Class 9 – Page 5 – HBSE Solutions

HBSE 9th Class Science Solutions Chapter 6 Tissues

September 19, 2022 / Class 9 / Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 6 Tissues Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 6 Tissues

HBSE 9th Class Science Tissues Intext Questions and Answers

Questions from Sub-section 6.1

Question 1.
What is a tissue?
Answer:
A tissue is a group of similar cells that performs similar function.

Question 2.
What is the utility of tissues in multicellular organisms ?
Answer:
They help in efficient working of the body of such organisms.

Questions from Sub-section 6.2.2 (i)

Question 1.
Which gas is needed for photosynthesis?
Answer:
Carbon dioxide.

Question 2.
State the role of transpiration in plants.
Answer:
In plants the role of transpiration is performed by leaves. There are many minute pores in the leaves of every plant. These pores are called stomata. There are guard cells on stomata which carry out the exchange of gases. These guard cells transpirate the water obtained from the earth.

Questions from Sub-section 6.2.2 (ii)

Question 1.
Name types of simple tissues.
Answer:
Simple tissues are of three types:
(1) Parenchyma
(2) Collenchyma
(3) Sclerenchyma.

Question 2.
Where is apical meristem found?
Answer:
It is located at the root and shoots tips.

Question 3.
Which tissue makes up the husk of coconut?
Answer:
Sclerenchymatous fibers make coconut husk.

Question 4.
What are the constituents of phloem?
Answer:
The constituents of phloem are:
1. sieve elements i. e., sieve cells, and sieve tubes
2. companion cells
3. phloem (bast) fibers
4. phloem parenchyma.

Questions from Sub-section 6.3.4

Question 1.
Name the tissue responsible for movement in our body.
Answer:
Muscular tissue.

Question 2.
What does a neuron look like?
Answer:
A neuron looks like a star having a tail.

Question 3.
Give three features of cardiac muscles.
Answer:
(i) Cardiac muscles are cylindrical, branched, and uninucleate.
(ii) They show faint cross striations.
(in) They are joined end to end by intercalated discs.

Question 4.
What are the functions of areolar tissue?
Answer:
(i) It binds muscles with skin.
(ii) It provides support to internal organs
(iii) It helps in repair of tissues.

HBSE 9th Class Science Tissues Textbook Questions and Answers

Question 1.
Define the term ‘tissue’.
Answer:
Tissue is a group of cells having a common origin, similar in structure and function.

Question 2.
How many types of elements together make up the xylem tissue?
Answer:
Four types of elements present in xylem are:
Tracheids, vessels, xylem fibres and xylem parenchyma.

Question 3.
How are simple tissues different from complex tissues in the tissue? Name them.
Answer:
Simple plant tissues are made up of only one type of cells, e.g. parenchyma, whereas complex tissues are made up of many types of cells, e.g. xylem.

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell wall.
Answer:

Parenchyma:
Cells are thin walled.

Collenchyma:
There is movement in cell wall.

Scierenchyma:
Cell walls are highly thickened due to lignin deposition.

Question 5.
What are the functions of the stomata?
Answer:
The main functions of stomata are transpiration and the exchange of gases.

Question 6.
Diagrammatically show the difference between the three types of muscle fibres.
Answer:
Three types of muscle fibres:

Straited muscle:
1. These are elongated, cylindrical, unbranched and multinucleate
2. These are voluntary.
3. These are attached to bones.

Untreated muscle:
1. These are spindle-shaped and uninucleate.
2. These are involuntary
3. These are found in eyelit, and lungs.

Cardiac muscle:
1. These are cylindrical branched and uninucleate.
2. These are involuntary.
3. These are found in cardiac walls.

Question 7.
What is the specific function of the cardiac muscle?
Answer:
They undergo rhythmic contraction and relaxation throughout life. They never get fatigued.

Question 8.
Differentiate between striated, unstriated, and cardiac muscles on the basis of their structure and site/location in the body.
Answer:

Cause of difference	Straited muscle	Unstraited muscle	Cardiac muscle
Site	These muscles work accordingly to our will.	These muscles do not work according to our will.	These muscles do their work themselves.
Location	They are attached to skeleton.	They are found in the walls of the organs of the body	They are found in the walls of heart.

