CBSE Class 9

NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 5 – The Fundamental Unit of Life solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 5 – The Fundamental Unit of Life Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

• NCERT Solutions for Class 9 Science
• HOTS Questions for Class 9 Science
• Value Based Questions in Science for Class 9
• NCERT Exemplar Solutions for Class 9 Science
• Previous Year Question Papers for CBSE Class 9 Science

NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
Who discovered cell, and how ?
Answer:
Robert Hooke (1665). He thinly pared a piece of cork and observed it under his self made primitive microscope. The scientist found that cork possesses a number of small box-like structures which he named cells (cellulae which later’abbreviated to cells). His work was published in the form of a book called Micrographia.

Question 2.
Why is cell called structural and functional unit of life ?
Answer:
Structural Unit: A living organism is made up of one or more cells. Therefore, cell is structural unit of life. Functional Unit. All life functions of an organism reside in its cells. Cells may also become specialised to perform specific functions like contraction in muscle cell or impulse transmission in nerve cell. Therefore, cells are functional units of life.

Question 3.
How do substances like CO₂ and water move into and out of the cell ? Discuss. (CCE 2011)
Answer:
CO₂ moves into and out of cells by diffusion while water does it through osmosis.
Diffusion. It is movement of particles of various substances from the region of their higher concentration to the region of their lower concentration,
(i) In a respiring cell, more CO₂ is produced internally. As a result its internal concentration rises. As concentration of CO₂ is lower in the outside medium, CO₂ passes out from cell into external medium,
(ii) In photosynthetic cell, CO₂ is being consumed in photosynthesis. Its intracellular concentration is lower than outside medium. Therefore, CO₂ diffuses from outside to inside of the cell. Osmosis. It is movement of water from the region of its higher concentration (pure water or dilute solution) to the region of its lower concentration (strong solution) when the two are separated by a semipermeable membrane. Plasma membrane functions as a semipermeable membrane. Cell sap functions as strong solution. Therefore, external water enters the cell (endosmosis) till wall pressure counter-balances this tendency. If the external medium has a very strong solution, water would pass out from the cell into the external medium. The phenomenon is called exosmosis.

Question 4.
Why is plasma membrane called selectively permeable membrane ? (CCE 2012)
Answer:
Cell membrane is semipermeable membrane for water. It permits the entry of gases through diffusion. Ions, sugar, amino acids, etc. pass through the plasma membrane by an active process. Plasma membrane is impermeable to certain other materials. Therefore, it is selectively permeable.

Question 5.
Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.

Answer:
2. (Left side), (i) Poorly defined due to absence of nuclear envelope (ii) Nucleoid.
4. (Right side). Membrane bound cell organelles are present.

Question 6.
Can you name the two organelles, we have studied that contain their own genetic material ?
Answer:
Yes. Mitochondria and plasdds.

Question 7.
If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen ?
Answer:
Lysosome will burst to release digestive enzymes. Digestive enzymes will cause breakdown of various cellular components causing destruction of the cell.

Question 8.
Why are lysosomes known as suicide bags ? (CCE 2011, 2012, 2013)
Answer:
Lysosomes contain digestive enzymes against all types of organic materials. If their covering membrane breaks as it happens during injury to cell, the digestive enzymes will spill over the cell contents and digest the same. As lysosomes are organelles which on bursting can kill the cells possessing them, they are called suicide bags.

Question 9.
Where are proteins synthesised inside the cell ?
Answer:
Proteins are synthesized over the ribosomes.

NCERT Exercise

Question 1.
Make a comparison to write down ways in which plant cells are different from animal cells.
Answer:

Question 2.
How is prokaryotic cell different from eukaryotic cell ?
Answer:

Question 3.
What would happen if the plasma membrane ruptures or breaks down ?
Answer:
There will be spilling of cytoplasm and cell organelles, bursting of lysosomes and digestion of cellular contents.

Question 4.
What would happen to the life of a cell if there was no Golgi apparatus ? (CCE 2011)
Answer:
There would not be any lysosome for intracellular digestion and cleansing, no complexing of molecules, no excretion and no formation of new plasma membrane.

Question 5.
Which organelle is known as power house of the cell ? Why ? (CCE 2011, 2013)
Answer:
Mitochondrion is known as power house of the cell because it produces most of the molecules of ATP (adenosine triphosphate) which are required for providing energy for synthesis of new chemicals, mechanical and other cellular functions.

Question 6.
Where do the lipids and proteins constituting the cell membrane get synthesised ?
Answer:
Proteins are synthesised over ribosomes of RER while lipids are synthesised over SER.

Question 7.
How does Amoeba obtain food ?
Answer:
Plasma membrane of Amoeba is flexible. With its help, Amoeba engulfs food particle. The engulfed food particle passes into the body of Amoeba as a phagosome. Phagosome combines with lysosome to produce digestive or food vacuole. Digestion occurs in food vacuole. The digested food passes into surrounding cytoplasm. The undigested matter is thrown out of the cell in exocytosis.

