyne

5. A force of 20N acting on a mass m1 produces an

acceleration of 4 ms�2. The same force is applied onmass m

2 then the acceleration produced is 0.5 ms�2.

What acceleration would the same force produce, whenboth masses are tied together ?

Sol. For mass m1: F = 20N, a = 4 ms�2

then m1 =

aF

= 4

20 = 5 kg

For mass m2 : F = 20N, a = 0.5 ms�2


then m2 =

aF

= 5.0

20 = 40 kg

When m1 and m2 are tied together :

Total mass = m1 + m

2 = 45 kg, F = 20N

then a = )mm(

F

21 =

4520

= 0.44 ms�2

IMPULSE OF FORCE

A large force acting for a short time to produce a finitechange in momentum is called impulsive force.The product of force and time is called impulse of force.i.e., Impulse = Force × Time

or I = Ft

The S.I. unit of impulse is Newton-second (N-s) andthe C.G.S unit is dyne- second (dyne-s)

Impulse and Momentum :

From Newton�s second law of motion

Force, F = tpp 12

or Ft = p

2 � p

1

i.e., Impulse = Change in momentum

This relation is called impulse equation or momentum-impulse theorem. It has an important application inour everyday life.

IMPULSE DURING AN IMPACT OR COLLISION

The impulsive force acting on the body produces achange in momentum of the body on which it acts. Weknow, Ft = mv � mu, therefore the maximum force

needed to produce a given impulse depends upontime. If time is short, the force required in a givenimpulse or the change in momentum is large and vice-versa.

NEWTON�S THIRD LAW

(a) Statement :

The law states that � To every action there is an equal

and opposite reaction�. Moreover, action and reaction

act on different bodies.

(b) Demonstration :

Two similar spring balances A and B joined by hook asshown in the figure. The other end of the springbalance B is attached to a hook rigidly fixed in a rigidwall.

Demonstration- Newton�s third law of motion

The other end of the spring balance A is pulled out tothe left. Both balances show the same reading (20 N)for the force.

The pulled balance A exerts a force of 20N on thebalance B. It acts as action, B pulls the balance A inopposite direction with a force of 20 N. This force isknown as reaction.We conclude that action-reaction forces are equal andopposite and act on two different bodies.

NO ACTION IS POSSIBLE WITHOUT REACTION

Examples :

(i) A nail cannot be fixed on a suspended wooden ball.

(ii) A paper cannot be cut by scissors of single blade.

(iii) A hanging piece of paper cannot be cut by blade.

(iv) Writing on a hanging page is impossible.

(v) Hitting on a piece of sponge does not producereaction. You do not enjoy hitting.

ACTION AND REACTION ARE NOT BALANCED

Action and reaction, though equal and opposite arenot balanced because they act on two different bodies.In case when they act on two different bodies forminga single system, they become balanced.

ANY PAIR OF EQUAL AND OPPOSITE FORCES IS

NOT AN ACTION�REACTION PAIR

Consider a book kept on a table. We have seen thatthe table pushes the book in the upward direction. Thenwhy does not the book fly up? It does not fly up becausethere is another force on the book pulling it down. Thisis the force exerted by the earth on the book, which wecall the weight of the book. So, there are two forces onthe book� the normal force, N acting upwards, applied

by the table and the force, W acting downwards,applied by the earth. As the book does not accelerate,we conclude that these two forces are balanced. Inother words, they have equal magnitudes but oppositedirections.

VVV

N

N=W

Can we call N the action and W the reaction ? Wecannot. This is because, although they are equal and


opposite, they are not forces applied by two bodies oneach other. The force N is applied by the table on thebook, its reaction will be the force applied by the bookon the table. Weight W is the force applied by the earthon the book, its reaction will be the force applied by thebook on the earth.So, although N and W are equal and opposite, they donot form an action�reaction pair.

PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM

By Newton�s second law, the rate of change of

momentum is equal to the applied force.

timemomentuminChange

= Force

Change in momentum = F × t

If F = 0 then,

Change in momentum = 0

If the force applied on the body is zero then itsmomentum will be conserved, this law is alsoapplicable on the system. If in a system the momentumof the objects present in the system are P

1, P

2, P

3...........

and external force on the system is zero, then�P

1 + P

2 + P

3 +................. = Constant

NOTE : If only internal forces are acting on the systemthen its linear momentum will be conserved.(a) The Law of Conservation of Linear

Momentum by Third Law of Motion :

Suppose A and B are two objects of masses m1 and

m2 are moving in the same direction with velocity u

1

and u2 respectively (u

1 > u

2). Object A collides with object

B and after time t both move in their original directionwith velocity v

1 and v

2 respectively.

The change in momentum of object A = m1v

1 � m

1u

1

u1 u2

m1 m2

before collision (u > u )1 2

The force on B by A is F1 =


F1 =

t

um�vm 1111 .............(1)

The change in momentum of object B = m2v

2 � m

2u

2

The force on A by B is F2 =


= t

um�vm 2222 .............(2)

v1 v2

m1 m2

after collision

By Newtons third law, F1 = �F

2

t

um�vm 1111 =

t

um�vm� 2222 m

1v

1 � m

1u

1

= �m2v

2 + m

2u

2

or m1u

1 + m

2u

2 = m

1v

1 + m

2v

2

or Initial momentum = Final momentum

6. Two ice hockey players, suitably padded collide directlywith each other and immediately become entangled.One has a mass of 120 kg and is travelling at 2 m/s,while the other has a mass of 80 kg and is travelling at4 m/s towards the first player at what speed do theytravel after they become entangled?

Sol. m1u

1 + m

2u

2 = m

1v

1 + m

2v

2

120 × 2 � 80 × 4 = (120 + 80)

240 � 320 = 200 V

� 80 = 200 V

V = � 52

m/s

SOME ILLUSTRATION ON CONSERVATION

OF MOMENTUM

(a) Recoil of Gun :

A loaded gun (rifle) having bullet inside it forming onesystem is initially at rest. The system has zeroinitial momentum.

vV

When the trigger (T) is pressed, the bullet is fired due tointernal force of explosion of powder in cartidge inside.The bullet moves forward with a high velocity and the gunmove behind (recoils) with a lesser velocity.Let the bullet and the gun have masses m and Mrespectively. Let the bullet move forward with velocity vand the gun recoils with velocity V.Then final momentum of the gun and bullet is MV + mvBy the law of conservation of momentum�Initial momentum of the system = Final momentum ofthe system.

0 = MV + mv

or V = � M

mv

Hence the recoil velocity of gun = M

mv

and the velocity of the gun is = � M

mv


(b) The Working of a Rocket :

the momentum of a rocket before it is fired is zero.When the rocket is fired, gases are produced. Thesegases come out of the rear of the rocket with high speed.The direction of the momentum of the gases comingout of the rocket is in the downward direction. Thus, toconserve the momentum of the system i.e., (rocket +gases), the rocket moves upward with a momentumequal to the momentum of the gases. So, the rocketcontinues to move upward as long as the gases areejected out of the rocket. Thus a rocket works on thebasis of the law of conservation of momentum.

7. A bullet of mass 0.01 kg is fired from a gun weighing

5.0 kg. If the speed of the bullet is 250 m/s, calculate

the speed with which the gun recoils.

Sol. V = 52501

= � 50 m/s

SYSTEM

Two or more than two objects which interact with each

other form a system.

Classification of forces on the basis of boundary of

system :

(a) Internal Forces : Forces acting with in a system

among its constituents.

(b) External Forces : Forces exerted on the

constituents of a system by the outside

surroundings are called as external forces.

FREE BODY DIAGRAM

A free body diagram consists of a diagrammaticrepresentations of single body or a subsystem ofbodies isolated from surroundings showing all theforces acting on it.

Steps for F.B.D.

Step 1 : Identify the object or system and isolate it from

other objects, clearly specify its boundary.

Step 2 : First draw non-contact external force in the

diagram, generally it is weight.

Step 3 : Draw contact forces which acts at the boundary

of the object of system. Contact forces are normal ,

friction, tension and applied force. In F.B.D, internal

forces are not drawn only external are drawn.

8. A block of mass �m� is kept on the ground as shown in

figure.

(i) Draw F.B.D. of block.

(ii) Are forces acting on block forms action- reactionpair.

(iii) If answer is no, draw action reaction pair.

Sol.(i) F.B.D. of block

(ii) �N� and mg are not action -reaction pair. Since pair

act on different bodies, and they are of same

nature.

(iii) Pair of �mg� of block acts on earth in opposite

direction.

mg earth

and pair of �N� acts on surface as shown in figure.

N

9. Two sphere A and B are placed between two verticalwalls as shown in figure. Draw the free body diagramsof both the spheres.

A

B

Sol.F.B.D. of sphere �A� :

F.B.D. of sphere �B� :

(exerted by A)

Note : Here NAB

and NBA

are the action - reaction pair

(Newton�s third law).


10. Draw F.B.D. for systems shown in figure below.

Sol.

TRANSLATORY EQUILIBRIUM

When several forces acts on a body simultaneously insuch a way that resultant force on the body is zero, i.e.,

F

= 0 with F

=

iF the body is said to be in translatory

equilibrium. Here it is worthy to note that :

(i) As if a vector is zero all its components must vanishi.e. in equilibrium as -

F

= 0 with F

=

iF = 0

xF = 0 ; y

F = 0 ; z

F = 0

So in equilibrium forces along x axes must balanceeach other and the same is true for other directions.If a body is in translatory equilibrium it will be either atrest or in uniform motion. If it is at rest, equilibrium iscalled static, otherwise dynamic.

Static equilibrium can be divided into following threetypes :

(a) Stable equilibrium :

If on slight displacement from equilibrium position abody has a tendency to regain its original position it issaid to be in stable equilibrium. In case of stableequilibrium potential energy is minimum and so centerof gravity is lowest.

O

(b) Unstable equilibrium : If on slight displacementfrom equilibrium position a body moves in the directionof displacement, the equilibrium is said to be unstable.In this situation potential energy of body is maximumand so center of gravity is highest.

O

(c) Neutral equilibrium : If on slight displacement fromequilibrium position a body has no tendency to comeback to its original position or to move in the directionof displacement, it is said to be in neutral equilibrium.In this situation potential energy of body is constantand so center of gravity remains at constant height.

.

(a) Newtons 2nd law of motion :

The rate of change of linear momentum of a body isdirectly proportional to the applied force and the changetakes place in the direction of the applied force.

In relation

F =

ma the force

F stands for the net

external force. Any internal force in the system is not to

be included in

F .

In S.I. the absolute unit of force is newton (N) andgravitational unit of force is kilogram weight or kilogramforce (kgf.)

Note : The absolute unit of force remains the sameeverywhere, but the gravitational unit of force variesfrom place to place because it depends on the value of g.

(b) Applications of Newton�s 2nd Law

(i) When objects are in equilibrium :Steps to solve problem involving objects inequilibrium :

Step 1 : Make a sketch of the problem.

Step 2 : Isolate a single object and then draw the free-

body diagram for the object. Label all external forces

acting on it.

Step 3 : Choose a convenient coordinate system and

resolve all forces into rectangular components along x

and Y direction.

Step 4 : Apply the equations 0Fx and 0Fy .

Step 5 : Step 4 will give you two equations with several

unknown quantities. If you have only two unknown

quantities at this point, you can solve the two equations


for those unknown quantities.

Step 6 : If step 5 produces two equations with more

than two unknowns, go back to step 2 and select

another object and repeat these steps. Eventually at

step 5 you will have enough equations to solve for all

unknown quantities.

11. A �block� of mass 10 kg is

suspended with string asshown in figure. Find tension in the string.(g = 10 m/s2).

Sol.F.B.D. of blockFor equilibrium of block along Y axis

0Fy

T � 10 g = 0 T = 100 N

12. The system shown in figure is in equilibrium. Find themagnitude of tension in each string ; T

1 , T

2, T

3 and T

4.

(g = 10 m/s2).

Sol.F.B.D. of 10 kg block

For equilibrium of block along Y axis.

0Fy

T0

10g

T0 = 10 g

T0 = 100 N

F.B.D. of point �A�

0Fy 30º

T2

T1x

T0

A

y

T2 cos 30º = T

0 = 100 N

T2 =

3

200N

0Fx

T1 = T

2 . sin 30º

= 3

200.

21

= 3

100N.

F.B.D. of point of �B�

60ºT4

T3x

B

y

30º

T2

yF = 0 T4 cos 60º = T

2 cos 30º

T4 = 200 N

and xF = 0 T3 + T

2 sin30º = T

4 sin 60º

T3 =

3

200N

13. Two blocks are kept in contact as shown in figure. Find :-

(a) forces exerted by surfaces (floor and wall) on

blocks.

(b) contact force between two blocks.

Sol. A : F.B.D. of 10 kg block

N1 = 10 g = 100 N .......(1)

N2 = 100 N .........(2)

F.B.D. of 20 kg block

N2 = 50 sin 30º + N

3

N3 = 100 � 25 = 75 N

& N4 = 50 cos 30º + 20 g

N4 = 243.30 N

14. Find magnitude of force exerted by string on pulley.

Sol B. F.B.D. of 10 kg block :

T = 10 g = 100 NF.B.D. of pulley :


Since string is massless, so tension in both sidesof string is same.So magnitude of force exerted by string on pulley

= 22 100100 = 100 2 N

Note : Since pulley is in equilibrium position, so netforces on it is zero.

(ii) Accelerating Objects :

Steps to solve problems involving objects that are in

accelerated motion :

Step 1 : Make a sketch of the problem.Step 2 : Isolate a single object and then draw the free- body diagram for that object. Label all external forcesacting on it. Be sure to include all the forces acting onthe chosen body, but be equally careful not to includeany force exerted by the body on some other body.Some of the forces may be unknown , label them withalgebraic symbols.Step 3 : Choose a convenient coordinate system, showlocation of coordinate axis explicitly in the free - bodydiagram, and then determine components of forceswith reference to these axis and resolve all forces intox and y components.

Step 4 : Apply the equations xF = max & yF = ma

y.

Step 5 : Step 4 will give two equations with severalunknown quantities. If you have only two unknownquantities at this point, you can solve the two equationsfor those unknown quantities.Step 6 : If step 5 produces two equations with morethan two unknowns, go back to step 2 and selectanother object and repeat these steps. Eventually atstep 5 you will have enough equations to solve for allunknown quantities.

15. A force F is applied horizontally on mass m1 as shown

in figure. Find the contact force between m1 and m

2.

Sol.Considering both blocks as a system to find thecommon acceleration.Common acceleration

a = 21 mmF

.......(1)

m1 m2F a

To find the contact force between �A� and �B� we draw

F.B.D. of mass m2.

F.B.D. of mass m2

xF = max

N = m2 . a

N = 21

2

mmFm

21 mmF

acesin

16. A 5 kg block has a rope of mass 2 kg attached to itsunderside and a 3 kg block is suspended from the otherend of the rope. The whole system is acceleratedupward at 2 m/s2 by an external force F

0.

(a) What is F0 ?

(b) What is the net force on rope ?

(c) What is the tension at middle point of the rope ?(g = 10 m/s2)

Sol.For calculating the value of F0.

F.B.D of whole system

(a) 2m/s2

F0

10 g = 100 N

F0 �100 = 10 × 2

F0 = 120 N ........(1)

(b) According to Newton�s second law, net force on

rope.

F = ma = 2 × 2

= 4N ............(2)

(c) For calculating tension at the middle point we draw

F.B.D. of 3 kg block with half of the rope (mass 1

kg) as shown.

T � 4 g = 4 . 2

T = 48 N

17. A block of mass 50 kg is kept on another block of mass

1 kg as shown in figure. A horizontal force of 10 N is

applied on the 1Kg block. (All surface are smooth).

Find : (g = 10 m/s2)

(a) Acceleration of blocks A and B.


(b) Force exerted by B on A.

50 kg 1 kg A

B

Sol.(a) F.B.D. of 50 kg

N2 = 50 g = 500 N

along horizontal direction, there is no force aB = 0

(b) F.B.D. of 1 kg block :

N1 N2

10 N

1g

along horizontal direction

10 = 1 aA.

aA

= 10 m/s2

along vertical direction

N1 = N

2 + 1g

= 500 + 10 = 510 N

18. One end of string which passes through pulley and

connected to 10 kg mass at other end is pulled by 100

N force. Find out the acceleration of 10 kg mass. (g

=9.8 m/s2)

Sol.Since string is pulled by 100 N force. So tension in the

string is 100 N

F.B.D. of 10 kg block

100 � 10 g = 10 a

100 � 10 × 9.8 = 10 a

a = 0.2 m/s2.

19. A man of mass m stands on a platform of equalmass m and pulls himself by two ropes passingover pulleys as shown. If he pulls each rope with aforce equal to half his weight, find his upwardacceleration ?

Sol. for (man + platform) system :2mg � 4T = 2m(a)

2mg � 4

2

mg = 2m (a) [ T =

2mg

]

a = 0

WEIGHING MACHINE

A weighing machine does not measure the weight but

measures the force exerted by object on its upper

surface.

20. A man of mass 60 Kg is

standing on a weighing

machine placed on ground.

Calculate the reading of

machine (g = 10 m/s2). weighing machine

Sol.For calculating the reading of weighing machine, we

draw F.B.D. of man and machine separately.

F.B.D of man

N

Mg

N = Mg

F.B.D of man taking mass of man as M

weighing machine N

N1Mg

F.B.D. of weighing machine


Here force exerted by object on upper surface is N

Reading of weighing machine

N = Mg

= 60 × 10

N = 600 N.

SPRING BALANCE

It does not measure the weight. It measures the force

exerted by the object at the hook. Symbolically, it is

represented as shown in figure.

A block of mass �m� is suspended at hook. When spring

balance is in equilibrium, we draw the F.B.D. of mass

m for calculating the reading of balance.

m

spring balance

hook

F.B.D. of �m�.

mg � T = 0

T = mgMagnitude of T gives the reading of spring balance.

21. A block of mass 20 kg is suspended through two lightspring balances as shown in figure . Calculate the :

(1) reading of spring balance (1).

(2) reading of spring balance (2).

Sol.For calculating the reading, first we draw F.B.D.of 20 kg

block.

F.B.D. 20 kg

T

20 g

mg � T = 0

T = 20 g = 200 NSince both the balances are light so, both the scaleswill read 200 N.

22. (i) A 10 kg block is supported by a cord that runs to aspring scale, which is supported by another cordfrom the ceiling figure (a). What is the reading onthe scale ?

(ii) In figure (b) the block is supported by a cord thatruns around a pulley and to a scale. The oppositeend of the scale is attached by cord to a wall. Whatis the reading of the scale.

(iii) In figure (c) the wall has been replaced with asecond 10 kg block on the left, and the assembly isstationary. What is the reading on the scale now ?

spring balance

hook

10 kg

T

T

(a)

T

TT

(b)

10kg

T

TT

T

10kg10kg

(c)

Sol. In all the three cases the spring balance reads 10 kg.To understand this let us cut a section inside the springas shown;


As each part of the spring is at rest, so F= T. As theblock is stationary, so T= 10g = 100N.

23. A block of mass m is placed on a weighing machineand it is also attached with a spring balance. Thewhole system is placed in a lift as shown in figure. Liftis moving with constant acceleration 10 m/s2 in upwarddirection. The reading of the weighing machine is 500N and the reading of spring balance is 300 N. Find theactual mass of the block in kg. (Take g = 10 m/s2)

Sol. m = mass of the blockApplying Newton's Law on the block in vertical direction

T + N � mg = ma ...............(1)

T = 300 N (given reading of spring balance)

N = 500 N (given reading of W.M.)a = 10 m/s2

m = agNT

=

20300500

m = 40 kg Ans.

24. Pull is easier than push

Push : Consider a block of mass m placed onrough horizontal surface. The coefficient of staticfriction between the block and surface is . Let apush force F is applied at an angle with thehorizontal.

As the block is in equilibrium along y-axis, so we have

;0Fy

or N = mg + F sin

To just move the block along x-axis, we have

F cos = N = (mg + F sin )

or F =

sin�cosmg

.......(i)

Pull : Along y-axis we have ;

;0Fy

N = mg � F sin

To just move the block along x-axis, we have

F cos = N = (mg � F sin )

or F =

sincosmg

. .......(ii)

It is clear from above discussion that pull force is

smaller than push force.

25. Discuss the direction of friction in the following cases :

(i) A man walks slowly, without change in speed.

(ii) A man is going with increasing speed.

(iii) When cycle is gaining speed.

(iv) When cycle is slowing down .

Sol. (i) Consider a man walks slowly without acceleration,

and both the legs are touching the ground as

shown in figure (a). The frictional force on rear leg

is in forward direction and on front leg will be on

backward direction of motion.

As a = 0,

Fnet

= 0 or f1 � f

2 = 0

f1 = f

2& N

1 = N

2.

N1 N2

(b)

Ground

f1 f2f1 f2

N1 N2

(ii) When man is gaining the speed : The frictionalforce on rear leg f

1 will be greater than frictional

force on front leg f2 (fig. b).

acceleration of the man, a = m

ff 21 .


(iii) When cycle is gaining speed : In this case torque

is applied on the rear wheel of the cycle by the

chain-gear system. Because of this the slipping

tendency of the point of contact of the rear wheel is

backward and so friction acts in forward direction.

The slipping tendency of point of contact of front

wheel is forward and so friction acts in backward

direction. If f1 and f

2 are the frictional forces on rear

and front wheel, then acceleration of the cycle a =

Mf�f 21 , where M is the mass of the cycle together

with rider (fig. a).

N1 N2

f1 f2

(a)

N1 N2

f1 f2

(b)

(iv) When cycle is slowing down : When torque is not

applied (cycle stops pedaling), the slipping

tendency of points of contact of both the wheels

are forward, and so friction acts in backward

direction (fig. b). If f1 and f

2 are the frictional forces

on rear and front wheel, then retardation

a = M

ff 21

26. A block of mass 25 kg is raised by a 50 kg man in two

different ways as shown in fig.. What is the action on

the floor by the man in the two cases ? If the floor yields

to a normal force of 700 N, which mode should the

man adopt to lift the block without the floor yielding.

50g

50g

Sol. The FBD for the two cases are shown in figure.

In Ist case, let the force exerted by the man on the floor is

N1. Consider the forces inside the dotted box, we have

N1 = T + 50 g.

Block is to be raised without acceleration, so

T = 25 g.

N1 = 25 g + 50 g

= 75 g = 75 × 9.8 = 735 N

In IInd case, let the force exerted by the man on the floor

in N2 . Consider the forces inside the dotted box, we

have

N2 = 50 g � T

and T = 25 g

N2 = 50 g � 25 g

= 25 g = 25 × 9.8 = 245 N.

As the floor yields to a downward force of 700 N, so the

man should adopt mode .

27. Figure shows a weighing machine kept in a lift is

moving upwards with acceleration of 5 m/s2. A block is

kept on the weighing machine. Upper surface of block

is attached with a spring balance. Reading shown by

weighing machine and spring balance is 15 kg and 45

kg respectively.

Answer the following questions. Assume that theweighing machine can measure weight by havingnegligible deformation due to block, while the springbalance requires larger expansion. (take g = 10 m/s2)

(i) Find the mass of the object in kg and the normalforce acting on the block due to weighing machine?

(ii) Find the acceleration of the lift such that weighingmachine shows its true weight ?

Sol. (i)

T + N � Mg = Ma

45 g + 15 g = M(g + a)

450 + 150 = M(10 + 5)

M = 40 kg

Normal force is the reaction applied by weighing

machine i.e. 15 × 10 = 150 N.


(ii)

T + N � Mg = Ma

45 g + 40 g = 40(g + a)

450 + 400 = 400 +40 a

a = 40

450=

445

m/s2

EXERCISE

1. Two blocks of masses 2 kg and 1 kg are placedon a smooth horizontal table in contact witheach other. A horizontal force of 3 newton isapplied on the first so that the blocks move witha constant acceleration. The force between theblocks would be :(A) 3 Newton (B) 2 Newton(C) 1 Newton (D) Zero

2. An object of mass 10 kg. moves at a constantspeed of 10ms�1. A constant force acts for 4 son the object and gives it a speed of 2 ms�1 inopposite direction. The force acting on the ob-ject is : (A) � 3N

(B) �30 N (C) 3 N(D) 30 N

3. A machine gun has a mass 5kg. It fires 50 grambullets at the rate of 30 bullets per minute at aspeed of 400 ms�1. What force is required tokeep the gun in position :(A) 10 N (B) 5 N(C) 15 N (D) 30N

4. A particle of mass m1 moving with velocity v collides

with a mass m2 at rest, then they get embedded.

Just after collision, velocity of the system :(A) Increases(B) Decreases(C) remains constant(D) becomes zero

5. When a car turns on a curved road, you are pushedagainst one of the doors of the car because of :

(IJSO/Stage-I/2012)(A) inertia(B) the centripetal force(C) the centrifugal force(D) the frictionaI force

6. What is the reading of the spring balance shownin the figure below? (IJSO/Stage-I/2012)

T

TT

0.2kg

(A) 0 (B) 2N(C) 4N (D) 6N

7. Two blocks are in contact on a frictionless table.

One has mass m and the other 2m.A force F is

applied on 2m as shown in the figure. Now the

same force F is applied from the right on m. In the

two cases respectively, the ratio of force of contact

between the two blocks will be :

(A) Same (B) 1 : 2

(C) 2 : 1 (D) 1 : 3

8. Two forces of 6N and 3N are acting on the two blocks of

2kg and 1kg kept on frictionless floor. What is the force

exerted on 2kg block by 1kg block ?:

2kg 1kg

6N3N

(A)1N (B) 2N

(C) 4N (D) 5N

9. There are two forces on the 2.0 kg box in the overheadview of figure but only one is shown. The second forceis nearly :

y

30º

F = 20 N1

a = 12 m/s2

x

(A) �20 j�N (B) � 20 i� + 20 j� N

(C) �32 i� � 12 3 j�N (D) �21 i� � 16 j�N

10. A dish of mass 10 g is kept horizontally in air by firingbullets of mass 5 g each at the rate of 100 per second.If the bullets rebound with the same speed, what is thevelocity with which the bullets are fired :(A) 0.49 m/s (B) 0.098 m/s(C) 1.47 m/s (D) 1.96 m/s


11. A block of metal weighing 2 kg is resting on africtionless plank. If struck by a jet releasing water at arate of 1 kg/s and at a speed of 5 m/s. The initialacceleration of the block will be :(A) 2.5 m/s2 (B) 5.0 m/s2

(C) 10 m /s2 (D) none of the above

12. A constant force F is applied in horizontal direction asshown. Contact force between M and m is N andbetween m and M� is N� then

(A) N= N� (B) N > N�(C) N�> N

(D) cannot be determined

13. STATEMENT-1 : Block A is moving on horizontal surfacetowards right under action of force. All surface aresmooth. At the instant shown the force exerted by blockA on block B is equal to net force on block B.

