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## Class 9 Science Chapter 3: Atoms and Molecules

NCERT Solutions for Class 9 Science Here we have given NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules.

## NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

INTEXT Questions

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Solution:
Total mass of reactants = mass of sodium carbonate + mass of ethanoic acid
= 5 – 3 g + 6 g = 11.3 g
Total mass of products = mass of sodium ethanoate + mass of carbon dioxide + mass of water
= 8.2 g + 2.2 g + 0.9 g = 11.3 g
Thus, the mass of reactants is equal to the mass of products, therefore the observations are in agreement with the law of conservation of mass.

Question 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:
∵ 1 g of hydrogen reacts with = 8 g of oxygen
∴ 3 g of hydrogen reacts with = 8 × 3 = 24 g of oxygen
Thus, 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Solution:
The postulate that “atoms can neither be created nor destroyed in a chemical reaction” is the result of the law of conservation of mass.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Solution:
The postulate that “A chemical compound always consists of the same elements combined together in the same proportion by mass” is the law of definite proportions.

Question 5.
Define the atomic mass unit.
Solution:
Atomic mass unit is defined as the mass unit equal to exactly one-twelfth (1/12th) of the mass of one atom of carbon-12. It is denoted by u (unified mass).
i.e. 1 u = 1.66 × 10-24 g

Question 6.
Why is it not possible to see an atom with naked eyes?
Solution:
It is not possible to see an atom with naked eye because of its extremely small size (atomic radius is of the order of 10-10 m).

Question 7.
Write down the formula of

1. sodium oxide’
2. aluminium chloride
3. sodium sulphide
4. magnesium hydroxide.

Solution:
Sodium oxide Aluminium chloride Sodium sulphide Magnesium hydroxide Question 8.
Write down the names of compounds represented by the following formulae:

1. AI2(S04)3
2. CaCI2
3. K2S04
4. KNO3
5. CaCO3

Solution:

(i) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate

Question 9.
What is meant by the term chemical formulae?
Solution:
The chemical formula of a compound is a symbolic representation of its composition. e.g., formula of calcium oxide. Thus, chemical formula of calcium oxide is CaO.

Question 10.
How many atoms are present in a
(i) H2S molecule and
(ii) ${ PO }_{ 4 }^{ 3- }$ ion?
Solution:
(i) 3 atoms because H2S molecule has two atoms of hydrogen and one atom of sulphur.
(ii) 5 atoms because ${ PO }_{ 4 }^{ 3- }$ ion has one atom of phosphorus and four atoms of oxygen.

Question 11.
Calculate the molecular masses of H2, O2 Cl2, CO2, CH4 C2H6, C2H4 NH3, CH3OH.
Solution: Question 12.
Calculate the formula unit masses of ZnO, Na2O, K2C03. Given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u,C = 12 u and,0= 16 u.
Solution:
Formula unit mass of ZnO = atomic mass of Zn + atomic mass of O Question 13.
If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Solution: Question 14.
Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given atomic mass of Na = 23 u, Fe = 56 u)?
Solution:

NCERT Exercises

Question 1.
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Solution:
We know that, % of any element in a Question 2.
When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.0 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Solution:

Total mass of reactants = total mass of products Hence, the law of conservation of mass is proved.
Further, it also shows carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3 : 8. Thus, it also proves the law of constant proportions. 3 g of carbon must also combine with 8 g of oxygen only. This means that (50 – 8) = 42 g of oxygen will remain un reacted.

Question 3.
What are polyatomic ions? Give examples.
Solution:
A polyatomic ion is a group of atoms carrying positive or negative charge. Question 4.
Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Solution:
(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate Question 5.
Give the names of the elements present in the following compounds:
(a) Quicklime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Solution:
(a) Quick lime is CaO. Elements present are calcium and oxygen.
(b) Hydrogen bromide is HBr. Elements present are hydrogen and bromine.
(c) Baking powder is NaHC03. Elements present are sodium, hydrogen, carbon and oxygen.
(d) Potassium sulphate is K2S04. Elements present are potassium, sulphur and oxygen.

