ix chem atoms &molecules

Class 9, NCERT (CBSE) Science (Chemistry) | Chapter 3, Atoms and Molecules - In Text Questions

Chapter 3, NCERT (CBSE) Science (Chemistry)

Atoms and Molecules  

In text Questions

NCERT Science Text Book Page 32

Q.1: In a reaction, 5.3 gm of sodium carbonate reacted with 6 gm of ethanoic acid. The products were 2.2 gm of carbon dioxide, 0.9 gm water and 8.2 gm sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.     

Ans: Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid

                                      = 5.3 + 6.0 = 11.3 gm

Mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium ethanoate

                            = 2.2 + 0.9 + 8.2 = 11.3 gm

The mass of products is equal to the mass of reactants. Thus, mass is neither created nor lost during the given chemical change which is in agreement with the law of conservation of mass.        

Q.2: Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 gm of hydrogen gas ?

Ans: Since hydrogen and oxygen combine in the ratio of 1:8 by mass, it means that x gm of hydrogen and 8x gm of oxygen will be required to form water.

So, Oxygen required to react with 3 gm of hydrogen to form water = 3 x 8 = 24 gm

Q.3: Which postulate of Dalton’s Atomic Theory is the result of the law ofconservation of mass ?  

Ans: “Atoms are indivisible particles, which can not be created nor destroyed in a chemical reaction.” The above postulate of Dalton’s Atomic Theory is the result of the law of conservation of mass.

Q.4: Which postulate of Dalton’s Atomic Theory can explain the law of definite proportions ?

Ans: “The relative number and kinds of atoms are constant in a given compound.” The above postulate of Dalton’s Atomic Theory can explain the law of definite proportions.

NCERT Science Text Book Page 35

Q.1: Define the Atomic Mass Unit.

Ans: The atomic mass of an element is the relative mass of its atom as compared with the mass of a particular atom of carbon isotope taken as 12 units. Thus the atomic mass of an element indicates the number of times one atom of the element is heavier than 1/12 th of a ¹²C isotope. For example the atomic mass of oxygen is 16 which indicates that an atom of oxygen is 16 times heavier than 1/12 th of a ¹²C isotope atom.

Thus, one Atomic Mass Unit (u) = 1/12 th mass of ¹²C isotope atom.     

Q.2: Why is it not possible to see an atom with naked eyes ?

Ans: The size of an atom is very small and is measured in nanometers. Therefore, it is not possible to see an atom with naked eyes. 

NCERT Science Text Book Page 39

Q.1: Write down the formulae of

  (i)    sodium oxide

 (ii)    aluminium chloride

(iii)    sodium sulphide

(iv)    magnesium hydroxide

Ans:

(i) Na₂O (ii) AlCl₃ (iii) Na₂S (iv) Mg(OH)₂

Q.2: Write down the names of compounds represented by the following formulae:

(i) Al₂(SO₄)₃, (ii) CaCl₂, (iii) K₂SO₄, (iv) KNO₃, (v) CaCO₃

Ans:

  (i)    aluminium sulphate

 (ii)    calcium chloride

(iii)    potassium sulphate

(iv)    potassium nitrate

(v)    calcium carbonate

Q.3: What is meant by the term chemical formula ?

Ans: The chemical formula of a compound is the symbolic representation of its composition. a chemical formula of a compound shows its constituent elements and the number of atoms of each combining element. For example, the chemical formula of hydrogen sulphide is H₂S. It indicates that in hydrogen sulphide two hydrogen atoms and one sulphur atom are chemically united.   

Q.4: How many atoms are present in a

(i) H₂S molecule and (ii) PO₄^3– ion ?

Ans: (i) Three atoms (ii) Five atoms

NCERT Science Text Book Page 40

Q.1: Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Ans:

Molecular mass of H₂  = 1 + 1 = 2u

Molecular mass of O₂ = 16 + 16 = 32u

Molecular mass of Cl₂ = 35.5 + 35.5 = 71u

Molecular mass of CO₂ = 12 + 32 = 44u

Molecular mass of CH₄ = 12 + 4 = 16u

Molecular mass of C₂H₆ = 24 + 6 = 30u

Molecular mass of C₂H₄ = 24 + 4 = 28u

Molecular mass of NH₃ = 14 + 3 = 17u

Molecular mass of CH₃OH = 12 + 3 + 16 + 1 =32u

Q.2: Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and O = 16u

Ans:

Formula unit mass of ZnO = 65 + 16 = 81u

Formula unit mass of Na₂O = 23 x 2 + 16 = 62u

Formula unit mass of K₂CO₃ = 39 x 2 + 12 x 1 + 16 x 3 = 138u. 

NCERT Science Text Book Page 42

Q.1: If one mole of carbon atoms weighs 12 gm, what is the mass (in gm) of 1 atom of carbon ?

Ans:

1 mole of carbon atoms = 6.022 x 10²³ atoms.

Now 6.022 x 10²³ atoms of carbon weighs = 12 gm

So, 1 atom of carbon weighs = 12 ÷ 6.022 x 10²³ = 1.99 x 10^–23 gm.

FURTHER STUDY ON THIS CHAPTER:

=> Atoms and Molecules | Class IX Science (Chemistry) | NCERT (CBSE) Textbook Exercise Solved   [Read]

=> Class IX, NCERT (CBSE) Guide | Chapter 3, Atoms and Molecules | Additional CCE type Sample Questions  [Read]

=> Class IX, NCERT (CBSE) | Chapter 3, Atoms and Molecules | Extrascore Multiple Choice Questions (MCQ) [Read]

http://cbse-ncert-solution.blogspot.in/2009/12/class-9-ncert-cbse-science-chemistry.html

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Saturday, 5 October 2013

CBSE Class 9th | Science | Chapter 11. Work And Energy | Solved Exercises

Intext Questions | Page 148 | Chapter 11. WORK AND ENERGY| CBSE Class 9th  Science 

Question 1. A force of 7 N acts on an object. The displacement is, say 8 m in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case ?
 Answer. Work done (W) on an object is equal to the product of magnitude of the force (F) and the distance moved by the body (s) in the direction of the applied force
Work done (W) = Force(F) × Displacement (s)
∴ W = 7 N × 8 m = 56 J

Intext Questions | Page 149 | Chapter 11. WORK AND ENERGY| CBSE Class 9th Science

Question 1. When do we say that work is done ?
 Answer. Whenever a force acting on an object, is able to move the object through some distance in its direction, a Work is said to be done .

Question 2. Write the expression for the work done when a force is acting on an object in the direction of its displacement. 

Answer. Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the body in the direction of the applied force.
Let F be the force acting on the object and s be displacement under went by the object in the direction of force, then from the above definition the work done is given by W = Fs

Question 3. Define 1 J of work.
 Answer. Work done on a body is said to be 1 J (Joule) ; if a force of 1 N (Newton) acting on a body, displaces it by a distance of 1 m along the line of action of the force..

Question 4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field ?

Answer. As we know, amount of work done in ploughing the length of the field will be given by the product of Force 'F' and distance 's'

I.e.	Work done = Fs

Here, as given values of Force and Distance , are : F = 140 N, s = 15 m

Hence

Work done = 140 N × 15 m = 2100 J

Intext Questions | Page 152 | Chapter 11. WORK AND ENERGY| CBSE Class 9th Science 

Question 1. What is kinetic energy of an object ? 

Answer. kinetic energy of an object, is the energy possessed by an object on account of its motion.
K.E. of an object of mass m, moving with a velocity v is given by

K.E. =	1	mv².
	2

Its SI unit is joule.

Example. Water falling down has K.E., due to which it can rotate the blades of a water turbine.

Question 2. Write an expression for kinetic energy of an object. 

Answer. Kinetic energy (K.E.) of an object of mass m moving with velocity v is given by

K.E =	1	mv²
	2

Question 3. The kinetic energy of an object of mass m, moving with a velocity of 5 ms^–1 is 25 J. What will be its kinetic energy when its velocity is doubled ? What will be its kinetic energy when its velocity is increased to three times ? 

Answer.

As given here, K.E.=25 J and velocity = 5 ms–1

Also as we know	K.E =	1	mv2	∴ 25 =	1	m × (5)²
		2			2

=>	1	m =	25	= 1	:. m = 2kg
	2		25

When velocity is doubled, i.e. v = 10 ms^–1

	K.E =	1	× 2 × (10)² = 100 J
		2

When velocity is increased three times, i.e. v = 15 ms^{– 1}

	K.E =	1	× 2 × (15)² = 15 × 15 = 225 J
		2

Intext Questions | Page 156 | Chapter 11. WORK AND ENERGY| CBSE Class 9th Science

Question 1. What is power ? 

Answer. Power is defined as the rate of doing work or the rate of transfer of energy. It is scalar  quantity. S.I. unit of power is watt. 1 W = 1 J/s. It is also measured in horsepower ( 1 HP = 746 watt).
If an agent does a work W in time t, then from the definition above, power is given by

	Power =	Work	=	W
		Time		t

	Power =	W	=	Joule
		t		Second

:.	1 W = 1 Js– 1 = 107 ergs–1

Question 2. Define 1 watt of power. 

Answer. The power is said to be one watt, When a work of 1 joule is done in 1 s,

:.	1 Watt = 1 Js– 1

Question 3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power ? 

Answer. Power is defined as the rate of doing work

=>	Power =	Work or Energy
		Time

As given, here,

Work 'W' = 1,000 J, time 't' = 10 s, Power 'P' = ?

:.	P =	1000 J	= 100 Watt
		10 s

Question 4. Define average power. 

Answer. Over a period of time, power of an object keeps on varying. When an object or machine performs different amounts of work or consumes different amount of energy in different intervals of time, the ratio between the total work done or total energy consumed to the total time taken is called average power.

:.	Average Power =	Total work done or energy consumed
		Total time taken

Solved Exercise

 Given at the end of Chapter 11. WORK AND ENERGY | CBSE Class 9th Science 

Question 1. Look at the activities listed below. Reason out whether or not work is d one in the light of your understanding of the term ‘work’.
(i) Suman is swimming in a pond.
(ii) A donkey is carrying a load in its back.
(iii) A wind mill is lifting water from a well.
(iv) A green plant is carrying out photosynthesis.
(v) An engine is pulling a train.
(vi) Food grains are getting dried in the sun.
(vii) A sail boat is moving due to wind energy.
 Answer. (i) Suman is swimming in a pond : Yes, Suman is doing work. By pushing water in backward direction, he is able to move in forward direction due to reactionary force offered by the water. Here, the work done is negative, as the displacement of Suman is in opposite direction .
(ii)A donkey is carrying a load in its back : Yes, Apparently, it seems,no work is being done in this case, as force is being applied in vertical direction but displacement is in horizontal direction. Which in real is not the case. The force applied by the donkey is inclined to ground. Vertical part of reaction force from the ground is supporting the weight and horizontal part of reaction force from the ground makes it to carry the wight horizontally
(iii) A wind mill is lifting water from a well : Yes, A wind mill is doing work in lifting water from a well, because the direction of displacement in terms of height as well as force are same.
(iv) A green plant is carrying out photosynthesis : No, in the absence of an applied force and a resulting displacement, apparent ally no work is being done in this case.
(v) An engine is pulling a train: Yes, an engine pulling a train is performing work as due to applied force by an engine, the displacement of train is taking place in the direction of applied force.
(vi) Food grains are getting dried in the sun : No, physically work is done , when food grains are getting dried in the sun. But looking this process in different way, answer is yes. The heat energy from the Sun, converts a part of grains in the form of moisture into steam rising above thus changing the position of water contents hence P.E.
(vii) A sail boat is moving due to wind energy : Yes, a sail boat is moving due to wind energy is example of work done as direction of applied force and displacement are the same .