Question 9.
Draw a labeled diagram of a neuron.
Answer:
Diagram of a neuron:

Question 10.
Name the following:
(a) Tissue that forms the inner lining of our mouth.
(b) Tissue that connects muscle to bone in humans
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Answer:
(a) Epithelial tissue (squamous epithelium).
(b) Tendon (a type of connective tissue).
(c) Sieve elements (phloem).
(d) Adipose tissue (a type of connective tissue).
(e) Vascular tissue (blood).
(f) Nervous tissue.

Question 11.
Identify the type of tissue in the following:
Skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:
Skin: Epithelial tissue.
Bark of tree: Cork.
Bone: Connective tissue.
Lining of kidney tubule: Epithelial tissue (cubodial epithelium).
Vascular bundle: Sclerenchyma.

Question 12.
Name the regions in which parenchyma tissue is present.
Answer:
Parenchyma is present in stem, root, leaves, flowers and fruits. It is the most common plant tissue.

Question 13.
What is the role of epidermis in plants ?
Answer:
It protects the parts of the plant. It helps in protections against excessive water loss. It helps the plant parts from mechanical injuries and attack of parasites (fungi, bacteria etc.).

Question 14.
How does the cork act as protective tissue ?
Answer:
Cork cells are highly thick walled, dead and protective in nature. They have deposition of suberin. They check water loss and prevent the entry of harmful microbes into plant body.

Question 15.
Complete the table:

Answer:

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HBSE 9th Class Science Solutions Chapter 5 The Fundamental Unit of Life

September 19, 2022 / Class 9 / Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 5 The Fundamental Unit of Life Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 5 The Fundamental Unit of Life

HBSE 9th Class Science The Fundamental Unit of Life Intext Questions and Answers

Questions from Sub-section 5.1

Question 1.
Who discovered cells, and how?
Answer:
Cells were first discovered by Robert Hooke in 1665. He observed the cells in a cork slice with the help of a primitive microscope.

Question 2.
Why is the cell called the structural and functional unit of life?
Answer:
All living organisms are made up of cells. A single cell may constitute a whole organism as in Amoeba, Paramoecium, Chlamydomonas, etc. These organisms are called unicellular organisms. On the other hand, many cells group together in a single body and form various body parts in multicellular organisms such as plants and animals. Each cell performs certain basic functions that are characteristic of all living forms. Due to these reasons, a cell is called the structural and functional unit of life.

All living organisms are made up of cells. A single cell may constitute a whole organism as in Amoeba, Paramoecium, Chlamydomonas, etc. These organisms are called unicellular organisms. On the other hand, many cells group together in a single body and form various body parts in multicellular organisms such as plants and animals. Each cell performs certain basic functions that are characteristic of all living forms. Due to these reasons, a cell is called the structural and functional unit of life.

Questions from Sub-section 5.2.1

Question 1.
How do substances like CO₂ and water move in and out of the cell ? Discuss.
Answer:
CO₂ and water can move across the cell membrane by a process called diffusion. We know that movement of substances is from a region of high concentration to a region where concentration is low. When the concentration is high in the cell due to the increasing in the quantity of CO₂ and the concentration of outer environment is relatively low, just then due to a difference of concentration of CO₂ inside and outside a cell, CO₂ moves out of the cell from a region of high concentration to a region of low concentration outside the cell by the process of diffusion. Similarly, molecule of water moves from outside (high concentration) to inside (low concentration) by the process of osmosis.

Question 2.
Why is the plasma membrane called a selectively permeable membrane?
Answer:
Plasma membrane is made up of protein and lipid molecules. It acts as selectively permeable membrane:
1. ft allows the substances to enter, which are required by cell.
2. It does not allow the substances to enter, which are not required by cell.
3. It allow the substances to move out, which are not required by cell.
4. It does not allows the substances to move out, which are required by cell.
Due to above stated properties, plasma membrane is called selectively permeable membrane.

Questions from Sub-sections 5.2.2-5.2.4

Question 1.
Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.

Answer:

Prokaryotic cell	Eukaryotic cell
I. Size: generally small. (1-10μm) 1μm = 10^-6m	1. Size: generally large. (5 – 100 μm)
2. Nuclear region: not well defined and is not surrounded by a nuclear membrane and is known as nucleoid.	2. Nuclear region: well defined and surrounded by a nuclear membrane.
3. Chromosome: single.	3. More than one chromosome.
4. Membrane-bound cell organelles absent.	4. Membrane-bound cell organelles present.

Questions from Sub-section 5.2.5

Question 1.
Can you name the two organelles we have studied that contain their own genetic material ?
Answer:
(i) Nucleus
(ii) Mitochondria.

