NEW SYLLABUS OF CBSE SCHOOLS
CBSE Syllabus 2021-22 for 9th, 10th, 11th, 12th: Subject-wise New CBSE Syllabus 2021-22 (PDF) or CBSE Curriculum 2021-22 for 9th, 10th, 11th & 12th has been officially released online at cbseacademic.nic.in. The new CBSE Syllabus is applicable for CBSE Academic Session 2021-22. The board has already instructed CBSE Schools to start new CBSE Academic Session 2021-22 from April and the board has released the new syllabus before April 2021. There is no reduction in CBSE Syllabus 2021-22 for 9th, 10th, 11th & 12th.
As this CBSE Syllabus is applicable for Academic Session 2021-22, so students preparing for CBSE board exams 2021 has nothing to do with this syllabus. They can access their syllabus from this link.
CBSE Syllabus 2021-22 (PDF): Class 9
With the links given above, students of CBSE Class 9th can check & download the latest CBSE Syllabus 2021-22 (PDF).
CBSE Syllabus 2021-22 (PDF): Class 10
With the links given above, students of CBSE Class 10th can check & download the latest CBSE Syllabus 2021-22 (PDF).
CBSE Syllabus 2021-22 for Class 11
With the links given above, students of CBSE Class 11th can check & download the latest CBSE Syllabus 2021-22 (PDF).
CBSE Syllabus 2021-22 for Class 12
chemistry
chapter1
Access Answers of NCERT Class 9 Science (Chemistry) Chapter 1 – Matter in Our Surroundings (All in text and Exercise Questions solved)
Exercise-1.1-1.2 Page: 3
1. Which of the following are matter?
Chair, air, love, smell, hate, almonds, thought, cold, lemon water, smell of perfume.
Solution:
The following substances are matter:
Chair
Air
Almonds
Lemon water
Smell of perfume (Smell is considered as a matter due to the presence of some volatile substances in air that occupy space & have mass.)
2. Give reasons for the following observation:
The smell of hot sizzling food reaches you several meters away, but to get the smell from cold
food you have to go close.
Solution:
Particles in the air, if fueled with higher temperatures, acquire high kinetic energy which aids them
to move fast over a stretch. Hence the smell of hot sizzling food reaches a person even at a distance
of several meters.
3. A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?
Solution:
The diver is able to easily cut through the water in the swimming pool because of the weak forces of
attraction between water molecules. It is this property of water that attributes to easy diving.
4. What are the characteristics of the particles of matter?
Solution:
The characteristics of particles of matter are:
(a) Presence of intermolecular spaces between particles
(b) Particles are in constant motion
(c) They attract each other
Exercise-1.3 Page: 6
1. The mass per unit volume of a substance is called density. (density=mass/volume). Arrange the following in the order of increasing density – air, exhaust from the chimneys, honey, water, chalk, cotton and iron.
Solution:
The following substances are arranged in the increasing density:
Air
Exhaust from chimney
Cotton
Water
Honey
Chalk
Iron
2. Answer the following.
a) Tabulate the differences in the characteristics of matter.
b) Comment upon the following: rigidity, compressibility, fluidity, filling a gas container,
shape, kinetic energy and density.
Solution:
(a) The difference in the characteristics of the three states of matter.
| Characteristics | Solid | Liquid | Gas |
| Shape | Fixed shape | No Fixed shape | No Fixed shape |
| Volume | Fixed volume | Fixed volume | No Fixed volume |
| Intermolecular force | Maximum | Less than solids | Very less |
| Intermolecular space | Very less | More than solids | maximum |
| Rigidity/Fluidity | Rigid/cannot flow | Can flow/not rigid | Can flow/not rigid |
| Compressibility | negligible | compressible | Highly compressible |
(b) (i) Rigidity: It is the propensity of a substance to continue to remain in its shape when treated
with an external force.
(ii) Compressibility: It is the attribute of the particles to contract its intermolecular space when
exposed to an external force thereby escalating its density.
(iii) Fluidity: It is the ability of a substance to flow or move about freely.
(iv) Filling the gas container: The particles in a container take its shape as they randomly vibrate in
all possible directions.
(v) Shape: It is the definite structure of an object within an external boundary
(vi) Kinetic energy: Motion allows particles to possess energy which is referred to as kinetic
energy. The increasing order of kinetic energy possessed by various states of matter are:
Solids < Liquids < Gases
Mathematically, it can be expressed as K.E = 1/2 mv2, where ‘m’ is the mass and ‘v’ is the velocity
of the particle.
(vii) Density: It is the mass of a unit volume of a substance. It is expressed as:
d = M/V, where ‘d’ is the density, ‘M’ is the mass and ‘V’ is the volume of the substance
3. Give reasons
a) A gas fills completely the vessel in which it is kept.
b) A gas exerts pressure on the walls of the container.
c) A wooden table should be called a solid.
d) We can easily move our hand in the air but to do the same through a solid block of wood we need a karate expert.
Solution:
a) Kinetic energy possessed by gas particles is very high which allows them to randomly move
across all directions when contained, hence the particles fills the gas vessel entirely.
b) Gas molecules possess high kinetic energy, due to which they are under constant motion inside
the container in random directions which causes them to hit the walls of the container and hence create vibrations. These collisions with the walls of the container generate pressure.
c) A wooden table should be called a solid as it possesses all the properties of a solid such as:
- Definite size and shape
- Intermolecular attraction between closely packed particles.
- It is rigid and cannot be compressed
d) Molecules in gases are loosely packed as compared to solid molecules which are densely packed.
Hence we are easily able to break the force of attraction when we move our hand through air but find it difficult to break through a solid (because of greater forces of attraction between molecules) which a karate expert is able to smash with the application of a lot of force.
4. Liquids generally have a lower density than solids. But you must have observed that ice floats on water. Find out why.
Solution:
Density of ice is less than the density of water. The low density of ice can be attributed to the small
pores it has which allows it to trap air hence ice floats on water.
Exercise-1.4 Page: 9
1. Convert the following temperature to Celsius scale:
a. 300K b. 573K
Solution:
a. 0°C=273K
300K= (300-273)°C = 27°C
b. 573K= (573-273)°C = 300°C
2. What is the physical state of water at:
a. 250°C b. 100°C ?
Solution:
(a) At 250°C – Gaseous state since it is beyond its boiling point.
(b) At 100°C – It is at the transition state as the water is at its boiling point. Hence it would be
present in both liquid and gaseous state.
3. For any substance, why does the temperature remain constant during the change of state?
Solution:
It is due to the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance hence the temperature stays constant.