Question 2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object ?
 Answer. Here the object uder the influence of force of gravity (F) acting in downward direction, exhibits a projectile motion
F = mg
Horizontal Displacement = s
Angle between force of gravity and displacement = 90⁰
We know, when force is inclined, then Work = Force × Displacement ×Cos Θ
∴ Workdone 'W' = Fs cos 90 ⁰ = 0
Therefor, the work done by the force of gravity on the object is Zero.

Question 3. A battery lights a bulb. Describe the energy changes involved in the process.

Answer. On connecting a lamp with battery, the chemical energy stored in the battery, is first converted into electric energy, which on passing through filament of lamp gets converted into heat energy and light energy.
The energy changes involved in the process can be shown as given below :
Chemical Energy => Electrical Energy => Heat Energy + Light Energy

Question 4. Certain force acting on a 20 kg mass changes its velocity from 5 ms^–1 to 2 ms^–1. Calculate the work done by the force.

Answer. Here as given,

Mass, m = 20 kg

Initial velocity, u = 5 ms–1

Final velocity, v = 2 ms– 1

Time, t = 1 s

We know that	v = u + at	=>	α =	v- u
				t

∴	α =	2 ms-1 - 5 ms-1	or	α = –3 ms–2
		1s

Also, as we know

v² – u² = 2as

:.	(2 ms–1)² – (5 ms–1)² = 2 × (–3 ms–2) × s

=>	–21 m²s–2 = –6 ms–2 × s	or	s =	7	m
				2

As

Work done = F × s

or

W = m × a × s

:.	W = 20 kg × (–3 ms–2) ×	7	m
		2

=>	Work done W = –210 Joules

Question 5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force ? Explain your answer. 

Answer.

As object is moved from point A to a point B, direction of its displacement AB is horizontal. The direction of gravitational force is vertically downwards or perpendicular to direction of displacement.

∴	W = FS cos 90° = 0

Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why ?

Answer. No. The law of conservation of energy is not violated as it states that the sum of the potential energy and kinetic energy of the object would be the same at all points.i.e. 'potential energy + kinetic energy = constant' The loss in potential energy results in an equal gain in kinetic energy of the object.

Question 7. What are the various energy transformations that occur when you are riding a bicycle ? 

Answer. First through bio-chemical reaction, the chemical energy stored in the form of synthesized food product is converted into heat and increased muscular force. On paddling, the muscular force gets transformed into mechanical energy resulting in motion of bicycle.

Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it ? Where is the energy you spend going ?

Answer. No, as there is no displacement of rock in the direction of applied force, so there is no energy transfer between the rock and our body because . The energy spend is being used up against the static inertia of the rock.

Question 9. A certain household has consumed 250 units of energy during a month. How much energy is this in joule ?
 Answer. As given, monthly household energy consumed = 250 units

= 250 kWh = 250 kW × 1 hr

= 250 × 1000 W × 3600 s

= 900,000,000 J or 9 × 108 Joules

Question 10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy ? If the object is allowed to fall, find its kinetic energy when it is half-way down.
 Answer.

As per the statement given

m = 40 kg, g = 9.8 ms–2 and h = 5 m

P. E. of the object at a height of 5 m  = mgh

= 40 × 9.8 × 5 = 1960 J

When the object is half-way down, i.e.,

s = 2.5 m

Let its velocity be v.

As	v² – u² = 2gs

:.

v² – 0² = 2 × 9.8 × 2.5

=>

v²= 49

:. K.E. of the object =	1	mv²=	1	×40 × 49 = 980 J
	2		2

Question 11. What is the work done by the force of gravity on a satellite moving round the earth ? Justify your answer. 

Answer. Work is done by the force of gravity on a satellite moving round the earth is zero. The reason being, as the satellite moves around the earth in a orbit, the direction of the force of gravity is along the radius of its orbit, where as direction of satellite motion is along the tangent to the orbital path at any point. Here as the direction of focre applied and resulting displacement are perpendicular, there for, the work done on the satellite is zero.(W = FS cos 90° = 0)

Question 12. Can there be displacement of any object in the absence of any external force ? 

Answer. Yes, there be displacement of any object in the absence of any external force. As per Newton's first law of motion, an object will remain in state of rest or motion in a straight line even in the absence of an external force .

Question 13. A person holds bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not ? Justify your answer.
 Answer. As on application of muscular force, there is no displacement of bundle of hay in the direction of applied force, so work done by him on hay is zero. However, in order to support the weight, the person has to do some work in straining the muscles, as they become loose on tiring.

Question 14. An electric heater is rated 1,500 W. How much energy does it use in 10 hours ?

Answer. As given above, Rated Power of heater = 1,500 W and Time = 10 hrs

As we know	Power =	Work or Energy
		Time

Or	Work or Energy used = Power × Time

:. Energy used by the heater = 1500 w × 10 h

=	1500 × 10	= 15 kWh
	1000

Question 15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually ? Is it a violation of the law of conservation of energy ? 

Answer. As per the law of conservation of energy, the sum of the potential energy and kinetic energy of the object would be the same at all points.i.e. 'potential energy + kinetic energy = constant' .

When we draw a pendulum bob to one side and allow it to oscillate. Its P.E. at side point say 'B' is mgh and K.E. is zero. As it swing to far end at 'C' through mean point 'A' Its whole P.E. which is mgh at B, gets converted into K.E. on reaching A. And its P.E. i.e. mgh becomes zero.

	1	mv² K.E.at point A  = mgh , P.E. at point B
	2

v = √2gh

As the bob overshoots A and reaches C' due to this speed. Its energy at C' is partly kinetic and partly potential but the sum of K.E. and P.E. is the same as K.E. at A and P.E. at B. So, when it reaches C, its whole energy is P.E.
Yes, the bob comes to rest after some time because of the resistance of the air to its motion and work done against friction at the point of oscillation O.
No, it is not a violation of the law of conservation of energy. The energy initially possessed by the bob is mostly converted to heat energy due to resistance of air and friction (at point of support). Also, a small fraction of energy is also converted into sound energy. The law of conservation of energy is not at all violated as sum total of all the energies is constant.

Question 16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest ?

Answer.

Here,

Mass = m

Constant velocity = u

As the body is in motion, its energy is given by

Energy possessed by the body =	1	mv²
	2

Thus, an equivalent amount of work should be done on the body in order to bring it to rest.

:. The work done on the object will be	1	mv²
	2

Question 17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.

Answer.

Here,

m = 1500 kg

v = 60 km/h =	60 ×1000m	=	50	m/s
	3600 s		3

Work required to be done to stop the car,

W = Change in K.E. of the car

=	1	mv² -	1	m (0)² =	1	mv²
	2		2		2

Putting the values, we get

W=	1	× 1500 ×	(	50	)	²	= 208333.3 J
	2			3

Question 18. In each of the following a force F is acting on an object of mass m. The direction of displacement is from west to east shown by the lower arrow. Observe the diagrams carefully and state whether the work done by the force F is negative, positive or zero.
Answer. (i) In the first case, the force is perpendicular to displacement.

:.	W = Fs cos 90° = 0

(ii) In the second case, F and s are in same direction.

:.	W = Fs cos 0° = Fs = a positive work

(iii) In the last case, as F and s are in opposite direction.

:.	W = Fs cos 180° = Fs (–1)

= a negative work

= –Fs

Question 19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her ? Why? 
Answer. Yes, I agree with her because an object, being acted upon by multiple forces can be in a state of rest due to cross cancellation of forces. Thus, will have zero velocity. As there is no change in velocity, it will have zero acceleration. On the other hand, When the object is moving with a constant velocity, its acceleration is zero. Even in this situation, multiple forces may act on the object , balancing each other.

Question 20. Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Answer. As power of each device = 500 W,
:. Total power of 4 devices = 4 × 500 = 2,000 W
:. Energy consumed in 10 hours = Power × Time

=>	E = 2,000 W × 10h = 20,000 Wh

or	E = 20 kWh.

Question 21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy ?
Answer. When a freely falling object eventually stops on reaching the ground, its kinetic energy is converted to heat and sound energy. On the impact with ground, there may be some deformation or destruction at the point of impact, followed by  a rise in temperature and production of some sound .

=================================================

Things to remember....

• Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the body in the direction of the applied force.
• Work done against friction is an example of negative work.
• The unit of work is joule.
  1 joule = 1 newton × 1 metre
• Living beings as well as machines need energy to do work.
• Work done by an object is an indicator of the energy possessed by it.
• The law of conservation of energy states that energy can neither be created nor destroyed. It can only change its form.
• The two main forms of mechanical energy are kinetic energy and potential energy.
• The spring of a watch contains elastic potential energy.
• The energy possessed by a body by virtue of its motion is called kinetic energy.
• Kinetic energy of a body of mass m and having a velocity v is given by

  K.E =	1	mv2
  	2

• The energy possessed by a body by virtue of its position or configuration is called potential energy.
• Work done on a body by an external forces is equal to the change in kinetic energy of the body.
• For a body falling freely the sum of kinetic and potential energies at any moment of its fall remains constant.
• The rate of doing work is called power. It is measured in watt and also in horse power (1 H.P. = 746 W).
• Power is given by P = W / t, where ‘W’ is the work done or energy consumed and ‘t’ is the time taken.

Friday, 27 September 2013

CBSE Class 10th Science | Chapter 5. Periodic Classification of Elements | Solved Exercises

Things to remember :

• From only 30 elements known around year 1800, today we have list of 114 known elements
• Elements are classified on the basis of similarities in their properties.
• Döbereiner grouped the elements into triads and Newlands gave the Law of Octaves.
• Mendeléev arranged the elements in increasing order of their atomic masses and according to their chemical properties.
• Mendeléev even predicted the existence of some yet to be discovered elements on the basis of gaps in his Periodic Table.
• Anomalies in arrangement of elements based on increasing atomic mass could be removed when the elements were arranged in order of increasing atomic number, a fundamental property of the element discovered by Moseley.
• Elements in the Modern Periodic Table are arranged in 18 vertical columns called groups and 7 horizontal rows called periods.
• Elements thus arranged show periodicity of properties including atomic size, valency or combining capacity and metallic and non-metallic character.

Intext Questions | Page 81 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. Did Döbereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.
Answer. Yes, Dobereiner's triads also exist in the columns of Newlands' Octave. For example, Li, Na, K is a triad of Dobereiner which also exists in the columns of Newlands' Octaves.

Question 2. What were the limitations of Döbereiner’s classification? 
Answer.

1. Dobereiner could not classify all the elements known at that time.
2. Dobereiner could identify only three triads from the elements known at that time.
3. So the system of classification into triads was not accepted.

Question 3. What were the limitations of Newlands’ Law of Octaves?
Answer.

1. Newlands' Law of Octaves was applicable only up to calcium. After calcium the eighth element did not possess properties similar to that of the first.
2. Newlands assume that only 56 elements existed in nature and no more elements would be discovered in the future.
3. Newlands' Octaves contains only few elements having similar properties.