Question 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, HCI
(e) Nitric acid, HN03
Solution:
(a) Molar mass of C2H2

(b) Molar mass of S = 8 × atomic mass of s = (8 × 32)u = 256 u Question 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 (Na2S03)?
Solution: Question 8.
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Solution: Question 9.
What is the mass of
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Solution:
(a) Mass of 1 mole of oxygen atom = 16 u Mass of 0.2 mole of oxygen atoms = (16 × 0.2) u = 3.2 u
(b) Mass of 1 mole of H20 molecule = Question 10.
Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur?
Solution:
Mass of 1 mole of S8 = 8 × 32 = 256

Question 11.

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
Solution:
Molar mass of Al2O3 = 2 × 27 + 3 × 16 # Chapter 4: Structure of the Atom

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

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

INTEXT Questions

Question 1.
What are canal rays?
Solution:
The beam of rays which travel in a direction away from anode towards cathode when a gas,taken in a discharge tube is subjected to the action of high voltage under low pressure are known as canal rays. It is also called anode rays. It was discovered by E. Goldstein in 1886.

Question 2.
If an atom contains one electron and one proton, will it carry any charge or not?
Solution:
No, the atom will not carry any charge because the electron has one negative charge (-1) and the proton has one positive charge (+1). They netutralise each other.

Question 3.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Solution:
According to Thomson’s model of an atom:

• An atom consists of a positively charged sphere and the electrons are embedded like the seeds in a water-melon.
• The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral.

Question 4.
On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom?
Solution:
Proton, a positively charged sub-atomic particle is present in the nucleus of an atom according to Rutherford’s model of atom.

Question 5.
Draw a sketch of Bohr’s model of an atom with three shells.
Solution:

Question 6.
What do you think would be the observation if the a-particle scattering experiment is carried out using a foil of a metal other than gold?
Solution:
On using the foil of heavy metals like gold (e.g., platinum, silver etc.) the observation will be same but if the foil is of light metal (e.g., sodium, magnesium etc.), the massive α-particles may push the nucleus aside and may not be deflected back.

Question 7.
Name the three sub-atomic particles of an atom.
Solution:
The three sub-atomic particles of an atom are :

• electron
• proton
• neutron.

Question 8.
Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Solution:
Mass number of helium = 4
Number of protons = 2
Number of neutrons (n)
= Mass number (A) – No. of protons (p) =4-2=2
Thus, no. of neutrons = 2

Question 9.
Write the distribution of electrons in carbon and sodium atoms.
Solution:
Atomic number of carbon = 6 Hence first shell (K-shell) have 2 electrons and the remaining 4 electrons will be present in the second shell, i.e. L-shell. Thus the distribution will be Atomic number of sodium = 11. Hence, first shell (K-shell) will have 2 electrons and second shell (L-shell) will have 8 electrons and third shell (M-shell) will have 1 electron. Thus, the distribution will be Question 10.
lf K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Solution:
The maximum number of electrons . present in a shell = 2n2 where n – shell number Value of n for K shell = 1
∴ Maximum number of electrons in K shell = 2n2 = 2(1)2 = 2 Value of n for L shell = 2
∴ Maximum number of electrons in L shell = 2(2)2 = 8 Thus, total no. of electrons = 2 + 8 = 10

Question 11.
How will you find the valency of chlorine, sulphur and magnesium?
Solution:
Valency of an atom is the number of electrons gained, lost or shared so as to complete the octet of electrons in the valence shell.
Valency of chlorine: It has electronic configuration = 2, 8, 7 Thus, one electron is gained to complete its octet and so its valency is 1. valency of sulphur: it it has electronic configuration = 2, 8, 6. Thus two electrons are gained to complete its octet and hence its valency = 2 Valency of magnesium : it has electronic configuration = 2, 8, 2. Thus, it can lose two electons to attain octet and hence its valency = 2.

Question 12.
If number of electrons in an atom is 8 and number of protons is also 8, then

1. what is the atomic number of the atom and
2. what is the charge on the atom?