Question 8.
What is osmosis ?
Answer:
Osmosis is diffusion of water from the region of its higher concentration (pure water or dilute solution) to the region of its lower concentration (strong solution) through a semipermeable membrane.

Question 9.
Carry out the following osmosis experiment.
Take four peeled potato halves and hollow each one out to make potato cups. One of these potato cups should be made from the boiled potato. Put each potato cup in the trough containing water. Now
(a) Keep cup A empty,
(b) Put one tea spoon sugar in cup B.
(c) Put one tea spoon of salt in cup C.
(d) Put one tea spoon sugar in boiled cup D.
Keep this set up for two hours. Then observe the four potato cups and answer the following :
(i) Explain why water gathers in the hollowed portion of B and C.
(ii) Why is potato A necessary for this experiment ?
(iii) Explain why water does not gather in the hollowed out portion of A and D.
Answer:
(i) Sugar and salt increases osmotic concentration which results in passage of water osmotically from the trough through the cells of potato B and C into its cavity.
(ii) Potato A functions as control experiment which indicates that the cavity of potato does not induce movement of water. Water does not gather in the hollowed out portion of A because it does not have a higher osmotic concentration than the cells of potato tuber.
(iii) Potato tuber D does not have living cells. Osmosis does not occur in dead cells. Therefore, despite presence of sugar in the cavity of D, no water passes from trough through dead potato cells into cavity of the tuber.

PRACTICAL BASED TWO MARKS QUESTIONS

Question 1.
Three students ‘A’, ‘B’ and ‘C were given five raisins each of equal mass. The raisins were soaked in distilled water at room temperature. ‘A’ soaked the raisins for 10 minutes, ‘B’ for overnight and ‘C for 60 minutes. They calculated the percentage of water absorbed by raisins. Now answer the following question :
(a) Name the student whose raisins will show the maximum percentage of water absorbed.
(b) Name the student whose raisins will show the minimum percentage of water absorbed. (CCE 2013)
Answer:
(a) Maximum Percentage ofWater Absorbed. Raisins of ‘B’ student.
(b) Minimum Percentage of Water Absorbed. Raisins of ‘A’ student.

Question 2.
A teacher soaked 10 g raisins in 35 ml of distilled water in a beaker A and a similar amount in beaker B. She maintained the temperature of beaker A at 20°C and beaker B at 40°C. After an hour, compare the percentage of water absorbed by raisins in beakers A and B. (CCE 2013, 2014)
Answer:
More water was absorbed in beaker B where temperature was 40°C. Permeability of cell membrane increases from 0°C to 40°C beyond which dénaturation sets in.

Question 3.
5g of raisins were placed in distilled water for 24 hours. The mass of water soaked raisins was found to be 7g. Calculate the percentage of water absorbed by raisins. (CCE 2013)
Answer:

Question 4.
A student puts five raisins in two beakers A and B. Beaker A contained 50 ml of distilled water at room temperature and beaker B had 50 ml of ice cold water. After some time what will be the observation of the student ? State reason for this observation.
(CCE 2013, 2015)
Answer:
Raisins in beaker A swell up. Those of beaker B do not. Low temperature of B water reduced membrane permeability as well as kinetic energy of water for osmosis.

Question 5.
A student recorded a mass of dry raisins as 6.0 g and mass of raisins after soaking them in water for about four hours as 10.5 g. Calculate the percentage of water absorbed by raisins. (CCE 2013)
Answer:

Question 6.
Write the mathematic equation used to determine the mass percentage of water absorbed by raisins.
Answer:
If initial weight is W₁ and final weight after soaking is W₂

Question 7.
Ravi took weight of five dry raisins and five swollen raisins of approximately equal size. If the weight of dry raisins was 7 g and weight of swollen raisins is ‘X’ g, then
(i) Write the formula to calculate the percentage of water absorbed by the raisins and
(ii) If the value of ‘X’ is 10.5 g, then what will be the percentage of water absorbed by the raisins ? (CCE 2014)
Answer:

Question 8.
Write four main steps of the method involved in an experiment “On determination of the percentage of water absorbed by raisins in the laboratory”. (CCE 2014)
Answer:
(i) Selection of raisins with intact stalks and their weight as W₁
(ii) Soaking the raisins in water at room temperature for at least 1-2 hours
(iii) Taking out wet raisins, wiping out water with blotting paper and their final weight as W₂

Question 9.
In the experiment of determining the percentage of water absorbed by raisins, two students Samya and Zahira used the following formula respectively

Which student used wrong formula ? What are W₁ and W₂ in the correct formula ? (CCE 2014)
Answer:
Student (i) used the wrong formula.
W₁ is initial weight of dry raisins. W₂ is final weight of swollen raisins.