STATEMENT-2 : From Newtons�s third law, the force

exerted by block A on B is equal in magnitude to forceexerted block B on A(A) statement-1 is true, Statement 2 is true, statement-2 is correct explanation for statement-1.(B) statement-1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1.(C) statement-1 is true, Statement 2 is false(D) statement-1 is False, Statement 2 is True

14. A certain force applied to a body A gives it an acceleration

of 10 ms�2 . The same force applied to body B gives it

an acceleration of 15 ms�2 . If the two bodies are joined

together and same force is applied to the combination,

the acceleration will be :

(IJSO/Stage-I/2011)

(A) 6 ms�2 (B) 25 ms�2

(C) 12.5 ms�2 (D) 9 ms�2

15. Four blocks are kept in a row on a smooth horizontaltable with their centres of mass collinear as shown inthe figure. An external force of 60 N is applied from lefton the 7 kg block to push all of them along the table.The forces exerted by them are :(IAO/Sr./Stage-I/2008)

7 kg 5 kg 2 kg 1 kg60N

P Q R S

(A) 32 N by P on Q (B) 28 N by Q on P(C) 12 N by Q on R (D) 4 N by S on R

16. A mass M is suspended by a rope from a rigid supportat A as shown in figure. Another rope is tied at the endB, and it is pulled horizontally with a force F. If the ropeAB makes an angle with the vertical inequilibrium,then the tension in the string AB is :

(A) F sin (B) F /sin (C) F cos (D) F / cos

17. In the system shown in the figure, the acceleration ofthe 1kg mass and the tension in the string connectingbetween A and B is :

(A) g4

downward, 8g7

(B) g4

upward, g7

(C) g7

downward, 67

g (D) g2

upward, g

18. A body of mass 8 kg is hanging from another body ofmass 12 kg. The combination is being pulled by astring with an acceleration of 2.2 m s�2. The tension T

1

and T2 will be respectively :(Use g =9.8 m/s2)

(A) 200 N, 80 N (B) 220 N, 90 N(C) 240 N, 96 N (D) 260 N, 96 N

19. Two masses M1 and M

2 are attached to the ends of a

light string which passes over a massless pulleyattached to the top of a double inclined smooth planeof angles of inclination and . If M

2 > M

1 then the

acceleration of block M2 down the inclined will be :

(A) 2

1 2

M (sin )

M M

g (B)

1

1 2

M g(sin )

M M

(C)

21

12

MMsinMsinM

g (D) Zero


20. Three masses of 1 kg, 6 kg and 3 kg are connected to

each other by threads and are placed on table as

shown in figure. What is the acceleration with which

the system is moving ? Take g = 10 m s�2:

(A) Zero (B) 1 ms�2

(C) 2 m s�2 (D) 3 m s�2

21. The pulley arrangements shown in figure are identical

the mass of the rope being negligible. In case I, the

mass m is lifted by attaching a mass 2m to the other

end of the rope. In case II, the mass m is lifted by

pulling the other end of the rope with a constant

downward force F= 2 mg, where g is acceleration due

to gravity. The acceleration of mass in case I is :

(A) Zero

(B) More than that in case II

(C) Less than that in case II

(D) Equal to that in case II

22. A 50 kg person stands on a 25 kg platform. He pullsmassless rope which is attached to the platform viathe frictionless, massless pulleys as shown in thefigure. The platform moves upwards at a steady velocityif the force with which the person pulls the rope is :

(A) 500 N (B) 250 N(C) 25 N (D) 50 N

23. Figure shows four blocks that are being pulled along a

smooth horizontal surface. The mssses of the blocks

and tension in one cord are given. The pulling force F is :

4kg 3kg 2kg 1kg60º

F30N

(A) 50 N (B) 100 N

(C) 125 N (D) 200 N

24. A10 kg monkey climbs up a massless rope that runs

over a frictionless tree limb and back down to a 15 kg

package on the ground. The magnitude of the least

acceleration the monkey must have if it is to lift the

package off the ground is :

(A) 4.9 m/s2 (B) 5.5 m/s2

(C) 9.8 m/s2 (D) none of these

25. Two blocks, each of mass M, are connected by a

massless string, which passes over a smooth

massless pulley. Forces F

act on the blocks as shown.

The tension in the string is :

(A) Mg (B) 2 Mg

(C) Mg + F (D) none of these

26. Two blocks of mass m each is connected with the

string which passes over fixed pulley, as shown in figure.

The force exerted by the string on the pulley P is :

(A) mg (B) 2 mg

(C) 2 mg (D) 4 mg


27. One end of a massless rope, which passes over a

massless and frictionless pulley P is tied to a hook C

while the other end is free. Maximum tension that rope

can bear is 360 N, with what minimum safe

acceleration (in m/s2) can a monkey of 60 kg move

down on the rope :

P

C

(A) 16 (B) 6

(C) 4 (D) 8

28. Which figure represents the correct F.B.D. of rod of

mass m as shown in figure :

(A) (B)

(C) (D) None of these

29. Two persons are holding a rope of negligible weighttightly at its ends so that it is horizontal. A 15 kg weightis attached to the rope at the mid point which now nolonger remains horizontal. The minimum tensionrequired to completely straighten the rope is :(A) 15 kg

(B) kg2

15

(C) 5 kg(D) Infinitely large (or not possible)

30. In the figure, the blocks A, B and C of mass each haveacceleration a

1 , a

2 and a

3 respectively . F

1 and F

2 are

external forces of magnitudes 2 mg and mgrespectively then which of the following relations iscorrect :

(A) a1 = a

2 = a

3(B) a

1 > a

2 > a

3

(C) a1 = a

2 , a

2 > a

3(D) a

1 > a

2 , a

2= a

3

31. A weight is supported by two strings 1.3 and 2.0 mlong fastened to two points on a horizontal beam 2.0m apart. The depth of this weight below the beam is :

(IAO/Jr./Stage-I/2007)

(A) 1.0 m (B) 1.23 m

(C) 0.77 m (D) 0.89 m

32. A fully loaded elevator has a mass of 6000 kg. Thetension in the cable as the elevator is accelerateddownward with an acceleration of 2ms�2 is (Take g = I0ms�2)

(A) 7·2 × 104 N (B) 4.8 × 104 N(C) 6 × 104 N (D) 1.2 × 104 N

33. A light string goes over a frictionless pulley. At its oneend hangs a mass of 2 kg and at the other end hangsa mass of 6 kg. Both the masses are supported byhands to keep them at rest. When the masses arereleased, they being to move and the string gets taut.(Take g = 10 ms�2) The tension in the string during themotion of the masses is :(A) 60 N (B) 30 N(C) 20 N (D) 40 N

34. In the given figure. What is the reading of the springbalance:

(A) 10 N (B) 20 N(C) 5 N (D) Zero

35. Two bodies of masses M1 and M

2 are connected to

each other through a light spring as shown in figure. Ifwe push mass M

1 with force F and cause acceleration

a1 in mass M

1 what will be the acceleration in M

2 ?

(A) F/M2

(B) F/(M1 + M

2)

(C) a1

(D) (F�M1a

1)/M

2

36. A spring balance is attached to 2 kg trolley and is usedto pull the trolly along a flat surface as shown in the fig.The reading on the spring balance remains at 10 kgduring the motion. The acceleration of the trolly is (Useg= 9.8 m�2) :

(A) 4.9 ms�2 (B) 9.8 ms�2

(C) 49 ms�2 (D) 98 ms�2


37. A body of mass 32 kg is suspended by a spring balancefrom the roof of a vertically operating lift and goingdownward from rest. At the instants the lift has covered20 m and 50 m, the spring balance showed 30 kg & 36kg respectively. The velocity of the lift is :(A) Decreasing at 20 m & increasing at 50 m(B) Increasing at 20 m & decreasing at 50 m(C) Continuously decreasing at a constant ratethroughout the journey(D) Continuously increasing at constant rate throughoutthe journey

38. A ship of mass 3 × 107 kg initially at rest is pulled by a

force of 5 × 104 N through a distance of 3m. Assume

that the resistance due to water is negligible, the speed

of the ship is :

(A) 1.5 m/s (B) 60 m/s

(C) 0.1 m/s (D) 5 m/s

39. When a horse pulls a cart, the force needed to move

the horse in forward direction is the force exerted by :

(A) The cart on the horse

(B) The ground on the horse

(C) The ground on the cart

(D) The horse on the ground40. A 2.5 kg block is initially at rest on a horizontal surface.

A 6.0 N horizontal force and a vertical force P

are appliedto the block as shown in figure. The coefficient of staticfriction for the block and surface is 0.4. The magnitudeof friction force when P = 9N : (g = 10 m/s2)

(A) 6.0 N (B) 6.4 N(C) 9.0 N (D) zero

41. The upper half of an inclined plane with inclination isperfectly smooth while the lower half is rough. A bodystarting from rest at the top will again come to rest at thebottom, if the coefficient of friction for the lower half is :(A) 2 tan (B) tan

(C) 2 sin (D) 2 cos

42. Minimum force required to pull the lower block is (takeg = 10 m/s2) :

(A) 1 N (B) 5 N(C) 7 N (D) 10 N

43. N bullets each of mass m are fired with a velocity v m/s at the rate of n bullets per sec., upon a wall. If thebullets are completely stopped by the wall, the reactionoffered by the wall to the bullets is :(A) N m v / n (B) n m v(C) n N v / m (D) n v m / N

44. A vehicle of mass m is moving on a rough horizontalroad with momentum P. If the coefficient of frictionbetween the tyres and the road be , then the stoppingdistance is :

(A) mg2P

(B) mg2

P2

(C) gm2

P2

(D) gm2

P2

2

45. What is the maximum value of the force F such that theblock shown in the arrangement, does not move :

F60º

12 3

m = 3kg

(A) 20 N (B) 10 N(C) 12N (D) 15 N

46. A bock of mass 5 kg is held against wall by applying ahorizontal force of 100N. If the coefficient of frictionbetween the block and the wall is 0.5, the frictionalforce acting on the block is : (g =9.8 m/s2)

5kg100N

(A)100 N (B) 50 N(C) 49 N (D) 24.9 N

47. A heavy roller is being pulled along a rough road asshown in the figure. The frictional force at the point ofcontact is : (IAO/Jr./Stage-I/2007)

F

(A) parallel to F (B) opposite to F(C) perpendicular to F (D) zero

48. When a motor car of mass 1500 kg is pushed on aroad by two persons, it moves with a small uniformvelocity. On the other hand if this car is pushed on thesame road by three persons, it moves with anacceleration of 0.2 m/s2. Assume that each person isproducing the same muscular force. Then, the force offriction between the tyres of the car and the surface ofthe road is : (IAO/Jr./Stage-I/2009)

(A) 300 N (B) 600 N(C) 900 N (D) 100 N


49. A block of mass M is at rest on a plane surface inclinedat an angle to the horizontal The magnitude of forceexerted by the plane on the block is :(A) Mg cos (B) Mg sin (C) Mg tan (D) Mg

50. A block of mass M rests on a rough horizontal table. A

steadily increasing horizontal force is applied such that

the block starts to slide on the table without toppling.

The force is continued even after sliding has started.

Assume the coefficients of static and kinetic friction

between the table and the block to be equal. The cor-

rect representation of the variation of the frictional

forces, �, exerted by the table on the block with time t is

given by :

(A) (B)

(C) (D)

51. A small child tries to move a large rubber toy placed on

the ground. The toy does not move but gets deformed

under her pushing force )F(

which is obliquely upward

as shown . Then

(A) The resultant of the pushing force )F(

, weight of

the toy, normal force by the ground on the toy and the

frictional force is zero.

(B) The normal force by the ground is equal and oppo-

site to the weight of the toy.

(C) The pushing force )F(

of the child is balanced by

the equal and opposite frictional force

(D) The pushing force )F(

of the child is balanced by

the total internal force in the toy generated due to

deformation

52. On a horizontal frictional frozen lake, a girl (36 kg) anda box (9kg) are connected to each other by means of arope. Initially they are 20 m apart. The girl exerts ahorizontal force on the box, pulling it towards her. Howfar has the girl travelled when she meets the box ?(A) 10 m(B) Since there is no friction, the girl will not move(C) 16 m(D) 4m

53. Which of the following does NOT involve friction ?(IJSO/Stage-I/2011)

(A) Writing on a paper using a pencil(B) Turning a car to the left on a horizontal road.(C) A car at rest parked on a sloping ground(D) Motion of a satellite around the earth.

54. In the two cases shown below, the coefficient of kineticfriction between the block and the surface is the same,and both the blocks are moving with the same uniformspeed. Then, (IAO/Sr./Stage-I/2008)

F1

F2

(A) F1 = F

2

(B) F1 < F

2

(C) F1 > F

2

(D) F1 = 2F

2 if sin = Mg/4F

2

55. The ratio of the weight of a man in a stationary lift andwhen it is moving downward with uniform acceleration�a� 3:2. The value of �a� is : (g = acceleration, due to

gravity)(A) (3/2)g (B) g(C) (2/3) g (D) g/3

56. A person standing on the floor of an elevator drops acoin. The coin reaches the floor of the elevator in timet1 when elevator is stationary and in time t2 if it is movinguniformly. Then(A) t1 = t2(B) t1 > t2(C) t1 < t2(D) t1 < t2 or t1 > t2 depending

57. STATEMENT-1 : A man standing in a lift which is moving

upward, will feel his weight to be greater than when

the lift was at rest.

STATEMENT-2 : If the acceleration of the lift is �a� upward

then the man of mass m shall feel his weight to be

equal to normal reaction (N) exerted by the lift given N

= m(g+a) (where g is acceleration due to gravity

(A) statement-1 is true, Statement 2 is true, statement-

2 is correct explanation for statement 1.

(B) statement-1 is true, Statement 2 is true, statement-

2 is NOT a correct explanation for statement-1.

(C) statement-1 is true, Statement 2 is false

(D) statement-1 is False, Statement 2 is True


58. A beaker containing water is placed on the platform of

a digital weighing machine. It reads 900 g. A wooden

block of mass 300 g is now made to float in water in

the beaker (without touching walls of the beaker). Half

the wooden block is submerged inside water. Now,

the reading of weighing machine will be :

(IAO/Jr./Stage-I/2009)

(A) 750 g (B) 900 g

(C) 1050 g (D) 1200 g

59. An object will continue accelerating until :(A) Resultant force on it begins to decreases(B) Its velocity changes direction(C) The resultant force on it is zero(D) The resultant force is at right angles to its directionof motion

60. In which of the following cases the net force is not zero ?(A) A kite skillfully held stationary in the sky(B) A ball freely falling from a height(C) An aeroplane rising upward at an angle of 45° with

the horizontal with a constant speed(D) A cork floating on the surface of water.

61. Figure shows the displacement of a particle going

along the X-axis as a function of time. The force acting

on the particle is zero in the region.

(A) AB (B) BC

(C) CD (D) DE

62. A 2 kg toy car can move along x axis. Graph shows force

Fx, acting on the car which begins to rest at time t = 0. The

velocity of the car at t = 10 s is :

(A) � i� m/s (B) � 1.5 i� m/s

(C) 6.5 i� m/s (D) 13 i� m/s

63. Figure shows the displacement of a particle going

along the x-axis as a function of time :

(A) The force acting on the particle is zero in the region AB

(B) The force acting on the particle is zero in the region BC

(C) The force acting on the particle is zero in the region CD

(D) The force is zero no where

64. A force of magnitude F1 acts on a particle so as to

accelerate if from rest to velocity v. The force F1 is then

replaced by another force of magnitude F2 which

decelerates it to rest.

(A) F1 must be the equal to F2

(B) F1 may be equal to F2

(C) F1 must be unequal to F2

(D) None of these

65. In a imaginary atmosphere, the air exerts a small forceF on any particle in the direction of the particle�s motion.

A particle of mass m projected upward takes a time t1in reaching the maximum height and t2 in the returnjourney to the original point. Then(A) t1 < t2(B) t1 > t2(C) t1 = t2(D) The relation between t1 and t2 depends on the massof the particle

66. A single force F of constant magnitude begins to act ona stone that is moving along x axis. The stone continuesto move along that axis. Which of the followingrepresents the stone�s position ?

(A) x = 5t � 3 (B) x = 5t2 + 8t � 3

(C) x = �5t2 + 5t � 3 (D) x = 5t3 + 4t2 � 3

67. Three forces act on a particle that moves with

unchanging velocity v = (3 i� � 4 j� ) m/s. Two of the

forces are 1F

= (3 i� + 2 j� � 4 k� ) N and 2F

= (�5 i� + 8 j�

+ 3 k� ) N. The third force is :

(A) (�2 i� + 10 j� � 7 k� ) N(B) (2 i� � 10 j� + k� ) N(C) (7 i� � 2 k� + 10 j� ) N(D) none of these

68. An 80 kg person is parachuting and experiencing a

downward acceleration of 2.5 m/s2 . The mass of the

parachute is 5.0 kg. The upward force on the open

parachute from the air is :

(A) 620 N (B) 740 N

(C) 800 N (D) 920 N


69. A block of mass m is pulled on the smooth horizontalsurface with the help of two ropes, each of mass m,connected to the opposite faces of the block. Theforces on the ropes are F and 2F. The pulling force onthe block is :

(A) F (B) 2F(C) F/3 (D) 3F/2

70. A body of mass 5 kg starts from the origin with an initialvelocity u

= 30 i� + 40 j� ms�1 . If a constant force

F

= �( i� + 5 j� ) N acts on the body, the time in which they-component of the velocity becomes zero is :(A) 5 s (B) 20 s(C) 40 s (D) 80 s

71. STATEMENT-1 :According to the newton�s third law of

motion, the magnitude of the action and reaction forceis an action reaction pair is same only in an inertialframe of reference.

STATEMENT-2 : Newton �s laws of motion are

applicable in every inertial reference frame.(A) statement-1 is true, Statement 2 is true, statement-2 is correct explanation for statement 1.(B) statement-1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1.(C) statement-1 is true, Statement 2 is false(D) statement-1 is False, Statement 2 is True

72. A body of mass 10 g moves with constant speed 2 m/s along a regular hexagon. The magnitude of changein momentum when the body crosses a corner is :

(IAO/Sr./Stage-I/2007)(A) 0.04 kg-m/s (B) zero(C) 0.02 kg-m / s (D) 0.4 kg-m/s

73. An object with uniform density is attached to a springthat is known to stretch linearly with applied force asshown below

When the spring object system is immersed in aliquid of density

1 as shown in the figure, the spring

stretches by an amount x1 ( >

1). When the experiment

is repeated in a liquid of density 2 <

1 . the spring is

stretched by an amount x2. Neglecting any buoyant force

on the spring, the density of the object is:

(A) 21

2211xx

xx

(B)

12

1221xx

xx

(C) 21

1221xx

xx

(D)

21

2211xx

xx

74. A body of 0.5 kg moves along the positive x - axis under

the influence of a varying force F (in Newtons) as shown

below :

3

3

1

F(N

)

0,0 2 4 6 8 10

x(m)

If the speed of the object at x = 4m is 3.16 ms�1 then its

speed at x = 8 m is :

(A) 3.16 ms�1 (B) 9.3 ms�1

(C) 8 ms�1 (D) 6.8 ms�1

75. A soldier with a machine gun, falling from an airplanegets detached from his parachute. He is able to resistthe downward acceleration if he shoots 40 bullets asecond at the speed of 500 m/s. If the weight of a bulletis 49 gm, what is the weight of the man with the gun ?Ignore resistance due to air and assume theacceleration due to gravity g = 9.8 m/s2 .(A) 50 kg (B) 75 kg(C) 100 kg (D) 125 kg

76. Blocks of mass M1 and M2 are connected by a cordwhich passes over the pulleys P1 and P2 as shownin the figure. If there is no friction, the accelerationof the block of mass M2 will be:

(A) )( 21

2

MM4

gM

(B) )( 21

2

MM4

gM2

(C) )M4M(gM2

21

1

(D) )( 21

1

MM

gM2

PAGE # 26

MOLE CONCEPT

ATOMS

All the matter is made up of atoms. An atom is thesmallest particle of an element that can take part in achemical reaction. Atoms of most of the elementsare very reactive and do not exist in the free state (assingle atom).They exist in combination with the atomsof the same element or another element.Atoms are very, very small in size. The size of an atomis indicated by its radius which is called "atomicradius" (radius of an atom). Atomic radius ismeasured in "nanometres"(nm).1 metre = 109 nanometre or 1nm = 10-9 m.Atoms are so small that we cannot see them underthe most powerful optical microscope.

Note :

Hydrogen atom is the smallest atom of all , having anatomic radius 0.037nm.

(a) Symbols of Elements :

A symbol is a short hand notation of an element whichcan be represented by a sketch or letter etc.Dalton was the first to use symbols to representelements in a short way but Dalton's symbols forelement were difficult to draw and inconvenient touse, so Dalton's symbols are only of historicalimportance. They are not used at all.

It was J.J. Berzelius who proposed the modernsystem of representing an element.The symbol of an element is the "first letter" or the"first letter and another letter" of the English name orthe Latin name of the element.

e.g. The symbol of Hydrogen is H.The symbol of Oxygen is O.There are some elements whose names begin withthe same letter. For example, the names of elementsCarbon, Calcium, Chlorine and Copper all begin withthe letter C. In such cases, one of the elements isgiven a "one letter "symbol but all other elements aregiven a "first letter and another letter" symbol of theEnglish or Latin name of the element. This is to benoted that "another letter" may or may not be the"second letter" of the name. Thus,The symbol of Carbon is C.The symbol of Calcium is Ca.The symbol of Chlorine is Cl.The symbol of Copper is Cu (from its Latin nameCuprum)

It should be noted that in a "two letter" symbol, thefirst letter is the "capital letter" but the second letter isthe small letter

English Name of the Element

Symbol

Hydrogen HHelium HeLithium LiBoron B

Carbon CNitrogen NOxygen OFluorine F

Neon NeMagnesium MgAluminium Al

Silicon SiPhosphorous P

Sulphur SChlorine ClArgon Ar

Calcium Ca

Symbol Derived from English Names

Symbols Derived from Latin Names

English Name of the Element

SymbolLatin Name of the Element

Sodium Na Natrium

Potassium K Kalium

(b) Significance of The Symbol of an

Element :

(i) Symbol represents name of the element.

(ii) Symbol represents one atom of the element.

(iii) Symbol also represents one mole of the element.That is, symbol also represent 6.023 × 1023 atoms ofthe element.

(iv) Symbol represent a definite mass of the element

i.e. atomic mass of the element.

Example :(i) Symbol H represents hydrogen element.

(ii) Symbol H also represents one atom of hydrogenelement.

(iii) Symbol H also represents one mole of hydrogenatom.(iv) Symbol H also represents one gram hydrogen

atom.

IONS

An ion is a positively or negatively charged atom orgroup of atoms.Every atom contains equal number of electrons


PAGE # 27

(negatively charged) and protons (positively charged).Both charges balance each other, hence atom iselectrically neutral.

(a) Cation :

If an atom has less electrons than a neutral atom,then it gets positively charged and a positivelycharged ion is known as cation.e.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc.A cation bears that much units of positive charge asthere are the number of electrons lost by the neutralatom to form that cation.

e.g. An aluminium atom loses 3 electrons to formaluminium ion, so aluminium ion bears 3 units ofpositive charge and it is represented as Al3+.

(b) Anion :

If an atom has more number of electrons than that ofneutral atom, then it gets negatively charged and anegatively charged ion is known as anion.e.g. Chloride ion (Cl

�

), oxide ion (O2-) etc.

An anion bears that much units of negative chargeas there are the number of electrons gained by theneutral atom to form that anion.e.g. A nitrogen atom gains 3 electrons to form nitrideion, so nitride ion bears 3 units of negative chargeand it is represented as N3-.

Note :Size of a cation is always smaller and anion is alwaysgreater than that of the corresponding neutral atom.

(c) Monoatomic ions and polyatomic ions :(i) Monoatomic ions : Those ions which are formedfrom single atoms are called monoatomic ions orsimple ions.e.g. Na+, Mg2+ etc.

(ii) Polyatomic ions : Those ions which are formedfrom group of atoms joined together are calledpolyatomic ions or compound ions.e.g. Ammonium ion (NH

4+) , hydroxide ion (OH�) etc.

which are formed by the joining of two types of atoms,nitrogen and hydrogen in the first case and oxygen andhydrogen in the second.

(d) Valency of ions :The valency of an ion is same as the charge presenton the ion.If an ion has 1 unit of positive charge, its valency is 1and it is known as a monovalent cation. If an ion has2 units of negative charge, its valency is 2 and it isknown as a divalent anion.