Activity 5.1 | Page 84 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. Looking at its resemblance to alkali metals and the halogen family, try to assign hydrogen a correct position in Mendeléev’s Periodic Table.
Answer. Mendeleev placed hydrogen with alkali metals while its some properties are similar to halogens. The position of hydrogen in Mendeleev's Periodic Table is correct because its properties are more similar to the alkali metals as it has the property of losing the electron, i.e., it is electropositive

Question 2. To which group and period should hydrogen be assigned?
Answer.First group and first period

Activity 5.2 | Page 85 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Consider the isotopes of chlorine, Cl-35 and Cl-37.
Question 1. Would you place them in different slots because their atomic masses are different?
Answer. No, the more fundamental base of classification is atomic number and not atomic mass
Question 2. Or would you place them in the same position because their chemical properties are the same?
Answer.Yes, both the isotopes are placed in same position because they have same chemical properties and same atomic number.

Intext Questions | Page 85 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. Use Mendeléev’s Periodic Table to predict the formulae for the oxides of the following elements:
K, C, AI, Si, Ba.
Answer.

K	------>	K2O
C	------>	CO2
Al	------>	Al2O3
Si	------>	SiO2
Ba	------>	BaO

Question 2. Besides gallium, which other elements have since been discovered that were left by Mendeléev in his Periodic Table? (any two)
Answer. (i) Scandium (ii) Germanium

Question 3. What were the criteria used by Mendeléev in creating his Periodic Table?
Answer. (i) He arranged the elements on the basis of their increasing atomic masses.
(ii) Second criteria was the similarity in the chemical properties
(iii) The formulae of hydides and oxides formed by an element were treated as one of the basic property of an element for the classification used by Mendeleev.

Question 4. Why do you think the noble gases are placed in a separate group?
Answer.Nobel gases are placed in separate group because:
(i) These gases discovered very late as they are very inert
(ii) These gases could be placed in a new group without disturbing the existing order

Activity 5.3 | Page 85 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. How were the positions of cobalt and nickel resolved in the Modern Periodic Table? 
Answer. Modern Periodic Table is based on the atomic number. Cobalt (27) is placed before Nickel (28) in Modern Periodic Table

Question 2. How were the positions of isotopes of various elements decided in the Modern Periodic Table? 
Answer.The isotopes have same atomic number so they are placed in the same group in Modern Periodic Table

Question 3. Is it possible to have an element with atomic number 1.5 placed between hydrogen and helium?
Answer.No, it is not possible because atomic number is a whole number

Question 4. Where do you think should hydrogen be placed in the Modern Periodic Table?
Answer.I think that the place of hydrogen in the Modern Periodic Table is correct

Activity 5.4 | Page 87 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. Look at the group 1 of the Modern Periodic Table, and name the elements present in it. 
Answer. The elements in group 1 are:
Hydrogen (H), Lithium (Li), Sodium (Na), Potassium (k), Rubidium (Rb), Cesium (Cs) and Francium (Fr).

Question 2. Write down the electronic configuration of the first three elements of group 1. 
Answer.
(i) H (1) --> 1
(ii) Li(3) -->2, 1
(iii) Na (11) --> 2, 8, 1

Question 3.What similarity do you find in their electronic configurations? 
Answer. All the elements have same number of valence electron, i.e. 1

Question 4. How many valence electrons are present in these three elements? 
Answer. One (1)

Activity 5.5 | Page 87 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. If you look at the long form of the Periodic Table, you will find that the elements Li, Be, B, C, N, O, F, and Ne are present in the second period. Write down their electronic configuration. 
Answer.

Li (3)	--->	2,1
Be (4)	--->	2, 2
B (5)	--->	2, 3
C (6)	--->	2, 4
N (7)	--->	2, 5
O (8)	--->	2, 6
F (9)	--->	2, 7
Ne (10)	--->	2, 8

Question 2. Do these elements also contain the same number of valence electrons? 
Answer. No

Question 3. Do they contain the same number of shells? 
Answer. Yes
The elements of same periods have same number of shells but not same number of valence electrons, which increases by 1

Activity 5.6 | Page 88 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. How do you calculate the valency of an element from its electronic configuration?
Answer.
Valency of metal : It is same as the number of valence electron i.e. 1, 2, and 3.
Valency of non-metal : Valency of non-metals can be calculated by subtracting number of valence electrons from 8 (i.e. 8- number of valence electrons).
For example :

8 - 4 = 4	8 - 5 = 3	8 - 6 = 2
8 - 7 = 1	8 - 8 = 0

Question 2.What is the valency of magnesium with atomic number 12 and sulphur with atomic number 16? 
Answer.Magnesium (12) --> 2, 8, 2
Sulphur (16) --> 2, 8, 6
The valency of Magnesium is same as valence electron, i.e. 2
The valency of Sulphur is 8-6 = 2 because it is non-metal.

Question 3. Similarly find out the valencies of the first twenty elements. 
Answer.

Sr No	Elements	Atomic No.	Configuration	Valency
1.	H	1	1	1
2.	He	2	2	0
3.	Li	3	2, 1	1
4.	Be	4	2, 2	2
5.	B	5	2, 3	3
6	C	6	2, 4	8-4 = 4
7.	N	7	2, 5	8-5 = 3
8.	O	8	2,6	8-6 = 2
9.	F	9	2, 7	8-7 = 1
10	Ne	10	2, 8	8-8 = 0
11.	Na	11	2, 8, 1	1
12.	Mg	12	2, 8, 2	2
13.	Al	13	2, 8, 3	3
14.	Si	14	2, 8, 4	8-4= 4
15.	P	15	2, 8, 5	8-5= 3
16.	S	16	2, 8, 6	8-6= 2
17.	Cl	17	2, 8, 7	8-7= 1
18.	Ar	18	2, 8, 8	8-8= 0
19.	K	19	2, 8, 8, 1	1
20.	Ca	20	2, 8, 8, 2	2

Question 4. How does the valency vary in a period on going from left to right? 
Answer. Valency first increases 1 to 4 then decreases from 4 to 0 (1, 2, 3, 4, 3, 2, 1, 0).

Question 5. How does the valency vary in going down a group?
Answer. Valency remains the same because valence electrons do not change on going down in a group.


Activity 5.7 | Page 88 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Atomic radii of the elements of the second period are given below:

Period II elements :	B	Be	O	N	Li	C
Atomic radius (pm) :	88	111	66	74	152	77

Answer.

Question 1. Arrange them in decreasing order of their atomic radii.
Answer.

Period II elements :	Li	Be	B	C	N	O
Atomic radius (pm) :	152	111	88	77	74	66

Question 2. Are the elements now arranged in the pattern of a period in the Periodic Table? 
Answer. Yes

Question 3.Which elements have the largest and the smallest atoms? 
Answer. Lithium (Li) has Largest atoms (152 pm)
Oxygen (O) has smallest atoms (77 pm)

Question 4. How does the atomic radius change as you go from left to right in a period?
Answer. The atomic radius decreases as we go left to right in a period
Activity 5.8 | Page 89 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. Study the variation in the atomic radii of first group elements given below and arrange them in an increasing order.

Group 1 Elements :	Na	Li	Rb	Cs	K
Atomic Radius (pm) :	186	152	244	262	231

Answer.

Group 1 Elements :	Na	Li	K	Rb	Cs
Atomic Radius (pm) :	86	152	231	244	262

Question 2. Name the elements which have the smallest and the largest atoms.
Answer. Na (Sodium) has smallest atom
Ca (Calcium) has largest atom.
Question 3. How does the atomic size vary as you go down a group? 
Answer.Atomic size increases as we go down a group


Activity 5.9 | Page 89 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. Examine elements of the third period and classify them as metals and non-metals.
Answer. Elements of third period are : Na(11), Mg (12), Al (13), Si (14), P (15), S (16), Cl (17), Ar (18)

Elements	Atomic No.	Configuration	Metals / Non Metals
Na	11	2, 8, 1	Metal
Mg	12	2, 8, 2	Metal
Al	13	2, 8, 3	Metal
Si	14	2, 8, 4	Non-Metal
P	15	2, 8, 5	Non-Metal
S	16	2, 8, 6	Non-Metal
Cl	17	2, 8, 7	Non-Metal
Ag	18	2, 8, 8	Non-Metal
The elements having 1,2,3 valence electrons are metals while the elements having 4,5,6,7,8 valence electrons are non- metals.

Question 2. On which side of the Periodic Table do you find the metals?
Answer.On the left side
Question 3. On which side of the Periodic Table do you find the non- metals? 
Answer. On the right side

Activity 5.10 | Page 89 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. How do you think the tendency to lose electrons will change in a group?
Answer. The tendency of lose electrons increases in a group on going down

Question 2. How will this tendency change in a period?
Answer. The tendency of lose electrons decreases as we go left to right in a period.


Activity 5.11 | Page 90 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1.How would the tendency to gain electrons change as you go from left to right across a period? 
Answer. The tendency of gaining electrons increases as we go left to right in a period upto 17th group. It decreases in 18th group.

Question 2. How would the tendency to gain electrons change as you go down a group? 
Answer. The tendency of gaining the electrons decreases as we go down a group.
====================================


Intext Questions | Page 90 | Chapter 5. Periodic Classification of Elements| CBSE Class 10th Science 

Question 1. How could the Modern Periodic Table remove various anomalies of Mendeléev’s Periodic Table?
Answer.(i) In Modern Periodic Table, the place of hydrogen is justify because it is electropositive and so it is placed in first group with metals.
(ii) In Modern Periodic Table, the elements are arranged on the basis of atomic number. So the place of isotopes is also justified as the atomic number remain same of all the isotopes.
(iii) The order of heavy and light elements is also corrected in the Modern Periodic Table
(iv) The position of inert gases is also justified.

Question 2. Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Answer. Calcium (ca) and Barium (Ba), as these two elements belong to same group as magnesium and have same valence electrons as magnesium has.

Question 3. Name :
(a) three elements that have a single electron in their outermost shells.
Answer. Lithium (li), Sodium (Na),Potassium (k), 
 
(b) two elements that have two electrons in their outermost shells.
Answer. Magnesium (mg), Calcium (ca), Barium (Ba) 
 
(c) three elements with filled outermost shells.
Answer. Helium (He), Neon (Ne), Argon (Ar).

Question 4.(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?
Answer. Lithium, sodium and potassium atoms have same number of electrons in their outermost shell and have same valency

Question 4.(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common? 
Answer. Helium and neon both have their outermost shell filled

Question 5. In the Modern Periodic Table, which are the metals among the first ten elements?
Answer. Lithium, Beryllium, Boron are the metals in Modern Periodic Table among the first ten elements.

Question 6. By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristic? 
Ga, Ge, As, Se, Be
Answer. Be has the maximum metallic characteristics because all other elements are situated at the right hand side in periodic table than Be. Due to the position their metallic characteristics decreases as we go from left to right.

=======================================================

Solved Exercises

Chapter 5. Periodic Classification of Elements| | CBSE Class 10th Science| Page 91-92)
 
Question 1. Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic Table. (a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer.(c) The atoms lose their electrons more easily.
(As we move from left to right across the periods of the periodic table, the non-metallic character increases. Hence, the tendency to lose electrons decreases.)

Question 2. Element X forms a chloride with the formula XCl₂, which is a solid with a high melting point. X would most likely be in the same group of the Periodic Table as :
(a) Na (b) Mg (c) AI (d) Si
Answer. (b) Mg, X would most likely be in the same group of the Periodic Table as magnesium (Mg) due to strong ionic bonding between Magnesium and Chlorine. _. 