Question 10.
(a) A student recorded the mass of dry raisins as 4.0 g and mass of raisins after soaking as 7.0 g. Calculate the percentage of water absorbed by raisins. (CCE 2014, 2016)
(b) Mention one application of the phenomenon of osmosis in plants. (CCE 2016)
Answer:

(b) Application of Osmosis. Turgidity of cells, root absorption.

Question 11.
(a) Ram while doing an experiment to find out the percentage of water absorbed by raisins measured the mass of dry raisins as 50 g. He soaked the raisins in water for four hours and again measured the mass as 80 g. Calculate the percentage of water absorbed by the raisins.
(b) He then placed raisins in concentrated salt solution. What will he observe ? (CCE 2015)
Answer:

(b) After Placing in Concentrated Salt Solution. Raisins will lose water and shrink to the maximum.

Question 12.
A student recorded the mass of dry raisins as 6.0 g and mass of raisins after soaking them in water for about four hours as 10.5 g. Calculate the percentage of water absorbed by raisins. Why do raisins get swelled up ? (CCE 2015)
Answer:

The raisins swell up due to absorption of water through osmosis.

Question 13.
A student recorded the following observations in an experiment for finding the percentage of water absorbed by raisins.
(i) Mass of water taken in beaker – 50 g
(ii) Mass of dry raisins before soaking water – 20 g
(iii) Mass of raisins after soaking water – 30 g.
(iv) Mass of remaining water in beaker after experiment – 40 g
Calculate the percentage of water absorbed by raisins. ( CCE 2015)
Answer:

Question 14.
If ‘X’ is the initial mass of the raisins and ‘Y’ is the final mass of raisins after soaking in water, calculate the percentage of water absorbed by raisins. Name the process due to which raisins absorb water. (CBSE 2015, 2016)
Answer:

Question 15.
A group of students selected 10 raisins with stalks and weighed them using digital balance. Then, they soaked them for a few hours. The weight of the swollen raisins was 9.2 g which was 4.6 g more than weight of dry raisins. Calculate the water imbibed by the raisins. ( CCE 2015)
Answer:

Question 16.
In the experiment “To determine the mass percentage of water imbibed by raisins”, the raisins absorb water when kept in water for 5-6 hours. Why does water absorption take place ? What is the phenomenon called ? (CCE 2015, 2016)
Answer:
Water moves into raisins due to endosmosis. Endosmosis is entry of water into a system, cell or organ due to presence of hypertonic solution in it and its separation from pure water or dilute solution by a semipermeable membrance. Skin of raisins function as semipermeable membrane. There is high concentration of sugar inside them. Therefore, external water passes into raisins and cause their swelling.

Question 17.
Before placing the raisins in water, the raisins weighed 10 g. The raisins were taken out of water, wiped well and then the weight was found to be 12.5 g. Determine the percentage of water absorbed by raisins. Define the process due to which raisins absorb water. (CCE 2016)
Answer:

Osmosis: Osmosis is the diffusion of water or solvent across a semipermeable membrane (which does not allow passage of solutes) from a region of its higher concentration to the region of its lower concentration. Plasma membrane functions as semipermeable membrane. There will be osmotic entry of water into cell if the external solution is pure water or dilute as compared to cell sap.

Question 18.
(a) In the experiment to determine the percentage of water absorbed by raisins, the raisins are wiped before weighing. Why ?
(b) While preparing a temporary stained mount of onion peel, Veena added a drop of glycerine. Why ? ( CCE 2016)
Answer:
(a) To remove unabsorbed water sticking to the surface of raisins.
(b) To prevent drying of onion peel.

Question 19.
A student took x gram water in a beaker and dipped p gram raisins in it. After keeping raisins in water for 2 hours, he measured the mass of soaked raisins as q grams. He also measured the mass of water left in the beaker which was y grams. On the
Answer:

SELECTION TYPE QUESTIONS

basis of his observations write correct formula to find the percentage of water absorbed by raisins. Mention the process due to which weight of raisins increased. (CCE 2016)

Alternate Response Type Questions
(True/False, Right/Wrong, Yes/No)

Question 1.
Cork comes from bark.
Question 2.
Robert Brown discovered protoplasm in 1831.
Question 3.
Amoeba has an everchanging shape.
Question 4.
Movement of a substance from the area of low concentration to an area of high concentration is called diffusion.
Question 5.
A dilute soludon is called hypertonic solution.
Question 6.
Lysosomes keep the cells clean by digesting foreign materials and worn out cell organelles.
Question 7.
SER detoxifies many poisons and drugs.
Question 8.
Central vacuole occupies 10-20% of cell volume.

Matching Type Questions :

Question 9.
Match the contents of the columns I and II (single matching)

Question 10.
Match the contents of columns I, II and III (double matching)

Question 11.
Which type of metabolism, anabolism (A) and catabolism (C) are performed by the following organelles (key or check list items) :

Question 12.
Match the stimulus with Appropriate Response.