LIST OF COMMON ELECTROVALENT POSITIVE RADICALS

LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS

Monovalent Electronegative Bivalent Electronegative

Trivalent Electronegative

Tetravalent Electronegative

1. Fluoride F� 1. Sulphate SO 42- 1. Nitride N3- 1. Carbide C4-

2. Chloride Cl� 2. Sulphite SO 32- 2. Phosphide P3-

3. Bromide Br� 3. Sulphide S2-3. Phosphite PO3

3-

4. Iodide I 4. Thiosulphate S2O32- 4. Phosphate PO4

3-

5. Hydride H� 5. Zincate ZnO22-

6. Hydroxide OH� 6. Oxide O2-

7. Nitrite NO2� 7. Peroxide O2

2-

8.Nitrate NO3� 8. Dichromate Cr2O7

2-

9. Bicarbonate or Hydrogen carbonate HCO3� 9. Carbonate CO3

2-

10. Bisulphite or Hydrogen sulphite HSO3� 10. Silicate SiO 3

2-

11. Bisulphide or Hydrogen sulphide HS�

12. Bisulphate or Hydrogen sulphate HSO4�

13. Acetate CH COO3�

Note :Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protons present inthem.

PAGE # 28

LAWS OF CHEMICAL COMBINATION

The laws of chemical combination are theexperimental laws which led to the idea ofatoms being the smallest unit of matter. The laws ofchemical combination played a significant role in the

development of Dalton�s atomic theory of matter.

There are two important laws of chemical combination.These are:

(i) Law of conservation of mass(ii) Law of constant proportions

(a) Law of Conservation of Mass or Matter :

This law was given by Lavoisier in 1774 . According tothe law of conservation of mass, matter can neitherbe created nor be destroyed in a chemical reaction.

OrThe law of conservation of mass means that in achemical reaction, the total mass of products is equalto the total mass of the reactants. There is no changein mass during a chemical reaction.Suppose we carry out a chemical reaction between Aand B and if the products formed are C and D then,A + B C + D

Suppose 'a' g of A and 'b' g of B react to produce 'c' g ofC and 'd' g of D. Then, according to the law ofconservation of mass, we have,

a + b = c + dExample :When Calcium Carbonate (CaCO

3) is heated, a

chemical reaction takes place to form Calcium Oxide(CaO) and Carbon dioxide (CO

2). It has been found

by experiments that if 100 grams of calcium carbonateis decomposed completely, then 56 grams of CalciumOxide and 44 grams of Carbon dioxide are formed.

Since the total mass of products (100g ) is equal tothe total mass of the reactants (100g), there is nochange in the mass during this chemical reaction.The mass remains same or conserved.

(b) Law of Constant Proportions / Law of

Definite Proportions :

Proust, in 1779, analysed the chemical composition(types of elements present and percentage ofelements present ) of a large number of compoundsand came to the conclusion that the proportion ofeach element in a compound is constant (or fixed).According to the law of constant proportions: Achemical compound always consists of the sameelements combined together in the same proportionby mass.

Note :The chemical composition of a pure substance isnot dependent on the source from which it is obtained.

Example :

Water is a compound of hydrogen and oxygen. It canbe obtained from various sources (like river, sea, welletc.) or even synthesized in the laboratory. Fromwhatever source we may get it, 9 parts by weight ofwater is always found to contain 1 part by weight ofhydrogen and 8 parts by weight of oxygen. Thus, inwater, this proportion of hydrogen and oxygen alwaysremains constant.

Note :The converse of Law of definite proportions that whensame elements combine in the same proportion, thesame compound will be formed, is not always true.

(c) Law of Multiple Proportions :

According to it, when one element combines with theother element to form two or more different compounds,the mass of one element, which combines with aconstant mass of the other, bears a simple ratio toone another.

Simple ratio means the ratio between small naturalnumbers, such as 1 : 1, 1 : 2, 1 : 3e.g.Carbon and oxygen when combine, can form twooxides that are CO (carbon monoxide), CO

2 (carbon

dioxide).In CO,12 g carbon combined with 16 g of oxygen.In CO

2,12 g carbon combined with 32 g of oxygen.

Thus, we can see the mass of oxygen which combinewith a constant mass of carbon (12 g) bear simpleratio of 16 : 32 or 1 : 2

Note :The law of multiple proportion was given by Dalton in1808.

Sample Problem :

1. Carbon is found to form two oxides, which contain42.8% and 27.27% of carbon respectively. Show thatthese figures illustrate the law of multiple proportions.

Sol. % of carbon in first oxide = 42.8% of oxygen in first oxide = 100 - 42.8 = 57.2% of carbon in second oxide = 27.27 % of oxygen in second oxide = 100 - 27.27 = 72.73

For the first oxide -Mass of oxygen in grams that combines with 42.8 gof carbon = 57.2 Mass of oxygen that combines with 1 g of carbon =

1.3442.857.2

g

For the second oxide -Mass of oxygen in grams that combines with 27.27 gof carbon = 72.73 Mass of oxygen that combines with 1 g of carbon =

2.6827.2772.73

g

Ratio between the masses of oxygen that combinewith a fixed mass (1 g) of carbon in the two oxides= 1.34 : 2.68 or 1 : 2 which is a simple ratio. Hence,this illustrates the law of multiple proportion.

PAGE # 29

(d) Law of Reciprocal Proportions :

According to this law the ratio of the weights of twoelement A and B which combine separately with afixed weight of the third element C is either the sameor some simple multiple of the ratio of the weights inwhich A and B combine directly with each other.e.g.

The elements C and O combine separately with thethird element H to form CH

4 and H

2O and they combine

directly with each other to form CO2.

H 24

CH4 H O2

C O16

12

12

CO2 32

In CH4, 12 parts by weight of carbon combine with 4

parts by weight of hydrogen. In H2O, 2 parts by weight

of hydrogen combine with 16 parts by weight ofoxygen. Thus the weight of C and O which combinewith fixed weight of hydrogen (say 4 parts by weight)are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8.Now in CO

2, 12 parts by weight of carbon combine

directly with 32 parts by weight of oxygen i.e. theycombine directly in the ratio 12 : 32 or 3 : 8 which isthe same as the first ratio.

Note :

The law of reciprocal proportion was put forward byRitcher in 1794.

Sample Problem :2. Ammonia contains 82.35% of nitrogen and 17.65%

of hydrogen. Water contains 88.90% of oxygen and11.10% of hydrogen. Nitrogen trioxide contains63.15% of oxygen and 36.85% of nitrogen. Show thatthese data illustrate the law of reciprocal proportions.

Sol

In NH3, 17.65 g of H combine with N = 82.35 g

1g of H combine with N = 17.65

82.35 g = 4.67 g

In H2O, 11.10 g of H combine with O = 88.90 g

1 g H combine with O = 11.10

88.90g = 8.01 g

Ratio of the weights of N and O which combinewith fixed weight (=1g) of H= 4.67 : 8.01 = 1 : 1.72

In N2O

3, ratio of weights of N and O which combine

with each other = 36.85 : 63.15 = 1 : 1.71

Thus the two ratio are the same. Hence it illustratesthe law of reciprocal proportions.

(e) Gay Lussac�s Law of Gaseous Volumes :

Gay Lussac found that there exists a definiterelationship among the volumes of the gaseousreactants and their products. In 1808, he put forwarda generalization known as the Gay Lussac�s Law of

combining volumes. This may be stated as follows :

When gases react together, they always do so in

volumes which bear a simple ratio to one another

and to the volumes of the product, if these are also

gases, provided all measurements of volumes are

done under similar conditions of temperature and

pressure.

e.g.Combination between hydrogen and chlorine to form

hydrogen chloride gas. One volume of hydrogen and

one volume of chlorine always combine to form two

volumes of hydrochloric acid gas.

H2 (g) + Cl

2 (g) 2HCl (g)

1vol. 1 vol. 2 vol.

The ratio between the volume of the reactants and

the product in this reaction is simple, i.e., 1 : 1 : 2.

Hence it illustrates the Law of combining volumes.

(f) Avogadro�s Hypothesis :

This states that equal volumes of all gases under

similar conditions of temperature and pressure

contain equal number of molecules.

This hypothesis has been found to explain elegantly

all the gaseous reactions and is now widely

recognized as a law or a principle known as Avogadro�s

Law or Avogadro�s principle.

The reaction between hydrogen and chlorine can be

explained on the basis of Avogadro�s Law as follows :

Hydrogen + Chlorine Hydrogen chloride gas 1 vol. (By experiment)1 vol. 2 vol.

n molecules. n molecules. 2n molecules.(By Avogadro's Law)

21 molecules. molecules. 1 molecules. (By dividing throughout by 2n)

1 Atom 1 Atom 1 Molecule (Applying Avogadro's hypothesis)

21

It implies that one molecule of hydrogen chloride gas

is made up of 1 atom of hydrogen and 1 atom of

chlorine.

(i) Applications of Avogadro�s hypothesis :

(A) In the calculation of atomicity of elementarygases.

e.g.

2 volumes of hydrogen combine with 1 volume of

oxygen to form two volumes of water vapours.Hydrogen + Oxygen Water vapours2 vol. 1 vol. 2 vol.

PAGE # 30

Applying Avogadro�s hypothesis

Hydrogen + Oxygen Water vapours2 n molecules n molecules 2 n molecules

or 1 molecule 2

1 molecule 1 molecule

Thus1 molecule of water contains 2

1 molecule of

oxygen. But 1 molecule of water contains 1 atom of

oxygen. Hence. 2

1 molecule of oxygen = 1 atom of

oxygen or 1 molecules of oxygen = 2 atoms of oxygeni.e. atomicity of oxygen = 2.

(B) To find the relationship between molecular massand vapour density of a gas.

Vapour density (V.D.) = hydrogenofDensitygas ofDensity

=

epressur and temp. same the at hydrogen of volume same the of Mass

gas the of volume certain a of Mass

If n molecules are present in the given volume of a gasand hydrogen under similar conditions of temperatureand pressure.

V.D. = hydrogen of molecules n of Massgas the of molecules n of Mass

= hydrogen of molecule 1 of Massgas the of molecule 1 of Mass

= hydrogen of mass Moleculargas the of mass Molecular

= 2

mass Molecular

(since molecular mass of hydrogen is 2)Hence, Molecular mass = 2 × Vapour density

ATOMIC MASS UNIT

The atomic mass unit (amu) is equal to one-twelfth(1/12) of the mass of an atom of carbon-12.The massof an atom of carbon-12 isotope was given the atomicmass of 12 units, i.e. 12 amu or 12 u.The atomic masses of all other elements are nowexpressed in atomic mass units.

RELATIVE ATOMIC MASS

The atomic mass of an element is a relative quantityand it is the mass of one atom of the element relativeto one -twelfth (1/12) of the mass of one carbon-12atom. Thus, Relative atomic mass

= atom12Coneofmass

12

1

elementtheofatomoneofMass

[1/12 the mass of one C-12 atom = 1 amu,1 amu = 1.66 × 10�24 g = 1.66 × 10�27 kg.]

Note :One amu is also called one dalton (Da).

GRAM-ATOMIC MASS

The atomic mass of an element expressed in grams

is called the Gram Atomic Mass of the element.

The number of gram -atoms

= elementtheofmassAtomicGram

graminelementtheofMass

e.g.

Calculate the gram atoms present in (i) 16g of oxygen

and (ii) 64g of sulphur.

(i) The atomic mass of oxygen = 16.

Gram-Atomic Mass of oxygen (O) = 16 g.

No. of Gram-Atom = 1616

= 1

(ii) The gram-atom present in 64 gram of sulphur.

= sulphurofMass AtomicGram64

= 32

64= 2

AtomicNumber Element Symbol

Atomicmass

1 Hydrogen H 12 Helium He 43 Lithium Li 74 Beryllium Be 95 Boron B 116 Carbon C 127 Nitrogen N 148 Oxygen O 169 Fluorine F 1910 Neon Ne 2011 Sodium Na 2312 Magnesium Mg 2413 Aluminium Al 2714 Silicon Si 2815 Phosphorus P 3116 Sulphur S 3217 Chlorine Cl 35.518 Argon Ar 4019 Potassium K 3920 Calcium Ca 40

RELATIVE MOLECULAR MASS

The relative molecular mass of a substance is themass of a molecule of the substance as comparedto one-twelfth of the mass of one carbon -12 atomi.e.,Relative molecular mass

= atom12Coneofmass

12

1

substancetheofmoleculeoneofMass

The molecular mass of a molecule, thus, representsthe number of times it is heavier than 1/12 of themass of an atom of carbon-12 isotope.

PAGE # 31

GRAM MOLECULAR MASS

The molecular mass of a substance expressed in

grams is called the Gram Molecular Mass of the

substance . The number of gram molecules

= ancetsubstheofmassmolecularGram

gramsintancesubstheofMass

e.g.

(i) Molecular mass of hydrogen (H2) = 2u.

Gram Molecular Mass of hydrogen (H2) = 2 g .

(ii) Molecular mass of methane (CH4) = 16u

Gram Molecular Mass of methane (CH4) = 16 g.

e.g. the number of gram molecules present in 64 g of

methane (CH4).

= 4CHofmassmolecularGram

64 =

16

64 = 4.

(a) Calculation of Molecular Mass :

The molecular mass of a substance is the sum of

the atomic masses of its constituent atoms present

in a molecule.

Ex.1 Calculate the molecular mass of water.

(Atomic masses : H = 1u, O = 16u).

Sol. The molecular formula of water is H2O.

Molecular mass of water = ( 2 × atomic mass of H)

+ (1 × atomic mass of O)

= 2 × 1 + 1 × 16 = 18

i.e., molecular mass of water = 18 amu.

Ex.2 Find out the molecular mass of sulphuric acid.

(Atomic mass : H = 1u, O = 16u, S = 32u).

Sol. The molecular formula of sulphuric acid is H2SO

4.

Molecular mass of H2SO

4

= (2 × atomic mass of H) + ( 1 × atomic mass of S)

+ ( 4 × atomic mass of O)

= (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98

i.e., Molecular mass of H2SO

4= 98 amu.

FORMULA MASS

The term �formula mass� is used for ionic compounds

and others where discrete molecules do not exist,

e.g., sodium chloride, which is best represented as

(Na+Cl�)n, but for reasons of simplicity as NaCl or

Na+Cl�. Here, formula mass means the sum of the

masses of all the species in the formula.

Thus, the formula mass of sodium chloride = (atomic

mass of sodium) + (atomic mass of chlorine)

= 23 + 35.5

= 58.5 amu

EQUIVALENT MASS

(a) Definition :

Equivalent mass of an element is the mass of theelement which combine with or displaces 1.008 partsby mass of hydrogen or 8 parts by mass of oxygen or35.5 parts by mass of chlorine.

(b) Formulae of Equivalent Masses of differentsubstances :(i) Equivalent mass of an element =

element the ofValency element the of wt.Atomic

(ii) Eq. mass an acid = acid the ofBasicity acid the of wt. Mol.

Basicity is the number of replaceable H+ ions fromone molecule of the acid.

(iii) Eq. Mass of a base = base the ofAcidity base the of wt. Mol.

Acidity is the number of replaceable OH� ions fromone molecule of the base

(iv) Eq. mass of a salt

= metal ofvalency atoms metal of Numbersalt the of wt. Mol.

(v) Eq. mass of an ion = ion the on Charge

ion the of wt.Formula

(vi) Eq. mass of an oxidizing/reducing agent

=

substance the ofatom/molecule

oneby gained or lost electrons of No.

wt At. or wt.Mol

Equivalent weight of some compounds are given inthe table :

S.No. CompoundEquivalent

weight

1 HCl 36.5

2 H2SO4 49

3 HNO3 63

4 45

5 .2H2O63

6 NaOH 40

7 KOH 56

8 CaCO3 50

9 NaCl 58.5

10 Na2CO3 53

COOH

COOH

PAGE # 32

In Latin, mole means heap or collection or pile. Amole of atoms is a collection of atoms whose totalmass is the number of grams equal to the atomicmass in magnitude. Since an equal number of molesof different elements contain an equal number ofatoms, it becomes convenient to express theamounts of the elements in terms of moles. A molerepresents a definite number of particles, viz, atoms,molecules, ions or electrons. This definite number iscalled the Avogadro Number (now called the Avogadroconstant) which is equal to 6.023 × 1023.

A mole is defined as the amount of a substance thatcontains as many atoms, molecules, ions, electronsor other elementary particles as there are atoms inexactly 12 g of carbon -12 (12C).

(a) Moles of Atoms :

(i) 1 mole atoms of any element occupy a mass whichis equal to the Gram Atomic Mass of that element.

e.g. 1 Mole of oxygen atoms weigh equal to GramAtomic Mass of oxygen, i.e. 16 grams.

(ii) The symbol of an element represents 6.023 x 1023

atoms (1 mole of atoms) of that element.

e.g : Symbol N represents 1 mole of nitrogen atomsand 2N represents 2 moles of nitrogen atoms.

(b) Moles of Molecules :

(i) 1 mole molecules of any substance occupy a masswhich is equal to the Gram Molecular Mass of thatsubstance.

e.g. : 1 mole of water (H2O) molecules weigh equal toGram Molecular Mass of water (H2O), i.e. 18 grams.

(ii) The symbol of a compound represents 6.023 x1023 molecules (1 mole of molecules) of thatcompound.

e.g. : Symbol H2O represents 1 mole of watermolecules and 2 H2O represents 2 moles of watermolecules.

Note :

The symbol H2O does not represent 1 mole of H2molecules and 1 mole of O atoms. Instead, itrepresents 2 moles of hydrogen atoms and 1 moleof oxygen atoms.

Note :The SI unit of the amount of a substance is Mole.

(c) Mole in Terms of Volume :

Volume occupied by 1 Gram Molecular Mass or 1mole of a gas under standard conditions oftemperature and pressure (273 K and 1atm.pressure) is called Gram Molecular Volume. Its valueis 22.4 litres for each gas.Volume of 1 mole = 22.4 litre (at STP)

Note :The term mole was introduced by Ostwald in 1896.

SOME IMPORTANT RELATIONS AND FORMULAE

(i) 1 mole of atoms = Gram Atomic mass = mass of6.023 × 1023 atoms

(ii) 1 mole of molecules = Gram Molecular Mass= 6.023 x 1023 molecules(iii) Number of moles of atoms

= elementofMassAtomicGram

gramsinelementofMass

(iv) Number of moles of molecules

= substanceofMassMolecularGram

gramsinsubstanceofMass

(v) Number of moles of molecules

= AN

N

numberAvogadro

elementofmolecules ofNo.

Ex.3 To calculate the number of moles in 16 grams ofSulphur (Atomic mass of Sulphur = 32 u).

Sol. 1 mole of atoms = Gram Atomic Mass.So, 1 mole of Sulphur atoms = Gram Atomic Mass ofSulphur = 32 grams.Now, 32 grams of Sulphur = 1 mole of SulphurSo, 16 grams of Sulphur= (1/32) x 16 = 0.5 molesThus, 16 grams of Sulphur constitute 0.5 mole ofSulphur.

6.023 × 10

(N ) Atoms

23

A

6.023 × 10

(N ) molecules

23

A

1 Mole

1 gram atomof element

1 gram molecule of substance

1 gram formula mass of substance

In terms of particles

In terms of mass

22.4 litre

In term ofvolume

PROBLEMS BASED ON THE MOLE CONCEPT

Ex.4 Calculate the number of moles in 5.75 g of sodium. (Atomic mass of sodium = 23 u)

Sol. Number of moles

= elementofMassAtomicGram

gramsinelementtheofMass =

23

5.75= 0.25 mole

or,1 mole of sodium atoms = Gram Atomic mass ofsodium = 23g.23 g of sodium = 1 mole of sodium.

5.75 g of sodium = 23

5.75mole of sodium = 0.25 mole

PAGE # 33

Ex.5 What is the mass in grams of a single atom ofchlorine ? (Atomic mass of chlorine = 35.5u)

Sol. Mass of 6.022 × 1023 atoms of Cl = Gram AtomicMass of Cl = 35.5 g.

Mass of 1 atom of Cl =23106.022

g35.5

= 5.9 × 10�23 g.

Ex.6 The density of mercury is 13.6 g cm�3. How manymoles of mercury are there in 1 litre of the metal ?(Atomic mass of Hg = 200 u).

Sol. Mass of mercury (Hg) in grams = Density(g cm�3)× Volume (cm3)= 13.6 g cm�3 × 1000 cm3 = 13600 g.

Number of moles of mercury

= mercuryofMassAtomicGramgramsinmercuryofMass

=200

13600 = 68

Ex.7 The mass of a single atom of an element M is3.15× 10�23 g . What is its atomic mass ? Whatcould the element be ?

Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms= mass of 1 atom × 6.022 × 1023

= (3.15 × 10�23g) × 6.022 × 1023

= 3.15 × 6.022 g = 18.97 g.

Atomic Mass of the element = 18.97uThus, the element is most likely to be fluorine.

Ex.8 An atom of neon has a mass of 3.35 × 10�23 g.How many atoms of neon are there in 20 g of thegas ?

Sol. Number of atoms

= atom1ofMass

massTotal = 23�103.35

02

= 5.97 × 1023

Ex.9 How many grams of sodium will have the samenumber of atoms as atoms present in 6 g ofmagnesium ?(Atomic masses : Na = 23u ; Mg =24u)

Sol. Number of gram -atom of Mg

= MassAtomicGramgramsinMgofMass

= 246

= 4

1

Gram Atoms of sodium should be = 4

1

1 Gram Atom of sodium = 23 g

4

1 gram atoms of sodium = 23 ×

4

1 = 5.75 g

Ex.10 How many moles of Cr are there in 85g of Cr2S

3 ?

(Atomic masses : Cr = 52 u , S =32 u)Sol. Molecular mass of Cr

2S

3=2 × 52 + 3 × 32 = 200 u.

200g of Cr2S

3 contains = 104 g of Cr.

85 g of Cr2S

3 contains =

200

85104 g of Cr = 44.2g

Thus, number of moles of Cr = 52

44.2 = 0.85 .

Ex.11 What mass in grams is represented by

(a) 0.40 mol of CO2,

(b) 3.00 mol of NH3,

(c) 5.14 mol of H5IO

6

(Atomic masses : C=12 u, O=16 u, N=14 u,

H=1 u and I = 127 u)

Sol. Weight in grams = number of moles × molecular

mass.

Hence,

(a) mass of CO2 = 0.40 × 44 = 17.6 g

(b) mass of NH3 = 3.00 × 17 = 51.0 g

(c) mass of H5IO

6 = 5.14 × 228 = 1171.92g

Ex.12 Calculate the volume in litres of 20 g of hydrogen

gas at STP.

Sol. Number of moles of hydrogen

= hydrogenofMassMolecularGram

grams in hydrogenofMass =

220

= 10

Volume of hydrogen = number of moles × Gram

Molecular Volume.

= 10 ×22.4 = 224 litres.

Ex.13 The molecular mass of H2SO

4 is 98 amu.

Calculate the number of moles of each elementin 294 g of H

2SO

4.

Sol. Number of moles of H2SO

4 =

98

294= 3 .

The formula H2SO

4 indicates that 1 molecule of

H2SO

4 contains 2 atoms of H, 1 atom of S and 4

atoms of O. Thus, 1 mole of H2SO

4 will contain 2

moles of H,1 mole of S and 4 moles of O atomsTherefore, in 3 moles of H

2SO

4 :

Number of moles of H = 2 × 3 = 6

Number of moles of S = 1 × 3 = 3

Number of moles of O = 4 × 3 = 12

Ex.14 Find the mass of oxygen contained in 1 kg ofpotassium nitrate (KNO

3).

Sol. Since 1 molecule of KNO3 contains 3 atoms of

oxygen, 1 mol of KNO3 contains 3 moles of

oxygen atoms. Moles of oxygen atoms = 3 × moles of KNO

3

= 3 × 101

1000 = 29.7

(Gram Molecular Mass of KNO3 = 101 g)

Mass of oxygen = Number of moles × Atomic

mass= 29.7 × 16 = 475.2 g

Ex.15 You are asked by your teacher to buy 10 moles ofdistilled water from a shop where small bottleseach containing 20 g of such water are available.How many bottles will you buy ?

Sol. Gram Molecular Mass of water (H2O) = 18 g

10 mol of distilled water = 18 × 10 = 180 g.

Because 20 g distilled water is contained in 1bottle,

180 g of distilled water is contained in =20

180

bottles = 9 bottles.

Number of bottles to be bought = 9

PAGE # 34

Ex.16 6.022 × 1023 molecules of oxygen (O2) is equal tohow many moles ?

Sol. No. of moles =

AN

N

moleculesofno.sAvogadro'

oxygenofmoleculesofNo. =

23

23

106.023

106.023

= 1

PERCENTAGE COMPOSITION

The percentage composition of elements in acompound is calculated from the molecular formulaof the compound.The molecular mass of the compound is calculatedfrom the atomic masses of the various elementspresent in the compound. The percentage by massof each element is then computed with the help of thefollowing relations.Percentage mass of the element in the compound

= massMolecular

elementtheofmassTotal× 100

Ex.17 What is the percentage of calcium in calciumcarbonate (CaCO

3) ?

Sol. Molecular mass of CaCO3 = 40 + 12 + 3 × 16

= 100 amu.Mass of calcium in 1 mol of CaCO

3 = 40g.

Percentage of calcium = 100

10040 = 40 %

Ex.18 What is the percentage of sulphur in sulphuric acid (H

2SO

4) ?

Sol. Molecular mass of H2SO

4 = 1 × 2 + 32 + 16 × 4 = 98 amu.

Percentage of sulphur = 98

10032 = 32.65 %

Ex.19 What are the percentage compositions ofhydrogen and oxygen in water (H

2O) ?

(Atomic masses : H = 1 u, O = 16 u)

Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.

H2O has two atoms of hydrogen.

So, total mass of hydrogen in H2O = 2 amu.

Percentage of H = 18

1002 = 11.11 %

Similarly,

percentage of oxygen = 18

10016 = 88.88 %

The following steps are involved in determining theempirical formula of a compound :(i) The percentage composition of each element isdivided by its atomic mass. It gives atomic ratio of theelements present in the compound.

(ii) The atomic ratio of each element is divided by theminimum value of atomic ratio as to get the simplestratio of the atoms of elements present in thecompound.

(iii) If the simplest ratio is fractional, then values ofsimplest ratio of each element is multiplied bysmallest integer to get the simplest whole numberfor each of the element.