Question 2.
If the organization of a cell is destroyed due to some physical or chemical influence, what will happen ?
Answer:
The cell will not be able to perform its functions like respiration, protein synthesis, excretion, nutrition, etc.

Question 3.
Why are lysosomes known as suicide bags?
Answer:
Lysosomes may digest their own cell under certain conditions when the cell is damaged. Therefore lysosomes are known as suicide bags.

Question 4.
Where are proteins synthesized inside the cell?
Answer:
Proteins are synthesized in ribosomes attached on the surface of rough endoplasmic reticulum (RER).

HBSE 9th Class Science The Fundamental Unit of Life Textbook Questions and Answers

Question 1.
Make a comparison and write down ways in which plants cells are different from animal cells.
Answer:
Differences between Plants and are as follows:

Plant Cells:
1. Cell wall present outside the plasma membrane.
2. Centrosome and Golgilbodies are absent.
3. Plastids (chloroplasts and chromoplasts) present in cytoplasm.
4. Reserve food material is in the form of starch.
5. These have definite size.
6. It is usually larger in size.

Animal Cells:
1. Cell wall absent. Bounded by a thin living plasma membrane only.
2. Centrosomes and Golgibodies are present.
3. Plastids absent in cytoplasm.
4. Reserve food material is in the form of glycogen.
5. These have indefinite size.
6. It is comparatively smaller in size.

Question 2.
How is a prokaryotic cell different from a eukaryotic cell?
Answer:
Differences between prokaryotic cell and eukaryotic cell are as follows:
Prokaryotic Cell:
1. Size is generally small (1-10 um).
2. Nuclear material (nucleoid) is not enclosed in a nuclear membrane.
3. Have a single chromosome.
4. Nucleolus absent.
5. Membrane-bound organelles absent.
6. Cell division occurs by fission or budding (no mitosis).

Eukaryotic Cell:
1. Size generally large (5-100 um).
2. Nuclear material is enclosed in a nuclear membrane.
3. Have more than one chromosome.
4. Nucleolus present.
5. Membrane-bound organelles present.
6. Cell division mitotic or meiotic.

Question 3.
What would happen if the plasma membrane ruptures or breaks down?
Answer:
Plasma membrane separates the contents of the cell from its external environment. It also controls entry or exit of materials in and out of the cell. In case, a plasma membrane ruptures or breaks down, the above functions cannot be performed and the cell may die.

Question 4.
What w ould happen to the life of a cell if there was no Golgi apparatus?
Answer:
Golgi apparatus performs the following important functions in the cell:
(i) It packages materials synthesised in E.R. and dispatches them to intercellular and extracellular targets.
(ii) It helps in the formation of cell plate in plant cells during cell division.
(iii) It is involved in synthesis of lysosomes and peroxisomes.
(iv) In some cases, complex sugars may be made from simple sugars in the Golgi apparatus.
The life of a cell will be affected for the above-mentioned functions if there is no Golgi apparatus.

Question 5.
Which organelle is known as the powerhouse of the cell? Why?
Answer:
Mitochondria are known as the powerhouses of the cell. They are called powerhouses of the cell because the energy required for various chemical activities needed for life is released by mitochondria in the form of ATP (Adenosine triphosphate) molecules. ATP is known as the energy currency of the cell.

Question 6.
Where do the lipids and proteins constituting the cell membrane get synthesized?
Answer:
Rough endoplasmic reticulum (RER) has particles called ribosomes attached to its surface. The ribosomes are the sites of protein synthesis. The smooth endoplasmic reticulum (SER) helps in the synthesis of fat molecules, or lipids. Some of these proteins and lipids help in the building of cell membranes.

Question 7.
How does an Amoeba obtain its food?
Answer:
The flexibility of the cell membrane also enables the cell to engulf in food and other materials from its external environment. Such processes are known as endocytosis. Amoeba acquires its food through such processes.

Question 8.
What is osmosis?
Answer:
Osmosis is the passage of water from a region of high water concentration through a semi- permeable membrane to a region of low water concentration.

Question 9.
Carry out the following osmosis experiment:
Take four peeled potato halves and scoos 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:
(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 portions of A and D.
Answer:
(i) Water gathers in the hollowed portions of B and C due to the process of osmosis. Potato membrane acts as a semipermeable membrane and the medium surrounding it has a higher water concentration than the cell (potato). Thus, water enters by osmosis.
(ii) Potato A necessary, because it acts as a control experiment.
(iii) Water does not gather in the hollowed out portions of A and D because of the following reasons:
1. Water does not enter in the hollowed out portion of A, because the cup is empty.
2. Water does not enter in the hollowed out portion of D, because the potato used in experiment is a boiled one.
On boiling, the cell dies and the selective permeability of the membrane is lost.