4. Suggest a method to liquefy atmospheric gases.
Solution:
It can be achieved by either increasing the pressure or decreasing the temperature which ultimately leads to the reduction of spaces between molecules.
Exercise-1.5 Page:
1. Why does a desert cooler cool better on a hot dry day?
Solution:
It is because the temperature is high and it is less humid on a hot dry day which enables better evaporation. High levels of this evaporation provide better cooling effects.
2. How does the water kept in an earthen pot (matka) become cool during summer?
Solution:
An earthen pot is porous in nature. These tiny pores facilitate penetration of water and hence their evaporation from the pot surface. The process of evaporation requires energy which is contributed by water in the pot as a result of which water turns cooler.
3.Why does our palm feel cold when we put on some acetone or petrol or perfume on it?
Solution:
Acetone, petrol, and perfume are volatile substances that get evaporated when they come in contact
with air. Evaporation is facilitated as it uses energy from palm hence leaving a cooling effect on our
palms.
4. Why are we able to sip hot tea or milk faster from a saucer rather than a cup?
Solution:
A saucer has a larger surface area than a cup which promotes quicker evaporation hence the tea or milk in a saucer cools down faster.
5. What type of clothes should we wear in summer?
Solution:
In summer, it is preferred to wear light-coloured cotton clothes because light colour reflects heat and cotton materials have pores that absorb sweat, facilitating their evaporation hence causing a cooling effect in the skin.
Exercise Page: 12
1. Convert the following temperature to Celsius scale.
(a) 293K (b) 470K
Solution:
0°C=273K
(a) 293K= (293 – 273)°C = 20°C
(b) 470K= (470 – 273)°C = 197°C
2.Convert the following temperatures to the Kelvin scale.
(a) 25°C (b) 373°C
Solution:
0°C = 273K
(a) 25°C = (25+273)K = 298K
(b) 373°C = (373+273)K = 646K
3. Give reason for the following observations:
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume while sitting several metres away.
Solution:
(a) At room temperature, naphthalene balls undergo sublimation wherein they directly get converted
from a solid to a gaseous state without having to undergo the intermediate state, i.e., the liquid state.
(b) Molecules of air move at a higher speed and have large intermolecular spaces. Perfumes comprise
of flavoured substances that are volatile which scatters quickly in air, becoming less concentrated over a distance. Hence we are able to smell perfume sitting several metres away.
4. Arrange the following in increasing order of forces of attraction between the particles – water, sugar, oxygen.
Solution:
Oxygen (gas) < water (liquid) < sugar (solid)
5. What is the physical state of water at –
(a) 25°C (b) 0°C (c) 100°C?
Solution:
(a) At 25°C, the water will be in liquid form (normal room temperature)
(b) At 0°C, the water is at its freezing point, hence both solid and liquid phases are observed.
(c) At 100°C, the water is at its boiling point, hence both liquid and gaseous state of water (water
vapour) are observed.
6.Give two reasons to justify –
(a) Water at room temperature is a liquid.
(b) An iron almirah is a solid at room temperature.
Solution:
(a) Transition in the states of matter of water occurs at 0°C and 100°C. At room temperature, water
is in the liquid state, thereby exhibiting all the properties of a liquid such as
- Water flows at this temperature
- It has a fixed volume and it takes the shape of its container
(b) The melting and boiling points of iron are as high as 1538°C and 2862°C respectively. The room
temperature is about 20-25 °C. Hence iron almirah is a solid at room temperature.
7. Why is ice at 273K more effective in cooling than water at the same temperature?
Solution:
Water at this temperature(273K) is less effective than ice as ice can readily form water through absorption of ambient heat energy as opposed to water which does not exhibit this property as it already possesses additional latent heat of fusion so does not require extra heat. Hence ice cools rapidly compared to water at the same temperature.
8. What produces more severe burns, boiling water or steam?
Solution:
Steam produces severe burns. It is because it is an exothermic reaction that releases a high amount of heat which it had consumed during vaporization.
9. Name A, B, C, D, E and F in the following diagram showing a change in its state.
Solution:
A: Melting (or) fusion (or) liquefaction
B: Evaporation (or) vaporization
C: Condensation
D: Solidification
E: Sublimation
F: Deposition
| Also Access |
| NCERT Exemplar Solutions for Class 9 Science Chapter 1 |
| CBSE Notes for Class 9 Science Chapter 1 |
NCERT Solutions for Class 9 Science Chapter 1 – Matter in Our Surroundings
Chapter 1 – Matter in Our Surroundings is a part of Unit 1: Matter – Its Nature and Behavior. According to the past trends and previous years question papers, this particular unit carries 23 marks out of 100. Therefore, it is quite important to ensure that this chapter is studied thoroughly.
The topics and Subtopics from NCERT Solutions Class 9 Science Chapter 1- Matter in Our Surroundings are given below:
- Physical nature of matter
- Matter is Made Up of Particles
- How Small Are These Particles Of Matter?
- Characteristic of particles of matter
- Particles of Matter Have Space between Them
- Particles of Matter Are Continuously Moving
- Particles of Matter Attract Each Other
- States of matter
- The Solid State
- The Liquid State
- The Gaseous State
- Can matter change its state?
- Effect of Change of Temperature
- Effect of Change o Pressure
- Evaporation
- Factors Affecting Evaporation
- How Does Evaporation Cause Cooling?
- Physical nature of matter
Matter is one of the fundamental constituents that make up everything in the universe – from minute sand particles on earth to the enigmatic black holes at the centre of many galaxies. Matter has a role to play in everything that we see around us, interacting to form new materials, some familiar and others exotic.
Explore how matter works and discover its molecular components. Also, learn how the term matter was coined and its significance in various fields of science. Find more important NCERT Solutions For Class 9 Science to aid your studies.
Exercises with Question count covered in NCERT Solutions for Class 9 Chapter 1:
Exercise 1.1 & 1.2, Page number 3 – Solution of 4 Questions
Exercise 1.3, Page number 6 – Solution of 4 Questions
Exercise 1.4, Page number 9 – Solution of 4 Questions
Exercise 1.5, Page number 10 – Solution of 5 Questions
Chapter Exercise, Page number 12 – Solution of 9 Questions
Key Features of NCERT Solutions for Class 9 Science Chapter 1 – Matter in Our Surroundings
- Content elaborated in detail, ensuring all jargons are explained
- Solutions have been written in an easy-to-understand language
- Crafted by qualified teachers and industry experts
- Includes questions from the latest prescribed syllabus
- Comprehensive analysis of previous year exam questions
- Explore additional learning tools such as sample papers and previous year question papers
More to Explore: NCERT Solutions Class 9
Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 1
Explain the different characteristics of state of matter covered in the Chapter 1 of NCERT Solutions for Class 9 Science.