Fill In the Blanks

Question 13. Cristae create a large surface area for …………… generating reactions.
Question 14. Plant cell wall is mainly composed of ……………..
Question 15. Same organelles perform …………… function in all organisms.
Question 16. Cell theory was proposed by …………. and ……………. .
Question 17. Cells were discovered by Robert Hooke in ………….. .

Answers:

SOME TYPICAL QUESTIONS

Question 1.
What is the junctional unit of life ? Define it.
Answer:
Cell is the functional unit of life. It can be defined as a tiny mass of protoplasm covered by plasma membrane which is capable of performing all functions of life.

Question 2.
What is the difference between plasma membrane and cell wall i Give the functions of each one. (CCE 2011)
Answer:
Plasma membrane is an elastic living membrane made up of lipids and proteins, whereas cell wall is a rigid non-living covering made up of cellulose.
Function of Plasma membrane. It acts as a selectively semipermeable membrane which allows only selective substances to pass through it.
Function of Cell Wall. It provides rigidity and protection to cell.

Question 3.
Differentiate between chromatin and chromosome.
Answer:
Chromatin. It is the nucleoprotein fine fibrous mass which stains strongly with basic dyes and is present as a network inside the nucleus.
Chromosome. Rod-like, stainable, condensed chromatin unit, visible at cell division and containing hereditary information in the form of genes.

Question 4.
Differentiate between RER and SER.
Answer:

Question 5.
Which type of ribosomes are found in prokaryotes and eukaryotes ?
Answer:
Prokaryotes have 70 S ribosomes and eukaryotes have 80 S ribosomes.

Question 6.
Which structure is called little nucleus ?
Answer:
Nucleolus.

Question 7.
Why is nucleus called director of the cell ?
Answer:
Nucleus controls and coordinates all the metabolic functions of the cell.

NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life

Hope given NCERT Solutions for Class 9 Science Chapter 5 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

Practical Based Questions for Class 9 Science Chemistry

Question 1.
Mention the temperature of the following in degree Celsius and Kelvin Scales

1. Ice and ice cold water
2. Boiling water and steam.

Answer:

1. 6°C and 273K
2. 100°C and 373K.

Question 2.
Rima took fine chalk powder, egg albumin, starch powder and alum powder in four test tubes. A, B, C and D respectively. After adding water to all the four test tubes, identify the test tubes as true solution, suspension and colloid. (CBSE 2013)
Answer:
Test tube A : Suspension; Test Tube B : Colloidal solution; Test Tube C : Colloidal solution; Test Tube D : True solution.

Question 3.
A student added water to sand and starch in different test tubes. How will you differentiate between the two on the basis of transparency ? (CBSE 2013)
Answer:
The solution of sand in water is a suspension and is opaque. On the other hand, the solution of starch in water is colloidal and is transleucent. This means that the light can pass partially through the second solution. (CBSE 2013)

Question 4.
List two precautions you must take while finding the melting point of ice.. (CBSE 2013)
Answer:

1. The bulb of the thermometer must not be dipping in ice. It must touch ice.
2. The constant stirring must be done during the process of melting.

Question 5.
Dipti was asked to prepare four separate mixtures in four beakers A, B, C and D by mixing sugar, fine sand, thin paste of starch and chalk powder respectively in water and then categorise each as stable or unstable. What will be correct categorization ?
Answer:

Question 6.
Identify two clear and transparent solutions from the following mixtures :
(a) Milk and water
(b) Sugar and water
(c) Starch powder and water
(d) Glucose and water.
(CBSE 2013)
Answer:
(b) Sugar and water
(d) Glucose and water.

Question 7.
In an experiment to determine the boiling point of water, state the reason for the following precautions :

1. The bulb of the thermometer should not touch the sides of the beaker.
2. While boiling water, pumice stone should be added. (CBSE 2013)

Answer:

1. The sides of the glass beaker are at higher temperature than the contents of the beaker. Therefore, the reading given by the thermometer touching the sides of the beaker does not give correct result.
2. Pieces of pumice stone are added to boiling water in order to check any bumping.

Question 8.
Four students A, B, C and D were given funnels, filter paper, test tubes, test tube stands, common salt, chalk powder, starch and glucose powder. They prepared the true solution, suspension and colloidal solutions. Test tubes were arranged as shown in the figure. Observe the filtrate obtained in the test tubes and residue on filter paper. Conclude about filtrate residue and type of solution. (CBSE 2013)
Answer:

Question 9.
If in the determination of melting point of ice, the ice is contaminated with some non-volatile impurity like common salt, how is the melting point of ice affected ? (CBSE 2013)
Answer:
Melting point of ice gets lowered. Impurities always lower the melting point temperature of solid.