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HBSE 9th Class Science Solutions Chapter 4 Structure of the Atom

September 19, 2022 / Class 9 / Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 4 Structure of the Atom Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 4 Structure of the Atom

HBSE 9th Class Science Structure of the Atom Intext Questions and Answers

Questions from Sub-section 4.1

Question 1.
What are canal rays?
Answer:
E. Goldstein in 1886 discovered the positively charged fluorescent radiations, which were renamed as canal rays.

Question 2.
If an atom contains one electron and one proton, will it carry any charge or not?
Answer:
If an atom contains one electron and one proton, it will possess no charge on it. Because proton and electron mutually balance the charges.

Questions from Sub-section 4.2

Question 1.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a w hole.
Answer:
According to Thomson’s model an atom is made up of positively charged sphere and electrons get embedded into it. Thus due to uniformity in the negative and positive magnitude, an atom as a whole is electrical neutral.

Question 2.
On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom ?
Answer:
According to Rutherford’s model of an atom, the nucleus of an atom consists of proton sub-atomic charged particle, since it deflects the a (alpha) particle.

Question 3.
Draw a sketch of Bohr’s model of an atom with three shells.
Answer:

Question 4.
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 ?
Answer:
Yes, the a-particie scattering experiment will be possible with any metal foil other than gold foil.

Questions from Subsection 4.2

Question 1.
Name the three sub-atomic particles of an atom.
Answer:
The three sub-atomic particles of an atom are electron, proton and neutron.

Question 2.
Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have ?
Answer:
Atomic mass of helium atom = 4u
Protons present in the nucleus of helium atom = 2u
Neutrons present in the nucleus of helium atom= Atomic mass – proton = 4 – 2 = 2

Questions from Subsection 4.3

Question 1.
Write the distribution of electrons in carbon and sodium atoms.
Answer:
(i) Carbon:
Mass number = 12
Atomic number = 6
Number of protons = 6
Number of electrons = 6
Number of neutrons = 12 – 6 = 6
Electron distribution = K = 2, L = 4

(ii) Sodium:
Mass Number = 23
Atomic number = 11
Number of protons = 11
Number of electrons = 11
Number of neutrons = 23 – 11 = 12
Electron distribution = K = 2 L = 8 M = 1

Question 2.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom ?
Answer:
Shell K = 2 electrons
Shell L = 8 electrons
Total electrons in the atom = 2 + 8 = 10 electrons

Questions from Sub-section 4.4

Question 1.
How will you find the valency of chlorine, sulphur and magnesium ?
Answer:
(1) The atomic number of chlorine is 17, therefore, its electron distribution will be as under:
K = 2 electrons
L = 8 electrons
M = 7 electrons
Therefore, to complete its octet, chlorine needs (8 – 7) = 1 electron.
Hence, valency of chlorine is 1.

(2) Atomic number of sulphur is 16, therefore, its electron distribution will be as under:
K = 2 electrons
L = 8 electrons M = 6 electrons
Therefore, to complete its octet, sulphur needs (8 – 6) = 2 electrons.
Hence, valency of sulphur is 2.

(3) Atomic number of magnesium is 12.
Therefore, its electron distribution will be as under:
K = 2 electrons .
L = 8 electrons
M = 2 electrons
Therefore, to complete its valence octet, it is easy in the case of magnesium to quit 2 electrons. Hence, the valency of magnesium is 2.

Questions from Subsection 4.5

Question 1.
If the number of electrons in an atom is 8 and the 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?
Answer:
Number of electrons in the atom = 8
Number of protons in atom = Number of electrons = 8
(i) Atomic number = Number of electrons = Number of Protons = 8
(ii) Electron distribution = K = 2, L = 6 .
To fulfil the outermost shell of atom 2 electrons are required. Therefore, the charge is -2.