Question 3. Which element has:
(a) two shells, both of which are completely filled with electrons? 
Answer. Neon (Ne) , Neon has two completely filled shells with 2 electrons in K shell and 8 electrons in L shell.
(b) the electronic configuration 2, 8, 2?
Answer. Magnesium (Mg)
(c) a total of three shells, with four electrons in its valence shell? 
Answer. Silicon (Si). Silicon has a total of three shells. K shell has 2 electrons, L has 8 and M i.e. valence shell has 4 electrons. .
(d) a total of two shells, with three electrons in its valence shell? 
Answer. Boron (B). It has a two shells, with 3 electrons in its L i.e. valence shell and 2 electrons in K shell
(e) twice as many electrons in its second shell as in its first shell? 
Answer. Carbon (C) has electronic configuration of 2 electrons in K shell and 4 electrons in L shell. Clearly, it has twice as many electrons in its second shell as in its first shell .

Question 4.(a) What property do all elements in the same column of the Periodic Table as boron have in common?
Answer. Both the elements are metals and show the following common properties :
(i) Both are good conductor of electricity.
(ii) Both show malleability

Question 4.(b) What property do all elements in the same column of the Periodic Table as fluorine have in common?
Answer. Both the elements are non-metal and show following common properties :
(i) Both are brittle
(ii) Both are bad conductor of electricity.


Question 5.An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
Answer. The atomic number of element is 17
(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)

N(7)

F(9)

P(15)

Ar(18)

Answer. It belongs chemically to F (9) because the electronic configuration of F is 2, 7 and having same valence electrons.

Question 6. The position of three elements A, B and C in the Periodic Table are shown below –

Group 16	Group 17
-	-
-	A
-	-
B	C

(a) State whether A is a metal or non-metal.
Answer. C is non-metal because it belongs to 17th group.

(b) State whether C is more reactive or less reactive than A.
Answer. C is less reactive than A because the reactivity of non-metal decreases from top to bottom.

(c) Will C be larger or smaller in size than B?
Answer. The size of C is smaller than B because B and C belong to the same period and the size decreases in a period on going left to right

(d) Which type of ion, cation or anion, will be formed by element A? 
Answer.A forms anion because C is non-metals for anion (Negative Ion)


Question 7. Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Answer. Atomic number of Nitrogen is 7 and electronic configuration is 2, 5.
Atomic number of Phosphorus is 15 and electronic configuration is 2,8, 5.
Phosphorus will be more electronegative because phosphorus and nitrogen both are non-metals. Phosphorus is situated in the lower side than Nitrogen. In non-metals, as we go top to bottom the electronegativity is increased.

Question 8. How does the electronic configuration of an atom relate to its position in the Modern Periodic Table? 
Answer. The electronic configuration is related to the position of element in periodic table. The numbers of electrons in outermost shell show the number of group and the number of shells show the number of periods.

Question 9.In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium? 
Answer. Atomic number of Calcium is 20 and its electronic configuration is 2, 8, 8, 2.
To see the resemblance with Calcium, first we have to check the Electronic Configuration of respective elements.

Elements with given Atomic Number	Electronic Configuration
Element with Atomic Number - 12	2, 8, 2
Element with Atomic Number - 19	2, 8, 8, 1
Element with Atomic Number - 21	2, 8, 8, 3
Element with Atomic Number - 38	2, 8, 18, 8, 2

From above, it is clear, elements with atomic number 12 and 38 has same electronic configuration of valence shell as that of Calcium, and therefor, will have physical and chemical properties resembling calcium

Question 10. Compare and contrast the arrangement of elements in Mendeléev’s Periodic Table and the Modern Periodic Table. 
Answer.
(i) Mendeleev's Periodic Table is based on atomic mass while Modern Periodic Table is based on atomic number.
(ii) In Medeleev's table there is no place for inert gases, while in modern Periodic Table they are classified.
(iii) In Modern Periodic Table all the anomalies of Mendleev's Periodic Table are removed.
(iv) There are 18 groups in Modern Periodic Table.

 ==================================================

Multiple Choice Questions (M.C.Q .)

CBSE Class 10th Science | Chapter 5. Periodic Classification of Elements

Question 1: The  elements of 3rd period will have _ _ _ _ _.
a. 1 shell
b. 3 shells
c. 2 shells
d. 0 shell

Answer : b. 3 shells

Question 2: Which element has zero electron affinity in the 3rd period ?
a. Ar
b. Al
c. P
d. S

Answer : a. Ar

Question 3: An element has configuration 2, 8, 1. It belongs to, _ _ _ _ _.
a. 17 group and 3rd period
b. 3 group and 1st period
c. 1 group and 8th period
d. 1 group and 3rd period

Answer : d. 1 group and 3rd period


Question 4: At the end of each period the valence shell is _ _ _ _ _.
a. incomplete
b. half filled
c. completely filled
d. singly occupied
Answer : c. completely filled

Question 5: The number of electrons in the valence shell is equal to its _ _ _
a. atomic mass
b. period number
c. group number
d. atomic volume
Answer : c. group number

Question 6: The non-metallic element present in the third period other than sulphur and chlorine is _ _ _ _ _.
a. phosphorus
b. oxygen
c. fluorine
d. nitrogen

Answer : a. phosphorus

Question 7: The family of elements to which potassium belongs is _ _ _ _ _.
a. halogens
b. alkaline earth metals
c. alkali metals
d. noble gases
Answer : c. alkali metals

Question 8: Which of the following factors does not affect the metallic character of an element?
a. Atomic size
b. Ionisation potential
c. Electronegativity
d. Atomic radius
Answer : c. Electronegativity

Question 9: Lanthanides and actinides are also called _ _ _ _ _.
a. normal elements
b. transition elements
c. noble gases
d. inner transition elements
Answer : d. inner transition elements

Question 10: The family of elements to which calcium belongs is _ _ _ _ _.
a. alkali metals
b. halogens
c. noble gases
d. alkaline earth metals
Answer : d. alkaline earth metals

Question 11: The modern periodic table is given by _ _ _ _ _.
a. Mendeleev
b. Einstein
c. Bohr
d. Mosley
Answer : d. Mosley

Question 12: The family of elements having seven electrons in the outermost shell is _ _ _ _ .
a. alkali metals
b. alkaline earth metals
c. halogens
d. noble gases

Answer : c. halogens

Question 13: A liquid non-metal is _ _ _ _ _.
a. phosphorous
b. mercury
c. bromine
d. nitrogen
Answer : c. bromine

Question 14: The first alkali metal is _ _ _ _ _.
a. hydrogen
b. lithium
c. sodium
d. francium

Answer : b. lithium

Question 15: A purple coloured solid halogen is _ _ _ _ _.
a. chlorine
b. bromine
c. iodine
d. astatine

Answer : c. iodine

Question 16: Six elements A, B, C, D, E and F have the following atomic numbers (A = 12, B = 17, C = 18, D = 7, E = 9 and F = 11). Among these elements, the element, which belongs to the 3rd period and has the highest ionization potential, is _ _ _ _ _.
a. C
b. A
c. B
d. F

Answer : a. C

Question 17: The least reactive element in group 17 is _ _ _ _ _.
a. fluorine
b. chlorine
c. bromine
d. iodine

Answer : d. iodine

Question 18: The statement that is not true about electron affinity is
a. It causes energy to be released
b. It causes energy to be absorbed
c. It is expressed in electron volts
d. It involves formation of an anion

Answer : b. It causes energy to be absorbed

Question 19: The valency of chlorine with respect to oxygen is _ _ _ _ _.
a. 1
b. 7
c. 3
d. 5

Answer : b. 7

Question 20: Elements belonging to groups 1 to 17 are called _ _ _ _ _.
a. noble gases
b. transition elements
c. normal elements
d. inner transition elements
Answer : c. normal elements

Question 21: Four elements along a period have atomic number (11, 13, 16 and 17). The most metallic among these has an atomic number of _ _ _ _ _.
a. 12
b. 16
c. 17
d. 11
Answer : d. 11

Question 22: A factor that affects the ionisation potential of an element is  _ _ _ _.
a. atomic size
b. electron affinity
c. electro-negativity
d. neutrons
Answer : a. atomic size

Question 23: The property of an element in the periodic table depends on its, _ _ _ _ _.
a. electronic configuration
b. atomic size
c. atomic mass
d. number of protons
Answer : a. electronic configuration

Question 24: The element, which has the highest electron affinity in the 3rd period is _ _ _ _ .
a. Na
b. Mg
c. Cl
d. Si
Answer : c. Cl

Question 25: Down a group, the electron affinity _ _ _ _ _.
1. decreases
2. increases
3. remains same
4. increases and then decreases

Answer : 1. decreases

Wednesday, 24 July 2013

CBSE Class IX (9th) Science | Chapter 10. GRAVITATION | Solved Exercises

Question 1.How does the force of gravitation between two objects change when the distance between them is reduced to half 


Answer. According to the law of gravitation , the force of attraction between any two objects of mass m1 and m2 is proportional to the product of their masses and inversely proportional to the square of the distance 'R' between them.

As given here	F =	Gm1 × m2
		R²

Here G is the gravitational constant. When the distance (R) is reduced to half .

Then	F' =	Gm1 × m2
		R²/2²

Or

F' =

4 F

Clearly, as the distance between the objects is reduced to half the force of gravitation becomes four times the original force.

Question 2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object ?
AnswerWeight of an object on surface of earth = mg
where 'm' is mass and 'g' is acceleration due to gravity .

Gravitational force acting on the object	F =	GM × m
		R²

Here M is mass of earth , R is radius of earth i.e. distance between the objects and G is gravitational constant

Weight of object 'mg' =

Gravitational force F acting on it

	mg =	GM × m
		R²

	g =	GM
		R²

From above expression it is clear , 'g' acceleration due to gravity is independent of mass of an object and all the objects- irrespective of being heavy or light experience the same acceleration due to gravity, hence fall with same speed from a given height.

Question 3. What is magnitude of gravitational force between the earth and a 1 kg object on its surface ? Take mass of earth to be 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m. G = 6.67 ×10^–11 Nm² kg^–2.

Answer. As given in the statement :
Gravitational Constant G = 6.67 × 10^–11 N m² kg^–2
Mass of the object m₁ = 1kg
Mass of the Earth m₂ = 6 × 10²⁴ kg ;
Radius of the earth R = 6.4 ×10⁶ m

As we know	F =	Gm1 × m2
		R²

	F =	6.67 × 10-11 ×1 × 6 ×1024
		(6.4 × 106)²

	=	6.67 × 6	× 10-11+24-12
		(6.4)²

=>	F = 0.977 × 10¹ = 9.77 N or F = 9.8 N Approximately

Question 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth ? Why ?
 
Answer. The universal law of gravitation states that the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.

As given here	F =	GM1 × M2
		R²

Where M₁ = Mass of earth, M₂ = Mass of moon, R = Distance between earth and moon.   G is the gravitational constant.
The above law applies to all objects anywhere in the universe. This is also true in case of force of attraction between Earth and moon. The magnitude of force (F) of attraction exerted by Earth on the moon, due to gravitation, is the same as that exerted by moon (F) on earth
These forces being equal and opposite, also in accordance with Newton's Third Law of Motion, which states that to every action there is an equal and opposite reaction.