Solution:
We know that,
No. of electrons (e) = No. of protons (p) = Atomic number (Z)

1. No. of electrons = 8 ∴ Atomic number = 8
2. The number of protons is equal to the number of electrons. So the atom will be neutral.

Question 13.
With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.
Solution:
Mass number of oxygen atom
= No. of protons + No. of neutrons = 8 + 8 = 16
Mass number of sulphur atom
= No. of protons + No. of neutrons = 16 + 16 = 32

Question 14.
Write the electronic configuration of any one pair of isotopes and isobars.
Solution:
Isotopes of chlorine: $_{ 17 }^{ 35 }{ CI },_{ 17 }^{ 37 }{ CI }$ Electronic configuration of each of them: 2,8, 7 Isobars : $_{ 20 }^{ 40 }{ Ca },_{ 18 }^{ 40 }{ CI }$
Electronic configuration of 20Ca → 2, 8, 8, 2 Electronic configuration of 18Ar → 2, 8, 8

Question 15.
For the symbol H, D and T tabulate three sub-atomic particles found in each of them.
Solution:

NCERT Exercises

Question 1.
Compare the properties of electrons, protons and neutrons.
Solution:

Question 2.
What are the limitations of J.J. Thomson’s model of the atom?
Solution:
The limitation of Thomson’s model are described below :

1. Thomson’s model of an atom considers an atom to lae a sphere of uniform positive charge. Later researches particularly, Rutherford’s a-particle scattering experiment showed that an atom has a positively charged ‘core’ at its centre.
2. According to Thomson’s atomic model, mass of an atom is considered to be uniformly distributed. Rutherford’s experiment showed that the entire mass of an atom is concentrated inside the core of the atom.

Question 3.
What are the limitations of Rutherford’s model of the atom?
Solution:
The Rutherford model suffers from the following drawbacks :

1. An electron revolving around the nucleus gets accelerated towards the nucleus. An accelerating charged particle must emit radiation, and lose energy. Thus, the electrons in an atom must continuously emit radiation and lose energy. Because of this loss of energy, the electron would slow down, and will not be able to withstand the attraction of the nucleus. As a result, the electron should follow a spiral path, and ultimately fall into the nucleus (see figure). If it happens, then the atom should collapse in about 10-8 second. But, this does not happen — atoms are stable. This indicates that there is something wrong in Rutherford’s nuclear model of atom.
2. Rutherford’s model of atom does not say anything about the arrangement of electrons in an atom.

Question 4.
Describe Bohr’s model of the atom.
Solution:
The Danish physicist Neils Bohr proposed the following postulates for revising the Rutherford’s model.

• Atom has central nucleus surrounded by electrons.
• An atom consists of small heavy positively charged nucleus in the centre and the electrons revolve around it, in circular paths called orbits or shells.
• Each orbit has fixed energy, so these orbits are called energy levels or energy shells.
• The order of the energy of these energy shells will be
K < L < M < N < O <…………. or, 1 < 2 < 3 < 4 < 5 <…………..
• As long as an electron remains in a particular orbit, it does not lose or gain energy.
• Energy is neither absorbed nor emitted when electron is moving in an orbit. But energy is absorbed when it jumps from lower orbit to higher orbit. Whereas energy is emitted when it jumps from higher orbit to lower orbit.

Question 5.
Compare all the proposed models of an atom given in this chapter.
Solution:

Question 6.
Summarise the rules for writing the distribution of electrons in various shells for the first eighteen elements.
Solution:
The distribution of elements in different orbits is governed by a scheme called Bohr- Bury scheme. There are following rules :

• The maximum number of electrons present in any shell is given by the formula 2n2. Where n = no. of orbit.
• The maximum number of electrons that can be accommodated in the outermost shell is 8.
• Electrons in an atom do not occupy a new shell unless all the inner shells are completely filled.