1. Shape
2. Volume
3. Rigidity or Fluidity
4. Intermolecular force
5. Intermolecular space
6. Compressibility
These concepts are briefly explained in the NCERT Solutions for Class 9 Science Chapter 1 curated by the experts at BYJU’S. The solutions are elaborated in a simple language to make it easier for the students while learning.
How many questions are present in each exercise of NCERT Solutions for Class 9 Science Chapter 1?
Exercise 1.1 & 1.2 – 4 Questions
Exercise 1.3 – 4 Questions
Exercise 1.4 – 4 Questions
Exercise 1.5 – 5 Questions
Chapter Exercise – 9 Questions
Is the NCERT Solutions for Class 9 Science Chapter 1 sufficient for the exam preparation?
Chapter 2
1. What is meant by a substance?
Solution:
It is a pure single form of matter. A substance has definite properties and compositions. Example – Iron
2. List the points of differences between homogeneous and heterogeneous mixtures.
Solution:
| Homogeneous mixture | Heterogeneous mixture |
| Particles are uniformly distributed throughout the mixture | All the particles are completely mixed and can be distinguished with the bare eyes or under a microscope. |
| Has a uniform composition | Irregular composition |
| No apparent boundaries of division | Noticeable boundaries of division. |
Exercise-2.2 Page: 18
1. Differentiate between homogenous and heterogeneous mixtures with examples.
Solution:
The following are the differences between heterogeneous and homogenous mixtures.
| Heterogeneous mixture | Homogeneous mixture |
| All the particles are completely mixed and can be distinguished with the bare eyes or under a microscope. | Particles are uniformly distributed throughout the mixture |
| Irregular composition | Has a uniform composition |
| Noticeable boundaries of division. | No apparent boundaries of division |
| Example: seawater, blood, etc. | Example: rainwater, vinegar, etc. |
2. How are sol, solution and suspension different from each other?
Solution:
| Attributes | Sol | Solution | Suspension |
| Type of Mixture | Heterogeneous | Homogeneous | Heterogeneous |
| Size of particles | 10-7 – 10-5 cm | Less than 1nm | More than 100nm |
| Tyndall effect | Exhibited | Not exhibited | May or may not be exhibited |
| Appearance | Usually glassy and clear | Unclouded and clear | Cloudy and opaque |
| Visibility | Visible with an ultramicroscope | Not visible | Visible with naked eye |
| Diffusion | Diffuses very slowly | Diffuses rapidly | Do not diffuse |
| Stability | Pretty stable | Highly stable | unstable |
| Settling | Get settled in centrifugation | Do not settle | Settle on their own |
| Example | Milk, blood, smoke | Salt solution, Sugar solution | Sand in water, dusty air |
3. To make a saturated solution, 36g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Solution:
Mass of solute (NaCl) = 36 g
Mass of solvent (H2O) = 100 g
Mass of solution (NaCl + H2O) = 136 g
Concentration = Mass of solute/Mass of solution x 100
Concentration = 36/136 x 100 = 26.47%
Hence, the concentration of the solution is 26.47%
Exercise-2.3 Page: 24
1. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?
Solution:
A technique known as simple distillation can be used to separate the mixture of miscible liquids, where the difference in boiling point is more than 25°C, to name a few – kerosene and petrol. The whole concept is established on the volatility property of substances. The following are the various steps in the process of simple distillation:
(a) In a distillation flask, take the mixture.
(b) Treat the mixture with heat while a thermometer is affix.
(c) We observe evaporation of petrol as it has a low boiling point.
(d) As the vapours advance towards the condenser, a dip in the temperature causes condensation of the vapours into liquid which can be accumulated in a flask.
(e) We notice that kerosene tends to remain in the flask in a liquid state due to comparatively higher boiling point.
(f) Consequently, the liquids are separated.
2. Name the techniques used to separate the following:
(a) Butter from curd.
(b) Salt from seawater
(c) Camphor from salt
Solution:
a) A process known as centrifugation is used to separate butter from curd. The process is governed on the principle of density.
b) We can use the simple evaporation technique to separate salt from seawater. Distillation causes water to evaporate leaving solid salt behind, hence the production of salt.
c) Sublimation can be used to separate camphor from salt as during the phase change, camphor does not undergo a liquid phase.
3. What type of mixtures are separated by the technique of crystallization?
Solution:
The technique of crystallization is used to separate solids from a liquid solution. It is linked to precipitation, but in this technique, the precipitate is achieved in a crystal form which exhibits extremely high levels of purity. The principle of crystallization can be applied to purify impure substances.
Exercise-2.4 Page: 24
1. Classify the following as physical or chemical changes:
- Cutting of trees
- Melting of butter in a pan
- Rusting of almirah
- Boiling of water to form steam
- Passing of electric current through water and water breaking into hydrogen and oxygen gases.
- Dissolving common salt in water
- Making a fruit salad with raw fruits, and
- Burning of paper and wood
Solution:
The following is the classification into physical and chemical change
| Physical change | Chemical change |
|
|
2. Try segregating the things around you as pure substances and mixtures.
Solution:
Listed below are the classifications based on pure substances and mixtures:
| Pure substance | Mixture |
| Water | Soil |
| Salt | Salad |
| Iron | Air |
| Diamond | Steel |
Exercise Page: 28
1. Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
Solution:
(a) In water, sodium chloride in its solution can be separated through the process of Evaporation.
(b) The technique of sublimation is apt as Ammonium chloride supports Sublimation.
(c) Tiny chunks of metal pieces in engine oil of car can be manually filtered.
(d) Chromatography can be used for the fine segregation of various pigments from an extract of flower petals.
(e) The technique of centrifugation can be applied to separate butter from curd. It is based on the concept of difference in density.
(f) To separate oil from water which are two immiscible liquids which vary in their densities, separating funnel can be an effective method.
(g) Tea leaves can be manually separated from tea using simple filtration methods.
(h) Iron pins can be separated from sand either manually or with the use of magnets as the pins exhibit strong magnetic quality which can be a key characteristic hence taken into consideration.
(i) The differentiating property between husk and wheat is that there is a difference in their mass. If treated with a small amount of wind energy, a remarkable variation in the moving distance is noticed. Hence to separate them, the sedimentation/winnowing procedure can be applied.
(j) Due to the property of water, sand or fine mud particles tends to sink in the bottom as it is denser provided they are undisturbed. Through the process of sedimentation/decantation water can be separated from fine mud particles as the technique is established on obtaining clear water by tilting it out.
2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate, and residue.
Solution:
(a) Into a vessel, add a cup of milk which is the solvent, supply it with heat.