Question 2.
With the help of Table 4.1. find out the mass number of oxygen and sulphur atom. Answer: According to the table,
(1) Atomic number of oxygen = 8 Number of protons in oxygen = 8
Number of neutrons in oxygen = 8
Mass number = Number of protons + Number of neutrons = 8 + 8 = 16

(2) Atomic Number of sulphur = 16 Number of protons in sulphur = ? 16
Number of neutrons in sulphur =16
Mass number = Number of protons + Number of neutrons = 16 + 16 = 32

Questions from Sub-section 4.6

Question 1.
For the symbols H, D and T tabulate three sub-atomic particles found in each of them.
Answer:
(1) Symbol H is the sign for Protium i.e. ₁H¹
∴ Atomic number = 1
Mass number = 1
Number of electrons = 1
Number of protons = 1
Number of neutrons = 1-1=0

(2) Symbol D is the sign for Deuterium i.e. ₁H²
Atomic number = 1
Mass number = 2
Number of electrons = 1
Number of protons = 1
Number of neutrons = 2 – 1 = 1

(3) Symbol T is the sign for Tritium i.e. ₁H³
Atomic number = 1
Mass number = 3
Number of electrons = 1
Number of protons = 1
Number of neutrons = 3 – 1 = 2

Question 2.
Write the electronic configuration of any one pair of isotopes and isobars.
Answer:
(i) Electronic configuration of the pair of isotope chlorine ₁₇Cl¹³⁵ and ₁₇Cl¹³⁷ will be as follows:

(ii) Electronic configuration of the pair of isobars calcium and argon will be as under:
(l) Calcium ₂₀Ca⁴⁰
e = 20
P = 20
N = 40 – 20 – 20

(2) Argon ₁₈Ar⁴⁰
e = 18
P = 18
N = 40 – 18 = 22

HBSE 9th Class Science Structure of the Atom Textbook Questions and Answers

Question 1.
Compare the properties of electrons, protons and neutrons.
Answer:
Properties of electrons, protons and neutrons can be compared as below:
(i) Charge: Electron is negatively charged and has an absolute charge of 1.602 x 10^-19 coulomb whereas proton is positively charged and has an absolute charge of 1.602 x 10^-19 coulomb. On the other hand, neutron is a neutral particle carrying no charge.

(ii) Mass: Electron has an absolute mass of 9.11 x 10^-31 kg which is equal to 1/1840 amu while proton
has an absolute mass of 1.672 x 10^-27kg which is equal to 1 amu. On the other hand, neutron has an absolute mass of 1.675 x 10^-27kg, i.e., neutron is slightly heavier than proton.

(iii) Location: Electrons are located outside the nucleus while protons and neutrons are located inside the nucleus.

(iv) Symbol: Electron is represented as _-1e⁰or e^– proton is represented as ₁p¹ or p⁺ and neutron is represented as ₀n¹ or n.

Question 2.
What are the limitations of J.J. Thomson’s model of the atom?
Answer:
Although Thomson’s model of the atom was able to explain the electrical neutrality of the atom but it failed to explain the results of experiments carried out by other scientists. For example, this model could not explain the results of scattering experiments carried out by Rutherford and was, therefore, rejected in favour of Rutherford’s model of the atom.

Question 3.
What are the limitations of Rutherford’s model of the atom?
Answer:
It was pointed by Neils Bohr that Rutherford’s atom should be highly unstable. He argued that if an electron (charged particle) moves around the nucleus in an orbit, it should be subjected to acceleration due to continuous change in its direction of motion. Therefore, the electrons should continuously emit radiations and lose energy.

Consequently, the orbit should become smaller and smaller and ultimately the electron should fall into the nucleus. In other words, the atom should collapse. Since the atoms do not collapse, therefore, there must be something wrong with Rutherford’s model of atom. Another serious drawback of Rutherford’s model of an atom is that it says nothing about the electronic structure of the atom, i.e., how electrons are distributed around the nucleus and what are the energies of these electrons.

Question 4.
Describe Bohr’s model of the atom.
Answer:
According to Neils Bohr’s model of the atom
(1) Electrons revolve or move in definite orbits which are known as discrete orbits of electrons.
(2) When electrons revolve in discrete orbits, they do not radiate energy. These orbits (or shells) are called energy levels. Energy levels in an atom are shown in the figure.

Question 5.
Compare all the proposed models of an atom given in this chapter.
Answer:
A comparison of all the proposed models of an atom given in this chapter is as follows:
1. According to Thomson’s model of an atom:
(1) An atom consists of a positively charged sphere and the electrons are embedded in it.
(2) The negative and positive charges are equal in magnitude. So, atoms are electrically neutral.

2. According to Rutherford’s Model of an atom:
(1) The centre in the atom is positively charged. About all the mass of an atom resides in the nucleus.
(2) The electrons revolve around the nucleus in some definite orbits.
(3) The size of the nucleus is very small as compared to the size of the atom.