(iv) To get the empirical formula, symbols of variouselements present are written side by side with theirrespective whole number ratio as a subscript to thelower right hand corner of the symbol.

(v) The molecular formula of a substance may bedetermined from the empirical formula if the molecularmass of the substance is known. The molecularformula is always a simple multiple of empiricalformula and the value of simple multiple (n) isobtained by dividing molecular mass with empiricalformula mass.

n = MassFormulaEmpiricalMassMolecular

Ex-20 A compound of carbon, hydrogen and nitrogencontains these elements in the ratio of 9:1:3.5 respectively.Calculate the empirical formula. If its molecular mass is108, what is the molecular formula ?

Sol.

Ele m e ntM as

sAtom i

cRe lative Num be r

Sim ple s t Ratio

Carbon 9 12

Hydrogen 1 1

Nitrogen 3.5 14

0.75129

111

0.25143.5

325.075.0

425.01

125.025.0

Empirical ratio = C3H

4N

Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54

n = MassFormulaEmpiricalMassMolecular

= 254

108

Thus, molecular formula of the compound

= (Empirical formula)2

= (C3H

4N)

2

= C6H

8N

2

Ex.21 A compound on analysis, was found to have thefollowing composition :(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen= 69.50%, (iv) Hydrogen = 6.22%. Calculate themolecular formula of the compound assuming thatwhole hydrogen in the compound is present as waterof crystallisation. Molecular mass of the compoundis 322.

Sol.

Element PercentageAtomic mass

Relative Number of atoms

Simplest ratio

Sodium 14.31 23 0.622

Sulphur 9.97 32 0.311

Hydrogen 6.22 1 6.22

Oxygen 69.50 16 4.34

20.3110.622

10.3110.311

200.3116.22

140.3114.34

23

31.14

1

22.6

16

50.69

3297.9

PAGE # 35

The empirical formula = Na2SH

20O

14

Empirical formula mass= (2 × 23) + 32 + (20 × 1) + (14 × 16)

= 322Molecular mass = 322Molecular formula = Na

2SH

20O

14

Whole of the hydrogen is present in the form of waterof crystallisation. Thus, 10 water molecules arepresent in the molecule.So, molecular formula = Na

2SO

4. 10H

2O

CONCENTRATION OF SOLUTIONS

(a) Strength in g/L :

The strength of a solution is defined as the amount ofthe solute in grams present in one litre (or dm3) of thesolution, and hence is expressed in g/litre or g/dm3.

Strength in g/L = litreinsolutionofVolume

graminsoluteofWeight

(b) Molarity :

Molarity of a solution is defined as the number ofmoles of the solute dissolved per litre (or dm3) ofsolution. It is denoted by �M�. Mathematically,

M = litreinsolutiontheofVolume

soluteofmolesofNumber

litreinsolutionofVolume

soluteofMassMoleculargram/GraminsoluteofMass

M can be calculated from the strength as given below :

M = solute of mass Molecular

litre per grams in Strength

If �w� gram of the solute is present in V cm3 of a givensolution , then

M = massMolecular

w ×

V

1000

e.g. a solution of sulphuric acid having 4.9 grams of itdissolved in 500 cm3 of solution will have its molarity,

M = massMolecular

w×

V

1000

M = 98

4.9 ×

500

1000 = 0.1

(c) Formality :

In case of ionic compounds like NaCl, Na2CO

3 etc.,

formality is used in place of molarity. The formality ofa solution is defined as the number of gram formulamasses of the solute dissolved per litre of thesolution. It is represented by the symbol �F�. The term

formula mass is used in place of molecular massbecause ionic compounds exist as ions and not asmolecules. Formula mass is the sum of the atomicmasses of the atoms in the formula of the compound.

litreinsolutionofVolume

soluteofMasslagram/FormuinsoluteofMass

(d) Normality :

Normality of a solution is defined as the number ofgram equivalents of the solute dissolved per litre (dm3)of given solution. It is denoted by �N�.Mathematically,

N = litre in solution the of Volume

solute of sequivalent gram of Number

N = litre in solution the of Volume

solute of weight equivalent/ graminsolute of Weight

N can be calculated from the strength as given below :

N = solute of mass Equivalent

litre per grams in Strength=

E

S

If �w� gram of the solute is present in V cm3 of a givensolution.

N = solute the of mass Equivalentw

× V

1000

e.g. A solution of sulphuric acid having 0.49 gram ofit dissolved in 250 cm3 of solution will have itsnormality,

N = solute the of mass Equivalentw

× V

1000

N = 49

0.49×

250

1000 = 0.04

(Eq. mass of sulphuric acid = 49).

Solution Seminormal

Decinormal

Centinormal

Normality101

1001

21

Some Important Formulae :

(i) Milli equivalent of substance = N × V

where , N normality of solutionV Volume of solution in mL

(ii) If weight of substance is given,

milli equivalent (NV) = E1000w

Where, W Weight of substance in gramE Equivalent weight of substance

(iii) S = N × E

S Strength in g/LN Normality of solutionE Equivalent weight

(iv) Calculation of normality of mixture :

Ex.22 100 ml of 10

NHCl is mixed with 50 ml of

5

NH

2SO

4 .

Find out the normality of the mixture.Sol. Milli equivalent of HCl + milli equivalent H

2SO

4

= milli equivalent of mixtureN

1 V

1 + N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

50

51

100101 N

3 × 150

N3 =

150

20 =

15

2= 0.133

PAGE # 36

Ex.23 100 ml of 10

NHCl is mixed with 25 ml of

5

NNaOH.

Find out the normality of the mixture.

Sol. Milli equivalent of HCl � milli equivalent of NaOH

= milli equivalent of mixtureN

1 V

1 � N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

25

51

�100101

= N3 × 125

N3 =

251

Note :

1 milli equivalent of an acid neutralizes 1 milliequivalent of a base.

(e) Molality :

Molality of a solution is defined as the number ofmoles of the solute dissolved in 1000 grams of thesolvent. It is denoted by �m�.Mathematically,

m = gram in solvent the of Weightsolute the of moles of Number

× 1000

�m� can be calculated from the strength as given below :

m = solute of mass Molecularsolvent of gram 1000 per Strength

If �w� gram of the solute is dissolved in �W� gram of the

solvent then

m = solute the of mass Mol.

w×

W1000

e.g. A solution of anhydrous sodium carbonate(molecular mass = 106) having 1.325 grams of it,dissolved in 250 gram of water will have its molality -

m = 250

1000106

1.325 = 0.05

Note :

Relationship Between Normality and Molarity of aSolution :Normality of an acid = Molarity × Basicity

Normality of base = Molarity × Acidity

Ex.24 Calculate the molarity and normality of a solutioncontaining 0.5 g of NaOH dissolved in 500 cm3

of solvent.Sol. Weight of NaOH dissolved = 0.5 g

Volume of the solution = 500 cm3

(i) Calculation of molarity :Molecular weight of NaOH = 23 + 16 + 1 = 40

Molarity =litre in solution of Volume

solute of weightmolecularsolute/ of Weight

= 500/1000

0.5/40 = 0.025

(ii) Calculation of normality :Normality

=litre in solution of Volume

solute of weightequivalentsolute/ of Weight

=500/1000

0.5/40 = 0.025

Ex.25 Find the molarity and molality of a 15% solution

of H2SO

4 (density of H

2SO

4 solution = 1.02 g/cm3)

(Atomic mass : H = 1u, O = 16u , S = 32 u)

Sol. 15% solution of H2SO

4 means 15g of H

2SO

4 are

present in 100g of the solution i.e.

Wt. of H2SO

4 dissolved = 15 g

Weight of the solution = 100 g

Density of the solution = 1.02 g/cm3 (Given)

Calculation of molality :

Weight of solution = 100 g

Weight of H2SO

4 = 15 g

Wt. of water (solvent) = 100 � 15 = 85 g

Molecular weight of H2SO

4 = 98

15 g H2SO

4 =

98

15= 0.153 moles

Thus, 85 g of the solvent contain 0.153 moles .

1000 g of the solvent contain= 85

0.153× 1000 = 1.8 mole

Hence ,the molality of H2SO

4 solution = 1.8 m

Calculation of molarity :

15 g of H2SO

4 = 0.153 moles

Vol. of solution = solution ofDensity

solution of Wt.

= 1.02

100 = 98.04 cm3

This 98.04 cm3 of solution contain H2SO

4

= 0.153 moles

1000 cm3 of solution contain H2SO

4

= 98.04

0.153 × 1000 = 1.56 moles

Hence the molarity of H2SO

4 solution = 1.56 M

(f) Mole Fraction :

The ratio between the moles of solute or solvent to

the total moles of solution is called mole fraction.

mole fraction of solute = Nn

n

solutionofMoles

soluteofMoles

= W/Mw/m

w/m

Mole fraction of solvent = Nn

N

solutionofMoles

solventofMoles

=

W/Mw/m

W/M

where,

n number of moles of solute

N number of moles of solvent

m molecular weight of solute

M molecular weight of solvent

w weight of solute

W weight of solvent

PAGE # 37

Ex.26 Find out the mole fraction of solute in 10% (by weight)urea solution.weight of solute (urea) = 10 gweight of solution = 100 gweight of solvent (water) = 100 � 10 = 90g

mole fraction of solute = solutionofMoles

soluteofMoles =

W/Mw/m

w/m

=

18/9060/10

60/10

= 0.032

Note :Sum of mole fraction of solute and solvent is alwaysequal to one.

STOICHIOMETRY

(a) Quantitative Relations in Chemical

Reactions :

Stoichiometry is the calculation of the quantities ofreactants and products involved in a chemicalreaction.

It is based on the chemical equation and on therelationship between mass and moles.N

2(g) + 3H

2(g) 2NH

3(g)

A chemical equation can be interpreted as follows -

1 molecule N2 + 3 molecules H

2 2 molecules

NH3(Molecular interpretation)

1 mol N2 + 3 mol H

2 2 mol NH

3

(Molar interpretation)

28 g N2 + 6 g H

2 34 g NH

3

(Mass interpretation)

1 volume N2 + 3 volume H

2 2 volume NH

3

(Volume interpretation)

Thus, calculations based on chemical equations aredivided into four types -

(i) Calculations based on mole-mole relationship.

(ii) Calculations based on mass-mass relationship.

(iii) Calculations based on mass-volume relationship.

(iv) Calculations based on volume -volumerelationship.

(i) Calculations based on mole-mole relationship :In such calculations, number of moles of reactantsare given and those of products are required.Conversely, if number of moles of products are given,then number of moles of reactants are required.

Ex.27 Oxygen is prepared by catalytic decompositionof potassium chlorate (KClO

3). Decomposition

of potassium chlorate gives potassium chloride(KCl) and oxygen (O

2). How many moles and how

many grams of KClO3 are required to produce

2.4 mole O2.

Sol. Decomposition of KClO3 takes place as,

2KClO3(s) 2KCl(s) + 3O

2(g)

2 mole KClO3 3 mole O

2

3 mole O2 formed by 2 mole KClO

3

2.4 mole O2 will be formed by

4.2

32

moles of KClO3 = 1.6 mole KClO

3

Mass of KClO3 = Number of moles × molar mass

= 1.6 × 122.5 = 196 g

(ii) Calculations based on mass-mass relationship:In making necessary calculation, following steps are

followed -

(a) Write down the balanced chemical equation.

(b) Write down theoretical amount of reactants and

products involved in the reaction.

(c) The unknown amount of substance is calculated

using unitary method.

Ex.28 Calculate the mass of CaO that can be prepared

by heating 200 kg of limestone CaCO3 which is

95% pure.

Sol. Amount of pure CaCO3 = 200

10095

= 190 kg

= 190000 g

CaCO3(s) CaO(s) + CO

2(g)

1 mole CaCO3 1 mole CaO

100 g CaCO3 56 g CaO

100 g CaCO3 give 56 g CaO

190000 g CaCO3 will give=

10056

× 190000 g CaO

= 106400 g = 106.4 kg

Ex.29 Chlorine is prepared in the laboratory by treatingmanganese dioxide (MnO

2) with aqueous

hydrochloric acid according to the reaction -

MnO2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

How many grams of HCl will react with 5 g MnO2 ?

Sol. 1 mole MnO2 reacts with 4 mole HCl

or 87 g MnO2 react with 146 g HCl

5 g MnO2 will react with =

87146

× 5 g HCl = 8.39 g HCl

Ex.30 How many grams of oxygen are required to burncompletely 570 g of octane ?

Sol. Balanced equation

2C H + 25O 8 18 2 16CO + 18H O2 2

2 mole2 × 114

25 mole25 × 32

First method : For burning 2 × 114 g of the octane,

oxygen required = 25 × 32 g

For burning 1 g of octane, oxygen required =1142

3225

g

Thus, for burning 570 g of octane, oxygen required

= 1142

3225

× 570 g = 2000 g

PAGE # 38

Mole Method : Number of moles of octane in 570grams

114570

= 5.0

For burning 2.0 moles of octane, oxygen required= 25 mol = 25 × 32 g

For burning 5 moles of octane, oxygen required

= 0.23225

× 5.0 g = 2000 g

Proportion Method : Let x g of oxygen be required forburning 570 g of octane. It is known that 2 × 114 g of

the octane requires 25 × 32 g of oxygen; then, the

proportion.

etanocg1142oxygeng3225

= etanocg570

x

x = 1142

5703225

= 2000 g

Ex.31 How many kilograms of pure H2SO

4 could be

obtained from 1 kg of iron pyrites (FeS2) according to

the following reactions ?4FeS

2 + 11O

2 2Fe

2O

3 + 8SO

2

2SO2 + O

2 2SO

3

SO3 + H

2O H

2SO

4

Sol. Final balanced equation,4FeS + 15O + 8H O2 2 2

2Fe O + 8H SO2 3 2 4

8 mole8 × 98 g

4 mole4 × 120 g

4 × 120 g of FeS2 yield H

2SO

4 = 8 × 98 g

1000 g of FeS2 will yield H

2SO

4 =

1204988

× 1000

= 1633.3 g

(iii) Calculations involving mass-volume relationship :In such calculations masses of reactants are givenand volume of the product is required and vice-versa.1 mole of a gas occupies 22.4 litre volume at STP.Mass of a gas can be related to volume according tothe following gas equation -PV = nRT

PV = mw

RT

Ex-32. What volume of NH3 can be obtained from 26.75 g

of NH4Cl at 27ºC and 1 atmosphere pressure.

Sol. The balanced equation is -

NH Cl(s) 4 NH (g) + HCl(g)3

1 mol1 mol53.5 g

53.5 g NH4Cl give 1 mole NH

3

26.75 g NH4Cl will give

5.531

× 26.75 mole NH3

= 0.5 molePV = nRT1 ×V = 0.5 × 0.0821 × 300

V = 12.315 litre

Ex-33 What quantity of copper (II) oxide will react with2.80 litre of hydrogen at STP ?

Sol. CuO + H 2 Cu + H O2

1 mol79.5 g

1 mol22.4 litre at NTP

22.4 litre of hydrogen at STP reduce CuO = 79.5 g2.80 litre of hydrogen at STP will reduce CuO

= 4.225.79

× 2.80 g = 9.93 g

Ex-34 Calculate the volume of carbon dioxide at STPevolved by strong heating of 20 g calcium carbonate.

Sol. The balanced equation is -

CaCO 3 CaO + CO 2

1 mol = 22.4 litre at STP

1 mol 100 g

100 g of CaCO3 evolve carbon dioxide = 22.4 litre

20 g CaCO3 will evolve carbon dioxide

= 100

4.22 × 20 = 4.48 litre

Ex.35 Calculate the volume of hydrogen liberated at 27ºC

and 760 mm pressure by heating 1.2 g of magnesiumwith excess of hydrochloric acid.

Sol. The balanced equation is

Mg + 2HCl MgCl + H2 2

1 mol 24 g

24 g of Mg liberate hydrogen = 1 mole

1.2 g Mg will liberate hydrogen = 0.05 mole

PV = nRT1 × V = 0.05 × 0.0821 × 300

V = 1.2315 litre

(iv) Calculations based on volume volumerelationship :These calculations are based on two laws :(i) Avogadro�s law (ii) Gay-Lussac�s Law

e.g.N (g) + 3H (g)2 2 2NH (g) (Avogadro's law)3

2 mol 2 × 22.4 L

1 mol1 × 22.4 L

3 mol3 × 22.4 L

(under similar conditions of temperature andpressure, equal moles of gases occupy equalvolumes)N (g) + 3H (g)2 2 2NH (g)3

1 vol 3 vol 2 vol(Gay- Lussac's Law)

(under similar conditions of temperature andpressure, ratio of coefficients by mole is equal to ratioof coefficient by volume).

Ex-36 One litre mixture of CO and CO2 is taken. This is

passed through a tube containing red hot charcoal.The volume now becomes 1.6 litre. The volume aremeasured under the same conditions. Find thecomposition of mixture by volume.

Sol. Let there be x mL CO in the mixture , hence, there willbe (1000 � x) mL CO

2. The reaction of CO

2 with red

hot charcoal may be given as -

CO (g) + C(s)2 2CO(g) 2 vol.2(1000 � x)

1 vol.(1000 �x)

Total volume of the gas becomes = x + 2(1000 � x)

x + 2000 � 2x = 1600

x = 400 mL

volume of CO = 400 mL and volume of CO2 = 600 mL

PAGE # 39

Ex-37 What volume of air containing 21% oxygen by volume

is required to completely burn 1kg of carbon containing

100% combustible substance ?

Sol. Combustion of carbon may be given as,

C(s) + O (g) 2 CO (g)2

1 mol12 g

1 mol32 g

12 g carbon requires 1 mole O2 for complete

combustion

1000 g carbon will require 1000121 mole O

2 for

combustion, i.e. , 83.33 mole O2

Volume of O2 at STP = 83.33 × 22.4 litre

= 1866.66 litre

21 litre O2 is present in 100 litre air

1866.66 litre O2 will be present in

21100

× 1866.66 litre air

= 8888.88 litre or 8.89 × 103 litre

Ex-38 An impure sample of calcium carbonate contains80% pure calcium carbonate 25 g of the impuresample reacted with excess of hydrochloric acid.Calculate the volume of carbon dioxide at STPobtained from this sample.

Sol. 100 g of impure calcium carbonate contains = 80 gpure calcium carbonate25 g of impure calcium carbonate sample will contain

= 10080

× 25 = 20 g pure calcium carbonate

The desired equation is -

CaCO + 2HCl 3 CaCl + CO + H O2 2 2

1 mol100 g

22.4 litre at STP

100 g pure CaCO3 liberate = 22.4 litre CO

2.

20 g pure CaCO3 liberate = 20

1004.22

= 4.48 litre CO2

VOLUMETRIC CALCULATIONS

The quantitative analysis in chemistry is primarily

carried out by two methods, viz, volumetric analysis

and gravimetric analysis.In the first method the mass

of a chemical species is measured by measurement

of volume, whereas in the second method it is deter-

mined by taking the weight.

The strength of a solution in volumetric analysis is

generally expressed in terms of normality, i.e., num-

ber of equivalents per litre but since the volume in the

volumetric analysis is generally taken in millilitres

(mL), the normality is expressed by milliequivalents

per millilitre.

USEFUL FORMULAE FOR

VOLUMETRIC CALCULATIONS

(i) milliequivalents = normality × volume in millilitres.

(ii) At the end point of titration, the two titrants, say 1

and 2, have the same number of milliequivalents,

i.e., N1V

1 = N

2V

2, volume being in mL.

(iii) No. of equivalents = 1000

.e.m.

(iv) No. of equivalents for a gas =

)STPat.eq1of.vol(volumeequivalentSTPatVolume

(v) Strength in grams per litre = normality × equivalent

weight.

(vi) (a) Normality = molarity × factor relating mol. wt.

and eq. wt.

(b) No. of equivalents = no. of moles × factor relat

ing mol. wt. and eq. wt.

Ex.39 Calculate the number of milli equivalent of H2SO

4

present in 10 mL of N/2 H2SO

4 solution.

Sol. Number of m.e. = normality × volume in mL =21

× 10 = 5.

Ex.40 Calculate the number of m.e. and equivalents of

NaOH present in 1 litre of N/10 NaOH solution.

Sol. Number of m.e. = normality × volume in mL

= 101

× 1000 = 100

Number of equivalents = 1000

.e.mof.no =

1000100

= 0.10

Ex.41 Calculate number of m.e. of the acids present in

(i) 100 mL of 0.5 M oxalic acid solution.

(ii) 50 mL of 0.1 M sulphuric acid solution.

Sol. Normality = molarity × basicity of acid

(i) Normality of oxalic acid = 0.5 × 2 = 1 N

m.e. of oxalic acid = normality × vol. in mL = 1 × 100

= 100.

(ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N

m.e. of sulphuric acid = 0.2 × 50 = 10

Ex.42 A 100 mL solution of KOH contains 10 milliequiva

lents of KOH. Calculate its strength in normality and

grams/litre.

Sol. Normality = mLinvolume.e.mof.no

= 1.010010

strength of the solution = N/10

Again, strength in grams/litre = normality × eq. wt.

= 56101 = 5.6 gram/litre.

56

156

acidity.wtmolecular

KOHof.wt.eq

PAGE # 40

Ex.43 What is strength in gram/litre of a solution of H2SO

4,

12 cc of which neutralises 15 cc of 10N

NaOH

solution ?

Sol. m.e. of NaOH solution = 101

× 15 = 1.5

m.e. of 12 cc of H2SO

4 = 1.5

normality of H2SO

4 =

125.1

Strength in grams/litre = normality × eq. wt.

= 12

5.1 × 49 grams/litre

= 6.125 grams/litre.

49

298

basicitywt.molecular

SOHofwt.eq. 42

Ex.44 What weight of KMnO4 will be required to prepare

250 mL of its 10N

solution if eq. wt. of KMnO4 is 31.6 ?

Sol. Equivalent weight of KMnO4 = 31.6

Normality of solution (N) = 101

Volume of solution (V) = 250 ml

1000NEV

W ; W = 1000

2506.31101 79.0

406.31 g

Ex.45 100 mL of 0.6 N H2SO

4 and 200 mL of 0.3 N HCl

were mixed together. What will be the normality of theresulting solution ?

Sol. m.e. of H2SO

4 solution = 0.6 × 100 = 60

m.e. of HCl solution = 0.3 × 200 = 60

m.e. of 300 mL (100 + 200) of acidic mixture= 60 + 60 = 120.

Normality of the resulting solution = .voltotal.e.m

= 300120

= 52

N.

Ex.46 A sample of Na2CO

3. H

2O weighing 0.62 g is added

to 100 mL of 0.1 N H2SO

4. Will the resulting solution

be acidic, basic or neutral ?

Sol. Equivalents of Na2CO

3. H

2O =

6262.0

= 0.01

62

2124

OH.CONaof.wt.eq 232

m.e. of Na2CO

3. H

2O = 0.01 × 1000 = 10

m.e. of H2SO

4 = 0.1 × 100 = 10

Since the m.e. of Na2CO

3. H

2O is equal to that of H

2SO

4,

the resulting solution will be neutral.

(a) Introduction :

Volumetric analysis is a method of quantitativeanalysis. It involves the measurement of the volumeof a known solution required to bring about thecompletion of the reaction with a measured volumeof the unknown solution whose concentration orstrength is to be determined. By knowing the volumeof the known solution, the concentration of the solutionunder investigation can be calculated. Volumetricanalysis is also termed as titrimetric analysis.

(b) Important terms used in volumetric

analysis :

(i) Titration : The process of addition of the knownsolution from the burette to the measured volume ofsolution of the substance to be estimated until thereaction between the two is just complete, is termedas titration. Thus, a titration involves two solutions:

(a) Unknown solution and (b) Known solution or stan-dard solution.

(ii) Titrant : The reagent or substance whose solu-tion is employed to estimate the concentration of un-known solution is termed titrant. There are two typesof reagents or titrants:

(A) Primary titrants : These reagents can beaccurately weighed and their solutions are not to bestandardised before use. Oxalic acid, potassiumdichromate, silver nitrate, copper sulphate, ferrousammonium sulphate, sodium thiosulphates etc., arethe examples of primary titrants.

(B) Secondary titrants : These reagents cannotaccurately weighed and their solutions are to bestandardised before use. Sodium hydroxide,potassium hydroxide, hydrochloric acid, sulphuricacid, iodine, potassium permanganate etc. are theexamples of secondary titrants.

(iii) Standard solution : The solution of exactly knownconcentration of the titrant is called the standardsolution.

(iv) Titrate : The solution consisting the substance tobe estimated is termed unknown solution. Thesubstance is termed titrate.

(v) Equivalence point : The point at which the reagent(titrant) and the substance (titrate) under investigationare chemically equivalent is termed equivalence pointor end point.

(vi) Indicator : It is the auxiliary substance used forphysical (visual) detection of the completion of titrationor detection of end point is termed as indicator.Indicators show change in colour or turbidity at thestage of completion of titration.

(c) Concentraion representation of solution

(A) Strength of solution : Grams of solute dissolvedper litre of solution is called strength of solution'

(B) Parts Per Million (ppm) : Grams of solutedissolved per 106 grams of solvent is calledconcentration of solution in the unit of Parts Per Million(ppm). This unit is used to represent hardness ofwater and concentration of very dilute solutions.

(C) Percentage by mass : Grams of solute dissolvedper 100 grams of solution is called percentage bymass.

(D) Percentage by volume : Millilitres of solute per100 mL of solution is called percentage by volume.For example, if 25 mL ethyl alcohol is diluted withwater to make 100 mL solution then the solution thusobtained is 25% ethyl alcohol by volume.

(E) Mass by volume percentage :Grams of solutepresent per 100 mL of solution is called percentagemass by volume.For example, let 25 g glucose is dissolved in water tomake 100 mL solution then the solution is 25% massby volume glucose.