(b) Add tea powder or tea leaves to the boiling milk, which acts as a solute. Continue to heat
(c) The solute i.e., the tea powder remains insoluble in the milk which can be observed while it is still boiling.
(d) At this stage, add some sugar to the boiling solution while stirring
(e) Sugar is a solute but is soluble in the solvent
(f) Continuous stirring causes the sugar to completely dissolve in the tea solution hence reaching saturation.
(g) Once the raw smell of tea leaves is vanished and tea solution is boiled enough, take the solution off the heat, filter or strain it to separate tea powder and the tea solution. The insoluble tea powder remains as a residue while the solute (sugar) and the solvent (essenced milk solution) strain through the filter medium which is collected as the filtrate.
3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of a substance dissolved in 100 grams of water to form a saturated solution).
| Substance dissolved | Temperature in K | ||||
| 283 | 293 | 313 | 333 | 353 | |
| Solubility | |||||
| Potassium nitrate | 21 | 32 | 62 | 106 | 167 |
| Sodium chloride | 36 | 36 | 36 | 37 | 37 |
| Potassium chloride | 35 | 35 | 40 | 46 | 54 |
| Ammonium chloride | 24 | 37 | 41 | 55 | 66 |
(a) What mass of potassium nitrate would be needed to produce a saturated solution of
potassium nitrate in 50 grams of water at 313K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the
solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this
temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Solution:
(a) Given:
Mass of potassium nitrate required to produce a saturated solution in 100 g of water at 313 K = 62g
To find:
Mass of potassium nitrate required to produce a saturated solution in 50 g of water =?
Required amount = 62 x 50/100 = 31
Hence 31 g of potassium nitrate is required.
(b) The solubility of potassium chloride in water is decreased when a saturated solution of potassium chloride loses heat at 353 K. Consequently, Pragya would observe crystals of potassium chloride which would have surpassed it solubility at low temperatures.
(c) Listed below is the solubility of each salt at 293 K:
- Solubility of Potassium nitrate —> 32/100
- Solubility of Sodium chloride —> 36/100
- Solubility of Potassium chloride —> 35/100
- Solubility of Ammonium chloride —> 37/100
It is observed that the ammonium chloride salt has the highest amount of solubility when compared to any other salt at 293 K.
(d) Effect of change of temperature on the solubility of salts:
The table clearly depicts that the solubility of the salt is dependent upon the temperature and increases with an increase in temperature. With this, we can infer that when a salt arrives at its saturation point at a specific temperature, there is a propensity to dissolve more salt through an increase in the temperature of the solution.
4. Explain the following giving examples.
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) suspension
Solution:
(a) Saturated solution: It is that state in a solution at a specific temperature when a solvent is no more soluble without an increase in the temperature. Example: Excess carbon leaves off as bubbles from a carbonated water solution saturated with carbon.
(b) Pure substance: A substance is said to be pure when it comprises of only one kind of molecules, atoms or compounds without adulteration with any other substance or any divergence in the structural arrangement. Example: Sulphur, diamonds
(c) Colloid: A Colloid is an intermediate between solution and suspension. It has particles of various sizes, that ranges between 2 to 1000 nanometers. Colloids can be distinguished from solutions using the Tyndall effect. Tyndall effect is defined as the scattering of light (light beam) through a colloidal solution. Example: Milk, gelatin.
(d) Suspension: It is a heterogeneous mixture that comprises of solute particles that are insoluble but are suspended in the medium. These particles that are suspended are not microscopic but visible to bare eyes and are large enough (usually larger than a micrometre) to undergo sedimentation.
5. Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
Solution:
The following is the classification of the given substances into homogenous and heterogenous mixture.
| Homogenous mixture | Heterogeneous mixture |
| Soda water | wood |
| vinegar | soil |
| Filtered tea | |
| Air |
6. How would you confirm that a colourless liquid given to you is pure water?
Solution:
We can confirm if a colourless liquid is pure by setting it to boil. If it boils at 100°C it is said to be pure. But if there is a decrease or increase in the boiling point, we infer that water has added impurities hence not pure.
7. Which of the following materials fall into the category of “pure substance”?
(a)Ice
(b)Milk
(c)Iron
(d)Hydrochloric acid
(e)Calcium oxide
(f)Mercury
(g)Brick
(e)Wood
(f)Air.
Solution:
Following substances from the above-mentioned list are pure substances:
- Iron
- Ice
- Hydrochloric acid
- Calcium oxide
- Mercury
8. Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water
Solution:
The following are the solutions from the above-mentioned list of mixture:
- Sea water
- Air
- Soda water
9. Which of the following will show the “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.
Solution:
Tyndall effect is exhibited by only milk and starch solution from the above-mentioned list of solutions.
10. Classify the following into elements, compounds and mixtures.
(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood.
Solution:
| Elements | Compounds | Mixture |
| Sodium | Calcium carbonate | Soil |
| Silver | Carbon dioxide | Sugar solution |
| Tin | Methane | Coal |
| Silicon | Air | |
| Blood | ||
| Soap |
11. Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of candle
Solution:
Out of the given, the following are chemical changes:
Growth of plant, rusting of iron, cooking of food, digestion of food and burning of candle.3
Chapter3
Class 9 Science Chapter 3 Exercise-3.1 Questions with Answer
Exercise-3.1 Page: 32
1. In a reaction, 5.3g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water
Solution:
Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water
5.3g 6g 8.2g 2.2g 0.9g
As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of
products
As per the above reaction, LHS = RHS i.e., 5.3g + 6g = 2.2g + 0.9 g + 8.2 g = 11.3 g
Hence the observations are in agreement with the law of conservation of mass.
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 g of hydrogen gas?
Solution:
We know hydrogen and water mix in the ratio 1: 8.
For every 1g of hydrogen, it is 8g of oxygen.
Therefore, for 3g of hydrogen, the quantity of oxygen = 3 x 8 = 24g
Hence, 24g of oxygen would be required for the complete reaction with 3g of hydrogen gas.
3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Solution:
The postulate of Dalton’s Atomic theory which is a result of the law of conservation of mass is,
“Atoms can neither be created nor destroyed”.
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Solution:
The postulate of Dalton’s atomic theory that can explain the law of definite proportions is – the
relative number and kinds of atoms are equal in given compounds.
Exercise-3.2 Page: 35
1. Define the atomic mass unit?
Solution:
An atomic mass unit is a unit of mass used to express weights of atoms and molecules where one
atomic mass is equal to 1/12th the mass of one carbon-12 atom.
2. Why is it not possible to see an atom with naked eyes?
Solution:
Firstly, atoms are miniscule in nature, measured in nanometers. Secondly, except for atoms of noble
gasses, they do not exist independently. Hence, an atom cannot be visible to the naked eyes.