Question 5. If the moon attracts the earth, why does the earth not move towards the moon ?
Answer. As per Newton’s third law of motion, the mutual forces of attraction due to gravitation, between earth and moon is same. The mass of the earth is 5.97x 10²⁴kg and that of moon is 7.349 x 10²² kg. Clearly the mass of earth is 82.23 times that of moon
Let F_e be the force of gravitational pull of earth on moon; F_m be the force of gravitational pull of moon on earth; m_e mass of earth ;m_m mass of moon ; g_e be the acceleration caused by earth on moon and g_m be the acceleration caused by moon on earth .

Fm	= mm × ge
Fe	= me × gm
As per, Newton's third law of motion,
Fm	= Fe
mm × ge	= me × gm
As given, me	= 83.23× mm
mm × ge	= 83.23× mm × gm
∴ ge	= 83.23× gm
Or gm	= ge/83.23 = 0 . 012 ge

Here clearly acceleration experienced by the earth due to gravitational pull of moon is very small, just .012 times ( 1.2 % ) of that experienced by the moon due to earth , which is not as effective to move the earth towards the moon.

Chapter 10. GRAVITATION |  CBSE Class IX (9th)  Science | Solved Exercises


Question 6. What happens to the force between two objects, if
(i) The mass of one object is doubled ?
(ii) The distance between the objects is doubled and tripled ?
(iii) The masses of both objects are doubled ?

Answer. According to the law of gravitation , the force of attraction between any two objects of mass m1 and m2 is proportional to the product of their masses and inversely proportional to the square of the distance 'R' between them.

i.e.	F =	Gm1 × m2
		R²

Here G is the gravitational constant. When the distance (R) is reduced to half .
(i) When the mass of one object say m1 is doubled, then

i.e.	F' =	G 2m1 × m2
		R²

i.e.	F' =	2 Gm1 × m2	= 2F
		R²

∴ As the mass of one object is doubled the force becomes 2 times.
(ii) When the distance between the bodies is doubled and tripled
when the distance is doubled

i.e.	F' =	Gm1 × m2	= F/4
		(R/2)²

The force is reduced to one fourth of the original force.
when the distance is tripled :

i.e.	F' =	Gm1 × m2	= F/9
		(R/3)²

The force is reduced to one ninth of the original force.
(iii) When the masses of both the objects are doubled, then

i.e.	F' =	G 2m1 × 2m2	= 4F
		R²

∴ When the masses of both the objects are doubled, then the force becomes four times the original force.

Question 7. What is importance of universal laws of gravitation ?
Answer. The Importance of Universal law of gravitation lies in the fact, that it was successful in explaining many phenomena such as.
(i) That, how does the different objects  in this universe,  affect  others.
(ii) That, how  gravity; the force of gravitation due to earth, is  responsible for the weight of a body and keeps us on the ground. And that why force of gravity decreases with altitude.
(iii) That, how does the lunar motion around the earth occur.
(v) That, how does the planetary motion of planets in our solar system as well as that of all other celestial objects take place
(vi) That how do the tidal waves originate, due to the gravitational pull of moon and the sun.

Question 8. What is the acceleration of free fall ?
Answer. Acceleration of free fall is the acceleration experienced by body falling freely towards earth under the influence of gravitation force of earth alone. It is denoted by g and its value on the surface of earth is 9.8 m s^–2.

Question 9. What do we call the gravitational force between an earth and an object ?
Answer.The gravitational force between an earth and an object Weight and it is equal to product of mass(m) and acceleration due to gravity(g).

Chapter 10. GRAVITATION |  CBSE Class IX (9th)  Science | Solved Exercises

Question 10. Amit buys a few gram (force) of gold at the poles as per instruction of one of his friends. He hands over the same when he meets him at equator. Will the friend agree with the weight of gold bought ? If not, why ?
Answer.The force of gravity decreases with altitude. It also varies on the surface of the earth, decreasing from poles to the equator.
 Also, The weight is equal to the product of mass and acceleration due to gravity.
i.e. W = mg, where ‘m’ is mass of the object ‘g’ is the acceleration due to gravity.
As value of g is higher at polar region than the equator, the weight of an object will also be more at polar region than at equtor. Therefore, his friend will not agree with weight of the gold bought at the poles when measured at equator.

Question 11. Why will a sheet of paper fall slower than one that is crumpled into a ball ?
Answer. Surface area of a Sheet which is crumpled into a ball, is much smaller than the surface area of a plain or flat sheet. Therefore, despite both experince same force of gravity, the plain or flat sheet of paper will have to face more air resistance than the crumpled ball, so it will fall slower than the sheet crumpled into a ball.

Question 12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton of a 10 kg object on the moon and on the earth ?
Answer. The mass of an object always being same.
 :. The mass of the object on the moon will be the same, as that on earth, m = 10 kg.
We know that, Acceleration due to gravity on earth (ge) = 9.81 ms^–2

Acc. due to gravity on moon(gm) =	9.81	ms–2
	6

:. Weight of the object on the surface of moon

= mgm

	= 10 ×	9.81	=16.35 N
		6

and Weight of the object on the surface of Earth = mg_e

= 10 × 9.81 = 98.1 N

Question 13. A ball is thrown vertically upwards with a velocity of 49 ms^–1. Calculate
(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.
Answer.
$$($$ i )  As per the statement given Initial velocity of the ball (u)            = 49 m s − 1 Final velocity of the ball (v)             = 0 m  s − 1 Downward gravity  ( g)                      = 9.8 m s − 2   Upward gravity  ( g )                           = −9.8 m s − 2 Max. Height attained by the ball  ( s )   =  ?  As we know :                     v 2   − u 2 = 2 gs = >                             ( 0 ) 2 − ( 49 ) 2 = 2 × ( −9.8 ) × s = >                                             s = − 49 × 49 − 2 × 9.8 Max. Height attained by the ball  ( s )   = 122.5 m ( ii )   Also, as we know :                   v  = u  +  gt = >                                              0  = 49 − 9.8 × t = >                                               t  =   49 9.8 = 5 s Time for upward journey of the ball will be the same as  time for downward journey i.e .t  =  5 s Total time taken by the ball to return to the surface of earth  = 2 × t = 2 × 5 = 10 s
$$$$
Question 14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer.As per given statement, initial velocity u = 0, g = – 9.8 ms^–2, height,  s = – 19.6 m

As we know that

u² – u² = 2 gs

:.	u² – 0² = 2 × (–9.8) × (–19.6)

or	v² = (19.6)²

=>	v = – 19.6 ms–1

The negative sign indicates that the velocity is in the downward direction.

Question 15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone ?
Answer. Here, initial velocity u = 40 ms^–1 and g = – 10 ms^–2
At the maximum height reached S , final velocity, v = 0

As we know that

v² – u² = 2 gs

:.	0² – 40² = 2 × (–10) S

=>	S =	40 ×40	= 80 m
		20

Also, the total distance covered = 2 × S = 2 ×= 160 m
As the stone returns back to the original position.

:.	Net displacement = S -( S ) =  80 – 80 = 0

Question 16. Calculate the force of gravitation between the earth and the sun, given that the mass of the earth = 6 × 10²⁴ kg and of the sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.
Answer.

As given in the statement,

Me = 6 × 1024 kg, Ms = 2 × 1030

r = 1.5 × 1011 m

As we know that	F = G	MeMs
		r²

:.	F =	6.67 × 10-11 × 6 ×1024 × 2 × 1030	N
		(1.5 × 1011)²

=>	F =	6.67 ×12 × 1021
		1.5 × 1.5

:.	F = 3.56 × 1022 N

Question 17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet ? 
Answer. Let the point at which, two stones meet after time t from the start, be at a height x from the ground.
Height of the tower = 100 m

Then distance covered by the stone allowed to fall from the top of the tower,

	100 – x = ut +	1	gt²
		2

	= 0 × t +	1	gt²
		2

	=	1	gt²	...(1)
		2

The distance covered by the stone thrown from the ground,

	x = ut -	1	gt²
		2

	= 25t –	1	gt²	...(2)
		2

Combining eq. (1) and (2), we get

100 = 25t

=>

t = 4s

:.	x = 25 × 4 –	1	× 9.8 × 4²
		2

= 100 – 78.4

= 21.6 m

Question 18. A ball thrown up vertically returns to the thrower after 6s. Find:
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches, and
(c) its position after 4s.
Answer. (a) The velocity with which ball was thrown up :
 Acceleration due to gravity, g = – 9.8 ms^–2
As the total time taken in upward and return journey by the ball is 6 s.
:. The upward journey, t = 6/2 s = 3 s

Final velocity, u = 0 ms–1

Initial velocity, u = ?

As we know, by the first equation of motion,

u = u + gt

=>	0 = u + (– 9.8) × 3

=>	0 = u – 29.4

=>	u = 29.4 ms–1

:. The velocity with which ball was thrown up = 29.4 ms^–1
(b) Max. Height the ball reaches (h) = Distance (s) = ?
As we know, by the second equation of motion,

	s = ut +	1	gt²
		2

:.	s = 29.4 × 3 +	1	(– 9.8) × 32
		2

=>	s = 88.2 m – 44.1 m

=>	s = 44.1 m

(c) The position of the ball after

4 seconds =Distance, s =  ?

              Here as given      Time, t = 4 s 
As we know by equation from laws of motion,

	s = ut +	1	gt²
		2

:.	s = 29.4 × 4 +	1	× (– 9.8) (4)²
		2

=>	s = 117.6 m – 78.4 m

=>	s = 39.2 m

Question 19. In what direction does the buoyant force on an object immersed in a liquid act? 

Answer. The direction of Buoyant force on an object immersed in a liquid acts is vertically upward towards the centre of buoyancy.

Question 20. Why does a block of plastic released under water come up to the surface of water?

Answer.All objects experience a force of buoyancy when they are immersed in a fluid. Objects having density less than that of the liquid in which they are immersed, float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed then it sinks in the liquid. Here the density of plastic block is less than that of water, there for, weight of water displaced by fully immersed plastic block is more than its own weight. Thus, the upward force acting on the plastic block due to buoyancy , is much more than the downward force due to the weight of the block. Due to this upward buoyant force, the block will be forced up tilll the weight of displaced liquid is equal to the weight of plastic block.

Question 21. The volume of 50 g of substance is 20 cm³. If the density of water is 1 g cm^–3, will the substance float or sink ?
Answer. As given in the statement: The Mass of substance = 50 g
The volume of substance = 20 cm³
density of water is 1 g cm^–3
$$Density\; of\; substance$$ =   Mass of substance volume of substance                                  = 50 20 = 2.5  g  cm − 3 
Clearly As the density of substance (2.5 g cm^-3 )is more than that of water (1 g cm^–3), so it will sink.
Question 22. The volume of a 500 g sealed packed is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm^–3 ? What will be the mass of water displaced by this packet ?
Answer.
As per the statement of question :
Mass of sealed packed M = 500g
Volume of sealed packed V= 350 cm³
Density of water = 1 g cm^–3
$$Density\; of\; sealed\; packet$$ = M V                                        = 500 350 = 107                                        = 1.43 g  cm − 3 As density of packet which is 1.43 g cm^-3 is more than that of water( 1 g cm^–3), it will sink in water.
The volume of water displaced is equal to the volume of packet V i.e., 350 cm³.
$$\therefore \; Mass\; of\; the\; water\; displaced$$ = Density  ×  Volume                                              = 1 g  cm − 3   × 350  cm 3                                              = 350 g 

Chapter 10. GRAVITATION |  CBSE Class IX (9th)  Science | Solved Exercises

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SATURDAY, SEPTEMBER 29, 2012

CBSE Class 9 - Science - CH 3 - Atoms and Molecules

Atoms and Molecules

Molecular Structure

NCERT Solution, MCQs, Study Notes, Q & A

Q1: Who established the two important laws of chemical combinations?