PAGE # 41

(d) Classification of reactions involved in

volumetric analysis

(A) Neutralisation reactions

The reaction in which acids and bases react to formsalt called neutralisation.

e.g., HCI + NaOH NaCI + H2O

H+(acid)

+ OH�

(base) H

2O (feebly ionised)

The titration based on neutralisation is calledacidimetry or alkalimetry.

(B) Oxidation-reduction reactions

The reactions involving simultaneous loss and gainof electrons among the reacting species are calledoxidation reduction or redox reactions, e.g., let usconsider oxidation of ferrous sulphate (Fe2+ ion) bypotassium permanganate (MnO

4� ion) in acidic

medium.

MnO4� + 8H+ + 5e� Mn2+ + 4H

2O

(Gain of electrons or reduction)5 [Fe2+ Fe3+ + e�](Loss of electrons or oxidation)

MnO4� + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H

2O

________________________________________________________________

In the above reaction, MnO4� acts as oxidising agent

and Fe2+ acts as reducing agent.

The titrations involving redox reactions are called redoxtitrations. These titrations are also called accordingto the reagent used in the titration, e.g., iodometric,cerimetric, permanganometric and dichromometrictitrations

(C) Precipitation reactions :A chemical reaction in which cations and anionscombine to form a compound of very low solubility (inthe form of residue or precipitate), is calledprecipitation.BaCl

2 + Na

2SO

4 BaSO

4+ 2NaCl

(white precipitate)

The titrations involving precipitation reactions arecalled precipitation titrations.

(D) Complex formation reactions :

These are ion combination reactions in which asoluble sl ightly dissociated complex ion orcompound is formed.Complex compounds retain their identity in thesolution and have the properties of the constituentions and molecules.e.g. CuSO

4 + 4NH

3 [Cu(NH

3)

4]SO

4

(complex compound)AgNO

3 + 2KCN K[Ag(CN)

2] + KNO

3

(complex compound)

2CuSO4 + K

4[Fe(CN)

6] Cu

2[Fe(CN)

6] + 2K

2SO

4

(complex compound)The titrations involving complex formation reactionsare called complexometric titrations.

The determination of concentration of bases bytitration with a standard acid is called acidimetry andthe determination of concentration of acid by titrationwith a standard base is called alkalimetry.The substances which give different colours withacids and base are called acid base indicators. Theseindicators are used in the visual detection of theequivalence point in acid-base titrations.The acid-base indicators are also called pHindicators because their colour change according tothe pH of the solution.

In the selection of indicator for a titration, followingtwo informations are taken into consideration :

(i) pH range of indicator(ii) pH change near the equivalence point in thetitration.

The indicator whose pH range is included in the pHchange of the solution near the equivalence point, istaken as suitable indicator for the titration.

(i) Strong acid-strong base titration : In the titrationof HCl with NaOH, the equivalence point lies in thepH change of 4�10. Thus, methyl orange, methyl red

and phenolphthalein will be suitable indicators.

(ii) Weak acid-strong base titration : In the titrationof CH

3COOH with NaOH the equivalence point lies

between 7.5 and 10. Hence, phenolphthalein (8.3�10) will be the suitable indicator.

(iii) Weak base-strong acid titration : In the titrationof NH

4OH (weak base) against HCl (strong acid) the

pH at equivalence point is about 6.5 and 4. Thus,methyl orange (3.1�4.4) or methyl red (4.2�6.3) will

be suitable indicators.

(iv) Weak acid-weak base titration : In the titration ofa weak acid (CH

3COOH) with weak base (NH

4OH)

the pH at the equivalence point is about 7, i.e., liesbetween 6.5 and 7.5 but no sharp change in pH isobserved in these titrations. Thus, no simple indica-tor can be employed for the detection of the equiva-lence point.

(v) Titration of a salt of a weak acid and a strongbase with strong acid:

H2CO

3 + 2NaOH Na

2CO

3 + 2H

2O

Weak acid Strong base

Na2CO

3 when titrated with HCl, the following two

stages are involved :Na

2CO

3 + HCl NaHCO

3 + NaCl (First stage)

pH = 8.3, near equivalence pointNaHCO

3 + HCl NaCl + H

2CO

3 (Second stage)

pH = 4, near equivalence point

PAGE # 42

For first stage, phenolphthalein and for second stage,methyl orange will be the suitable indicator.Titration of mixture of NaOH, Na2CO3 and NaHCO3

by strong acid like HCl

In this titration the following indicators are mainly used :

(i) Phenolphthalein (weak organic acid) : It showscolour change in the pH range (8 � 10)

(ii) Methyl orange (weak organic base) : It shows

colour change in the pH range (3.1 � 4.4). Due to

lower pH range, it indicates complete neutralisation

of whole of the base.

Let for complete neutralisation of Na2CO

3, NaHCO

3 and NaOH, x,y and z mL of standard HCl are required. The

titration of the mixture may be carried by two methods as summarised below :

Mixture Phenlphthalein Methyl orange Methyl orange from from after first end beginning beginning point

1. NaOH z + (x/2) (x + z) x/2 (for remaining 50% + Na CO

Na CO )

2. NaOH z + 0 (z + y) y (for remaining 100% + NaHCO NaHCO

3. Na CO (x/2) + 0 (x + y) x/2 + y (for remaining 50% + NaHCO of Na CO and 100% NaHCO are indicated)

2 3 2 3

3 3

2 3

3 2 3

3

Volume of HClused with

An indicator is a substance which is used to determine the end point in a titration. In acid-base titrations organicsubstances (weak acids or weak bases) are generally used as indicators. They change their colour within a certainpH range. The colour change and the pH range of some common indicators are tabulated below:

PAGE # 43

________________________________________Indicator pH range Colour

change________________________________________

Methyl orange 3.2 � 4.5 Orange to red

Methyl red 4.4 � 6.5 Red to yellow

Litmus 5.5 � 7.5 Red to blue

Phenol red 6.8 � 8.4 Yellow to red

Phenolphthalein 8.3 � 10.5 Colourless to pink________________________________________

Theory of acid-base indicators : Two theories havebeen proposed to explain the change of colour ofacid-base indicators with change in pH.

1. Ostwald's theory:According to this theory

(a) The colour change is due to ionisation of the acid-base indicator. The unionised form has differentcolour than the ionised form.

(b) The ionisation of the indicator is largely affected inacids and bases as it is either a weak acid or a weakbase. In case, the indicator is a weak acid, itsionisation is very much low in acids due to commonions while it is fairly ionised in alkalies. Similarly if theindicator is a weak base, its ionisation is large inacids and low in alkalies due to common ions.

Considering two important indicators phenolphtha-lein (a weak acid) and methyl orange (a weak base),Ostwald's theory can be illustrated as follows:Phenolphthalein: It can be represented as HPh. Itionises in solution to a small extent as:

HPh H+ + Ph�

Colourless Pink

Applying law of mass action,

K = ]HPh[

]Ph][H[

The undissociated molecules of phenolphthalein arecolourless while ph� ions are pink in colour. In pres-ence of an acid, the ionisation of HPh is practicallynegligible as the equilibrium shifts to left hand sidedue to high concentration of H+ ions. Thus, the solu-tion would remain colourless. On addition of alkali,hydrogen ions are removed by OH� ions in the form ofwater molecules and the equilibrium shifts to righthand side. Thus, the concentration of ph� ions in-creases in solution and they impart pink colour to thesolution.

Let us derive Henderson's equation for an indicator

HIn + H2O H

3+O + In�

'Acid form' 'Base form'

Conjugate acid-base pair

KIn =

]HIn[]OH][In[ 3

KIn = Ionization constant of indicator

[H3

+O] = KIn ]In[

]HIn[

pH = �log10

[H3

+O] = �log10

[KIn] � log

10 ]In[

]HIn[

pH = pKIn + log

10

]HIn[]In[

(Henderson's equation for

indicator)At equivalence point ;[In�] = [HIn] and pH = pK

In

Methyl orange : It is a very weak base and can berepresented as MeOH. It is ionised in solution to giveMe+ and OH� ions.

MeOH Me+ +OH�

Orange Red

Applying law of mass action,

K = ]MeOH[

]OH][Me[

In presence of an acid, OH� ions are removed in theform of water molecules and the above equilibriumshifts to right hand side. Thus, sufficient Me+ ions areproduced which impart red colour to the solution. Onaddition of alkali, the concentration of OH� ions in-creases in the solution and the equilibrium shifts toleft hand side, i.e., the ionisation of MeOH is practi-cally negligible. Thus, the solution acquires the colourof unionised methyl orange molecules, i.e. orange.This theory also explains the reason why phenol-phthalein is not a suitable indicator for titrating a weakbase against strong acid. The OH� ions furnished bya weak base are not sufficient to shift the equilibriumtowards right hand side considerably, i.e., pH is notreached to 8.3. Thus, the solution does not attainpink colour. Similarly, it can be explained why methylorange is not a suitable indicator for the titration ofweak acid with strong base.

SOLUBILITY

The solubility of a solute in a solution is alwaysexpressed with respect to the saturated solution.

(a) Definition :

The maximum amount of the solute which can bedissolved in 100g (0.1kg) of the solvent to form asaturated solution at a given temperature.Suppose w gram of a solute is dissolved in W gramof a solvent to make a saturated solution at a fixedtemperature and pressure. The solubility of the solutewill be given by -

W

w× 100 =

solventtheofMass

solutetheofMass× 100

For example, the solubility of potassium chloride inwater at 20ºC and 1 atm. is 34.7 g per 100g of water.

This means that under normal conditions 100 g ofwater at 20ºC and 1 atm. cannot dissolve more than

34.7g of KCl.

PAGE # 44

(b) Effect of Temperature and Pressure on

Solubility of a Solids :

The solubility of a substance in liquids generallyincreases with rise in temperature but hardly changeswith the change in pressure. The effect of temperaturedepends upon the heat energy changes whichaccompany the process.

Note :If heat energy is needed or absorbed in the process,it is of endothermic nature. If heat energy is evolvedor released in the process, it is of exothermic nature.

(i) Effect of temperature on endothermic dissolutionprocess : Most of the salts like sodium chloride,potassium chloride, sodium nitrate, ammoniumchloride etc. dissolve in water with the absorption ofheat. In all these salts the solubility increases withrise in temperature. This means that sodium chloridebecomes more soluble in water upon heating.

(ii) Effect of temperature on exothermic dissolutionprocess : Few salts like lithium carbonate, sodiumcarbonate monohydrate, cerium sulphate etc.dissolve in water with the evolution of heat. Thismeans that the process is of exothermic nature. Inthese salts the solubility in water decreases with risein temperature.

Note :

1. While expressing the solubility, the solution mustbe saturated but for expressing concentration (masspercent or volume percent), the solution need not tobe saturated in nature.

2. While expressing solubility, mass of solvent isconsidered but for expressing concentration themass or volume of the solution may be taken intoconsideration.

(c) Effect of Temperature on the Solubility

of a Gas

(i) The solubility of a gas in a liquid decreases withthe rise in temperature.

(ii) The solubility of gases in liquids increases onincreasing the pressure and decreases on decreas-ing the pressure.

SAMPLE PROBLEMS

Ex.47 12 grams of potassium sulphate dissolves in75 grams of water at 60ºC. What is the solubility

of potassium sulphate in water at that temperature ?

Sol. Solubility = 100solventofmasssoluteofmass

= 75

12×100 = 16 g

Thus, the solubility of potassium sulphate inwater is 16 g at 60ºC.

Ex.48 4 g of a solute are dissolved in 40 g of water toform a saturated solution at 25ºC. Calculate the

solubility of the solute.

Sol. Solubility = solventofMass

soluteofMass× 100

Mass of solute = 4 gMass of solvent = 40 g

Solubility = 404

× 100 = 10 g

Ex.49 (a) What mass of potassium chloride would beneeded to form a saturated solution in 50 g ofwater at 298 K ? Given that solubility of the salt is46g per 100g at this temperature.

(b) What will happen if this solution is cooled ?

Sol. (a) Mass of potassium chloride in 100 g of waterin saturated solution = 46 gMass of potassium chloride in 50 g of water insaturated solution.

= 10046

× 50 = 23 g

(b) When the solution is cooled, the solubility ofsalt in water will decrease. This means, that uponcooling, it will start separating from the solutionin crystalline form.

It is defined as the charge (real or imaginary ) which

an atom appears to have when it is in combination.

In the case of electrovalent compounds, the oxidation

number of an element or radical is the same as the

charge on the ion. This is the real charge and is

developed by the loss and gain of electron or

electrons.

For example, in the electrovalent compound , sodium

chloride (NaCl), the charge on sodium and chlorine

is +1 and �1, respectively.

The oxidation numbers of atoms in covalent

compounds can be derived by assigning the electrons

of each bond to the more electronegative atom of the

bonded atoms. For a molecule of HCl both the

electrons of the covalent bond are assigned to the

chlorine atom since it is more electronegative than

hydrogen.

H Cl×

Thus, chlorine atom has one more electron than the

neutral chlorine atom which brings one unit negative

charge on chlorine. The oxidation number of chlorine

in the compound is �1. The hydrogen atom has lost

the only electron possessed by it, thus acquiring one

unit positive charge. The oxidation number of

hydrogen is , therefore , +1 in this compound.

PAGE # 45

Counting of electrons in this fashion is not convenient

in many molecules and therefore the following

operational rules are followed which are helpful and

convenient in determining the oxidation numbers :

(i) The oxidation number (Ox. no.) of an atom in free

elements is zero, no matter how complicated the

molecule is, hydrogen in H2, or O

3,all have zero value

of oxidation numbers.

(ii) The fluorine, which is the most electronegative

element, has oxidation number �1 in all of its

compounds.

(ii i) Oxidation number of oxygen is �2 in all

compounds except in peroxides, superoxides and

oxygen fluorides.

In peroxides (O2

2�) oxygen has oxidation number �1,

in superoxide (O2�) oxygen has oxidation number �½

and in OF2, the oxygen has an oxidation number +2.

(iv) The oxidation number of hydrogen is +1 in all of

its compounds except in metallic hydrides like NaH,

BaH2, etc. hydrogen is in �1 oxidation state in these

hydrides.

(v) The oxidation number of an ion is equal to the

electrical charge present on it.

(vi) The oxidation number of IA elements (Li, Na, K,

Rb, Cs and Fr) is +1 and the oxidation number of IIA

elements (Be, Mg, Ca, Sr, Ba and Ra) is +2.

(vii) For complex ions, the algebraic sum of oxidation

numbers of all the atoms is equal to the net charge

on the ion.

(viii) In the case of neutral molecules, the algebraic

sum of the oxidation numbers of all the atoms present

in the molecule is zero.

Ex.50 What is the oxidation number of Mn in KMnO4 and of

S in Na2S

2O

3 ?

Sol. Let the ox. no. of Mn in KMnO4 be x

We known that ox. no. of K = +1

ox. no. of O = �2

so ox. no. K + ox. no. Mn + 4 (ox. no. O) = 0

or +1 + x + 4(�2) = 0

or +1 + x � 8 = 0

or x = 8 � 1 = + 7

Hence, ox. no. of Mn in KMnO4 is + 7.

Similarly for S in Na2S2O3,

2 (ox. no. Na) +2 (ox. no. S) +3(ox. no. O)= 0

x = +2

Hence ox. no. of S in Na2S

2O

3 = +2

Ex.51 What is the oxidation number of Cr in K2Cr

2O

7 ?

Sol. Let the Ox.no. of Cr in K2Cr

2O

7 be x.

We known that Ox. no. of K = + 1

Ox. no. of O = 2

So, 2(ox. no. K) + 2(ox. no. Cr) + 7 (ox. no. O) = 0

2(+1) + 2(x) + 7(�2) = 0

or +2 + 2x � 14 = 0

or 2x = + 14 �2 = + 12

or x = + 2

12= + 6

Hence, oxidation number of Cr in K2Cr

2O

7 is + 6

Ex.52 Find the oxidation number of

(a) S in SO4

2� (b) S in HSO3�

(c) Pt in [Pt Cl6]2� (d) Mn in (MnO

4)� ion

Sol. (a) Let the oxidation number of S be x.

we known that ox. no. of O = � 2

so ox. no. S + 4 (ox. no. O) = � 2

or x + 4 (�2) = � 2

or x � 8 = � 2

or x = + 8 � 2 = + 6

The oxidation number of S in SO4

2� ions is +6.

(b) Let the oxidation number of S be x in HSO3� ion.

we known that

ox. no. of H = + 1

ox. no. of O = � 2

so ox. no. H + ox. no. S + 3 (ox. no. O) = � 1

or +1 + x + 3(�2) = � 1

or +1 + x � 6 = � 1

x � 5 = � 1

or x = + 5 � 1 = + 4

The oxidation number of S in HSO3� ion is + 4.

(c) Let oxidation number of Pt be x.

We know that Ox. no. of Cl = � 1

So ox. no. Pt + 6(ox. no. Cl) = � 2

or x + 6(�1) = � 2

or x � 6 = �2

or x = + 6 � 2 = + 4

The oxidation number of Pt in [PtCl6]2� ions is + 4.

(d) Let oxidation number of Mn be x.

We known that ox. no. of O = � 2.

So Ox. no. Mn + 4 (ox. no. O) = � 1

or x + 4(�2) = � 1

or x � 8 = � 1

or x = +8 � 1 = + 7

The oxidation number of Mn in [MnO4�] ion is +7.

PAGE # 46

Note : (i) Oxidation state of chromium in CrO5.

CrO5 has butterfly structure having two peroxo bonds.

Peroxo oxygen has (�1) oxidation state

Let oxidation state of chromium be �x�

x + 4(�1) + (�2) = 0

x = + 6

(ii) Oxidation state of chlorine in bleaching powder :

Bleaching powder has two chlorine atoms having

different oxidation states.

(iii) Fractional values of oxidation numbers are

possible as in Na2S

4O

6, Fe

3O

4, N

3H, etc.

(iv) Fe0.94

O (Oxidation state of iron is to be determined) :

0.94x � 2 = 0

x = 2/0.94 = 200/94

(v) Na2[Fe(CN)

5NO] : In iron complex NO lies in NO+

state ; thus oxidation state of �Fe� may be determined

as :

+2 + x � 5 + 1 = 0

x = + 2

(vi) [Fe(NO)(H2O)

5]SO

4 :

x + 1 + 5(0) � 2 = 0

x = + 1

Balancing oxidation reduction reactions by

oxidation number method :In a balanced redox reaction, total increase in oxidationnumber must be equal to the decrease in oxidationnumber. This equivalence provides the basis forbalancing redox reaction. This method is applicableto both molecular and ionic equation. The generalprocedure involves the following steps:

(i) Write the skeleton equation (if not given, frame it)representing the chemical change.

(ii) Assign oxidation number to the atoms in theequation and find out which atoms are undergoingoxidation and reduction. Write separate equations forthe atoms undergoing oxidation and reduction.

(iii) Find the change in oxidation number in eachequation. Make the change equal in both theequations by multiplying with suitable integers. Addboth the equations.

(iv) Complete the balancing by inspection. Firstbalance those substance which have undergonechange in oxidation number and then other atomsexcept hydrogen and oxygen. Finally balancehydrogen and oxygen by putting H

2O molecules

wherever needed.The final balanced equation should be checked toensure that there are as many atoms of each elementon the right as there are on the left.

(v) In ionic equations the net charges on both sidesof the equations must be exactly the same. Use H+

ion/ions in acidic reactions and OH� ion/ions in basicreactions to balance the charge and number ofhydrogen and oxygen atoms.The following examples illustrate the above rules.

Ex.53 Balance the following equation by oxidation numbermethod :Cu + HNO

3 Cu(NO

3)

2 + NO

2 + H

2O

Sol. Writing the oxidation number of all the atoms.

Change in ox. no. has occurred in copper and nitrogen.

Increasing in ox. no. of copper = 2 unit per moleculeCudecrease in ox. no. of nitrogen = 1 unit per moleculeHNO

3

To make increase and decrease equal eq. (ii) ismultiplied by 2.Cu + 2HNO

3 Cu(NO

3)

2 + 2NO

2 + H

2O

Balancing nitrate ions, hydrogen and oxygen, thefollowing equation is obtained.Cu + 4HNO

3 Cu(NO

3)

2 + 2NO

2 + 2H

2O

This is the balanced equation.

PAGE # 47

Ion - electron method for balancing redoxreactions:

The method for balancing redox reactions by ionelectron method was developed by Jette and LaMevinvolves the following steps :(i) Write down the redox reaction in ionic form.(ii) Split the redox reaction into two half reactios, onefor oxidation and the other for reduction.(iii) Balance each half reaction for the number ofatoms of each element . For this purpose.(a) Balance the atoms other than H and O for eachhalf reaction using simple multiples.(b) Add water molecules to the side deficient in oxygenand H+ to the side deficient in hydrogen. This is donein acidic or neutral solutions.(c) In alkaline solution, for each excess of oxygen,add one water molecule to the same side and twoOH� ions to the other side. If hydrogen is stillunbalanced, add one OH� ion for each excesshydrogen on the same side and one water moleculeto the other side.(iv) Add electrons to the side deficient in electrons asto equalise the charge on both sides.(v) Multiply one or both the half reactions by a suitablenumber so that the number of electrons becomeequal in both the equations.(vi) Add the two balanced half reaction and cancel anyterm common to both sides.The following solved problems illustrate the varioussteps of ion electron method :

Ex.54 Balance the following equations by ion electronsmethod :(a) MnO

4� + Fe2+ + H+ Mn2+ + Fe3+ + H

2O

(b) Cl2 + lO

3� + OH� IO

4� + Cl� + H

2O

Sol. (a) MnO4� + Fe2+ + H+ Mn2+ + Fe3+ + H

2O

Ist step : splitting into two half reactions.MnO

4� Mn2+ ; Fe2+ Fe3+

(Reduction half reaction) (oxidation half reaction)

2nd step : In first reaction add 4H2O in R.H.S. to

balance oxygen.MnO

4� Mn2+ + 4H

2O

3rd Step : Adding hydrogen ions to the side deficientin hydrogen.MnO

4� + 8H+ Mn2+ + 4H

2O

4th steps : Adding electrons to the side deficient inelectrons.MnO

4� + 8H+ + 5e� Mn2+ + 4H

2O

Fe2+ Fe3+ + e�

5th step : Balancing electrons in both half reaction.MnO

4� + 8H+ + 5e� Mn2+ + 4H

2O

5Fe2+ 5Fe3+ + 5e�

6th steps : Adding both the half reaction.MnO

4� + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H

2O

Sol. (b) Cl2 + IO

3� + OH� IO

4� + Cl� + H

2O

Ist step : Splitting into two half reactions,IO

3� IO�

4; Cl

2 Cl�

(oxidation half reaction (Reduction half reaction)

2nd step : Adding 2OH� ions and H2O to balance H

and OIO

3� + 2OH� IO

4� + H

2O

3rd step : Adding electrons to the sides deficient inelectrons,IO

3� + 2OH� IO

4� + H

2O + 2e�

Cl2 + 2e� 2Cl�

4th step : Adding both the half reaction.IO

3� + 2OH� + Cl

2 IO

4� + 2Cl� + H

2O

EXERCISE

1. The solubility of K2SO

4 in water is 16 g at 50ºC. The

minimum amount of water required to dissolve 4 g

K2SO

4 is -

(A) 10 g (B) 25 g

(C) 50 g (D) 75 g

2. Molarity of H2SO

4 (density 1.8g/mL) is 18M. The

molality of this solution is -

(A)36 (B) 200

(C) 500 (D) 18

3. 8g of sulphur are burnt to form SO2, which is oxidised

by Cl2 water. The solution is treated with BaCl

2

solution. The amount of BaSO4 precipitated is -

(A) 1.0 mole (B) 0.5 mole

(C) 0.75 mole (D) 0.25 mole

4. In a compound AxB

y -

(A) Mole of A = Mole of B = mole of AxB

y

(B) Eq. of A = Eq. of B = Eq. of AxB

y

(C) X × mole of A = y × mole of B = (x + y) × mole of AxB

y

(D) X × mole of A = y × mole of B

5. The percentage of sodium in a breakfast cereal

labelled as 110 mg of sodium per 100 g of cereal is -

(A) 11% (B) 1.10%

(C) 0.110% (D) 110%

6. Two elements A (at. wt. 75) and B (at. wt. 16) combine

to yield a compound. The % by weight of A in the

compound was found to be 75.08. The empirical

formula of the compound is -

(A) A2B (B) A

2B

3

(C) AB (D) AB2

7. No. of oxalic acid molecules in 100 mL of 0.02 N

oxalic acid are -

(A) 6.023 × 1020 (B) 6.023 × 1021

(C) 6.023 × 1022 (D) 6.023 × 1023

PAGE # 48

8. Which of the following sample contains the maximum

number of atoms -

(A) 1 mg of C4H

10(B) 1 mg of N

2

(C) 1 mg of Na (D) 1 mL of water

9. The total number of protons, electrons and neutrons

in 12 g of C126 is -

(A) 1.084 × 1025 (B) 6.022 × 1023

(C) 6.022 × 1022 (D) 18

10. 4.4 g of CO2 and 2.24 litre of H

2 at STP are mixed in a

container. The total number of molecules present in

the container will be -

(A) 6.022 × 1023 (B) 1.2044 × 1023

(C) 2 mole (D) 6.023 × 1024

11. Which is not a molecular formula ?

(A) C6H

12O

6(B) Ca(NO

3)

2

(C) C2H

4O

2(D) N

2O

12. The hydrated salt, Na2SO

4. nH

2O undergoes 55.9%

loss in weight on heating and becomes anhydrous.

The value of n will be -

(A) 5 (B) 3

(C) 7 (D) 10

13. Which of the following mode of expressing

concentration is independent of temperature -

(A) Molarity (B) Molality

(C) Formality (D) Normality

14. The haemoglobin of the red blood corpuscles of most

of the mammals contains approximately 0.33% of

iron by weight. The molecular weight of haemoglobin

is 67,200. The number of iron atoms in each molecule

of haemoglobin is (Atomic weight of iron = 56) -

(A) 2 (B) 3

(C) 4 (D) 5

15. An oxide of metal have 20% oxygen, the eq. wt. of

metal oxide is -

(A) 32 (B) 40

(C) 48 (D) 52

16. How much water is to be added to dilute 10 mL of 10

N HCl to make it decinormal ?