Exercise-3.3-3.4 Page: 39
1. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Solution:
The following are the formulae:
(i) sodium oxide – Na2O
(ii) aluminium chloride – AlCl3
(iii) sodium sulphide – Na2S
(iv) magnesium hydroxide – Mg (OH)2
2. Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3.
Solution:
Listed below are the names of the compounds for each of the following formulae
(i) Al2(SO4)3 – Aluminium sulphate
(ii) CaCl2 – Calcium chloride
(iii) K2SO4 – Potassium sulphate
(iv) KNO3 – Potassium nitrate
(v) CaCO3 – Calcium carbonate
3. What is meant by the term chemical formula?
Solution:
Chemical formula is the symbolic representation of a chemical compound. For example: The chemical formula of hydrochloric acid is HCl.
4. How many atoms are present in a
(i) H2S molecule and
(ii) PO43- ion?
Solution:
The number of atoms present are as follows:
(i) H2S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in totality.
(ii) PO43- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in totality.
Exercise-3.5.1-3.5.2 Page: 40
1. Calculate the molecular masses of H2 , O2 , Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Solution:
The following are the molecular masses:
The molecular mass of H2 – 2 x atoms atomic mass of H = 2 x 1u = 2u
The molecular mass of O2 – 2 x atoms atomic mass of O = 2 x 16u = 32u
The molecular mass of Cl2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u
The molecular mass of CO2 – atomic mass of C + 2 x atomic mass of O = 12 + ( 2×16)u = 44u
The molecular mass of CH4 – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u
The molecular mass of C2H6– 2 x atomic mass of C + 6 x atomic mass of H = (2 x 12) +
(6 x 1)u=24+6=30u
The molecular mass of C2H4– 2 x atomic mass of C + 4 x atomic mass of H = (2x 12) +
(4 x 1)u=24+4=28u
The molecular mass of NH3– atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u= 17u
The molecular mass of CH3OH – atomic mass of C + 3x atomic mass of H + atomic mass of O + atomic mass of H = (12 + 3×1+16+1)u=(12+3+17)u = 32u
2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u,
Na = 23 u, K=39u, C = 12u, and O=16u.
Solution:
Given:
Atomic mass of Zn = 65u
Atomic mass of Na = 23u
Atomic mass of K = 39u
Atomic mass of C = 12u
Atomic mass of O = 16u
The formula unit mass of ZnO= Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u
The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u
The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u
Exercise-3.5.3 Page: 42
1. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?
Solution:
Given: 1 mole of carbon weighs 12g
1 mole of carbon atoms = 6.022 x 1023
Molecular mass of carbon atoms = 12g = an atom of carbon mass
Hence, mass of 1 carbon atom = 12 / 6.022 x 1023 = 1.99 x 10-23g
2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23u, Fe = 56 u)?
Solution:
Given: Atomic mass of Na=23u, Atomic mass of Fe= 56u
To calculate the number of atoms in 100g of sodium:
23g of Na contains = 6.022 x 1023 atoms
1g of Na contains = 6.022 x 1023 atoms / 23
100g of Na contains = 6.022 x 1023 atoms x 100 / 23
= 2.6182 x 1024 atoms
To calculate the number of atoms in 100g of sodium:
56g of Fe contains = 6.022 x 1023 atoms
1g of Fe contains = 6.022 x 1023 atoms / 56
100g of Fe contains = 6.022 x 1023 atoms x 100 / 56
= 1.075 x 1024 atoms
Hence, through comparison, it is evident that 100g of Na has more atoms.
Exercise Page: 43
1. A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Solution:
Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g
To calculate percentage composition of the compound:
Percentage of boron = mass of boron / mass of the compound x 100
= 0.096g / 0.24g x 100 = 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 = 60%
2. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Solution:
11.00g of carbon dioxide is formed when 3.00g carbon is burnt in 8.00g of oxygen.
Carbon and oxygen are combined in the ratio 3:8 to give carbon dioxide using up all the carbon and
oxygen
Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon is formed, the
left oxygen is unused i.e., 50-8=42g of oxygen is unused.
This depicts the law of definite proportions – The combining elements in compounds are present in
definite proportions by mass.
3. What are polyatomic ions? Give examples.
Solution:
Polyatomic ions are ions that contain more than one atom but they behave as a single unit
Example: CO32-, H2PO4–
4. Write the chemical formula of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Solution:
The following are the chemical formula of the above-mentioned list:
(a) Magnesium chloride – MgCl2
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)2
(d) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3
5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Solution:
The following are the names of the elements present in the following compounds:
(a) Quick lime – Calcium and oxygen (CaO)
(b) Hydrogen bromide – Hydrogen and bromine (HBr)
(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3)
(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4)
6. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
Solution:
Listed below is the molar mass of the following substances:
(a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g
(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8 x 32 = 256g
(c) Molar mass of Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g
(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g
(e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+
3×16 = 63g
7. 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)?
Solution:
The mass of the above mentioned list is as follows:
(a) Atomic mass of nitrogen atoms = 14u
Mass of 1 mole of nitrogen atoms= Atomic mass of nitrogen atoms
Therefore, mass of 1 mole of nitrogen atom is 14g
(b) Atomic mass of aluminium =27u
Mass of 1 mole of aluminium atoms = 27g
1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g
(c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3 x Mass of O = (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g
Therefore, mass of 10 moles of Na2SO3 = 10 x 126 = 1260g
8. Convert into mole.
(a) 12g of oxygen gas
(b) 20g of water
(c) 22g of carbon dioxide
Solution:
Conversion of the above-mentioned molecules into moles is as follows:
(a) Given: Mass of oxygen gas=12g
Molar mass of oxygen gas = 2 Mass of Oxygen = 2 x 16 = 32g
Number of moles = Mass given / molar mass of oxygen gas = 12/32 = 0.375 moles
(b) Given: Mass of water = 20g
Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g
Number of moles = Mass given / molar mass of water
= 20/18 = 1.11 moles
(c) Given: Mass of carbon dioxide = 22g
Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g
Number of moles = Mass given/ molar mass of carbon dioxide = 22/44 = 0.5 moles
9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Solution:
The mass is as follows:
(a) Mass of 1 mole of oxygen atoms = 16u, hence it weighs 16g
Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2u
(b) Mass of 1 mole of water molecules = 18u, hence it weighs 18g
Mass of 0.5 moles of water molecules = 0.5 x 18 = 9u
10. Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.