Question 10.
A mixture of sand, powdered glass and common salt is dissolved in water and then filtered. Name the substance left on the filter paper. Name the substance in the filtrate. (CBSE 2013, 2016)
Answer:
Sand and powdered glass are left as residue on the filter paper. Common salt (sodium chloride) solution in water constitutes the filtrate.

Question 11.
In an experiment to determine the boiling point of water, mention two important precautions. (CBSE 2013)
Answer:

1. Constant stirring must be done so that heating may be uniform.
2. Bulb of the thermometer must not dip in water.

Question 12.
Four students A, B, C and D are asked to prepare colloidal solutions. The following diagrams show the preparation done by them. Name the student who will be able to prepare colloidal solution. Write two properties of colloidal solutions. (CBSE 2014)

Answer:
Only student A’ has prepared the colloidal solution. Egg as such does not mix with water. Only white of an egg forms colloidal solution on stirring in water. Both sugar and common salt form true solution in water. The two properties of colloidal solution are :

1. Colloidal solutions are of heterogeneous nature consisting of dispersed phase and dispersion medium.
2. Colloidal solutions show Tyndall effect.

Question 13.
Rima took fine chalk powder, egg albumin, starch powder and alum powder in four test tubes A, B, C and D respectively. After adding water to all the four test tubes, identify the test tubes as true solution, suspension and colloid. (CBSE 2014)
Answer:
Test tube (A) contains a suspension
Test tube (B) contains a colloidal solution
Test tube (C) contains a colloidal solution
Test tube (D) contains a true solution.

Question 14.
What is a suspension ? Give two characteristics of suspension ? (CBSE 2015)
Answer:
A suspension may be defined as a heterogeneous mixture in which the solid particles are spread throughout the liquid without dissolving in it. They settle as precipitate if the suspension is left undisturbed for sometime.
Two characteristics: 

1. A suspension is of heterogeneous nature. There, are two phases. The solid particles represent one phase while the liquid in which these are suspended or distributed forms the other phase.
2. The particle size in a suspension is more than 100 nm .

Question 15.
List two properties of a true solution. How would you prepare a true solution. List two different solutes which will form a true solution. (CBSE 2015)
Answer:

1. A true solution is always homogeneous in nature.
2. A true solution is transparent in nature.

A true solution can be prepared by dissolving water soluble solute such as sugar/sodium chloride in water. Both sugar and sodium chloride form true solution when dissolved separately in water.

Question 16.
While doing an experiment to determine the melting point of ice, state the role of glass stirrer and mention the correct position of bulb of the thermometer. (CBSE 2015)
Answer:
Stirring the contents of the beaker with a glass stirrer keeps the heating uniform. The position of the thermometer should be such the bulb is just touching the ice cubes.

Question 17.
Rima took fine chalk powder, egg albuminm starch powder and alum powder in four test tubes A, B, C and D respectively. After adding water to all the four test tubes, identify the test tubes as true solution, suspension and colloid. (CBSE 2015)
Answer:
True solution is formed in test tube D. Test tube A contains suspension while colloidal solutions are formed in test tubes B and C.

Question 18.
While determining the melting point of ice, it was observed that even when ice cubes were being moderately heated using the gas burner, the temperature did not rise for sometime till the whole ice melted. Give the possible reason.  (CBSE 2015)
Answer:
Once the melting of ice, starts upon heating with the help of a gas burner, the temperature becomes constant till the entire ice melts to form water. In fact the heat energy now absorbed is used up to overcome the intermoleular foces or is used as latent heat of fusion. Therefore, the temperature does not rise.

Question 19.
In an experiment to verify the law of Conservation of Mass in a chemical reaction, four students A, B, C and D noted down the following observations for the difference in the mass of apparatus before and after the chemical reaction;
A – 4 g ; B- 8 g ; C – zero g ; D – 10 g
Which student has made the correct observation and why ?
Answer:
Student C has made the correct observation since according to the law, there is no change in mass of reactants and products taking part in a chemical reaction.

Question 20.
When 15g of copper sulphate react with 15g of sodium hydroxide, 20g of sodium sulphate along with copper hydroxide is formed. What is the mass of copper hydroxide formed ?
Answer:
The chemical reaction taking place is :

On the basis law of conservation of mass :
Mass of copper hydroxide (x) = (15 g + 15 g) – (20 g) = 10 g

Question 21.
To verify the law of conservation of mass in a chemical reaction, four students A, B, C and D perfomed the following chemical reactions in the school laboratory.
(A) Added zinc granules to dil hydrochloric acid
(B) Heated lead nitrate (solid) in a test tube
(C) Dipped Mg ribbon in copper sulphate solution
(D) Added barium chloride solution to sodium sulphate solution.
Which student according to you is likely to obtain the best results and why ?
Answer:
Student (D) is likely to obtain the best results because the reaction is completed immediately and a white precipitate is formed.