Question 7.
Define valency by taking examples of silicon and oxygen.
Solution:
The number of electrons gained, lost or shared so as to complete the octet of electrons in valence shell is called valency.
Valency of silicon : It has electronic configuration → 2, 8, 4 Thus, 4 electrons are shared with other atoms to complete the octet and so its valency = 4
Valency of oxygen : It has electronic configuration → 2, 6 Thus, It will gain 2 electrons to complete its octet. So its valency = 2

Question 8.
Explain with examples

1. Atomic number,
2. Mass number,
3. Isotopes and
4. Isobars. Give any two uses of isotopes.

Solution:

1. Atomic number : The number of protons present in the nucleus of an atom is called atomic number. It is denoted by Z. e.g., in $_{ 20 }^{ 40 }{ Ca }$, atomic number = 20
2. Mass number : The sum of the number of protons and neutrons present in the nucleus of an atom is called mass number. It is denoted by A. e.g., in $_{ 20 }^{ 40 }{ Ca }$, mass number = 40
3. Isotopes : The atoms of the same elements having same atomic number but different mass numbers are called isotopes, e.g., $_{ 17 }^{ 37 }{ Cl }$ and $_{ 17 }^{ 35 }{ Cl }$
4. Isobars : The atoms of the different elements having same mass number but different atomic numbers are called isobars. e.g., $_{ 20 }^{ 40 }{ Ca }$ and $_{ 18 }^{ 40 }{ Ar }$

Uses of isotopes :

• As nuclear fuel : An isotope of uranium (U – 235) is used as a nuclear fuel.
• In medical field : An isotope of cobalt is used in the treatment of cancer.

Question 9.
Na+ has completely filled K and L shells. Explain.
Solution:
Atomic number of sodium (Na) = 11 No. of electrons in Na = 11 No. of electrons in Na+ = 11 — 1 = 10 Electronic configuration of Na+ → 2, 8 K L
For K – shell; 2n2 = 2 × l2 = 2
For L – shell; 2n2 = 2 × 22 = 8
Thus, in Na+, K and L shells are completely filled.

Question 10.
If bromine atom is available in the form of, say two isotopes $_{ 35 }^{ 79 }{ Br }$ (49.70%) and $_{ 35 }^{ 81 }{ Br }$ (50.30%), calculate the average atomic mass of bromine atom.
Solution:

Question 11.
The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes $_{ 8 }^{ 16 }{ X }$ and $_{ 8 }^{ 18 }{ X }$ in the sample?
Solution:

Question 12.
If Z = 3, what would be the valency of the element? Also, name the element.
Solution:
The electronic configuration of (Z) = 2,1 Thus, outermost shell has 1 electron. So, its valency = 1 Atomic number (Z) = 3, So name of the element is lithium.

Question 13.
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 = No. of protons +
No. of neutrons = 6 + 6 = 12
Mass number of Y = 6 + 8 = 14 The species X and Y are isotopes because their atomic numbers are same and their mass numbers are different i.e. $_{ 6 }^{ 12 }C$ and $_{ 6 }^{ 14 }C$.

Question 14.
For the following statements, write T for True and F for False.

1. J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
2. A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
3. The mass of an electron is about times that of proton. $\cfrac { 1 }{ 2000 }$
4. An isotope of iodine is used for making tincture iodine, which is used as a medicine.

Solution:

1. F : Because it was not proposed by J.J. Thomson.
2. F : Because neutron is an independent sub-atomic particle.
3. T : Because it is a fact known from experiments.
4. F : Because tincture iodine is a solution of ordinary iodine in alcohol.

Question 15.
Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) atomic nucleus
(b) electron
(c) proton
(d) neutron.
Solution:
(a)

Question 16.
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)

Question 17.
Number of valence electrons in Cl ion are :
(a) 16
(b) 8
(c) 17
(d) 18
Solution:
(b) : Electronic configuration of Cl (Z = 17) = 2, 8, 7. Thus, it has 7 valence electrons. Cl gains 1 electron to form CL ion. So, number of valence electrons = 7 + 1 = 8.