Solution:
To calculate molecular mass of sulphur:
Molecular mass of Sulphur (S8) = 8xMass of Sulphur = 8×32 = 256g
Mass given = 16g
Number of moles = mass given/ molar mass of sulphur
= 16/256 = 0.0625 moles
To calculate the number of molecules of sulphur in 16g of solid sulphur:
Number of molecules = Number of moles x Avogadro number
= 0.0625 x 6.022 x 10²³ molecules
= 3.763 x 1022 molecules
11. Calculate the number of aluminium ions present in 0.051g 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 = 27u)
Solution:
To calculate the number of aluminium ions in 0.051g of aluminium oxide:
1 mole of aluminium oxide = 6.022 x 1023 molecules of aluminium oxide
1 mole of aluminium oxide (Al2O3) = 2 x Mass of aluminium + 3 x Mass of Oxygen
= (2x 27) + (3 x16) = 54 +48 = 102g
1 mole of aluminium oxide = 102g = 6.022 x 1023 molecules of aluminium oxide
Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 1023 / 102
= 3.011 x 1020 molecules of aluminium oxide
One molecule of aluminium oxide has 2 aluminium ions, hence number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 1020 molecules of aluminium oxide
= 6.022 x 1020
Chapter 4
Exercise-4.1 Page: 47
1. What are the canal rays?
Solution:
The radiations that are positively charged are canal rays. This discovery was crucial in the discovery
of another subatomic particle that was positively charged – proton.
2. If an atom contains one electron and one proton, will it carry any charge or not?
Solution:
Since a proton is a positively charged particle and an electron is a negatively charged particle, the net
charge becomes neutral as both the particles neutralizes each other.
Exercise-4.2 Page: 49
1. On the basis of Thompson’s model of an atom, explain how the atom is neutral as a whole.
Solution:
As per Thompson’s model of an atom,
(i) An atom contains a positively charged sphere in which the negatively charged electrons are implanted.
(ii) Electrons and protons are equal in magnitude, hence an atom on the whole is electrically neutral.
2. On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an atom?
Solution:
As per Rutherford’s model of an atom, the positively charged protons are the ones that are present in the atom.
3. Draw a sketch of Bohr’s model of an atom with three shells.
Solution:
4. What do you think would be the observation if the ∝– particle scattering experiment is carried out using a foil of a metal other than gold?
Solution:
In the ∝ – particle scattering experiment, when any other metal foil is used instead of gold, the observation would remain the same. This is because the structure of an atom when considered individually remains the same.
Exercise-4.2.4 Page: 49
1. Name the three subatomic particles of an atom.
Solution:
An atom consists of three subatomic particles:
- Protons – positively charged
- Electrons – negatively charged
- Neutrons – neutral in nature ( no charge )
2. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Solution:
Given: Atomic mass of helium atom = 4u, 2 protons in helium nucleus
Atomic mass = number of protons + number of neutrons
4 = 2 + number of neutrons
Number of neutrons = 4 – 2 = 2
Hence, Helium has 2 neutrons.
Exercise-4.3 Page: 50
1. Write the distribution of electrons in Carbon and Sodium atoms.
Solution:
Distribution of electrons in Carbon atoms:
The atomic number of Carbon is 6
Number of electrons is equal to the number of protons in carbon atom i.e., 6
The distribution of electrons in carbon atom is K – 2, L – 4
Distribution of electrons in sodium atoms:
The atomic number of Sodium is 11
Number of electrons is equal to the number of protons in sodium atom i.e., 11
The distribution of electrons in sodium atom is K – 2, L – 8, M – 1
2. If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Solution:
K shell can hold 2 electrons
L shell can hold 8 electrons
Hence, when both the shells are full, the total number of electrons present in the atom = 2+8 = 10 electrons.
Exercise-4.4 Page: 52
1. How will you find the valency of chlorine, sulphur and magnesium?
Solution:
The definite combining capacity of the atoms of each element, wherein electrons are lost, gained or shared to make the octet of electrons present in the outermost shell is defined as valency. To measure valency, we can figure out the number of electrons that are required to complete the shell in which it is contained or losing excess electrons if present, once the filling is complete.
To find the valency of chlorine:
The atomic number of chlorine is 17
Number of electrons is equal to the number of protons in chlorine i.e., 17
The distribution of electrons in chlorine atom is K – 2, L – 8, M – 7
Hence, from the distribution of chlorine it is clearly evident that to fill the M shell only one electron is required. Therefore its valency is -1. i.e, one electron less
To find the valency of sulphur:
The atomic number of sulphur is 16
Number of electrons is equal to the number of protons in sulphur i.e., 16
The distribution of electrons in sulphur atom is K – 2, L – 8, M – 6
Hence, from the distribution of sulphur it is clearly evident that to fill the M shell two more electrons are required. Therefore its valency is -2, i.e., two electrons lesser.
To find the valency of magnesium:
The atomic number of magnesium is 12
Number of electrons is equal to the number of protons in magnesium i.e., 12
The distribution of electrons in magnesium atom is K – 2, L – 8, M – 2
Hence, from the distribution of magnesium it is clearly evident that to fill the M shell six more electrons are required. But M shell has two electrons only. It possesses lesser electrons than needed to fill the shell.
Thus, we say that the magnesium atom is not stable as the M shell has 2 electrons. Its valency is +2, meaning it has 2 electrons in excess.
Exercise-4.5 Page: 52
1. If the number of electrons in an atom is 8 and number of protons is also 8, then
(i) What is the atomic number of the atom? and
(ii) What is the charge on the atom?
Solution:
Given: Number of electrons = 8
Number of protons = 8
(i) The atomic number of an atom is the same as the number of protons in that atom, hence its atomic number is 8.
(ii) In an atom, the number of protons is equal to the number of electrons. Hence both the charges – positive and negative neutralize each other. Therefore, the atom does not possess any charge.