Question 22.
16.8 g of sodium hydrogen carbonate are added to 12.0 g of acetic acid. The residue left weighed 20.0 g. What is the mass of CO₂ escaped in the reaction ?
Answer:
The chemical reaction taking place is :

Mass of CO₂ released = (16.8 g +12.0 g ) – (20.0 g) = 8.8 g.

Question 23.
While studying the properties of cathode rays in a discharge tube, a student placed a mica wheel mounted on an axle in the path of the rays. What would happen to the. wheel ? What conclusions can be drawn from it ? (CBSE 2016)
Answer:
The mica wheel would start rotating around the axle. This shows that the cathode rays are made of material particles (electrons).

NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

Multiple Choice Questions

Question 1.
Which of the following correctly represents the electonic distribution in the Mg atom ?
(a) 3, 8, 1         (b) 2, 8, 2
(c) 1, 8, 3         (d) 8, 2, 2.
Correct Answer:
(b).

More Resources

• NCERT Exemplar Solutions for Class 9 Science
• NCERT Solutions for Class 9 Science
• Value Based Questions in Science for Class 9
• HOTS Questions for Class 9 Science
• Previous Year Question Papers for CBSE Class 9 Science

Question 2.
Rutherfords alpha (a) particles scattering experiment resulted in to discovery of :
(a) Electron
(b) Proton
(c) Nucleus in the atom
(d) Atomic mass.
Correct Answer:
(c).

Question 3.
The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element ?

Correct Answer:
(a) Mass no. = no. of electrons (protons) + no. of neutrons = 15 + 16 = 31.

Question 4.
Daltons atomic theory successfully explained :
(i) Law of conservation of mass
(ii) Law of constant composition
(iii)Law of radioactivity
(iv) Law of multiple proportion
(a) (i), (ii) and (iii)         (b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)       (d) (i), (ii) and (iv).
Correct Answer:
(d) Except for the law of radioactivity, Dalton’s atomic theory explained all other laws which have been listed.

Question 5.
Which of the following statements about Rutherford’s model of atom are correct ?
(i) Considered the nucleus as positively charged
(ii) Established that the α-particles are four times as heavy as a hydrogen atom
(iii) Can be compared to solar system
(iv) Was in agreement with Thomson’s model
(a) (i) and (iii)         (b) (ii) and (iii)
(c) (i) and (iv)         (d) only (i).
Correct Answer:
(a) The statements (i) and (iii) are both correct.

Question 6.
Which of the following are true for an element ?
(i) Atomic number = number of protons + number of electrons
(ii) Mass number = number of protons + number of neutrons
(iii) Atomic Mass = number of protons + number of neutrons
(iv) Atomic number = number of protons = number of electrons
(a) (i) and (ii)           (b) (i) and (iii)
(c) (ii) and (iii)         (d) (ii) and (iv).
Correct Answer:
(d) The statements (it) and (iv) are both correct.

Question 7.
In the Thomson’s model of atom, which of the following statements are correct ?
(i) The mass of the atom is assumed to be uniformly distributed over the atom
(ii) The positive charge is assumed to be uniformly distributed over the atom
(iii) The electrons are uniformly distributed in the positively charged sphere
(iv) The electrons attract each other to stabilise the atom
(a) (i), (ii) and (iii)             (b) (i) and (iii)
(c) (i) and (iv)                   (d) (i), (iii) and (iv).
Correct Answer:
(a) Except for the statements (iv), all other statements are correct.

Question 8.
Rutherford’s α-particle scattering experiment showed that
(i) electrons have negative charge
(ii) the mass and positive charge of the atom is concentrated in the nucleus
(iii) neutron exists in the nucleus
(iv) most of the space in atom is empty Which of the above statements are correct ?
(a) (i) and (iii)        (b) (ii) and (iv)
(c) (i) and (iv)        (d) (iii) and (iv).
Correct Answer:
(b) The statements (ii) and (iv) are correct.

Question 9.
The ion of an element has 3 positive charges. Mass , number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion ?
(a) 13      (b) 10
(c) 14      (d) 16.
Correct Answer:
(b) The ion has 13 protons (27-14) and 10 electrons. It is Al³⁺ ion.

Question 10.
Identify the Mg²⁺ ion from the given figure where n and p represent the number of neutrons and protons respectively

Correct Answer:
(d) Since the ion has 10 electrons, it is Mg²⁺ ion.

Question 11.
In a sample of methyl ethanoate (CH₃COOCH₃) the two oxygen atoms have the same number of electrons but different number of neutrons. Which of the following is the correct reason for it ?
(a) One of the oxygen atoms has gained electrons
(b) One of the oxygen atoms has gained two neutrons
(c) The two oxygen atoms are isotopes
(d) The two oxygen atoms are isobars.
Correct Answer:
(c) The two oxygen atoms are the isotopes. These are the different atoms of the same element which have same number of electrons (or atomic number) but different number of neutrons.