Question 18.
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) : Atomic number of sodium (Na) = 11 Its electronic configuration = 2, 8,1

Question 19.
Complete the following table.
Solution:

First Row : Mass number = Atomic number (9)+ No. of neutrons (10) = 19 No. of protons = Atomic number = 9 No. of electrons = Atomic number = 9 Name of the species = Fluorine (F)

Second Row :
No. of neutrons = Mass number (32) – Atomic number (16) = 16 No. of protons = Atomic number = 16 No. of electrons = Atomic number = 16

Third Row :
Atomic number = No. of protons = 12 No. of neutrons = Mass number (24) – Atomic number (12) = 12
No. of electrons = Atomic number = 12 Name of the species = Magnesium (Mg)

Fourth Row :
Atomic number=No. of protons=1 No. of neutrons = Mass number (2) – Atomic number (1) = 1
No. of electrons = Atomic number = 1 Name of the species = Deuterium (D).

Fifth Row :
Atomic number = No. of protons = 1 Name of the species = Protium or Hydrogen (H) The complete table can also be represented as:

# Chapter 5: The Fundamental Unit of Life

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

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

INTEXT Questions

Question 1.
Who discovered cells, and how ?
Solution:
Robert Hooke discovered the cell in 1665. He examined a thin slice of cork through a self designed primitive microscope and saw that the cork resembled the structure of a honey comb consisting of many tiny compartments. He called them cells which is derived from the Latin word “cellula” meaning ‘a little room’.

Question 2.
Why is the cell called the structural and functional unit of life ?
Solution:
All living organisms are made up of cells. Thus, cell is the structural unit of life. Each cell acquires distinct structure and function due to the organisation of its membrane and cytoplasmic organelles in the specific way. Such an organisation enables the cells to perform basic functions such as respiration, obtaining nutrition, clearing of waste material, forming new proteins etc. The cell is, therefore, the basic functional unit of living organisms.

Question 3.
How do substances like CO2 and water move in and out of the cell? Discuss.
Solution:
CO2 moves in and out of the cells by the process of diffusion which involves movement of molecules from higher concentration to lower concentration across the cell membrane.

Water moves in and out of the cells by osmosis. Osmosis is the movement of water or solvent through a semi-permeable membrane from a solution of lower concentration of solutes to a solution of higher concentration of solutes.

Question 4.
Why is the plasma membrane called a selectively permeable membrane?
Solution:
Plasma membrane permits the entry and exit of only selected materials in the cell. It also prevents movement of selected materials. Therefore, the plasma membrane is called a selectively permeable membrane.

Question 5.
Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells Solution:
The differences between prokaryotic cell and eukaryotic cell are as follows: Question 6.
Can you name the two organelles we have studied that contain their own genetic material?
Solution:

• Mitochondrion
• Plastid

Question 7.
If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?
Solution:
If the organisation of a cell gets destroyed due to any reason then lysis of the whole cell will occur due to the enzymes released by lysosomes.

Question 8.
Why are lysosomes known as suicide bags?
Solution:
Lysosomes contain digestive enzymes that are capable of digesting the whole cell. During breakdown of cell structure, when the cell gets damaged, lysosomes digest their own cells by releasing own enzymes. Therefore, they are known as suicide bags of a cell.

Question 9.
Where are proteins synthesised inside the cell?
Solution:
The ribosomes attached to the endoplasmic reticulum as well as present freely in the cytoplasm of all active cells are the site for the synthesis of proteins.

NCERT Exercises

Question 1.
Make a comparison and write down ways in 2. which plant cells are different from animal cells.
Solution:
Differences between plant cells and animal cells are as follows:

Question 2.
How is a prokaryotic cell different from a eukaryotic cell?
Solution:
Differences between prokaryotic and eukaryotic cells are as follows:

Question 3.
What would happen if the plasma membrane ruptures or breaks down?
Solution:
The plasma membrane acts as a mechanical barrier to the protoplasm which regulates transport of materials into and out of cell maintaining the identity of the cell. In case of rupturing of the plasma membrane, the protoplasmic contents will get dispersed in the surrounding medium and thus cell will disintegrate.