2. With the help of given Table, find out the mass number of oxygen and sulphur atom.
Table: Composition of Atoms of the First Eighteen Elements with Electron Distribution in Various Shells.
| Name of Element | Symbol | Atomic number | Number of Protons | Number of Neutrons | Number of electrons | Distribution of electrons K L M N | Valency | |||
| Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium | H He Li Be B C N O F Ne Na Mg | 1 2 3 4 5 6 7 8 9 10 11 12 | 1 2 3 4 5 6 7 8 9 10 11 12 | – 2 4 5 6 6 7 8 10 10 12 12 | 1 2 3 4 5 6 7 8 9 10 11 12 | 1 2 2 2 2 2 2 2 2 2 2 2 | – – 1 2 3 4 5 6 7 8 8 8 | – – – – – – – – – – 1 2 | – – – – – – – – – — – – | 1 0 1 2 3 4 3 2 1 0 1 2 |
| Aluminium Silicon Phosphorus Sulphur Chlorine Argon | Al Si P S Cl Ar | 13 14 15 16 17 18 | 13 14 15 16 17 18 | 14 14 16 16 18 22 | 13 14 15 16 17 18 | 2 2 2 2 2 2 | 8 8 8 8 8 8 | 3 4 5 6 7 8 | – – – – – | 3 4 3,5 2 1 0 |
Solution:
(a) To find the mass number of Oxygen:
Number of protons = 8
Number of neutrons = 8
Atomic number = 8
Atomic mass number = Number of protons + number of neutrons = 8 + 8 = 16
Therefore, mass number of oxygen = 16
(b) To find the mass number of Sulphur:
Number of protons = 16
Number of neutrons = 16
Atomic number = 16
Atomic mass number = Number of protons + number of neutrons = 16 + 16 = 32
Exercise-4.6 Page: 53
1. For the symbol H, D and T, tabulate three subatomic particles found in each of them.
Solution:
The following table depicts the subatomic particles in Hydrogen (H), Deuterium (D), and Tritium(T).
| Isotope | Symbol | Mass no. | Atomic no. | No. of electrons | No. of protons | No. of neutrons |
| Hydrogen | H | 1 | 1 | 1 | 1 | 0 |
| Deuterium | D | 2 | 1 | 1 | 1 | 1 |
| Tritium | T | 3 | 1 | 1 | 1 | 2 |
2. Write the electronic configuration of any one pair of isotopes and isobar.
Solution:
(a) Isotopes: Isotopes are atoms which have the same number of protons but the number of neutrons differs. This leads to the variation in mass number too.
Example: Carbon molecule exists as 6C12 and 6C14 but when their electronic configuration is noticed, both have K-2; L-4
(b) Isobars: Isobars are atoms which have the same mass number but differ in the atomic number. Electronic configuration of an isobar pair is as follows,
Example: Electronic configuration of 20Ca40 – K-2; L-8; M-8; N- 2
Electronic configuration of 18Ar40 – K-2; L-8; M-8
Exercise Page: 54
1. Compare the properties of electrons, protons and neutrons.
Solution:
| Property | Electrons | Protons | Neutrons |
| Charge | Negatively charged | Positively charged | No charge. |
| Location | Located outside the nucleus | Located within the nucleus | Located inside the nucleus of an atom |
| Weight | Mass is negligible | 1 a.m.u | 1 a.m.u |
| Affinity | Attracted towards positively charged | Attracted towards negatively charged | Do not get attracted to any charged particle |
2. What are the limitations of J.J.Thomson’s model of the atom?
Solution:
The following are the limitations of the J.J. Thomson’s model of an atom.
- The model failed to explain the outcome of alpha particle scattering which was conducted by Rutherford. The model failed to depict why majority of these alpha particles pass through gold foil while some are diverted through small and big angles, while some others rebound completely, returning back on their path.
- It did not provide any experimental evidence and was established on imagination.
3. What are the limitations of Rutherford’s model of the atom?
Solution:
Following are the limitations of Rutherford’s model of the atom:
- There is no expected stability in the revolution of the electron in a circular orbit
- Charged particles radiate energy when accelerated thus causing the revolving electrons to lose energy and would fall into the nucleus
- Hence atoms must be highly unstable. Matter would not exist in their known form which clearly is an assumption as atoms are highly stable.
4. Describe Bohr’s model of the atom.
Solution:
- An atom holds the nucleus at the centre.
- Negatively charged electrons revolve around the nucleus.
- The atoms in it contains distinct orbits of electrons.
- Electrons do not radiate energy when they are in their orbits.
- The distinct orbits are named as K, L, M, N orbits. Numbers used to denote them are n=1, 2, 3, 4
5. Compare all the proposed models of an atom given in this chapter.
Solution:
| Thomson | Rutherford | Bohr |
| ● Sphere is positively charged ●Electrons are negatively charged and scattered all through the inside of the sphere. ● Positively charged = negatively charged ● The net charge in the atom is zero. | ● The nucleus is at the centre and is positively charged holding the entire mass. ● Electrons are negatively charged revolving in a well-defined path ●In comparison with the nucleus, the size of the atom is very large. | ●Nucleus is present at the centre and is positively charged ● Electrons are negatively charged, revolving around but do not radiate energy. ● The distinct orbits are labelled as K, L, M, N |
6. Thomson’s Model of Atom
7. Rutherford’s Model of Atoms
8. Bohr’s model of the atom
Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Solution:
- Maximum number of electrons that can be accommodated in a shell is given by the formula: 2n2, where n= 1, 2, 3…
- Maximum number of electrons in different shells are:
K shell – n=1 ; 2n2 = 2(1)2 = 2
L shell – n=2 ; 2n2 = 2(2)2 = 8
M shell – n=3 ; 2n2 = 2(3)2 = 18
N shell- n=4 ; 2n2 = 2(4)2 = 32
- The outermost orbit can be accommodated with 8 electrons at the maximum.
- The electrons are not taken in unless the inner shells are filled which are filled step-wise, hence the highest element has K-2; L-8 ; M-8 distribution of electrons.
9. Define valency by taking examples of silicon and oxygen.
Solution:
The definite combining capacity of the atoms of each element, wherein electrons are lost, gained or shared to make the octet of electrons present in the outermost shell is defined as valency. To measure valency, we can figure out the number of electrons that are required to complete the shell in which it is contained or losing excess electrons if present, once the filling is complete.
Example : To find the valency of silicon:
The atomic number of silicon is 14
Number of electrons is equal to the number of protons in silicon i.e., 14
The distribution of electrons in silicon atom is K – 2, L – 8, M – 4
Hence, from the distribution of silicon it is clearly evident that to fill the M shell 4 electrons are required. Therefore its valency is 8-4=4.
To find the valency of oxygen:
The atomic number of oxygen is 8
Number of electrons is equal to the number of protons in oxygen i.e., 8
The distribution of electrons in oxygen atom is K – 2, L – 6
Hence, from the distribution of oxygen it is clearly evident that to fill the M shell 6 more electrons are required. Therefore its valency is 8-6=2.
10. Explain with examples
(i) Atomic number,
(ii) Mass number,
(iii) Isotopes and
(iv) Isobars.
Give any two uses of isotopes.
Solution:
(i) The number of positively charged protons present in the nucleus of an atom is defined as the atomic number and is denoted by Z. Example: Hydrogen has one proton in its nucleus, hence its atomic number is one.
(ii) The total number of protons and neutrons present in the nucleus of an atom is known as the mass number. It is denoted by A. 20Ca40 . Mass number is 40. Atomic number is 20.