(A) 990 mL (B) 1010 mL

(C) 100 mL (D) 1000 mL

17. The pair of compounds which cannot exist in solution is -

(A) NaHCO3 and NaOH

(B) Na2SO

3 and NaHCO

3

(C) Na2CO

3 and NaOH

(D) NaHCO3 and NaCl

18. If 250 mL of a solution contains 24.5 g H2SO

4 the

molarity and normality respectively are -

(A) 1 M, 2 N (B) 1M,0.5 N

(C) 0.5 M, 1N (D) 2M, 1N

19. The mole fraction of NaCl, in a solution containing 1

mole of NaCl in 1000 g of water is -

(A) 0.0177 (B) 0.001

(C) 0.5 (D) 0.244

20. 3.0 molal NaOH solution has a density of 1.110 g/

mL. The molarity of the solution is -

(A) 2.9732 (B) 3.05

(C) 3.64 (D) 3.0504

21. How many atoms are contained in a mole of Ca(OH)2 -

(A) 30 × 6.02 × 1023 atoms/mol

(B) 5 × 6.02 × 1023 atoms/mol

(C) 6 × 6.02 × 1023 atoms/mol

(D) None of these

22. Insulin contains 3.4% sulphur. The minimum

molecular weight of insulin is -

(A) 941.176 u (B) 944 u

(C) 945.27 u (D) None of these

23. Number of moles present in 1 m3 of a gas at NTP are -

(A) 44.6 (B) 40.6

(C) 42.6 (D) 48.6

24. Weight of oxygen in Fe2O

3 and FeO is in the simple

ratio of -

(A) 3 : 2 (B) 1 : 2

(C) 2 : 1 (D) 3 : 1

25. 2.76 g of silver carbonate on being strongly heated

yield a residue weighing -

(A) 2.16g (B) 2.48 g

(C) 2.32 g (D) 2.64 g

26. How many gram of KCl would have to be dissolved in

60 g of H2O to give 20% by weight of solution -

(A) 15 g (B) 1.5 g

(C) 11.5 g (D) 31.5 g

27. When the same amount of zinc is treated separately

with excess of H2SO

4 and excess of NaOH, the ratio

of volumes of H2 evolved is -

(A) 1 : 1 (B) 1 : 2

(C) 2 : 1 (D) 9 : 4

28. Amount of oxygen required for combustion of 1 kg of a

mixture of butane and isobutane is -

(A) 1.8 kg (B) 2.7 kg

(C) 4.5 kg (D) 3.58 kg

29. Rakesh needs 1.71 g of sugar (C12

H22

O11

) to sweeten

his tea. What would be the number of carbon atoms

present in his tea ?

(A) 3.6 × 1022 (B) 7.2 × 1021

(C) 0.05 × 1023 (D) 6.6 × 1022

30. The total number of AlF3 molecule in a sample of AlF

3

containing 3.01 × 1023 ions of F� is -

(A) 9.0 × 1024 (B) 3.0 × 1024

(C) 7.5 × 1023 (D)1023

PAGE # 49

31. The volume occupied by one molecule of water(density 1 g/cm3) is -(A) 18 cm3 (B) 22400 cm3

(C) 6.023 × 10�23 (D) 3.0 × 10�23 cm3

32. 224 mL of a triatomic gas weigh 1 g at 273 K and 1atm. The mass of one atom of this gas is -(A) 8.30 × 10�23 g (B) 2.08 × 10�23 g(C) 5.53 × 10�23 g (D) 6.24 × 10�23 g

33. The percentage of P2O

5 in diammonium hydrogen

phosphate is -(A) 77.58 (B) 46.96(C) 53.78 (D) 23.48

34. The mole fraction of water in 20% (wt. /wt.) aqueoussolution of H

2O

2 is -

(A) 68

77(B)

77

68

(C) 80

20(D)

20

80

35. Which of the following has the maximum mass ?(A) 25 g of Hg(B) 2 moles of H

2O

(C) 2 moles of CO2

(D) 4 g atom of oxygen

36. Total mass of neutrons in 7mg of 14C is -(A) 3 × 1020 kg (B) 4 × 10�6 kg(C) 5 × 10�7 kg (D) 4 × 10�7 kg

37. Vapour density of a metal chloride is 66. Its oxidecontains 53% metal. The atomic weight of metal is -(A) 21 (B) 54(C) 26.74 (D) 2.086

38. The number of atoms in 4.25 g NH3 is approximately -

(A) 1 × 1023 (B) 1.5 × 1023

(C) 2 × 1023 (D) 6 × 1023

39. The modern atomic weight scale is based on -(A) C12 (B) O16

(C) H1 (D) C13

40. Amount of oxygen in 32.2g of Na2SO

4. 10H

2O is -

(A) 20.8 g (B) 22.4 g(C) 2.24 g (D) 2.08 g

41. Which of the followings does not change on dilution ?(A) Molarity of solution(B) Molality of solution(C) Millimoles and milli equivalent of solute(D) Mole fraction of solute

42. Equal masses of O2, H

2 and CH

4 are taken in a

container. The respective mole ratio of these gases

in container is -

(A) 1 : 16 : 2 (B) 16 : 1 : 2

(C) 1 : 2 : 16 (D) 16 : 2 : 1

43. The number of molecules present in 11.2 litre CO2 at

STP is -

(A) 6.023 × 1032 (B) 6.023 × 1023

(C) 3.011 × 1023 (D) None of these

44. 250 ml of 0.1 N solution of AgNO3 are added to 250

ml of a 0.1 N solution of NaCl. The concentration of

nitrate ion in the resulting solution will be -

(A) 0.1N (B) 1.2 N

(C) 0.01 N (D) 0.05 N

45. Amount of BaSO4 formed on mixing the aqueous

solution of 2.08 g BaCl2 and excess of dilute H

2SO

4 is -

(A) 2.33 g (B) 2.08 g

(C) 1.04 g (D) 1.165 g

46. 2g of NaOH and 4.9 g of H2SO

4 were mixed and

volume is made 1 litre. The normality of the resulting

solution will be -

(A) 1N (B) 0.05 N

(C) 0.5 N (D) 0.1N

47. 1g of a metal carbonate neutralises completely 200

mL of 0.1N HCl. The equivalent weight of metal

carbonate is -

(A) 25 (B) 50

(C) 100 (D) 75

48. 100 mL of 0.5 N NaOH were added to 20 ml of 1N

HCl and 10 mL of 3 N H2SO

4. The solution is -

(A) acidic (B) basic

(C) neutral (D) none of these

49. 1M solution of H2SO

4 is diluted from 1 litre to 5 litres ,

the normality of the resulting solution will be -

(A) 0.2 N (B) 0.1 N

(C) 0.4 N (D) 0.5 N

50. The volume of 7g of N2 at S.T.P. is -

(A) 11.2 L (B) 22.4 L

(C) 5.6 L (D) 6.5 L

51. One mole of calcium phosphide on reaction with

excess of water gives -

(A) three moles of phosphine

(B) one mole phosphoric acid

(C) two moles of phosphine

(D) one mole of P2O

5

52. Mg (OH)2 in the form of milk of magnesia is used to

neutralize excess stomach acid. How many moles of

stomach acid can be neutralized by 1 g of Mg(OH)2 ?

(Molar mass of Mg(OH)2 = 58.33)

(A) 0.0171 (B) 0.0343

(C) 0.686 (D) 1.25

PAGE # 50

53. Calcium carbonate decomposes on heating

according to the following equation -


2(g)

How many moles of CO2 will be obtained by

decomposition of 50 g CaCO3 ?

(A) 23

(B) 25

(C) 21

(D) 1

54. Sulphur trioxide is prepared by the following two

reactions -

S8(s) + 8O

2(g) 8SO

2(g)

2SO2(g) + O

2(g) 2SO

3(g)

How many grams of SO3 are produced from 1 mole

of S8 ?

(A) 1280 (B) 640

(C) 960 (D) 320

55. PH3(g) decomposes on heating to produce

phosphorous and hydrogen. The change in volume

when 100 mL of such gas decomposed is -

(A) + 50 mL (B) + 500 mL

(C) � 50 mL (D) � 500 mL

56. What amount of BaSO4 can be obtained on mixing

0.5 mole BaCl2 with 1 mole of H

2SO

4 ?

(A) 0.5 mol (B) 0.15 mol

(C) 0.1 mol (D) 0.2 mol

57. In the reaction , CrO5 + H

2SO

4 Cr

2(SO

4)3 + H

2O + O

2

one mole of CrO5 will liberate how many moles of O

2 ?

(A) 5/2 (B) 5/4

(C) 9/2 (D) None

58. Calcium carbonate decomposes on heating

according to the equation -


2(g)

At STP the volume of CO2 obtained by thermal

decomposition of 50 g of CaCO3 will be -

(A) 22.4 litre (B) 44 litre

(C) 11.2 litre (D) 1 litre

59. When FeCl3 is ignited in an atmosphere of pure

oxygen, the following reaction takes place-

4FeCl3(s) + 3O

2(g) 2Fe

2O

3(s) + 6Cl

2(g)

If 3 moles of FeCl3 are ignited in the presence of 2

moles of O2 gas, how much of which reagent is

present in excess and therefore, remains unreacted ?

(A) 0.33 mole FeCl3 remains unreacted

(B) 0.67 mole FeCl3 remains unreacted

(C) 0.25 mole O2 remains unreacted

(D) 0.50 mole O2 remains unreacted

60. The volume of CO2 (in litres) liberated at STP when

10 g of 90% pure limestone is heated completely, is-(A) 22.4 L (B) 2.24 L(C) 20.16 L (D) 2.016 L

61. A metal oxide has the formula Z2O

3. It can be reduced

by hydrogen to give free metal and water. 0.1596 g ofthe metal requires 6 mg of hydrogen for completereduction. The atomic mass of the metal is -(A) 27.9 (B) 159.6(C) 79.8 (D) 55.8Question number 62, 63, 64 and 65 are based on thefollowing information :

Q. Dissolved oxygen in water is determined by using aredox reaction. Following equations describe theprocedure -

I 2Mn2+(aq) + 4OH�(aq) + O2 (g) 2MnO2(s) + 2H2O( )

II MnO2(s)+2I�(aq)+4H+(aq) Mn2+(aq)+I2(aq) + 2H2O( )

III �232OS2 (aq) + I2(aq) �2

64OS2 (aq) + 2I�(aq)

62. How many moles of �232OS are equivalent to each

mole of O2 ?

(A) 0.5 B) 1(C) 2 (D) 4

63. What amount of I2 will be liberated from 8 g dissolved

oxygen ?(A) 127 g (B) 254 g(C) 504 g (D) 1008 g

64. 3 × 10�3 moles O2 is dissolved per litre of water, then

what will be molarity of I� produced in the givenreaction ?(A) 3 × 10�3 M (B) 4 × 3 × 10�3 M

(C) 2 × 3 × 10�3 M (D) 3�10321

M

65. 8 mg dissolved oxygen will consume -(A) 5 × 10�4 mol Mn+2

(B) 2.5 × 10�4 mol Mn2+

(C) 10 mol Mn2+

(D) 2 mol Mn2+

66. 2 g of a base whose eq. wt. is 40 reacts with 3 g of anacid. The eq. wt. of the acid is-(A) 40 (B) 60(C) 10 (D) 80

67. Normality of 1% H2SO

4 solution is nearly -

(A) 2.5 (B) 0.1(C) 0.2 (D) 1

68. What volume of 0.1 N HNO3 solution can be prepared

from 6.3 g of HNO3 ?

(A) 1 litre (B) 2 litres(C) 0.5 litre (D) 4 litres

69. The volume of water to be added to 200 mL of

seminormal HCl solution to make it decinormal is -

(A) 200 mL (B) 400 mL

(C) 600 mL (D) 800 mL

PAGE # 51

70. 0.2 g of a sample of H2O

2 required 10 mL of 1N KMnO

4

in a titration in the presence of H2SO

4. Purity of H

2O

2 is-

(A) 25% (B) 85%

(C) 65% (D) 95%

71. Which of the following has the highest normality ?

(A) 1 M H2SO

4(B) 1 M H

3PO

3

(C) 1 M H3PO

4(D) 1 M HNO

3

72. The molarity of 98% H2SO

4(d = 1.8g/mL) by wt. is -

(A) 6 M (B) 18.74 M

(C) 10 M (D) 4 M

73. 0.7 g of Na2CO

3 . xH

2O is dissolved in 100 mL. 20 mL

of which required to neutralize 19.8 mL of 0.1 N HCl.

The value of x is -

(A) 4 (B) 3

(C) 2 (D) 1

74. 0.45 g of an acid of molecular weight 90 was

neutralised by 20 mL of 0.5 N caustic potash. The

basicity of the acid is -

(A) 1 (B) 2

(C) 3 (D) 4

75. 1 litre of 18 molar H2SO

4 has been diluted to 100

litres. The normality of the resulting solution is -

(A) 0.09 N (B) 0.18

(C) 1800 N (D) 0.36

76. 150 ml of 10N

HCl is required to react completely with

1.0 g of a sample of limestone. The percentage purity

of calcium carbonate is -

(A) 75% (B) 50%

(C) 80% (D) 90%

77. 50 ml of 10N

HCl is treated with 70 ml 10N

NaOH.

Resultant solution is neutralized by 100 ml of

sulphuric acid. The normality of H2SO

4 -

(A) N/50 (B) N/25

(C) N/30 (D) N/10

78. 200 mL of 10N

HCl were added to 1 g calcium car-

bonate, what would remain after the reaction ?

(A) CaCO3

(B) HCl

(C) Neither of the two (D) Part of both

79. Equivalent mass of KMnO4, when it is converted to

MnSO4 is -

(A) M/5 (B) M/3

(C) M/6 (D) M/2

80. A 3 N solution of H2SO

4 in water is prepared from

Conc. H2SO

4 (36 N) by diluting

(A) 20 ml of the conc. H2SO

4 to 240 ml

(B) 10 ml of the conc. H2SO

4 to 240 ml

(C) 1 ml of the conc. H2SO

4 to 36 ml

(D) 20 ml of the conc. H2SO

4 to 36 ml

81. The solubility curve of KNO3 as a function of tempera-

ture is given below

0 20 40 60 80 100

0

50

100

150

200

250

Temperature (°C)

Sol

ubili

ty (

g/10

0 m

l wat

er)

The amount of KNO3 that will crystallize when a satu-

rated solution of KNO3 in 100 ml of water is cooled

from 90°C to 30 °C, is

(A) 16 g (B) 100 g

(C) 56 g (D)160 g

82. The volume of 0.5 M aqueous NaOH solution required

to neutralize 10 ml of 2 M aqueous HCl solution is :

(A) 20ml (B) 40ml

(C) 80ml (D) 120ml

83. 3.01×1023 molecules of elemental Sulphur will react

with 0.5 mole of oxygen gas completely to produce

(A) 6.02 × 1023 molecules of SO3

(B) 6.02 × 1023 molecules of SO2

(C) 3.01 × 1023 molecules of SO3

(D) 3.01 x 1023 molecules of SO2

84. The solubility of a gas in a solution is measured in

three cases as shown in the figure given below where

w is the weight of a solid slab placed on the top of the

cylinder lid. The solubility will follow the order :

gas

solution

w

gas

solution

w w

gas

solution

w w w

(A) a > b > c (B) a < b < c

(C) a = b = c (D) a >b < c

PAGE # 52

85. The density of a salt solution is1.13 g cm�3 and itcontains 18% of NaCI by weight. The volume of thesolution containing 36.0 g of the salt will be :(A) 200 cm3 (B) 217 cm3

(C) 177 cm3 (D) 157cm3

86. One mole of nitrogen gas on reaction with 3.01 x 1023

molecules of hydrogen gas produces -(A) one mole of ammonia(B) 2.0 x 1023 molecules of ammonia(C) 2 moles of ammonia(D) 3.01 × 1023 molecules of ammonia

87.Solubility g/I

20 40 60 80 100

KNO3

KCl

Temperature (ºC)

50

100

150

200

250

Given the solubility curves of KNO3 and KCl, which of

the following statements is not true ?(A) At room temperature the solubility of KNO

3 and

KCI are not equal(B) The solubilities of both KNO

3 and KCI increase

with temperature(C) The solubility of KCI decreases with temperature(D) The solubility of KNO

3 increases much more com-

pared to that of KCl with increase in temperature

88. 10 ml of an aqueous solution containing 222 mg ofcalcium chloride (mol. wt. = 111) is diluted to 100 ml.The concentration of chloride ion in the resulting so-lution is -(A) 0.02 mol/lit. (B) 0.01 mol/lit.(C) 0.04 mol/lit (D) 2.0 mol/lit.

89. Aluminium reduces manganese dioxide to manga-nese at high temperature. The amount of aluminiumrequired to reduce one gram mole of manganesedioxide is -(A) 1/2 gram mole (B) 1 gram mole(C) 3/4 gram mole (D) 4/3 gram mole

90. One mole of oxalic acid is equivalent to -(A) 0.5 mole of NaOH (B) 1 mole of NaOH(C) 1.5 mole of NaOH (D) 2 mole of NaOH

91. 8 Grams of oxygen at NTP contain[IJSO-Stage-I/2012]

(A) 1.5 × 1023 molecules(B) 3.0 × 1023 molecules(C) 6.023 × 1023 molecules(D) 1.5 × 1022 molecules

92. When 1g of CaCO3 reacts with 50 ml of 0.1 M HCI, the

volume of CO2 produced is - [IJSO-Stage-I/2012]

(A) 11.2 mL (B) 22.4 mL(C) 112 mL (D) 56 mL

93. Molality of a solution is the number of -[IJSO-Stage-I/2012]

(A) moles of the solute per 1000 mL of the solution.(B) moles of the solute per 1000 mL of the solvent.(C) moles of the solute per 1000 g of the solvent.(D) moles of the solute per 100g of the solvent.

94. The oxidation number of chlorine in CaOCI2 is -

[IJSO-Stage-I/2012]

(A) 0 (B) �1

(C) +1 (D) +3

SUBJECTIVE QUESTIONS

95. (a) Sachin was suffering from problem of acidity , sohe visited a physician who advised him to take 0.0025dm3 of milk of magnesia for a fast relief. He exactlyfollowed what the doctor told him to do. Out of curios-ity he saw the label on milk of magnesia bottle andhe found that there were different ingredients writtenon it and the concentration of milk of magnesia men-tioned was 29 ppm. Assuming, the volume of milk ofmagnesia required for neutralization of acid is equalto intake of milk of magnesia, help Sachin to find outthe following

(i) How many moles of acid was produced in Sachin�sstomach ?

(ii) Write down the neutralization reaction of thisprocess.

(iii)Calculate the concentration of acid produced inmol/dm3

53PAGE # 53

(a) Definition :

An algebraic expression f(x) of the form

f(x) = a0 + a

1x + a

2x2 + ..........+ a

nxn, Where a

0 ,a

1, a

2.....a

n

are real numbers and all the indices of x are non

negative integers is called a polynomial in x and the

highest index n is called the degree of the polynomial,

if an 0. Here a

0 , a

1x, a

2x2 .....,a

nxn are called the terms

of the polynomial and a0, a

1, a

2, ...... a

n are called various

co-efficients of the polynomial f(x). A polynomial in x is

said to be in its standard form when the terms are

written either in increasing order or decreasing order

of the indices of x in various terms.

EXAMPLES :

(i) 2x3 + 4x2 + x + 1 is a polynomial of degree 3.

(ii) x7 + x5 + x2 + 1 is a polynomial of degree 7.

(iii) x3/2 + x2 + 1 is not a polynomial as the indices of x

are not all non negative integer

(iv) x2 + 2 x + 1 is a polynomial of degree 2.

(v) x�2 + x + 1 is not a polynomial as �2 is not non

negative.

(a) Polynomial Based on Degree :

There are five types of polynomials based on degree.

(i) Constant polynomial :

A polynomial of degree zero is called a zero degree

polynomial or constant polynomial.

e.g. f(x) = 4 = 4x0

(ii) Linear polynomial :

A polynomial of degree one is called a linear polynomial.

The general form of a linear polynomial is ax + b, where

a and b are any real numbers and a 0

e.g. 4x + 5, 2x + 3, 5x + 3 etc.

POLYNOMIALS

(iii) Quadratic polynomial :

A polynomial of degree two is called a quadratic

polynomial. The general form of a quadratic polynomial

is ax2 + bx + c where a 0

e.g. x2 + x + 1, 2x2 + 1, 3x2 + 2x + 1 etc.

(iv) Cubic polynomial :

A polynomial of degree three is called a cubic

polynomial. The general form of a cubic polynomial is

ax3 + bx2 + cx + d, where a 0

e.g. x3 + x2 + x + 1, x3 + 2x + 1, 2x3 + 1 etc.

(v) Biquadratic polynomial :

A polynomial of degree four is called a biquadratic or

quartic polynomial. The general form of biquadratic

polynomial is ax4 + bx3 + cx2 + dx + e where a 0

e.g. x4 + x3 + x2 + x + 1 , x4 + x2 + 1 etc.

NOTE :

A polynomial of degree five or more than five does not

have any particular name. Such a polynomial is usually

called a polynomial of degree five or six or ..... etc.

(b) Polynomial Based on Terms :

There are three types of polynomial based on number

of terms.

(i) Monomial : A polynomial is said to be a monomial if

it has only one term.

For example, x, 9x2, � 5x2 are all monomials

(ii) Binomial : A polynomial is said to be a binomial if it

contains two terms.

For example 2x2 + 3x, 3 x + 5x4, � 8x3 + 3 etc are all

binomials.

(iii) Trinomial : A polynomial is said to be a trinomial if

it contains three terms.

For example 3x3 � 8x +2

5, 7 x10 + 8x4 � 3x2 etc are all

trinomials.

REMARKS :

(i) A polynomial having four or more than four terms

does not have any particular name. They are simply

called polynomials.

(ii) A polynomial whose coefficients are all zero is called

a zero polynomial, degree of a zero polynomial

is not defined.


54PAGE # 54

(a) Value of a Polynomial :

The value of a polynomial f(x) at x = is obtained by

substituting x = in the given polynomial and is

denoted by f().

Consider the polynomial f(x) = x3 � 6x2 + 11x � 6,

If we replace x by � 2 everywhere in f(x), we get

f(� 2) = (� 2)3 � 6(� 2)2 + 11(� 2) � 6

f(� 2) = � 8 � 24 � 22 � 6

f(� 2) = � 60 0.

So, we can say that value of f(x) at x = � 2 is � 60.

(b) Zero or root of a Polynomial :

The real number is a root or zero of a polynomial

f(x), if f( = 0.

Consider the polynomial f(x) = 2x3 + x2 � 7x � 6,

If we replace x by 2 everywhere in f(x), we get

f(2) = 2(2)3 + (2)2 � 7(2) � 6

= 16 + 4 � 14 � 6 = 0

Hence, x = 2 is a root of f(x).

Let �p(x)� be any polynomial of degree greater than or

equal to one and a be any real number and If p(x) is

divided by (x � a), then the remainder is equal to p(a).

Ex.1 Find the remainder, when f(x) = x3 � 6x2 + 2x � 4 is

divided by g(x) = 1 � 2x.

Sol. f(x) = x3 � 6x2 + 2x � 4

Let, 1 � 2x = 0

2x = 1

x = 21

Remainder =

21

f

21

f = 4�21

221

6�21

23

= 4�123

�81

= 8

35�

832�812�1

.

FACTOR THEOREM

Let p(x) be a polynomial of degree greater than or equal

to 1 and �a� be a real number such that p(a) = 0, then

(x � a) is a factor of p(x). Conversely, if (x � a) is a factor

of p(x), then p(a) = 0.

Ex.2 Show that x + 1 and 2x � 3 are factors of

2x3 � 9x2 + x + 12.

Sol. To prove that (x + 1) and (2x � 3) are factors of

2x3 � 9x2 + x + 12 it is sufficient to show that p(�1) and

23

p both are equal to zero.

p (� 1) = 2 (� 1)3 � 9 (� 1)2 + (� 1) + 12

= � 2 � 9 � 1 + 12

= � 12 + 12 = 0.

and

23

p = 122

3

2

39�

2

32

23

= 1223

481

�4

27

= 4

48681�27

=4

8181�

= 0.

Hence, (x + 1) and (2x � 3) are the factors

2x3 � 9x2 + x + 12.

Ex. 3 Find the values of a and b so that the polynomials

x3 � ax2 � 13x + b has (x � 1) and (x + 3) as factors.

Sol. Let f(x) = x3 � ax2 � 13x + b

Because (x � 1) and (x + 3) are the factors of f(x),

f(1) = 0 and f(� 3) = 0

f(1) = 0

(1)3 � a(1)2 � 13(1) + b = 0

1 � a � 13 + b = 0

� a + b = 12 .... (i)

f(�3) = 0

(� 3)3 � a(� 3)2 � 13(� 3) + b = 0

� 27 � 9a + 39 + b = 0

� 9a + b = �12 ...(ii)

Subtracting equation (ii) from equation (i)

(� a + b) � (� 9a + b) = 12 + 12

� a + 9a = 24

8a = 24

a = 3.

Put a = 3 in equation (i)

� 3 + b = 12

b = 15.

Hence, a = 3 and b = 15.

55PAGE # 55

ALGEBRAIC IDENTITIES

Some important identities are

(i) (a + b)2 = a2 + 2ab + b2

(ii) (a � b)2 = a2 � 2ab + b2

(iii) a2 � b2 = (a + b) (a � b)

(iv) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2 bc + 2ca

(v) a3 + b3 = (a + b) (a2 � ab + b2)

(vi) a3 � b3 = (a � b) (a2 + ab + b2)

(vii) (a + b)3 = a3 + b3 + 3ab (a + b)

(viii) (a � b)3 = a3 � b3 � 3ab (a � b)

(ix) a3 + b3 + c3 � 3abc = (a + b + c) (a2 + b2 + c2 � ab � bc � ac)

Special case : if a + b + c = 0 then a3 + b3 + c3 = 3abc.