Question 4.
What would happen to the life of a cell if there was no Golgi apparatus?
Solution:
In the absence of Golgi apparatus, the following problems will arise:

• The secretory activities. of the cell will cease to occur.
• The broken membranes like those of lysosomes, cell wall, plasma membrane, etc. will not get repaired.
• In case of sperms, acrosome formation will not take place, causing inability of sperms to enter the egg.

Question 5.
Which organelle is known as the power house of the cell? Why?
Solution:
Mitochondrion is known as the power house of the cell. It is because the mitochondrion is the site of cellular respiration where energy in the form of ATP (adenosine triphosphate) is generated as a consequence of oxidation of carbohydrates and fats (lipids).

Question 6.
Where do the lipids and proteins constituting the cell membrane get synthesised?
Solution:
Smooth endoplasmic reticulum (SER) helps in the manufacturing of lipids which are important for the formation of cell membrane. Ribosomes are the site for protein synthesis. The manufactured proteins are then sent to different places in the cell depending upon the need by the endoplasmic reticulum (ER).

Question 7.
How does an Amoeba obtain its food?
Solution:
Amoeba acquires its food through endocytosis. Endocytosis involves invagina-tion of a small region of the plasma membrane and ultimately forming an intra cellular membrane bound vesicle. This process is generally involved in the ingestion of food materials. Intake of solid particles by a cell through its cell membrane is called phagocytosis or cell eating. In this process, cell membrane of Amoeba puts up protoplasmic processes around the food particle. The processes join and fuse to form phagosome.

Question 8.
What is osmosis?
Solution:
Osmosis involves the passive flow of water or any other solvent from a region of higher water, concentration to a region of lower water concentration through a semi- permeable membrane.

Question 9.
Carry out the following osmosis experiment. Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now
(a) Keep cup A empty
(b) Put one teaspoon sugar in cup B
(c) Put one teaspoon salt in cup C
(d) Put one teaspoon sugar in the boiled potato cup D.
Keep these for two hours. Then observe the four potato cups and answer the following

1. Explain why water gathers in the hollowed portion of B and C.
2. Why is potato A necessary for this experiment?
3. Explain why water does not gather in the hollowed out portions of A and D.

Solution:

1. Water gathers in B and C because in both the situations there is difference in the concentration of water in the trough and water in the potato cup. Hence endosmosis (i.e. water enters into the cell) takes place as the potato cells act as a semi-permeable membrane.
2. Potato cup A is necessary in the experiment as a ‘control’ for providing comparison with situations created in potato cups B, C and D. It indicates that the potato cavity alone does not induce any movement of water.
3. Water does not gather in hollowed out portions of A. This can be explained as follows. For osmosis to take place, a concentration gradient is necessary to develop between the two solutions on either side of a semi- permeable membrane (in this case potato strip). Here water is present only on one side i.e., convex side of potato cup whereas concave side of the cup is empty. Water will fill in the concavity only on account of endosmosis that can take place when some solution having concentration higher than that of water is present it. Hence, the hollowed out portion of A remained empty. In cup D cells become dead due to boiling, hence semi permeability of membrane is lost so no osmosis will take place in it.

# Chapter 6 Tissues

## NCERT Solutions for Class 9 Science Chapter 6 Tissues

INTEXT Questions

Question 1.
What is a tissue?
Solution:
A tissue may be defined as a group or collection of similar or dissimilar cells that perform dr help to perform a common function and have common orgin.

Question 2.
What is the utility of tissues in multicellular organism?
Solution:
The utility of tissues in multicellular organism is given below:

• Division of labour: Tissues bring about division of labour in multicellular organ-isms. It increases efficiency.
• Higher organisation: Tissues become organised to form organs and organ systems.
• Individual cells: Work load of individual cells has decreased.
• Higher survival: Because of division of labour, higher efficiency,and organisation, the multicellular organisms have high survival.

Question 3.
Name the types of simple tissues.
Solution:
There are three types of simple tissues. These are parenchyma, collenchyma and sclerenchyma.

Question 4.
Where is apical meristem found?
Solution:
Apical meristem occurs at root and stem tips.

Question 5.
Which tissue makes up the husk of coconut?
Solution:
Sclerenchyma (fibres) makes up the husk of coconut.