Answer: Antoine L. Lavoisier and Joseph L. Proust

Q2: What are the laws of chemical combinations?

Answer: These are group of laws which describe how different elements and compounds combine together to form new compounds. Most of these were established based on chemical experimental results. These laws are:

1. Law of Conservation of Mass
2. Law of Constant composition (or definite proportions)
3. Law of Multiple proportions
4. Law of reciprocal proportions
5. Law of Combining Volumes (Gay Lussac's law of Gaseous Volumes)
6. Avogadro Law 

[Note: CBSE Class 9 syllabus covers first two laws only. In class 11, you will revisit all of these.]

Q3: What is the law of conservation of mass?

Answer: The law states "During any physical change or chemical change, the total mass of the products remains equal to the total mass of the reactants."
It means during a chemical reaction matter is neither created nor destroyed. It is also called the Law of indestructibility of matter.

Q4: Give an example to show Law of conservation of mass applies to physical change also.

Answer: When ice melts into water, is a physical change. Take a piece of ice in small flask, cork it and weight it (say W_ice gm). Heat the flask gently and ice (solid) slowly melts into water (liquid). Weigh the flask again (W_water gm). It is found there is no change in the weight i.e. W_ice= W_water.

                              Heat(Δ)

                        Ice  ----------➜   Water

This shows law of conservation of mass holds true for physical changes.

Q5: Explain with example that law of conservation of mass is valid for chemical reactions.

Answer:According to this law, total mass of the products (chemicals produced) is equal to the total mass of the reactants (chemicals take part in reaction). e.g. Carbon combines with Sulphur to form Carbon disulphide. The mass of reactants i.e. carbon and sulphur is same mass of products (carbon disulphide).

                 Carbon + Sulphur -----➜   Carbon DiSulphide

                   C    +   S     -----➜     C2S  

                   1g   +  5.34g    =       6.34 g 

                       LHS          =     RHS

Q6: Is there any exception to law of conservation of mass?

Answer: Later after atoms were discovered, it was found during nuclear reactions this law does not hold good. In a nuclear reaction, some of the mass gets converted into energy, as given by famous Einstein's mass-energy relationship (E = mC²). The law was rechristened as Law of conservation of mass and energy.

Lavosier and "Reign of Terror"

Remember French Revolution and Robespierre's Reign of Terror. Lavosier was declared as traitor. Within a day, he was tried, convicted and executed. Later French Government admitted her mistake for the false conviction of him.

Lavosier made important contributions in the field of chemistry and biology. He was a pioneer of analytical chemistry. He discovered the gas oxygen and gave its name. He concluded air is mixture of gases and oxygen supports breathing and combustion.

Q7(NCERT): In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
       sodium carbonate + ethanoic acid  → sodium ethanoate + carbon dioxide + water

Answer: Given, The reaction is
       sodium carbonate + ethanoic acid  → sodium ethanoate + carbon dioxide + water

Mass of  sodium carbonate (Na₂CO₃) = 5.3 g
Mass of ethanoic acid (CH₃COOH) = 6g
Mass of carbon dioxide (CO₂) = 2.2g
Mass of Water (H₂O) = 0.9 g
Mass of sodium ethanoate (CH₃COONa) = 8.2 g

Total mass of reactants = 5.3g + 6g = 11.3g
Total mass of products = 2.2g + 0.9 g + 8.2g = 11.3g

Since, Total mass of reactants = Total mass of products, the observations in this reaction are in agreement with the law of conservation of mass.

Q8(CBSE 2011): 2.8 g of nitrogen gas was allowed to react with 0.6 g of hydrogen gas to produce 3.4 g of ammonia. Show that these observations are in agreement with the law of Conservation of mass.

Answer: Given,
mass of nitrogen (N₂) gas (reactant) = 2.8g
mass of hydrogen (H₂) gas (reactant) = 0.6g
mass of ammonia gas (NH₃) (product) = 3.4 g

The reaction is

 nitrogen + hydrogen  → ammonia

Total mass of reactants = 2.8g + 0.6g = 3.4g
Total mass of product = 3.4g

Since Total mass of reactants = Total mass of products, the observations in this reaction are in agreement with the law of conservation of mass.

Q9(CBSE 2011): If 12 g of carbon is burnt in the presence of 32 g of oxygen, how much carbon dioxide will be formed? 

Answer: Given, the reaction is

carbon + oxygen → carbon dioxide

Mass of carbon = 12g
Mass of oxygen = 32g
Mass of carbon dioxide (CO₂) = xg

J L Proust

Since the reaction is in agreement with law of conservation of mass, then
                      Total mass of reactants = Total mass of products
i.e. 12g + 32g = x
i.e. x = 44g
∴ mass of carbon dioxide (CO₂) formed = 44g

Q10: Who proposed Law of Definite Proportions (or Law of Constant Composition)?

Answer: J.L. Proust 1799 (French Chemist)

Q11: State Law of constant proportions. Explain with an example.

OR

Q(CBSE 2011): (i) State the law of constant proportions.
(ii) Show that water illustrates the law of constant proportions.

Answer: In a chemical substance the elements are always present in definite proportions by
mass, no matter how that chemical compound is prepared.

Example: Water always contains Hydrogen and Oxygen in the same proportion  i.e. the ratio of
the mass of hydrogen to the mass of oxygen is always 1:8, whatever the source of water. Whether water comes from melting of ice or by condensation of steam (physical changes) or by it is produced through chemical reactions.



Q12(NCERT): Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Answer: Given that the ratio of hydrogen and oxygen by mass to form water is 1 : 8.
⇒ the mass of oxygen gas required to react completely with 1 g of hydrogen gas =  8 g of O₂
∴ 3g of hydrogen (H₂) gas requires = 3 × 8 = 24g of O₂

Q13(NCERT): A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer: Given, mass of boron = 0.096 g
    mass of the oxygen = 0.144g
mass of the sample compound (boron+oxygen) = 0.24 g
According to Law of definite proportions, boron and oxygen are in fixed ratio in the given compound.= 0.096:0.0144  = 2:3

⇒ % of boron = mass of boron × 100 / mass of compound = 0.096 × 100/0.24 = 40%and % of oxygen = mass oxygen  × 100 / mass of compound = 0.144 × 100/0.24 = 60%

Q14(NCERT): When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Answer: The given reaction is

                Carbon + Oxygen ➜ Carbon Dioxide

                  3g   +  8g    =   11g 

⇒ Total mass of reactants = Total mass of products
∴ Law of conservation of mass is obeyed.

It also shows that carbon dioxide (CO₂) contains carbon and oxygen in fixed ratio i.e. 3:8 which follows Law of Constant proportion.

⇒ 3g of carbon (C) will react with 8g of oxygen (O) to give 11g of carbon dioxide (CO₂).
⇒ Remaining oxygen (50 -8 = 42g) will not participate in the reaction.

Q14(CBSE 2011): Magnesium and oxygen combine in the ratio of 3 : 2 by mass to form magnesium oxide. How much oxygen is required to react completely with 12 g of magnesium?

Answer: Magnesium (Mg) and Oxygen (O) combine in the ratio = 3:2 to form compound magnesium oxide (MgO). It follows Law of Constant proportions.
⇒ 3g of Mg combines with = 2g of O.
∴ 12g of Mg combines with = 2 × 12/3 = 8g of O.




Q15: What are the postulates of Dalton's atomic theory?


Answer: Dalton's theory states that matter, whether an element, a compound or a mixture is composed of small particles called atoms. 
The  postulates are:

1. All matter is made of very tiny particles called atoms.
2. Atoms are indivisible particles i.e. can neither be created or destroyed in a chemical reaction. (Law of Conservation of mass)
3. Atoms of a give element have same mass and identical chemical properties.
4. Atoms of different elements have different masses and chemical properties.
5. Atoms combine in the ratio of small whole numbers to form compounds. A chemical reaction is a re-arrangement of atoms. (Law of Multiple Proportions)
6. The relative number and kinds of atoms are constant in a given compound. (Law of Definite Proportions)

Q16(NCERT): Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer: The postulate of Dalton’s atomic theory i.e. "Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction" is the result of law of conservation of mass.

Q17(NCERT): Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer: The postulate of Dalton's atomic theory that explains the law of definite proportions is:
The relative number and kinds of atoms are constant in a given compound.  

Boltzman

Suicide of Ludwig Boltzman (1844-1906) 

When Dalton proposed his theory (1804), it was not well accepted by many scientists, because there was no experiment to prove atoms exist. However few scientists believed in his theory. Ludwig Boltzman, famous Austrian physicist, was firm the believer in atoms and used this concept to explain kinetic theory of gases. Unfortunately it did not go well with other scientists (Ostwald and Mach) who criticized it badly. This had a grave impact on Boltzman. His health deteriorated and eventually he committed suicide. Within couple of years after his death, scientists confirmed experimentally and convinced the world that atoms do exist. Later, J.J. Thomson was the first one to propose a model for the structure of an atom.    

Q18(MCQ): Which of the following statements is NOT true about an atom?
(a) Atoms are the building blocks from which molecules and ions are formed.
(b) Atoms cannot exist independently.
(c) Atoms are neutral in nature
(d) Atoms combine together to form matter that we can see, feel or touch.

Answer: (b) Atoms cannot exist independently.

Q19:  What is an atom?

Answer: An atom is the smallest particle of an element which takes part in chemical reactions. An atom is considered to be building blocks of matter.

STM Image of Gold Atoms
credits:wikicommons

Q20: Why is it not possible to see an atom with naked eyes?

Answer: The size of an atom is so small that it cannot be seen by naked eye or by simple optical instruments. e.g. Considering an atom a sphere, the radius of Hydrogen atom is approximately 10nano-metres (1 nm = 10^-9m).

To see atoms of different elements, a special type of microscope called scanning tunnelling microscope  (STM) is used.

Nano-science and Nano-technology:
With the invention of scanning tunneling microscope, a new branch of science and engineering emerged called Nano-science and Nano-technology. It deals in designing functional systems at atomic and molecular scale i.e. of size in nano-meters. At nano-scale, laws of Newton fail. It follows principles of Quantum Physics.

Q20: Who proposed the chemical notation based on first two letters of the name of the element?

Answer: Berzilius

Q21: Name the international organization who approves names of elements.

Answer: IUPAC (International Union of Pure and Applied Chemistry)

Q22: What is the chemical symbol for iron?

Answer: Fe

Q23: Name five elements have single letter chemical symbol.