Question 12.
Elements with valency 1 are :
(a) always metals
(b) always metalloids
(c) either metals or non-metals
(d) always non-metals.
Correct Answer:
(c) Elements with valency 1 may be metals (+1 valency) or non-metals (-1 valency)

Question 13.
The first model of an atom was given by :
(a) N. Bohr
(b) E. Goldstein
(c) Rutherford
(d) J.J. Thomson.
Correct Answer:
(d).

Question 14.
An atom with 3 protons and 4 neutrons will have a valency of :
(a) 3         (b) 7
(c) 1         (d) 4.
Correct Answer:
(c) The atom has also 3 electrons with electronic configuration 2, 1. It has valency of 1.

Question 15.
The electronic distribution in an aluminium atom is :
(a) 2, 8, 3         (b) 2, 8, 2
(c) 8, 2, 3         (d) 2, 3, 8.
Correct Answer:
(d).

Question 16.
Which of the following do not represent Bohr’s model of an atom correctly ?

(a) (i) and (ii)           (b) (ii) and (iii)
(c) (ii) and (iv)          (d) (i) and (iv).
Correct Answer:
(c) The first shell (K) cannot have more than two electrons in it.

Question 17.
Which of the following statements is always correct ?
(a) An atom has equal number of electrons and protons.
(b) An atom has equal number of electrons and neutrons.
(c) An atom has equal number of protons and neutrons.
(d) An atom has equal number of electrons, protons and neutrons.
Correct Answer:
(a) The number of electrons or protons in an atom are always equal.

Question 18.
Atomic models have been improved over the years. Arrange the following atomic models in the order of their chronological order :
(i) Rutherford’s atomic model
(ii) Thomson’s atomic model
(iii) Bohr’s atomic model
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (i)
(c) (ii), (i) and (iii)
(d) (iii), (ii) and (i).
Correct Answer:
(c). It is the correct order.

Short Answer Questions

Question 19.
An element has one proton, one electron and no neutron. Name the element. How will you represent it ?
Answer:
The element is called protium or hydrogen (H). It can be represented as ¹₁H.

Question 20.
Write any two observations to show that atoms are divisible.
Answer:
The discovery of electrons and protons led to the opinion that the atoms are divisible. Even neutrons were also discovered at a later stage.

Question 21.
Will ³⁵Cl and ³⁷Cl have different valencies ? Justify your answer.
Answer:
The atomic number (Z) of the element (Cl) is 17. Its electronic distribution is 2, 8, 7. It has valency (8 – 7) = 1 which is the same for both the species, which act as isotopes.

Question 22.
Why did Rutherford select a gold foil in his α-scattering experiment ?
Answer:
Gold is a highly malleable. It could be converted into very fine foils or leaves. Moreover, the nucleus of the element is very heavy. It could not be displaced by the impact of fast moving α-particles.

Question 23.
Find out the valency of the atoms represented by the Fig. (a) and (b).
Answer:
(a) The electronic configuration of the atom is 2, 8, 8. It has completely filled K, L, M shells. Its valency is zero. The atom belongs to the element Argon (Ar).
(b) The electronic configuration of the atom is 2, 7. It has seven electrons in the valence shell. Its valency is (8 – 7) equal to one. The atom belongs to the element fluorine (F).

Question 24.
One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell ?
Answer:
Since the atom of the element X has lost one electron from the outermost shell, it has +1 unit charge.

Question 25.
Write down the electron distribution of chlorine atom. How many electrons are there in the L shell ? (Atomic number of chlorine is 17).
Answer:
The electronic distribution in chlorine (Z = 17) is K(2), L(8) and M(7). The L-shell has eight electrons.

Question 26.
In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed ?
Answer:
With six (6) electrons in the outermost shell, the atom of the element needs two (2) electrons more to have eight (8) electrons in the outermost shell. The ion (x^2-) has -2 units charge.

Question 27.
What information do you get from the following electronic distribution about the atomic number, mass number and valency of atoms X, Y and Z ? Give your answer in a tabular form.

Answer:

• In atom X, valency is equal to number of valence electrons (2, 3) = 3
• Atom Y has electronic configuration 2, 6. Valency is eight – No. of valence electrons (8-6)
• In atom Z, valency is equal to number of valence electrons (2, 8, 5), or eight – no. of valence electrons
  (8 – 5) = 3.

Question 28.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement ? Justify your answer.
Answer:
The statement is wrong. In an atom, number of protons and electrons are always equal. These can be different only in an ion (positive or negative).

Question 29.
Calculate the number of neutrons present in the nucleus of an element ³¹₁₅X which is represented as .
Answer:
Mass no. of the element (A) = 31
Atomic no. of the element (Z) =15
No. of neutrons (n) = A – Z = 31 – 15 = 16.