Value Form :

(i) a2 + b2 = (a + b)2 � 2ab, if a + b and ab are given.

(ii) a2 + b2 = (a � b)2 + 2ab, if a � b and ab are given.

(iii) a + b = ab4ba 2 , if a � b and ab are given.

(iv) a � b = ab4ba 2 , if a + b and ab are given.

(v) a2 + 2a

1 =

2

a

1a

� 2, if a +

a1

is given.

(vi) a2 + 2a

1 =

2

a1

a

+ 2, if a �

a1

is given.

(vii) a3 + b3 = (a + b)3 � 3ab (a + b), if (a + b) and ab aregiven.(viii) a3 � b3 = (a � b)3 + 3ab (a � b), if (a � b) and ab aregiven.

(ix) 3

3

a

1a =

3

a1

a

� 3

a1

a , if a +a1

is given.

(x) 3

3

a

1a =

3

a1

a

+ 3

a1

a , if a � a1

is given.

(xi) a4 � b4 = (a2 + b2) (a2 � b2) = [(a + b)2 � 2ab](a + b) (a � b).

Ex.4 Expand :

(i)2

x3

1x2

(ii) 22 y5x3

(iii) y3x2 y3x2 (iv) 2

1b21

a41

Sol. (i) 2

x31

x2

= (2x)2 � 2(2x)

x31

+ 2x3

1= 4x2 �

34

+ 2x9

1.

(ii) (3x2 + 5y)2

= (3x2)2 + 2(3x2)(5y) + (5y)2

= 9x4 + 30x2y + 25y2

(iii) ( 2 x � 3y)( 2 x + 3y)

= ( 2 x)2 � (3y)2

= 2x2 � 9y2

(iv) 2

1b2

1a

4

1

= 2

a4

1

+

2

b2

1

+ (1)2 + 2

a

41

b

21

+ 2

b

21

(1) + 2(1)

a

41

= 161

a2 + 41

b2 + 1 � 4

ab � b +

2a

.

Ex. 5 Simplify :

(i)

44

22

x

1x

x

1x

x1

xx1

x

(ii) 22 yx4 yx2 yx2

Sol.

x1

x

x1

x

22

x

1x

44

x

1x

=

22

x

1x

22

x

1x

44

x

1x

=

2

222

x

1)x(

44

x

1x

=

44

x

1x

44

x

1x

= (x4)2 �

2

4x

1

= x8 � 8x

1.

(ii) (2x + y)(2x � y)(4x2 +y2)= [(2x)2 � (y)2](4x2 + y2)= (4x2 � y2)(4x2 + y2)= (4x2)2 � (y2)2 = 16x4 � y4.

Ex.6 Find the value of x � y when x + y = 9 & xy = 14:

Sol. x + y = 9On squaring both sides

x2 + y2 + 2xy = 81Putting value of xy = 14x2 + y2 + 28 = 81x2 + y2 = 81 � 28 = 53 ...(i)(x � y)2 = x2 + y2 � 2xy

Putting xy = 14 and (i)(x � y)2 = 53 � 2 (14) = 53 � 28

(x � y)2 = 25

x � y = ± 25 = ±5

56PAGE # 56

Ex.7 If x2 + 2x

1 = 23, find the values of

x1

x ,

x1

x and

44

x

1x .

Sol. x2 + 2x

1 = 23 �(i)

x2 + 2x

1 + 2 = 25 [Adding 2 on both sides of (i)]

(x2) +

2

x1

+ 2 x

x1

= 25

2

x1

x

= (5)2

x + x1

= 5

2

x1

x

= x2 + 2x

1� 2

2

x1

x

= 23 � 2 = 21

x1

x = 21 .

2

22

x

1x

=

44

x

1x + 2

44

x

1x =

2

22

x

1x

� 2

44

x

1x = (23)2 � 2 = 529 � 2

44

x

1x = 527.

Ex. 8 Find the value of

aba

b�a

b5ab6�a

ab5�a2

22

22

2

.

Sol. 22

2

b5ab6�a

ab5�a

×

aba

b�a2

22

= )b5�a)(b�a()b5�a(a

× )ba(a)ba)(b�a(

= 1

Ex. 9 Find the value of 96.5

02.202.2�98.798.7

Sol.96.5

02.202.2�98.798.7

= 96.5

)02.2�98.7)(02.298.7( =

96.596.510

= 10.

Ex.10 Simplify :

(i) (3x + 4)3 � (3x � 4)3 (ii) 33

x2

xx2

x

Sol. (i)(3x + 4)3 � (3x � 4)3

= [(3x)3 + (4)3 + 3 (3x) (4) (3x + 4)] � [(3x)3 � (4)3 � 3 (3x)

(4) (3x � 4)]

= [27x3 + 64 + 36x (3x + 4)] � [27x3 � 64 � 36x (3x � 4)]

= [27x3 + 64 + 108x2 + 144x] � [27x3 � 64 � 108x2 + 144x]= 27x3 + 64 + 108x2 + 144x � 27x3 + 64 + 108x2 � 144x

= 128 + 216x2.

(ii) 3

x2

x

+

3

x2

x

= x3 + 3

x2

+ 3(x)

x2

x2

x + x3 �

3

x2

� 3(x)

x2

x2

x

= x3 + 3x

8 + 6x +

x12

+ x3 � 3x

8 � 6x +

x12

= 2x3 + x

24.

Ex.11 Evaluate :(i) (1005)3 (ii) (997)3

Sol. (i) (1005)3 = (1000 + 5)3

= (1000)3 + (5)3 + 3 (1000) (5) (1000 + 5)= 1000000000 + 125 + 15000 (1000 + 5)= 1000000000 + 125 + 15000000 + 75000= 1015075125.

(ii) (997)3 = (1000 � 3)3

= (1000)3 � (3)3 � 3 × 1000 × 3 × (1000 � 3)

= 1000000000 � 27 � 9000 × (1000 � 3)

= 1000000000 � 27 � 9000000 + 27000

= 991026973.

Ex.12 If x � x1

= 5, find the value of x3 � 3x

1.

Sol. We have, x � x1

= 5 ...(i)

3

x1

x

= (5)3 [Cubing both sides of (i)]

x3 � 3x

1 � 3x

x1

x1

x = 125

x3 � 3x

1� 3

x1

x = 125

x3 � 3x

1� 3 × 5 = 125 [Substituting

x1

x = 5]

x3 � 3x

1� 15 = 125

x3 � 3x

1 = (125 + 15) = 140.

57PAGE # 57

Ex.13 Find the products of the following expression :(i) (4x + 3y) (16x2 � 12xy + 9y2)(ii) (5x � 2y) (25x2 +10xy + 4y2)

Sol. (i) (4x + 3y) (16x2 � 12 xy + 9y2)= (4x + 3y) [(4x)2 � (4x) × (3y) + (3y)2]= (a + b) (a2 � ab + b2) [Where a = 4x, b = 3y ]= a3 + b3

= (4x)3 + (3y)3 = 64x3 + 27y3.

(ii) (5x � 2y) (25x2 + 10xy + 4y2)= (5x � 2y) [(5x)2 + (5x) × (2y) + (2y)2]= (a � b) (a2 + ab + b2) [Where a = 5x, b = 2y]= a3 � b3

= (5x)3 � (2y)3

= 125x3 � 8y3.

Ex. 14 If a + b + c = 9 and ab + bc + ac = 26, find the value ofa3 + b3 + c3 � 3abc.

Sol. We have a + b + c = 9 ...(i) (a + b + c)2 = 81 [On squaring both sides of (i)] a2 + b2 + c2 + 2(ab + bc + ac) = 81 a2 + b2 + c2 + 2 × 26 = 81 [ ab + bc + ac = 26]

a2 + b2 + c2 = (81 � 52)

a2 + b2 + c2 = 29.Now, we have a3 + b3 + c3 � 3abc

= (a + b + c) (a2 + b2 + c2 � ab � bc � ac)

= (a + b + c) [(a2 + b2 + c2) � (ab + bc + ac)]

= 9 × [(29 � 26)]

= (9 × 3) = 27.

Ex.15 Simplify : 333

322322322

accbba

accbba

.

Sol. Here, 0accbba 222222

322322322 accbba

= 222222 ac cb ba3

Also, 0accbba

333 accbba = accbba3

Given expression

=

ac cb ba3ac cb ba3 222222

=

ac cb ba3ac ac cb cb ba ba3

= accbba .

Ex.16 Find the value of (28)3 � (78)3 + (50)3.Sol. Let a = 28, b = � 78, c = 50

Then, a + b + c = 28 � 78 + 50 = 0

a3 + b3 + c3 = 3abc.

So, (28)3 + (�78)3 + (50)3 = 3 × 28 × (�78) × 50

= � 327600.

1. Which of the following is a polynomial :

(A) x2 + 2 x + 3 (B) x2 + x2 + 3

(C) x3/2 + 2 x + 3 (D) 5x2 + x

2 + 3 x

2. The remainder obtained when t6 +3t2 + 10 is divided byt3 + 1 is :(A) t2 � 11 (B) 3t2 + 11(C) t3 � 1 (D) 1 � t3

3. If a2 �b2 =21 and a2 + b2 = 29, which of the followingcould be the value of ab ?

I. �10 II. 5 2 III. 10(A) I only (B) II only(C) III only (D) I and III only

4. If c1

b1

a1

and ab =c, what is the average (arithmetic

mean) of a and b ?

(A) 0 (B) 21

(C) 1 (D) c2ba

5. On simplifying (a + b)3 + (a � b)3 + 6a(a2 � b2) we get :(A) 8a2 (B) 8a2b(C) 8a3b (D) 8a3

6. Find the value of 222

333

c�b�a�cabcab

abc3�cba

, when

a = � 5, b = � 6, c = 10.

(A) 1 (B) �1

(C) 2 (D) �2

7. If (x + y + z) = 1, xy + yz + zx = �1, xyz = �1, then value of

x3 + y3 + z3 is :(A) �1 (B) 1(C) 2 (D) �2

8. If (x + a) is a factor of x2 + px + q and x2 + mx + n then thevalue of a is :

(A) q�np�m

(B) p�mq�n

(C) pmqn

(D) qn

pm

9. If x2 � 4 is a factor of 2x3 + ax2 + bx + 12, where a and bare constant. Then the values of a and b are :(A) � 3, 8 (B) 3, 8(C) �3, � 8 (D) 3, � 8

10. If 31

31

31

zyx = 0 then which one of the following

expression is correct :

(A) x3 + y3 + z3 = 0

(B) x + y + z = 3 31

31

31

zyx

(C) x + y + z = 3xyz(D) x3 + y3 + z3 = 3xyz

58PAGE # 58

11. If x51 + 51 is divided by (x + 1) the remainder is :

(A) 0 (B) 1

(C) 49 (D) 50

12. If a4 + 4a

1 = 119, then find the value of a3 �

3a

1.

(A) 11 (B) 36

(C) 33 (D) 12

13. Evaluate : )c�b)(b�a(

)a�c()a�c)(b�a(

)c�b()a�c)(c�b(

)b�a( 222

.

(A) 0 (B) 1

(C) 2 (D) 3

14. The polynomials ax3 + 3x2 � 3 and 2x3 � 5x + a when

divided by (x � 4) leaves remainders R1 & R

2 respectively

then value of �a� if 2R1 � R

2 = 0.

(A) 12718

� (B) 12718

(C) 12717

(D) 12717

�

15. A quadratic polynomial is exactly divisible by (x + 1) &(x + 2) and leaves the remainder 4 after division by(x + 3) then that polynomial is :(A) x2 + 6x + 4 (B) 2x2 + 6x + 4(C) 2x2 + 6x � 4 (D) x2 + 6x � 4

16. If x2 � 4 is a factor of 2x3 + ax2 + bx + 12, where a and bare constant. Then the values of a and b are :(A) � 3, 8 (B) 3, 8(C) �3, � 8 (D) 3, � 8

17. The value of 24.024.024.076.076.076.024.024.024.076.076.076.0

is :

(A) 0.52 (B) 1(C) 0.01 (D) 0.1

18. If x + y = 3 and xy = 2, then the value of x3 � y3 is equal to(A) 6 (B) 7(C) 8 (D) 0

19. If x = 2

1, then the value of x +

x1

1

11

1

is :

(A) 4

5(B)

5

4

(C) 4

3(D) None of these

20. If (a2 + b2)3 = (a3 + b3)2 then ba

+ ab

=

(A) 32

(B) 23

(C) 65

(D) 56

21. m5 + m4 + m3 + m2 + m + 1 = (m3 + 1) × _______

(A) m5 + m4 + m2 + m (B) m2 + m3

(C) m3 + m3 + m + 1 (D) m2 + m + 1

22. If x = 22 , then x4 + 4x

4 is :

(A) 2(3 � 2 ) (B) 6 2 � 2

(C) 6 � 2 (D) 12

23. If 4x � 5z = 16 and xz = 12, 64x3 � 125z3 =

(A) 14512 (B) 15676(C) 25833 (D) 15616

24. 13213

33

xy)xy(yx

yx

(A) x + y (B) y � x

(C) x1�

y1

(D) x1

+ y1

25. If

b�aab4b�a

2

= 35

, then the value of a : b is :

(A) 1 : 16 (B) 1 : 4(C) 4 : 1 (D) 16 : 1

26. If x = 0.50, then the value of the expression

x1x

)xx1(3

2 is :

(A) 4 (B) 2(C) 1.50 (D) 1

27. If p = 22/3 + 21/3, then :(A) p3 � 6p + 6 = 0 (B) p3 � 3p � 6 = 0

(C) p3 � 6p � 6 = 0 (D) p3 � 3p + 6 = 0

28. The polynomial p(x) = 2x4 � x3 � 7x2 + ax + b is divisibleby x2 � 2x � 3 for certain values of a and b. The value of

(a + b), is :(A) � 34 (B) � 30

(C) � 26 (D) � 18

29. When the polynomial (6x4 + 8x3 + 17x2 + 21x + 7) isdivided by (3x2 + 4x + 1), the remainder is (ax � b).

Therefore : [IJSO-2011](A) a = 1, b = 2 (B) a = 1, b = � 2

(C) a = 2, b = 1 (D) a = �1, b = � 2

30. If 22x�1 + 21�2x = 2, then the value of x is : [IJSO-2011](A) 0.5 (B) �0.5

(C) 1 (D) 0

31. Given that a (a+b) = 36 and b (a + b) = 64, where a andb are positive, (a � b) equals : [IJSO-2011](A) 2.8 (B) 3.2(C) �2.8 (D) �2.5

32. If a, b, c are positive, cbca

is : [IJSO-2011]

(A) always smaller than ba

(B) always greater than ba

(C) greater than ba

only if a > b.

(D) greater than ba

only if a < b.

59PAGE # 59

33. Find x2 + y2 + z2 if x2 + xy + xz = 135, y2 + yz + yx = 351and z2 + zx + zy = 243. [IJSO-2012](A) 225 (B) 250(C) 275 (D) 300

34. lf a + b + c = 1, a2 + b2 + c2 = 21 and abc = 8 then findthe value of (1� a)(1� b) (1� c). [IJSO-2012](A) � 10 (B) � 18

(C) � 24 (D) � 30

60BIOLOGY_IJSO_PAGE # 60

NATURAL RESOURCES

� It indicates the potential wealth of a country. The varietyof substances that man gets from earth and nature tomeet his basic needs are called natural resources.The word resource means a source of supplying amaterial generally held in reserve.

� Natural resources are both living and non � living.

Some of these resources are found in abundance,while others are found in limited quantities & that tooin some restricted parts of our land. For this reason,the natural resources have to be wisely used.However, in reality it is not so. They are being usedindiscriminately.

� Types of Natural Resources :

(i) Based on availability : The natural resources arecategorized into two types i.e.

(a) nexhaustible natural resources

(b) Exhaustible natural resources

Natural Resources

(i)

(ii)

Resources that are in unlimited quantity.

Resources that are not likely to be exhausted by human activity or their use.Examples : Air, Water &Solar Radiations. (i)

(ii)

Resources that are in limited quantity.

Resources that are likely to be exhausted by human activities.

Renewable (i)

(ii)

(iii)

Can replenish themselves by quickrecycling and replacement within a reasonable time.

Not likely to be exhausted.

Examples : Soil, Forests and Wild life.

(i)

(ii)

(iii)

Cannot replenish themselves by recycling & replacement.

These may be exhausted.

Example : Minerals, Fossil fuels

Non�Renewable

Exhaustible

Inexhaustible

(ii) Based on origin : On the basis of their originresources may be biotic (organic) or abiotic(inorganic). Biotic resources are obtained from thebiosphere. Forest and forest products, crops, birds,animal, fish and other marine life forms are examplesof biotic resources. Coal and mineral oil also belongto this category since they originate from organicmatter. Some biotic resources like forest and livestockare renewable, whereas coal and oil arenon�renewable. Resources composed of non�living

inorganic matter are called abiotic resources. Land,water and minerals like iron, copper, lead and goldare abiotic resources.

(iii) Based on utility : Every resource has some utility.For example, some are used as food, some as rawmaterials and others as sources of energy.

AIR OR ATMOSPHERE

� The multilayered, transparent and protective envelopeof gases surrounding the planet earth is calledatmosphere.

� In other words atmosphere is the layer of air abovethe earth�s surface and air is a mixture of several

gases. About 95% of total air is present up to the heightof 20 km above earth's surface, remaining 5% is up tothe height of 280 km.

(a) Composition of Air :

Gas Relative percentage / volumeNitrogen 78.08 %Oxygen 20.94 %Argon 0.9 %Carbon dioxide 0.03 %He, Ne, Kr, Xe, in trace amounts.

� Besides these gaseous components air alsopossesses water vapour, industrial gases, dust,smoke particles, microorganisms, pollen grains,fungal spores etc.

(b) The Different Zones of Atmosphere or Air

(i) Troposphere : t is the basal part that extends about8-16 km above the earth's surface. (upto 8 km onpoles). In this layer important climatic events occurlike cloud formation, lightning, thundering etc.In thisregion air temperature gradually decreases withheight. Its upper limit is called Tropopause

(ii) Stratosphere : t lies next to troposphere and isup to 50 km high. In this layer temperature rises. Thereis a formation of ozone layer in this region which canabsorb the harmful ultra violet rays coming from sun.

(iii) Mesosphere : t lies next to stratosphere and isup to 80km in height. Temperature decreases in thisregion.

NATURAL RESOURCES



(iv) Ionosphere : t lies upto 400 km above earth's surface. In this layer gaseous components become ionized dueto sun's energy and remain there as ions.

S.No. Region Range of distance Density of air Importance1. Troposphere 11 kms from the

surface of earthHighest Most of the atmospheric air is

present here. It is a medium forlocomotion of flying animals,helps in dispersal of seeds andfruits; region of cloud formation.

2. Stratosphere 50 kms from the surface of earth Less than troposphere Contains ozone layer that trapsmost of UV rays and cosmic raysof the Sun.

3. Mesosphere 80 kms from the surface of earth Low --

4. Thermosphere 100 kms upwards Extremely low Reflect radio waves back toearth, artificial satellites arepresent here.

Different Regions of Atmosphere

ATMOSPHERIC REGIONS BASED ON TEMPERATURE

150

100

50

HE

TE

RO

SP

HE

RE

HO

MO

SP

HE

RE

TEMPERATUREINCREASINGWITH HEIGHT

THERMOSPHERE

MESOSPHERE

STRATOSPHERE

TROPOSPHERE

TEMPERATURE DECREASINGWITH HEIGHT

TEMPERATURE INCREASINGWITH HEIGHT

OZONOSPHERE

IONOSPHERE

00 100 200 300

HE

IGH

T IN

KIL

OM

ET

ER

S

TYPICAL TEMPERATURE IN DEGREE CELSIUS

Different Regions of Atmosphere

(c) Role of Air or Atmosphere :

� It acts as medium for movement of insects, birds etc.� It protects the life on earth from harmful ultra violet rays.� It is a source of oxygen, carbon dioxide and nitrogen

required for various metabolic activities of living beings.

� It helps in dispersal of spores, pollen grains etc.� t maintains temperature on earth required for life.� It transmits sound for communication.� Ionosphere reflects the radio waves back to earth for

long distance communication due to presence of ionsand free electrons.


� Burning (combustion) takes place in presence ofoxygen and produces carbon dioxide.

� Specific climatic conditions and water cycle ismaintained due to circulation of air.

� Eukaryotic cells and many prokaryotic cells require O2

for break down of glucose to get energy throughrespiration and release CO

2.

(d) The Role of Atmosphere in Climate

Control :

� Climate is an average weather of an area.Temperature, light and rainfall are important factorsthat determine climate of an area. Atmosphere playsa crucial role in its control :

� It acts like a blanket covering the whole earth.� It keeps the temperature of earth steady.

� It acts as bad conductor of heat thus prevents thesudden increase in temperature during the day aswell as slows down the escape of heat into the outerspace during night.The temperature range on thesurface of moon �190ºC to 110ºC.

(e) Wind :

� Air in motion is called wind. Speed of wind can bedetermined by :

� Heating of air� Formation of water vapour

� Atmosphere can be heated from below by radiations,such radiations are reflected back.

� Convection currents appear in air on being heated.

� Factors controlling movement of air

(A) Rotation of earth.

(B)n the path of wind, mountain ranges may comeacross.

� The general pattern of winds over earth is known asgeneral circulation and specific winds are named forthe direction from which they originate (e.g. windblowing from west to east is westerly).

� Wind speeds are often classified according toBeaufort scale.

(f) Rain :

� The warm, moist and rising air cools and forms cloudsin the sky.

� This happens due to heating of water bodies duringday time which get mixed with atmosphere.

� The air rises, it expands and cools.� Cool air in the atmosphere sinks towards the ground.� Due to cooling water vapours present in air get

facilitated.� These tiny droplets become bigger and bigger due to

condensation.� When they become heavy, they fall down in the form of rain.

POLLUTION

Any undesirable change in physical, chemical orbiological characteristics in the air, water and landwhich is harmful to the men directly or indirectlythrough animals, plants, industrial units or rawmaterials is called as pollution.

� Pollution is man made. But it can also be natural.� 99.95 % of pollution is natural only 0.05 % pollution is

manmade.

(a) Pollutants :

Any material or act of man, or nature which leads topollution is called as pollutants.

� The pollution is usually brought about by the additionto the environment of waste products of human activity.

� When the waste products are not efficientlyassimilated, decomposed or other wise removed bynatural, biological and physical processes (recycling)and the system is unable to utilize them properly, sothat the balance of the system breaks down.

� Therefore such type of pollutants can stimulate orinhibit the biological reactions or change in theircapacity. Therefore changes also take place in theecosystem.

� The amount, number and type of pollutants areincreasing with the growth of the population.

Difference between primary and secondary pollutants

Primary pollutants Secondary pollutants`

Pollutants which persist in the environment in the form it is produced.

Pollutant formed from a primary one through change or reaction.(No)x and hydrocarbons react photochemically to produce PAN (Peroxyacetyl nitrate) and O3 .

These include particulate matter, CO, CO2, SO2, (NO)x and hydrocarbons.

Photochemical smog, O3 and peroxy acetyl nitrate are secondary pollutants.

These are less toxic.These are more toxic than the primary pollutants & this phenomenon is called synergism.

(b) Air Pollution :

� Air pollution is caused due to the addition of theunwanted substances or gases.

� The atmospheric pollution is mainly caused by theactivities of man and concentrated to the inhabitedand the industrial complexes in cities.

� There are two main categories of air pollutants.

(i) Gaseous : The gaseous materials include variousgases and vapours of volatile substances or thecompound with a boiling point below 200ºC.

(ii) Particulate : Dust particles , carbon particles,particles of other metals etc.


(c) Major air Pollutants and Their Effects :

(i) Carbon monoxide (CO) : This is the main airpollutant.

� Carbon monoxide is highly toxic & it is colourless andodourless in nature.

� It combines with haemoglobin of the blood and blocksthe transportation of oxygen.Thus, it impairsrespiration and it causes death.

(ii) Unburnt hydrocarbons : Out of them 3, 4 �benzpyrene is the main pollutant. This causes cancerin lungs.

(iii) Ethylene : The falling of leaves without particularreason, falling buds etc. effects are seen in plants aredue to ethylene.

(iv) Oxides of nitrogen : These oxides of nitrogen formphotochemical smog in the atmosphere and releaseozone.

� Ozone causes harm to mucilagenous membrane.� The oxide pollutants of nitrogen are nitric oxide (NO),

and nitrogen di oxide (NO2).

� These oxides and ozone are very harmful for theplants.

� The entry of these pollutants causes various diseasesin animals like-respiratory trouble such asemphysema, bronchitis, swelling of lungs and lungcancer etc.

(v) Smoke : Many constituents are present in smokesuch as Sulphur dioxide (SO

2), Sulphur trioxide

(SO3),Sulphuric acid (H

2SO

4), Ozone (O

3), Carbon

dioxide (CO2), PAN (peroxyacetyl nitrate), Arsenic and

Fluoride etc.

� The distribution area of lichen and mosses are theindicators of SO

2 pollution because lichen and mosses

cannot grow in the industrial regions or the regionscontaining SO

2 pollutants.

� The higher concentration of ozone produces harmfuleffects.

� Ozone layer absorbs U.V. rays which are harmful forthe living beings.

(vi) Aerosol : The aerosol like C.F.C. (Chloro fluorocarbon) released into the atmosphere from therefrigerators, air conditioners and jet planes, depleteor reduce the ozone layer.

� This thin layer of ozone is also known as ozone holeresults in the increase in temperature of the earth.

� Secondary effect of Air pollution :� Green -House Effect : Usually carbon dioxide is not

considered as pollutant, but its higher concentrationforms the thick layer above the earth surface whichchecks the radiation of the heat from the earth surface.Because of this the temperature of the earth surfaceincreases.This is called as"Green house effect".