Answer: Hydrogen (H), Oxygen(O), Boron(B), Sulphur(S), Carbon(C) and Nitrogen (N)

Q24: Name the element having following Latin names
(i)   Stibium
(ii)  Cuprum
(iii) Argentum
(iv) Natrium
(v)  Stannum
(vi) Wolfram
(vii) plumbum
(viii) Kalium

Answer:
(i)   Stibium - Antimony (Sb)
(ii)  Cuprum - Copper (Cu)
(iii) Argentum - Silver(Ag)
(iv) Natrium - Sodium (Na)
(v)  Stannum - Tin (Sn)
(vi) Wolfram - Tungeston (W)
(vii) plumbum - Lead (Pb)
(viii) Kalium - Potassium (K)

Q25: Write the chemical symbols of  the following:
(i) Gold
(ii) Iron
(iii) Chlorine
(iv) Mercury

Answer:
(i) Gold - Au
(ii) Iron - Fe
(iii) Chlorine - Cl
(iv) Mercury - Hg

Q26: How will you define chemical symbol?

Answer: A chemical symbol of an element is a shorthand chemical notation for the full name of an element.

Q27: What is the significance of a chemical symbol?

Answer: The symbol of an element has the following significance:

1. symbol stands for the name of the element in while representing chemical formula and chemical equations.
2. it stands for one atom of the element.
3. it represents quantity of the element in terms of mass, moles or gram-atomic mass etc. while computing chemical reactions.

Q28: Can atoms of an element exist independently? Give examples of elements which exist in atomic form. Give examples of elements that do not exist in atomic form.

Water Molecule

Answer:  Atoms of most elements do not to exist independently. There are few elements like Helium (He), Argon (Ar), Krypton(Kr), Radon(Rn), Neon(Ne) and Xenon(Xe) - collectively called noble gases, exist at atomic form. Atoms of other elements combine together and form molecules and
ions. These molecules or ions aggregate in large numbers to form the matter.  e.g.

• Hydrogen (H₂), Oxygen(O₂), Nitrogen(N₂) etc. atoms of the same element combine together and form molecules.
• Water (H₂O), Carbon Dioxide(CO₂), Common Salt(NaCl) etc. atoms of different elements combine together and form molecules.
• Na⁺, Cl^-, Cu²⁺ etc. atoms of elements form ions in aqueous solutions. 

Q29: Why do atoms of the most of the elements not exist independently?

Answer: Atoms of the most of the elements are reactive by nature. They are not able to exist independently and form molecules and ions.

Q30: Which element has the smallest atom in size?

Answer: Hydrogen.

Q31(NCERT): What is the atomic mass unit?

Answer: The mass equal to 1/12th of the mass of  ¹²₆C atom is called the one atomic mass unit. It is 1.66 × 10^-27 kg.  It is represented by symbol 'u' (older symbol 'amu').

            1 u  = 1/12th of the mass of carbon-12 atom.


Q32: Magnesium is two times heavier than C-12 atom, what shall be the mass of Mg atom in terms of atomic mass units? (Given mass of C-12 atom = 12u)


Answer:  1 u = 1/12th of C-12 atom
 mass of C-12 atom = 12u
 mass of Mg is twice of C-12 atom = 2 ×12u  = 24u

Q33(CBSE 2011): What is relative atomic mass  of an element? How it is related to atomic mass unit?

Answer: The relative atomic mass (A_r) of an element is the average relative mass of an element of the element with respect to an atom of Carbon-12 taken as 12u.

                                      Average mass of 1 atom of an element 

Relative Atomic mass of an element = -------------------------------------- 

                                      1/12th of Mass of one C-12 atom

The relative atomic mass is a pure number and has no units. It indicates number of times one atom of an element is heavier than 1/12th of C-12 atom. 

Relative atomic mass is also called average atomic mass.  

Atomic mass is expressed in amu or u while relative atomic mass is a number.

e.g. Atomic mass of one Hydrogen atom is 1u and its relative atomic mass is 1.


Q34: Define molecule. What are its important properties?

Answer: The smallest particle of matter (element or compound) which can exist in a free state is called a molecule.

 Atoms of the same element or of different elements can join together to form molecules.

Properties:

1. Molecules of any given substance are similar.
2. Molecules of different substances are different.
3. The properties of a substance are the properties of the molecule of that substance.
4. Each molecule is described by its chemical formula.

Buckminsterfullerene consists
of carbon atoms only
(credits:wikipedia)

Q34: Based on type of substance, how molecules are classified?

Answer:

1. elementary molecules: molecules formed by atoms of the same element. e.g. Chlorine (Cl₂), Hydrogen (H₂), Oxygen(O₂), Nitrogen(N₂) gases, Buckminsterfullerene (C₆₀) etc.
2. compound molecules: molecules formed by atoms of different elements e.g. (H₂O), Carbon Dioxide(CO₂), Glucouse (C₆H₁₂O₆), Aspirin (C₉H₈O₄)

Aspirin Molecule

Q35: What is atomicity?

Answer: The number of atoms contained in a molecule of a substance (element or compound) is called its atomicity.

Q36: Based on atomicity, how molecules are categorized?

Answer:

1. Monoatomic molecules (1 atom molecule)
2. Diatomic molecules (2 atoms molecule)
3. Triatomic molecules (3 atoms molecule)
4. Tetraatomic molecules (4 atoms molecule)
5. Polyatomic molecules (more than 4 atoms molecule)

Q37: Give three examples of monoatomic molecules.

Answer: Noble gases. e.g. Helium (He), Neon(Ne) and Argon(Ar).

Q38: Give four examples of diatomic molecules.

Answer:  Chlorine (Cl₂), Hydrogen (H₂),Carbon Monoxide (CO), Sodium Chloride (NaCl)

Q39: (i) What is the chemical formula of Water molecule?
(ii) What is its atomicity?
(iii) Calculate the ratio of masses of atoms of elements present in water molecule.
(iv) Calculate the ratio by number of atoms of elements present in water molecule.

Answer: 
(i) Water molecule:  (H₂O)
(ii) It is triatomic i.e. 2 atoms of H + 1 atom of O = 3 atoms of water molecule.
(iii) Mass of 1 H-atom = 1 u
       Mass of two H-atoms = 2u
       Mass of O atom = 16u
       Mass ratio of H:O = 2u:16u = 1:8
(iv) Number of atoms ration H:O = 2:1


Q40: What is an ion?

Answer: An ion is positively or negatively charged atom or group of atoms. There are two types of ions:

1. cations (positively charged ions) e.g.Na⁺, K⁺, Cu²⁺
2. anions (negatively charge ions) e.g. Cl^-, SO₄^2-

Q41(NCERT): What are polyatomic ions? Give examples?

Answer: A group of atoms carrying a charge is known as a polyatomic ion.
Examples are: Sulphate anion (SO₄^2-), carbonate anion (CO₃^2-), Ammonium cation (NH₄¹⁺) etc.

Q42: Give examples of triatomic molecules.

Answer: Ozone (O₃), Carbon Dioxide(CO₂), Nitrogen Dioxide (NO₂)

Q43: What is valency of an element?

Answer: The valency of an element is defined as its capacity to combine with other elements. It can be be used to find out how the atoms of an element will combine with the atom(s) of another element to form a chemical compound.

Elements may have valency: 1(monovalent), 2 (divalent), 3(trivalent) and 4 (tetravalent) respectively. Some elements show variable valency in different compounds.
e.g. In red oxide of Copper called Copper (I) Oxide (Cu₂O), Cu has valency=1. While in black oxide of Copper i.e. Copper(II) Oxide (CuO), Cu is divalent (valency=2).

Q44(NCERT): What is meant by the term chemical formula?

Answer:  A chemical formula is a representation of chemical compound using a set of chemical symbols that forms the compound. It may be an empirical formula or a molecular formula. The molecular formula represents the actual number of atoms of elements present in the compound. While empirical formula tells the whole number ratio of atoms of elements present in the compound.

e.g. The empirical formula of Water is HO which tells it consists of hydrogen and oxygen atoms. The chemical (molecular) formula of Water is : H₂O, which tells 2 Hydrogen atoms combine with one atom of oxygen.
The chemical formula of a molecular compound is determined by the valency of each element.

Q45(NCERT): Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide

Answer:
(i) sodium oxide: Na⁺ + O^2- ➜Na₂O
(ii) aluminium chloride: Al³⁺ + Cl^- ➜ AlCl₃
(iii) sodium suphide: Na⁺ + S^2- ➜ Na₂S
(iv) magnesium hydroxide: Mg²⁺ + OH^- ➜ Mg(OH)₂

Q46(NCERT): Write down the names of compounds represented by the following formulae:

 (i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃

Answer:
(i) Al₂(SO₄)₃: Aluminium Sulphate
(ii) CaCl₂: Calcium Chloride
(iii) K₂SO₄:Potassium Sulphate
(iv) KNO₃: Potassium Nitrate
(v) CaCO₃ : Calcium Carbonate

Q47(NCERT): Write the chemical formulae of the following. Also identify the ions present.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

Answer:
(a) Magnesium chloride: MgCl₂ (ions present are: Mg²⁺, Cl^1- )
(b) Calcium oxide: CaO (ions present are: Ca²⁺, O^2- )
(c) Copper nitrate: Cu(NO₃)₂ (ions present are: Cu²⁺, NO₃^1- )
(d) Aluminium chloride: AlCl₃ (ions present are: Al³⁺, Cl^1- )
(e) Calcium carbonate: CaCO₃ (ions present are: Ca²⁺, CO₃^2- )

Q48(NCERT): Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

Answer:
(a) Quick lime: Calcium Oxide (CaO), Elements Present: Calcium(Ca), Oxygen(O)
(b) Hydrogen bromide: HBr, Elements Present: Hydrogen (H), Bromine(Br)
(c) Baking powder: Sodium Bicarbonate (NaHCO₃) Elements Present: Sodium(Na), Hydrogen(H), Carbon(C) and Oxygen(O).
(d) Potassium sulphate: K₂SO₄Elements Present:Potassium(K), Sulphur(S), Oxygen(O)

Q49(NCERT): How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄^3- ion?

Answer:
(i) H₂S molecule: Three atoms (2 H-atoms + 1 S atom)
(ii) PO₄^3- ion: Five atoms (1 P-atom + 4 O-atoms)

Q50(CBSE 2011): (a) Write a chemical formula of a compound using zinc ion and phosphate ion.
(b) Calculate the ratio by mass of atoms present in a molecule of carbon dioxide.
(Given C =12, O =16)

Answer: (a) Zinc ion: Zn²⁺, Phosphate ion: PO₄^3-
⇒ Zinc Phosphate: Zn₃(PO₄)₂

(b) Carbon DiOxide (CO₂) = 1 C-atom + 2 O-atoms
Atomic mass of 1 C-atom = 12u
Atomic mass of 1 O atom = 16u
Atomic mass of 2 O-atoms = 32u
Ratio by mass of C:O atoms = 12u:32u = 3:8

Q51: What is the molecular mass of a substance?

Answer: The average mass of a molecule of a substance expressed in atomic mass units is called its molecular mass.
It is obtained by adding together the atomic masses of all the atoms present in one molecule of the substance.

Q52: What is Formula Unit Mass? How it is different from molecular mass?

Answer:   The formula unit mass of a substance is a sum of the atomic masses of all atoms in a
formula unit of a compound. Formula unit mass is calculated in the same manner as we calculate the molecular mass.
The difference is that we use the word formula unit for ionic compounds.