Question 30.
Match the names of the Scientists given in column A with their contributions towards the understanding of the atomic structure
as given in column B.

Answer:
(a) — (iii) ;
(b) — (iv);
(c) — (i) ;
(d) — (ii)
(e) — (vi) ;
(f) — (vii) ;
(g) — (v).

Question 31.
The atomic numbers of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements ?
Answer:
The elements having same mass number but different atomic numbers are known as isobars.

Question 32.
Complete the following table on the basis of information available in the symbols given below :

Answer:

Question 33.
Helium atom has 2 electrons in its valence shell but its valency is not 2. Explain.
Answer:
Helium (He) atom with atomic number (Z) equal to two has only one shell (K-shell). It can have a maximum of two electrons which it has. Therefore, the valence shell is complete and valency of the atom is zero.

Question 34.
Fill in the blanks in the following statements :
(a) Rutherfords α-particle scattering experiment led to the discovery of the .
(b) Isotopes have same but different
(c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be
and respectively.
(d) The electronic configuration of silicon is and that of sulphur is
Answer:
(a) nucleus
(b) atomic numbers, mass numbers
(c) zero and one
(d) 2, 8, 4 and 2, 8, 6.

Question 35.
An element X has a mass number 4 and atomic number 2. Write the valency of this element.
Answer:
The element X with mass number 4 and atomic number 2 is helium (He). It has 2 electrons (maximum possible) in its only shell which is K-shell. Therefore, the valency of the element is zero.

Long Answer Questions

Question 36.
Why do Helium, Neon and Argon have zero valency ?
Answer:
All the three elements have completely filled valence shells, according to Bohr Bury scheme. Therefore, their valencies are zero.

Question 37.
The ratio of the radii of hydrogen atom and its nucleus is 10⁵. Assuming the atom and the nucleus to be spherical,
(i) what will be the ratio of their sizes ?
(ii) If atom is represented by planet earth ‘R_e’ = 6.4 x 10⁶ m, estimate the radius of the nucleus.
Answer:

Question 38.
Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment.
Answer:
He made the following conclusions from the experiment.

1. As most of the alpha particles passed through undeflected, this means that they did not come across any obstruction in their path. Thus, most of the space in an atom is expected to be empty.
2. As a few alpha particles suffered minor deflections and a very few major deflections, this means that these must have met with some obstructions in their path.
3. This obstruction must be :
   1. Very small : Only a few particles were obstructed by it.
   2. Massive : Each alpha particle has 4 u mass and is quite heavy. It could easily pass through a light obstruction by pushing it aside.

Question 39.
In what way is the Rutherfords atomic model different from Thomsons atomic model ?
Answer:
According to Thomsons model, an atom may be regarded as a positively charged sphere containing protons in which the negatively charged protons are supposed to be studied or embedded. He gave no clue about the nucleus and extranuclear portion. This was given for the first time by Rutherfords model atom with the help of α-ray scattering experiment.

Question 40.
What were the drawbacks of Rutherfords model of an atom ?
Answer:
There are infact, two main drawbacks of Rutherford Model atom

1. Rutherford model of atom could not explain the stability of the atom.
2. Rutherford model of atom could not explain as to how the electrons are distributed in the extra nuclear portion in an atom.

Question 41.
What are the postulates of Bohr’s model of an atom ?
Answer:
The main postulates of the theory are listed :

1. In the extra nuclear portion of an atom, the electrons revolve in well defined circular paths known as orbits.
2. These circular orbits are also known as energy levels or energy shells.
3. These have been designated as K, L, M, N, O, … (or as 1, 2, 3, 4, 5, …) based on the energy present.
4. The order of the energy of these energy shells is :
   K<L<M<N<0 <…. or 1< 2< 3 < 4<5 <….
5. While revolving in an orbit, the electron is not in a position to either lose or gain energy. In other words, its energy remains stationary. Therefore, these energy states for the electrons are also known as stationary states.

Question 42.
Show diagramatically the electronic distribution in a sodium atom and a sodium ion and also give their atomic number.
Answer:
The atomic number of sodium (Na) is 11. The electronic distribution in the atom is K(2) L(8) and L(1). Sodium
ion (Na⁺) is formed by removal of one electron from the atom. Its electronic distribution is K(2) and L(8). This means that Na⁺ ion has the elecronic configuration of Ne atom which is inert gas atom. The electronic distribution of the atom and ion are shown as follows :

Question 43.
In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherfords model of an atom, 1.00% of the α-particles were found to deflect at angles > 50°. If one mole of α-particles were bombarded on the gold foil, compute the number of α-particles that would deflect at angles less than 50°.
Answer:
Percentage (%) of particles deflected at an angle more than 50° = 1%
Percentage (%) of α-particles deflected at an angle less than 50° = 100 – 1 = 99%

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.