(iii) The atoms which have the same number of protons but different number of neutrons are referred to as isotopes. Hence the mass number varies.
Example: The most simple example is the Carbon molecule which exists as 6C12 and 6C14
(iv) Isobars: Isobars are atoms which have the same mass number but differ in the atomic number.
Examples are, 20Ca40and 18Ar40
Uses of isotopes:
- The isotope of Iodine atom is used to treat goitre and iodine deficient disease.
- In the treatment of cancer, an isotope of cobalt is used.
- Fuel for nuclear reactors is derived from the isotopes of the Uranium atom.
11. Na+ has completely filled K and L shells. Explain.
Solution:
The atomic number of sodium is 11. It has 11 electrons in its orbitals wherein the number of protons is equal to the number of electrons. Hence, its electronic configuration is K-2 ; L-8 ; M-1 ; The one electron in the M shell is lost and it obtains a positive charge since it has one more proton than electrons, and obtains a positive charge, Na+ . The new electronic configuration is K-1 ; L-8 which is the filled state. Hence it is very difficult to eliminate the electron from a filled state as it is very stable.
12. If bromine atom is available in the form of, say, two isotopes 35Br79 (49.7%) and 35Br81 (50.3%), calculate the average atomic mass of Bromine atom.
Solution:
The atomic masses of two isotopic atoms are 79 (49.7%) and 81 (50.3%).
Thus, total mass = (79 * 49.7 / 100) + (81 * 50.3 / 100) = 39.263 + 40.743 = 80.006 u
13. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 8X16 and 8X18 in the sample?
Solution:
Let the percentage of 8X16 be ‘a’ and that of 8X18 be ‘100-a’.
As per given data,
16.2u = 16 a / 100 + 18 (100-a) /100
1620 = 16a + 1800 – 18a
1620 = 1800 – 2a
a = 90%
Hence, the percentage of isotope in the sample 8X16 is 90% and that of
8X18 = 100-a = 100- 90=10%
14. If Z=3, what would be the valency of the element? Also, name the element.
Solution:
Given: Atomic number, Z = 3
The electronic configuration of the element = K-2; L-1, hence its valency = 1
The element with atomic number 3 is Lithium.
15. Composition of the nuclei of two atomic species X and Y are given as under
X Y
Protons = 6 6
Neutrons = 6 8
Give the mass numbers of X and Y. What is the relation between the two species?
Solution:
Mass number of X: Protons + neutrons = 6+6 = 12
Mass number of Y: Protons + neutrons = 6+8 = 14
They are the same element as their atomic numbers are the same.
They are isotopes as they differ in the number of neutrons and hence their mass numbers.
16. For the following statements, write T for true and F for false.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore it is neutral.
(c) The mass of an electron is about 1/2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Solution:
(a) Statement is False
(b) Statement is False
(c) Statement is True
(d) Statement is False
17. Put a tick(✓) against correct choice and cross(x) against wrong choice in questions 15, 16 and 17.
Rutherford’s alpha – particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron
Solution:
(a) Atomic nucleus
Isotopes of an element have
(a) The same physical properties
(b) Different chemical properties
(c) Different number of neutrons
(d) Different atomic numbers.
Solution:
(c) Different number of neutrons
18. Number of valence electrons in Cl– ion are:
(a) 16
(b) 8
(c) 17
(d) 18
Solution:
(b) 8
Electronic distribution of Cl is K-2, L-8, M-7. Valence electrons are 7, hence chlorine gains one electron for the formation of Cl–. Therefore, its valency is 8.
19. Which one of the following is a correct electronic configuration of Sodium?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1
Solution:
(d) 2, 8, 1
Complete the following table.
| Atomic number | Mass number | Number of neutrons | Number of Protons | Number of electrons | Name of the atomic species |
| 9 16 – – – | – 32 24 2 1 | 10 – – – 0 | – – 12 1 1 | – – – – 0 | – Sulphur – – – |
Solution:
The following table depicts the missing data:
Atomic number(Z) =Number of protons
Mass number = Number of neutrons + atomic number
(or)
Mass number(A) = Number of neutrons + number of neutrons
| Atomic number | Mass number | Number of neutrons | Number of Protons | Number of electrons | Name of the atomic species |
| 9 16 12 1 1 | 19 32 24 2 1 | 10 16 12 1 0 | 9 16 12 1 1 | 9 16 12 1 0 | Fluorine Sulphur Magnesium Deuterium Hydrogen |
| Also Access |
| NCERT Exemplar Solutions for Class 9 Science Chapter 4 |
| CBSE Notes for Class 9 Science Chapter 4 |
NCERT Solutions for Class 9 Science Chapter 4 – Structure Of The Atom
NCERT Class 9 Science Chapter 4 – Structure of the atom is categorized under Unit I – Matter – Its nature and behaviour which fundamentally deals with the physical nature of matter and what constitutes the particles that make up for matter. The whole unit of Matter makes up for 23 marks out of which Structure of the atom chapter is allocated 10 marks. In order to score maximum marks, students practise NCERT Class 9 Science Solutions.
This chapter has a good weightage. Thorough knowledge of all the concepts covered in this chapter will help you achieve up to 10 marks from this chapter alone. Around 2 questions with 5 marks each appear in the examination from this chapter and are expected to appear in the following year too.
List of subtopics covered in Class 9 Science Chapter 4 – Structure Of The Atom:
| Number | Subtopic |
| 4.1 | Charged particles in matter |
| 4.2 | The structure of an atom |
| 4.2.1 | Thomson’s model of an atom |
| 4.2.2 | Rutherford’s model of an atom |
| 4.2.3 | Bohr’s model of an atom |
| 4.2.4 | Neutrons |
| 4.3 | How are electrons distributed in different orbits(shells)? |
| 4.4 | Valency |
| 4.5 | Atomic number and mass number |
| 4.5.1 | Atomic number |
| 4.5.2 | Mass number |
| 4.6 | Isotopes |
| 4.6.1 | Isobars |
List of Exercises in Class 9 Science Chapter 4
Number 4.1 – Charged particles in matter 2 Question ( 2 short)
Number 4.2 – The structure of an atom 4 Question ( 4 short)
Number 4.2.4 – Neutrons 2 Question ( 2 short)
Number 4.3 – How are electrons distributed
indifferent orbits(shells)? 2 Question ( 1 long, 1 short)
Number 4.4 – Valency 1 Question ( 1 long)
Number 4.5 – Atomic number and mass number 2 Question ( 2 long)
Number 4.6 – Isotopes 2 Question ( 1 long, 1 short)
Exercise Solutions – 19 Question ( 6 long, 9 short, 4 MCQ)
0 Comments