Q53(NCERT): Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Answer:
(i) molecular mass of H₂ = 2 × Atomic mass of H = 2 × 1u = 2u
(ii) molecular mass of O₂ = 2 × Atomic mass of O = 2 × 16u = 32u
(iii) molecular mass of Cl₂ = 2 × Atomic mass of Cl = 2 × 35.5u = 71u
(iv) molecular mass of  CO₂ = 1 × Atomic mass of C + 2 × Atomic mass of O
       = 1 × 12u + 2 × 16u = 12u + 32u = 44u
(v) molecular mass of CH₄= 1 × Atomic mass of C + 4 × Atomic mass of H
       = 1 × 12u + 4 ×1u = 12u + 4u = 16u
(vi) molecular mass of C₂H₆ =  2 × Atomic mass of C + 6 × Atomic mass of H
      =  2 × 12u + 6 ×1u = 24u + 6u = 30u
(vii) molecular mass of C₂H₄ =  2 × Atomic mass of C + 4 × Atomic mass of H
      =  2 × 12u + 4 ×1u = 24u + 4u = 28u
(viii) molecular mass of NH₃ = 1 × Atomic mass of N + 3 × Atomic mass of H
      = 1  × 14u + 3 × 1u = 14u + 3u = 17u
(ix) molecular mass of CH₃OH =  1 × Atomic mass of C + 4 × Atomic mass of H + 1 × Atomic mass of O = 1 × 12u + 4 × 1u + 1 × 16u = 12u + 4u + 16u = 32u

Q54(NCERT): Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

Answer:
(i) Formula unit mass of  ZnO = 1 × Atomic mass of Zn + 1 × Atomic mass of O = 1 × 65u + 1 ×16 u
        = 65u + 16u = 81u
(ii) Formula unit mass of  Na₂O =  2 × Atomic mass of Na + 1 × Atomic mass of O
       = 2 × 23u + 1 × 16u = 46u + 16u = 62u
(iii) Formula unit mass of  K₂CO₃= 2 × Atomic mass of K + 1 × Atomic mass of C + 3 × Atomic mass of O
= 2 × 39u + 1 × 12u + 3 × 16u = 78u + 12u + 48u = 138u

Q55: What are ionic compounds?

Answer: Compounds which are made up of ions (cation + anion) are called ionic compound. In ionic compounds (e.g. NaCl),  the positively charged ions (cations, e.g. Na¹⁺) and negatively charged ions (anions, e.g. Cl^1-) combine together due to strong forces of attraction between them.

Q56(CBSE 2011): How do we know the presence of atoms if they do not exist independently for most of the elements?

Answer: Matter consists of tiniest particles which cannot be further divided. Elements combine together to form compounds and Law of Combinations prove that elements combine in definite proportions to form a particular compound. It implies atoms of various elements combine together and form molecules. Each molecule can be described by a unique chemical formula.

Q57(CBSE 2011): An element 'Z' forms the following compound when it reacts with hydrogen, chlorine, oxygen and phosphorus.
                        ZH₃, ZCl₃, Z₂O₃ and ZP
(a) What is the valency of element Z?
(b) Element 'Z' is metal or non-metal?

Answer: The valency of element Z is 3. It is a metal, it combines with non-metals like Hydrogen(H), Chlorine(Cl), Oxygen and Phosphorous(P).

Q58(CBSE 2011): Name one element each which forms diatomic and tetra atomic molecule.

Answer: Diatomic: Oxygen, Tetra Atomic: Phosphorus

Q59: Name one element which forms diatomic and triatomic molecule.

Answer: Oxygen (forms diatomic oxygen O₂ gas and triatomic ozone O₃ gas molecules).

Q60: What is gram-atomic mass of an element?

Answer: When atomic mass of an element is expressed in grams, it is called the gram-atomic mass of the element.
e.g. Atomic mass of Oxygen (O) = 16u
Gram-atomic mass of oxygen = 16g

Q61: What is gram-molecular mass of a substance?

Answer: The molecular mass of substance expressed in grams is called the gram-molecular mass of the substance.

e.g. molecular mass of Hydrogen (H₂)molecule = 2u
Gram-molecular mass of Hydrogen (H₂)molecule = 2g

Q62: Define mole. What is its significance?

Answer: A mole (or mol) is the amount of a substance which contains the same number of chemical units (atoms, molecules or ions) as there are atoms in exactly 12 grams of pure carbon-12.

Since 12g of  carbon-12 is found to contain 6.023 × 10²³ atoms of C-12. Therefore, a mole represents a collection of 6.023 × 10²³ chemical units (atoms, molecules or ions). This number 6.023 × 10²³ is also called Avagadro's Number (N_A or N_O).

Significance of mole: A mole represents the following

1. It represents quantity in number i.e. 6.023 × 10²³ chemical units.
2. The mass of 1 mole  of an element is equal to 6.023 × 10²³ atoms of the element.
3. 1 mol of a substance represents one gram-formula mass of the substance i.e.
   1 mole of atoms of an element = Gram atomic mass of the element.
   1 mole of molecules of a substance = Gram-molecular mass of the substance.
   e.g. 1 mole of Oxygen (O₂) molecules = 6.023 × 10²³ molecules = 32g of O₂
4. One mole of gas at standard temperature and pressure occupies 22.4 litres volume (Avagadro's Law).
   
   

Q63(CBSE 2011): What is molar mass?

Answer: Mass of 1 mole of any substance is called its molar mass (M). It is measured as grams/mol  (g/mol) or kg/mol. For an element, its molar mass is equal to its gram atomic mass. For compounds (or molecules), its molar mass is equal to its gram-molecular mass.
e.g. Gram-molecular mass of  Hydrogen (H₂)  = 2g
⇒ Molar mass of  Hydrogen (H₂) = 2g/mol.

Q64: Who introduced the term 'mole' in chemistry?

Answer: Wilhelm Ostwald in 1896.

Q65: When 'mole' was chosen internationally standard way to express larger number of chemical units?

Answer: 1967

Q66: How many moles are there in 4.6 gms of Sodium(Na)?

Answer: Number of moles = mass in gm ÷ atomic mass
Atomic mass of Na = 23.
 moles = 4.6g/23 = 2/10 = 0.2 moles

Q67(NCERT): If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

Answer:  1 mol of Carbon atoms = 12g
1 mol of atoms = N_o atoms = 6.023 × 10²³ atoms⇒ 6.023 × 10²³ carbon atoms = 12g
∴ mass of 1 carbon atom = 12g ÷ N_o = 12g / 6.023 × 10²³ = 1.992 × 10^-23 gram

Q68(NCERT): Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)? 

Answer: We know that, 1 mol of atoms of an element has mass = gram atomic mass.
∴ mass of 1 mol of Na atoms = 23g
Or 23g of Na has = 1 mol of atoms
∴ 100g of Na = 100/23 = 4.35 moles

Similarly, 56g of Fe has = 1 mol of atoms
∴ 100g of Fe has = 100/56 = 2.17 moles
⇒ 100g of Na has more number of atoms than 100g of Fe.

Q69(NCERT): What is the mass of
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na₂SO₃)?

Answer: (a) 1 mol of nitrogen (N) atoms = gram atomic mass of N = 14g

(b) mass = molar mass × number of moles
1 mol of Aluminium (Al) = gram atomic mass of Al = 27g
mass of 4 mols of Al = 4 × 27 = 108g

(c) mass = molar mass × number of moles
molecular mass of sodium sulphite (Na₂SO₃) =
    =  2 × Atomic mass of Na + 1 × Atomic mass of S  + 3 × Atomic mass of O
    = 2 × 23u + 1 × 32u + 3 × 16u = 46u + 32u + 48u
    = 126u
mass of 1 mol of sodium sulphite (Na₂SO₃) = gram molecular mass or molar mass = 126g
mass of 10 moles of Na₂SO₃ = 10 × 126g = 1260g

Q70(NCERT): Convert into moles:

(a) 12 g of oxygen gas 

(b) 20 g of water 

(c) 22 g of carbon dioxide

Answer: 

(a) 32g of Oxygen = 1 mol

12g of Oxygen = 12/32 = 0.375 mol

(b) 20g of water

Molar mass of Water (H₂O) = 2 × 1 + 16 = 18g

18g of H₂O = 1 mol

20g of  H₂O = 20/18 = 1.1 mol

(c) Molar mass of  carbon dioxide(CO₂) = 12g + 2 × 16g = 12g + 32g = 44g

44g of CO₂ = 1 mol

22g of CO₂ = 22/44 = 0.5 mol

Q71(NCERT): What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Answer:

(a)  mass = molar mass × number of moles
molar mass of Oxygen atoms = 16g
mass of 0.2 mol of Oxygen atoms = 16 × 0.2 = 3.2g

(b) molar mass of water molecule (H₂O) = 2 × 1 + 16 = 18g
mass of 0.5 mole of Water = 18 × 0.5 = 9g

Q72: Find out number of atoms in 15 moles of He.

Answer: 1 mole of He has = 6.023 × 10²³ atoms
∴ 15 moles of He have =  15 × 6.023 × 10²³ = 9.034 × 10²⁴ atoms

Q73(NCERT): Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.

Answer: Gram Atomic mass of Sulphur = 32g
Molar mass of sulphur (S₈) = 8 × 32g = 256g
∴ 256g of sulphur (S₈) = 6.023 × 10²³ molecules
16g of  sulphur (S₈) = 6.023 × 10²³ × 16 / 256 = 3.76 × 10²² molecules

Q74(NCERT): Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Answer: Molar mass of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102g
⇒ 102g of  aluminium oxide (Al₂O₃) contains = 1 mole of  molecules = 6.023 × 10²³molecules
The molecular formula of  aluminium oxide = Al₂O₃
⇒ 1 molecule of aluminium oxide = 2 atoms of aluminum
∴ 102g of  aluminium oxide (Al₂O₃) contains = 2 × 6.023 × 10²³ atoms of Al
∴ 0.051g of aluminium oxide (Al₂O₃)will have = 2 × 6.023 × 10²³× 0.051 / 102
    =  6.023 × 10²⁰ atoms of Al

Q75: Calculate the mass percentage of Carbon(C) , Hydrogen (H) and Oxygen (O) in one molecule of glucouse (C₆H₁₂O₆).
(Atomic mass of C = 12u, H = 1u and O = 16u)

Answer:  Molecular mass of glucouse (C₆H₁₂O₆) = 6 × 12u + 12 × 1u + 6 × 16u
                = 72u + 12u + 86u = 180u
⇒ Molar mass of  glucouse (C₆H₁₂O₆) = 180 g/mol

% of C = Total mass of C/ Molar mass of glucouse = 6 × 12g × 100/ 180g = 7200/180 = 40%
% of H = Total mass of H/ Molar mass of glucouse =12 × 1g × 100/ 180g = 1200/180 = 6.67%
% of O = Total mass of O/ Molar mass of glucouse =6 × 16g × 100/ 180g = 9600/180 = 53.33%


Q76: Calculate the number of molecules of phosphorus (P₄) present in 31 gram of phosphorus.
Answer:  Atomic mass of phosphorus = 31 u
  Molecular mass of P₄ = 31× 4 = 124 u
 ⇒ Molar mass of P4 = 124 g/mol.

⇒ Mass of 1 mole ( 6.023 × 10²³ ) P₄ molecules = 124 g

∴ 124 g of  P₄ =  6.023 × 10²³ molecules of P₄

 1g of  P₄ = ( 6.023 × 10²³ )/(124) = 0.25 × 10²³ molecules
 31g  P₄ =  0.25 × 10²³ × 31 = 1.5 × 10²³ molecules                           .... answer

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