Question 6.
What are the constituents of the phloem?
Solution:
The constituents of the phloem are sieve tubes, companion cells, phloem parenchyma and phloem fibres.

Question 7.
What are the functions of areolar tissue?
Solution:
Areolar tissue fills the space inside the organs, supports internal organs and helps in repair of tissues. It binds the skin with underlying parts.

Question 8.
Name the tissue responsible for movement in our body.
Solution:
Muscular and skeletal tissue are responsible for movement in our body.

Question 9.
What does a neuron look like?
Solution:
Neuron is a thread or hair like structure, with palm leaf shaped structure at one end.

Question 10.
Give three features of cardiac muscle.
Solution:
The three features of cardiac muscle are given below:

• Cardiac muscles show morphological characteristics of both striated and unstriated muscles.
• Intercalated discs are present in the cardiac muscle fibres.
• The muscles show rhythmic contractions, without fatigue.

NCERT Exercises

Question 1.
Define the term “tissue”.
Solution:
A tissue may be defined as a group or collection of similar or dissimilar cells that perform or help to perform a common function and have common orgin.

Question 2.
How many types of elements together make up the xylem tissue? Name them.
Solution:
Four types of elements together make up the xylem tissue. These elements are-

• tracheids,
• vessels,
• xylem parenchyma and
• xylem fibres.

Question 3.
How are simple tissues different from complex tissues in plants?
Solution:
Simple tissues are made up of only one type of cells, which look like each other. On the other hand, complex tissues are made up of more than one type of cells. Parenchyma, collenchyma and sclerenchyma are the examples of simple plant tissue whereas xylem and phloem are complex tissues.

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell wall.
Solution:
On the basis of the cell wall, differences between parenchyma, collenchyma and sclerenchyma are-

Question 5.
What are the functions of the stomata?
Solution:
Stomata are necessary for exchanging gases with the atmosphere. Transpiration (loss of water in the form of water vapour) also

Question 6.
Diagrammatically show the difference among three types of muscle fibres.
Solution:

Question 7.
What is the specific function of the cardiac muscle?
Solution:
The specific function of the cardiac muscle is rhythmic contraction and relaxation simultaneously throughout life without getting fatigued.

Question 8.
Differentiate among striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.
Solution:
The differences among striated, unstri-ated and cardiac muscles are as follows:

Question 9.
Draw a labelled diagram of a neuron.
Solution: Question 10.
Name the following:

1. Tissue that forms inner lining of our mouth.
2. Tissue that connects muscle to bone in humans.
3. Tissue that transports food in plants.
4. Tissue that stores fat in our body.
5. Connective tissue with a fluid matrix.
6. Tissue present in the brain.

Solution:

1. Squamous epithelium, an epithelial tissue
2. Tendon
3. Phloem
5. Fluid connective tissue i.e., blood and lymph
6. Nervous tissue

Question 11.
Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Solution:
Skin—Squamous epithelium Bark of tree—Epidermal tissue/cork Bone—Supportive connective tissue or
skeletal tissue Lining of kidney tubule—Ciliated epithelium Vascular bundle—Xylem and phloem (conducting tissue).

Question 12.
Name the regions in which parenchyma tissue is present.
Solution:
In small herbaceous plants, parenchyma makes up the bulk of the plant body. It is mainly found in the cortex, pith, ground tissue leaf, mesophyll and also in vascular bundles.

Question 13.
What is the role of epidermis in plants?
Solution:
It protects internal tissues against mechanical injury, parasites and also cold or heat. Thick cuticle, wax, epidermal hair and epidermis reduce loss of water from internal tissue. Epidermal cells of roots have hairs that greatly increase the absorptive surface areas for the absorption of water and nutrients.

Question 14.
How does the cork act as a protective tissue?
Solution:
The walls of cork cells are heavily thickened by the deposition of suberin. This structural characteristic helps the cork to protect and prevent from infection and mechanical injury. It also prevents desiccation, by preventing loss of water from the plant body.

Question 15.
Complete the table: Solution:

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