Sun light

Green House Gases

HeatEarth Surface

Atmosphere

Heat

Radiantheat trappedby CO2

Green House Effect

� The various green house gases are CO2 (Warming

effect 60% ), CH4 (warming effect 20%) , chlorofluoro

carbon or CFCs (14%) and nitrous oxide (N2O 6%)

� Even 2�3º C rise in temperature will lead to melting of

glaciers & ice caps of polar regions and consequentlycauses floods in rivers, rise in sea level and changesin cycle of rain. Islands may be emerged in sea water.

� Global Warming : Global warming is the increase inaverage global temperature due to increase in amountof GHGs in earth�s atmosphere.

� Consequences of global warming :(i) Increase in the sea level : Global warming willmelt polar ice caps.

� If all the ice on the earth will melt water would beadded to surface of all oceans. Thus low lying coastalcities like Shanghai, Kolkata, Bangkok, Dhaka, Venice,etc. will be inundated.

(ii) Increase in global temperature : f present inputof GHGs will be continued, the earth �s global

temperature will rise.

(iii) Effect on agriculture : Grain production will bereduced. India�s annual monsoon rains may even

cease together. One third of global forest might beswept away. Deserts are likely to increase

(iv) Chances of hurricanes, cyclones and floods willbe more.


(v) Increased temperature and humidity caused byglobal warming will lead to spread of diseases likemalaria, filariasis etc. due to spread of vectors.Incidences of respiratory and skin diseases are likelyto increase.

� Acid rain : The term was coined by Robert August(1872). Acid rain is rainfall and other forms ofprecipitation with a pH of less than 5. pH of normalrain is 5.6 � 6.5. The most acidic rain has occured

over West Virginia U.S.A. with a pH of 1.5. Acids fromatmosphere are deposited over earth in two forms,wet and dry. Wet deposition occurs through rain, snowand fog. Dry deposition is settling down of wind blownacidic gases and particles over trees, various articles

and soil. About 50% of acidity is passed to earth asdry deposition. Rainfall will wash it down from treesand other articles. Acid rain is caused by large scaleemission of acidic gases into the atmosphere fromthermal power plants, industries and automobiles.The common ones are sulphur dioxide, nitrogenoxides (NO

X), volatile organic carbons (VOCs) and

hydrogen chloride. NOX are also produced in

atmosphere through lightning. Sulphur dioxide andnitrogen oxides are changed in the atmosphere intosulphuric acid and nitric acid by combining with oxygenand water.2SO

2 + O

2 2SO

3 ; SO

3 + H

2O H

2SO

4

2NO + [O] N2O

5 ; N

2O

5 + H

2O 2HNO

3

Surface run off

Air PollutantsSO , NOx

Acid Precursors2

Transport and change,Complex oxidation

reactions

Acid rain, snow or fog

Buildingsand

monuments

Urban

areas,

Pollution sourceDirection of wind flow and acid rain path

H SO HNO2 4 , 3

Acid rain, snow or fog

River

ecosystemLake ecosystem

Forest ecosystemUrban area,

power plants,

vehicles, etc

Clouds Clouds

Acid rain formationAcid rain damages plants by direct effect on foliageand growing points (eg � chlorosis, necrosis,

defoliation, dieback.) It causes leaching of essentialminerals of soil. Toxic minerals left in the soil furtherkill the plants. 50% of natural forests have beendestroyed by acid rain in Germany, Sweden, north east,U.S.A., Romania, Poland, etc. Acid rain has also ruinedfresh water reservoirs of most industrialised countries.Acidity dissolves toxic metals like Hg, Pb, Zn, Al. Bothacidity and toxic metals kill all types of aquatic lifeexcept some algae and fungi. Acid rain corrodesmetals, marble, painted surfaces, slate, stone, etc.The phenomenon is called stone leprosy.

OZONE DEPLETION

� Between 20 and 26 km above the sea level it occursozone layer and the part of atmosphere containing itis called ozonosphere (Stratosphere).

� This layer is established due to an equilibriumbetween photo dissociation of ozone by UV �radiations and regeneration of ozone.

� The thickness of this ozonosphere averages 5 km.� The ozone layer acts as a shield and absorbs the

harmful UV�radiations of the sunlight so protects the

earth�s biota from the harmful effects of strong UV�

radiations. So this layer is very important for the survivaland existence of life on earth.

(a) Causes of Thinning of Ozone Layer :

� The decline in spring � layer thickness is called ozone

hole.� Ozone hole is largest over Antarctica and was just

short of 27 million sq. km. during Spetember 2003.� Main chemicals responsible for destruction of ozone

� layer are : chlorofluorocarbons (CFCs), halogens

(used in fire extinguishers) methane and nitrous oxide.Out of these, most damaging is the effect of CFCs


which are a group of synthetic chemicals and are usedas coolants in refrigerators and air conditioners; ascleaning solvents, propellants and sterilants etc.

� These CFCs produce �active chlorine� (Cl and ClO

radicals) in the presence of UV�radiations. These

active chlorine radicals catalytically destroy ozone andconvert it into oxygen.

Cl + O3 ClO + O

2ClO + O

3 Cl + 2O

2

� Nitrous oxide : It is produced by industrial processes,forest fires, solid waste disposal , spraying ofinsecticides and pesticides, etc. Methane and nitrousoxide also cause ozone destruction.

(b) Effects of Ozone Depletion :

The thinning of ozone layer results in an increase in

the UV radiations (in the range of 290 � 320 nm)

reaching the earth�s surface. It is estimated that a 5

per cent loss of ozone results in a 10 per cent increase

in UV � radiations. These UV � radiations :

(i) Increase incidences of cataract and skin cancer.

(ii) Decrease the functioning of immune system : due

to killing of melanin � producing cells of the skin.

(iii) nhibit photosynthesis in most of phytoplanktons

so adversely affect the food chains of aquatic

ecosystems.

(iv) Damage nucleic acids of the living organisms.

Ozoneshield

Atmosphere

Hole over Antarctica

Ultraviolet radiationfrom sun

(c) Measures to Control Air Pollution :

� Barium compounds should be mixed with petrol whichreduce the smoke.

� It is also very essential to check the quality of gasesreleased from the factories.

� Industries should not be established at one place.� The smoke should be released into the atmosphere

after filtration and purification (by cyclone collector orelectrostatic precipitators).

WATER (HYDROSPHERE)

� It is a renewable resource which is essential forsustainance of life.

� It covers 3/4th of the earth�s surface.

� Of the total water present in hydrosphere 97% ispresent in oceans which is not utilizable by livingbeings.

� Only 3% water is fresh water. Among this 3%, 02.0%in Ice caps, 0.68% in Ground Water, 0.009% inFreshwater lakes, 0.009% in Salt lakes and 0.0019%in Atmosphere is present.

(a) Types of Water Resources :

(i) Fresh water resource : t consists of ponds, lakes,large rivers. t can be recycled. It is essential for life onearth as well as for survival. It can be obtained bythree different types of natural resources.

(A) Rain water : ndia receives 3 trillion m3 of waterfrom rainfall or precipitation. Its intensity is different indifferent zones, on this basis zones are classified as :

� Wet zone : with very high rainfall� ntermediate zone : with heavy rainfall� Semi arid zone : with moderate rainfall� Arid zone : with low rainfall.

(B) Surface water :

� These are major river systems with plenty of lakes &ponds etc.

(C) Ground water :

� t is the water which percolate into the ground. Thereis a certain level below the soil surface where therocks are saturated with water and this level is knownas the zone of saturation.


� The upper level of the zone of saturation is called thewater table. However, the vertical distance from thesurface of a region to the water table is called thewater level.

(ii) Salt water resource : t consists of oceans, seasetc. It cannot be used by living beings for drinking.

(b) Role of Water or Hydrosphere :

� Water is the main constituent of protoplasm.� It is the universal solvent. Through which mineral salts

are transported from one part of the plant to the other.� Various metabolic reactions take place in the medium

containing water.� It acts as a reactant in numerous metabolic reactions.� During photosynthesis, water releases oxygen.� Turgidity of the growing cells is maintained with water.� Various movements of plant organs like movements in

sensitive plant (touch-me-not) are controlled by water.� The growth of the cells during elongation phase mainly

depends upon absorption of water.� Metabolic end product of respiration is water.� It acts as a temperature buffer as its specific heat is

highest (only exception - liquid ammonia).� It shows the properties of cohesion and adhesion

which account for the capillary action of water.

(c) Water Pollution :

� The water pollution is the addition of organic andinorganic chemicals as well as the biologicalmaterials which change the physical and chemicalproperties of water.

� The water pollution is caused by many sources suchas sewage matter, industrial wastage, agriculturalwastage, domestic wastage, hot water of thermalplants and nuclear reactors etc. Water pollution canbe caused by the following man made sources :

(i) Household detergents : The household detergentsinclude the compounds of phosphate, nitrate,ammonium and alkylbenzene sulphonate etc. harmfulsubstances which are gathered in water. Alkyl benzenesulphonate (ABS) is not degradable, so that itsconcentration increases which is harmful for aquaticlife.

� Control measures : For the control of this pollutionlime, ferric chloride etc. are used to precipitate thephosphate. Zirconium is considered best for thispurpose.

(ii) Sewage : Sewage contains highest amount oforganic materials and biological materials.

� These organic materials increase the number ofdecomposers like bacteria and fungus.

� The acceleration of microbial activity increases theBOD of water. BOD is very less in pure water.

� The higher BOD (Biological oxygen demand) is theindication of water pollution and the water of pollutedreservoir can not be utilized and produces a very badsmell spreading around the locality.

� The infection or disease also takes place. Daphniaand some fishes are sensitive to water pollution andshow the intensity of water pollution.

� Control measures : To control the water pollution ofsewage water it should be left into reservoir after theprimary and secondary treatment.

� The big particles are mainly separated in primarytreatment through floatation and sedimentation.

� Micro organisms are used for secondary treatmentsuch as oxidation chamber or activated sludgeprocess.

� Oxidation chamber is a shallow reservoir in whichthe sewage is stored.

� Algae and bacteria grow very well because of thehigher amount of organic materials in it.

� Bacteria decomposes the organic materials andproduce CO

2 which is utilized by the algae in

photosynthesis.� Oxygen released by photosynthesis protects the water

pollution. Therefore oxidation pond is the goodexample of symbiosis in between algae and bacteria.

� The infectious bacteria are destroyed during theactivity (reactions) in the oxidation pond. So that thesimple substances are left after decomposition oforganic matter.

(iii) Industrial wastes : The wastes of industries aredischarged into the running water, rivers and canals,Industrial wastes mainly contain inert suspendedparticles such as dust, coal, toxins like acid, base,phenols, cyanides, mercury, zinc etc., inorganicmaterials like � ferrous salts, sulphides, oils and other

residues of organic material and hot water.

� The water polluted by mercury, lead etc. causesdisorganization of nervous system.

� It means it produces insanity. The minamata diseasecaused in Japan by eating of mercury polluted fishes.So many people died because of this disease.

� Control measures : The industrial wastes and toxiccomponents should be made pure before releasinginto rivers, lakes, ponds or sea .So that the waterpollution of industrial effluents can be controlled bysuitable treatment to remove the pollutants.

� Bioaccumulation of pesticides : Pesticides like DDTare poisonous chemicals sprayed on crops to protectthem from pests and diseases.

� This increase in concentration of harmful non-biodegradable chemical substances in the body ofliving organisms at each trophic level of a food chainis called biological magnification.

� Eutrophication : The discharge of sewage water anddetergents in water bodies promotes excessive growthof phytoplanktons (minute aquatic algae).

� This excessive growth causes reduction in oxygenlevel of water. The excessive growth of phytoplanktonsbrings about a reduction in dissolved oxygen whichaffects other aquatic organisms. Consequentlypotential sources of food are highly reduced.


LITHOSPHERE

Lithosphere is the main life supporting system. Toplayer of earth is called soil. It is the main naturalresource essential for survival and development.

(a) Structure and Formation of Soil :

Soil is formed due to interaction between weatheringof rocks, rain, wind, temperature (physicalcomponents) and plants, animals and microbes(biological components).

� It is formed by combined action of climatic factors suchas temperature, rainfall, light etc. and biotic factorssuch as plants and microbes on earth crust.

(b) Constituents of Soil :

� Soil contains :(a) Inorganic constituents of parent rocks(b) Organic products of living organisms(c) Living organisms including microorganisms(d) Air in the pores.

� There are four important components of soil. Theyare

(i) Mineral matter 50�60%

(ii) Organic matter 10%(A) Living organisms(B) Decomposed matter(iii) Soil water 25 � 35 %

(iv) Soil air 15�25 %

(c) Types of Soil :

On the basis of its nature and composition, soil ismainly of six types �

(i) Alluvial soil : rich in loam and clay.(ii) Black soil � which has clay.

(iii) Red soil : which is sandy to loam.(iv) Mountain soil � which is a stony and sandy soil.

(v) Desert soil � which is sandy.

(vi) Laterite soil � which has porous clay.

� Outer most layer of earth is called crust. Many types ofminerals are found in crust. They provide many typesof nutrients to living beings.

(d) Factors / Processes Responsible for

Formation of Soil :

(i) Sun : Rocks get expanded due to heat produced bysun during day time. At night, the rocks cool down andcontract. Due to this unequal expansion andcontraction of rocks, cracks in rocks appear. This leadsto formation of smaller pieces of rocks.

(ii) Water : Due to continuous movement of rain andfast flowing river water, rock pieces collide and breakdown in still finer particles due to their abrasive effect.

(iii) Wind : Wind has abrasive effect on rocks. Finerrock particles are blown away and get deposited atother distant places.

(iv) Living organisms : The step of weathering isbrought about by plants and animals. Lichens arefirst to appear on bare rocks. They produce acidswhich corrode the rocky surface to produce fineparticles. Now plants like mosses can appear on it. Insuch type of soil, certain microbes, algae, insects and

worms appear and die. Organic matter getsaccumulated. Roots of some plants grow into thecervices of rocks.

(e) Soil Pollution :

Soil is also polluted through the polluted water and air.

� These pollutants are mixed into the soil through therainy water. Such as H

2SO

4 acid is formed by mixing of

SO2 with rainy water in the air.

� The fertilizers are used to increase yield of thecrops.Various types of pesticides and weedicides etc.are sprayed over the crops. All these mixed with soilto produce harmful effects.

� The growth of plants inhibited or reduced due to thistype of pollution and sometimes death also takesplace. Excluding to these soil pollution is also causedby the disposal of house hold detergents, sewage,flowing oils, radioactive substances and hot water etc.

� The main substances of pesticides in soil pollutantsare D.D.T. and weedicides 2, 4 � D (2, 4 di�

chlorophenoxy acetic acid) 2, 4, 5�T (2, 4, 5, tri�

chlorophenoxy acetic acid).

� Control measures : Soil pollution can be controlledthrough biological degradation of waste materials.

� The various carbonic materials are of agriculturalwaste, cattle dung etc.which can be minimized by theuse of biogas plants which can produce energy also.

� Inspite all measures pesticides and weedicidesshould be used in limited quantity only when they arerequired.

� Bhopal GasTragedy is the best example of humanhazard which took the life of many persons the tank ofmethyl isocyanate burst during the manufacturing ofSavin insecticide on 3rd December 1984.

(f) Soil erosion :

� Fertility of soil depends on(A) Presence of organic matter (humus) and nutrients,(B) Capacity of soil to retain water and air.A loamy soilis the best - suited for plant growth.

� The fertility of soil is threatened due to various activitiesof humans. The main threat to the fertility of soil isfrom soil erosion, which is the loss of soil due to windor water flow.

� Methods of Preventing Soil Erosion : Prevention ofsoil erosion can be brought about by controlling thefactors which cause soil erosion. The methods wouldthus be as follows :

� Deforestation should be stopped, rather, trees shouldbe planted (afforestation). Afforestation should beundertaken not only in areas already cut , but additionalareas should be brought under plantation.

� To reduce the effect of strong wind in the fields, theboundaries of the fields should be planted with treesin two or three rows.

� To maintain the soil in its natural condition, it isadvisable to grow different crops. Crop rotation helpsto maintain the fertility of the soil. The water - holdingcapacity of the soil is also maintained by this method.

� Proper drainage and irrigation arrangements shouldbe made in the fields.

� On the sloping areas in hills, strip cropping (meansthe planting of crops in rows or strips to check flow ofwater). should be practised, thereby reducing thesteepness of the slopes and checking soil erosion.


BIOGEOCHEMICAL CYCLES

These are the cyclic pathways through which chemicalelements move from environment to organisms andback to the environment. The earth and itsenvironment, with reference to these elements, areconsidered as closed system and there is no inflowof such elements from outside the earth and theiramount is limited.

� Two types of biogeochemical cycles are :

1. Gaseous cycles 2. Sedimentary cycles

Characters Gaseous cycles Sedimentary

Cycles

Reservoir pool Air or water Rocks

Speed Faster Slower

ExamplesCarbon, nitrogen & oxygen cycles

Calcium, phosphorous

& sulphur cycles.

Differences between gaseous & sedimentary cycles.

(a) Water Cycle :

Water is the most abundant (60�90%) component of

protoplasm.It acts as a habitat for hydrophytes & manyaquatic animals, a good ionizer, good solvent,temperature, buffer and perform transportation ofmaterials. It also helps in digestion of organiccompounds & in photosynthesis of plants.

Water Cycle

� Types of water cycles are :

(A) Global water cycle : It does not involve livingorganisms and involves the interchange of waterbetween the earth�s surface and the atmosphere via

the processes of precipitation and evaporation.

� Ocean is the biggest store house of water.� Evaporation involves the conversion of liquid and solid

forms of water into vapours and later form the clouds.� Precipitation involves the rainfall, hail, snow, etc.

Energy for global water cycle is provided by sunlight.

(B) Biological water cycle : t is the interchange ofwater between abiotic and biotic components ofenvironment

� e.g. the plants absorb water from water bodies andsoil while loose most of the water by the process oftranspiration, animals consume water from waterbodies or the food ingested, while release water viathe processes of respiration and excretion.

(b) Nitrogen Cycle :

Nitrogen is an essential component of amino acids,proteins, enzymes and nucleic acids of theprotoplasm.Reservoir pool of nitrogen is atmospherewhich contains about 78.08% of nitrogen in gaseousstate. But it cannot be used directly and is changedinto nitrites and nitrates and then utilized.


� Steps of nitrogen cycle are :

(A) Nitrogen fixation : It involves the conversion of free diatomic nitrogen (N2) into nitrites and nitrates.t occurs in

three ways :

� Physical nitrogen fixation :� Atmospheric nitrogen fixation in the presence of

photochemical and electrochemical reactions inducedby thundering and lightning.

� Industrial nitrogen fixation in the industries at hightemperature and high pressure.

� Biological nitrogen fixation : � Biological nitrogen fixation occurs in the presence of

certain living organisms such as. � Rhizobium bacterium in the root nodules of legumes. � Azotobacter bacterium in the soil.� Anabaena (blue green algae) in water in the paddy

fields. � Azospirillum bacterium in loose association with the

roots of maize, sorghum, etc.

(B) Ammonification : t involves the decomposition ofproteins of dead plants and animals to ammonia inthe presence of ammonifying bacteria like Bacillusramosus.

(C) Nitrification : It involves the oxidation of ammoniato nitrites (NO

2�) and nitrates (NO

3�) in the presence of

nitrifying bacteria like Nitrosomonas (Ammonia tonitrite), Nitrobacter (Nitrite to nitrate), etc.Plants absorbthe nitrites and nitrates from the soil through their rootsand convert them into organic compounds (e.g.proteins) of protoplasm by the process called nitrogenassimilation.

(D) Denitrification : t involves reduction of ammoniumcompounds, nitrites and nitrates to molecularnitrogen in the presence of denitrifying bacteria likeThiobacillus denitrificans.

� Carbon Cycle : Carbon is the basic component of allthe organic compounds like carbohydrates, proteins,lipids, enzymes and nucleic acid of the protoplasm.

� In atmosphere, it is present as carbon dioxide. � It involves two types of processes, one involving CO

2

utilization and another involving CO2 production.

� They are expressed as follows :


(i) CO2 utilization : Carbon dioxide is utilized by thephotosynthetic organisms like green plants,photosynthetic bacteria, diatoms and blue green algaein the process of photosynthesis, t occurs in thepresence of chlorophyll and radiant energy ofsunlight.Glucose synthesized in photosynthesis isused to synthesize other organic compounds.

(ii) CO2 production :

� CO2 is released during respiration of both producers

and consumers. � During decomposition of organic compounds of dead

bodies. � During burning of fossil fuels like wood, coal, petro

leum, etc. � Volcanic eruptions and hot springs. � During weathering of rocks by acids produced by

microorganisms and roots of higher plants.

(d) Oxygen Cycle :

Oxygen is present in water and forms 20% of air inatmosphere.

� All living beings need it for respiration. Oxygen contentof atmosphere has remained constant for the lastseveral million years.

� Most of O2 is replenished by photosynthesis. During

photosynthesis CO2

is used by plants to form foodalong with release of oxygen.

� The oxides can be reduced both chemically andbiologically to produce oxygen. Microbial oxidation canalso occur.

� Due to burning of materials oxygen form carbondioxide.

� When oxygen combines with nitrogen, it forms oxidesof nitrogen, amino acids, proteins etc.


OBJECTIVE DPP - 16.1

1. Soil is a part of(A) atmosphere (B) lithosphere(C) hydrosphere (D) ionosphere

2. Maximum air in which we breath is present at(A) troposphere (B) stratosphere(C) ionosphere (D) mesosphere

3. Corrosion of statues and monuments occurs due to :(A) Photochemical smog (B) CO(C) Acid rain (D) Methane

4. Lichens indicate pollution by :(A) O

3(B) SO

2

(C) NO2

(D) CO

5. Pollutant released by jet planes is :(A) Fog (B) Aerosol(C) Smog (D) Colloid

6. Thickness of ozonosphere is -(A) 3km. (B) 7km.(C) 5cm. (D) 5km.

7. Soil erosion can be prevented by �(A) deforestation(B) afforestation(C) overgrazing(D) removal of vegetation

8. A renewable source of energy is �(A) petroleum (B) coal(C) nuclear fuel (D) trees

9. Percentage of nitrogen in air is �(A) 77.02 % (B) 78.09 %(C) 76.08% (D) 74.09%

10. Ozone layer is present in atmosphere in �(A) troposphere (B) stratosphere(C) mesosphere (D) thermosphere

11. Nodules in the roots of legume plants contain(A) nitrogen fixing bacteria(B) sulphur fixing bacteria(C) potassium fixing bacteria(D) none of the above

12. Which gas is mainly responsible for the depletion ofozone layer ?(A) Oxygen (B) CFC(C) Nitrogen dioxide (D) All of the above

13. Acid rain mainly contains �(A) nitric acid (B) hydrochloric acid(C) sulphuric acid (D) (A) and (C) both

14. Plants and animals are known as �(A) biotic resources (B) abiotic resources(C) machines (D) none of these

15. Coal is an / a �(A) exhaustible resource(B) inexhaustible resource(C) potential resource(D) none of these

16. Ozonosphere occurs at height of �(A) 8 � 10 km above poles

(B) 8 � 10 km above equator

(C) 20 � 26 km above the earth surface

(D) 11 � 16 km above equator

17. Biosphere is made of �(A) living beings and their remains(B) living beings, lithosphere, hydrosphere andatmosphere(C) living beings and lithosphere(D) living beings, lithosphere and hydrosphere.

18. Free living bacteria involve in nitrogen fixation �(A) Rhizobium (B) Azotobacter(C) Anabaena (D) Azospirillum

19. Which one of the following is renewable resource ?(A) Water (B) Metals(C) Fossil fuel (D) All of these

20. Which gas is responsible for the global warming �(A) O

2(B) N

2

(C) H2

(D) CO2

21. Biogeochemical cycles are also known as(A) sedimentary cycles (B) gaseous cycles(C) material cycles (D) cycles of water

22. Which of the following is a free living nitrogen fixingbacteria present in soil ?(A) Azotobacter (B) Nitrosomonas(C) Rhizobium (D) Pseudomonas

23. CO2 and O

2 balance in atmosphere is due to

(A) photosynthesis(B) respiration(C) leaf anatomy(D) photorespiration

24. Nitrogen fixation is -(A) Nitrogen Nitrite(B) Nitrogen nitrates(C) Nitrogen Amino acid(D) Both A and B

25. Ozone depletion in stratosphere would result in :(A) Forest fires(B) Increased incidence of skin cancer(C) Global warming(D) None of the above

72ANSWER KEY IJSO_PAGE # 72

ANSWER KEY

Force and Newton�s law of motion

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. C D A C A B B C C B A B D A ACDQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. B C C C C C B D A A A C C D BQue. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. B B B A D C B C B A A C B D AQue. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. C A B D A A D D CD D A A D C BQue. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75Ans. A,C C A,B B A B B A C C D C B D CQue. 76Ans. A

MOLE CONCEPT(CHEMISTRY)

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. B C D B C B A D A B B D B C B

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. A A A A A B A A D A A A D A D

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45Ans. D C C B C B C D A B C A C D A

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. B B C C C C B C B A A D C A D

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75Ans. C D A B A B C A D B C B C B D

Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90Ans. A A C A A D B D B C B C C D D

Ques. 91 92 93 94Ans. A D C A

POLYNOMIALS (MATHEMATICS)

Q. 1 2 3 4 5 6 7 8 9 10

Ans. A B D B D A B B C B

Q. 11 12 13 14 15 16 17 18 19 20

Ans. D B D B B C B B A A

Q. 21 22 23 24 25 26 27 28 29 30

Ans. D D D B D B C A B A

Q. 31 32 33 34

Ans. C D C B

NATURAL RESOURCES(BIOLOGY)

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20A. B A C B B D B D B B A B D A A C B B A DQ. 21 22 23 24 25A. C A A D B


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