3. According to Bohr’s model of an atom:
(1) The electrons can revolve only in certain definite orbits which are known as discrete orbits of electrons.
(2) When electrons revolve in discrete orbits, then they do not radiate energy. These orbits are called energy levels.

Question 6.
Summarise the rules for the writing of distribution of electrons in various shells for the first eighteen elements.
Answer:
Bohr and Bury proposed identical schemes regarding the arrangement of electrons in various orbits. The main rules of the Bohr-Bury scheme are:

(1) The maximum number of electrons which can be accommodated in any orbit or shell is equal to 2r? where n is the number of orbit or shells.
(2) The maximum capacity of the outermost shell is of 8 electrons and that of the penultimate shell (next to the outermost) is of 18 electrons.
(3) It is not necessary that a shell should be completed to its maximum capacity before another starts. In fact, a new shell always starts when the outermost shell attains 8 electrons.
(4) The outermost shell cannot have more than 2 electrons and the penultimate shell cannot have more than 9 electrons unless the next innermost shell has received the maximum number of electrons as required by rule (i).

Question 7.
Define valency by taking examples of silicon and oxygen.
Answer:
Valency of an element is the combining capacity of the element and is equal to the number of electrons that take part in chemical reaction. The electronic configuration of silicon is :
K L M
2 8 4
Since it has 4 electrons in its valence shell, therefore,
Valency of silicon = 8 – Number of valence electrons = 8 – 4 = 4
The electronic configuration of oxygen is:
K L
2 6
Since it has 6 electrons in its valence shell, therefore,
Valency of oxygen = 8 – Number of valance electrons = 8 – 6 = 2.

Question 8.
Explain with examples
(i) Atomic number
(ii) Mass number
(iii) Isotopes
(iv) Isobars. Give any two uses of isotopes.
Answer:
(i) Atomic Number:
The total number of protons present in the nucleus of an atom of an element is known as atomic number (Z). for example, the atomic number of oxygen is 8 and that of carbon is 6.

(ii) Mass Number:
The total number of protons and neutrons present in an atom of an element is known as its Mass number (A). For example, the mass number of oxygen and carbon is 16u and 12u respectively.

(iii) Isotopes:
Atoms of the same element having same atomic number but different mass numbers are known as isotopes. For example, Protium (₁H¹), deueterium (₁H²), and tritium (₁H³) are three isotopes of hydrogen and ₆C¹² and ₆C¹⁴ are two isotopes of carbon.

(iv) Isobars: Atoms of different elements which have same mass number but different atomic numbers are known as isobars. For example, calcium (₂₀Ca⁴⁰) and argon (₁₈Ar⁴⁰) are isobars.

Uses of Isotopes:
(1) Isotope of uranium is used as a fuel in nuclear reactors.
(2) Isopote of cobalt is used in the treatment of cancer.

Question 9.
Na⁺ has completely filled K and L shells. Explain.
Answer:
The atomic number ofNa (sodium) is 11, so an atom of Na contains 11 electrons. The arrangement of electrons in Na atom will be:
K L M
2 8 1
Now, Na⁺ ion is formed by loss of 1 valence shell electron, therefore, the remaining 10 electrons are arranged as:
K L
2 8
According to Bohr and Bury rule (2n² formula), the K and L shells can accommodate 2 and 8 electrons respectively. This explains that the K and L shells in Na⁺ ions are completely filled.

Question 10.
If the bromine atom is available in the form of, say, two isotopes, ₃₅⁷⁹Br(49.7%) and ₃₅⁸¹Br (50.3%), calculate the average atomic mass of the bromine atom.
Answer:
Average atomic mass of bromine atom
\(\left(79 \times \frac{49.7}{100}+81 \times \frac{50.3}{100}\right)\)
\(\left(\frac{79 \times 497}{1000}+\frac{81 \times 503}{1000}\right)\)
\(\left(\frac{39263}{1000}+\frac{40743}{1000}\right)\) = 39.263 + 40.743 = 80.006 = 80u

Question 11.
The average atomic mass of a sample of element X is 16.2 u. What are the percentages 16 18 of isotopes ₈¹⁶X and ₁₈⁸X in the sample?
Answer:
Let the percentage of isotope ₁₆⁸X be x. Then the percentage of ₁₈⁸X is (100 – x).
Now,
Average atomic mass = \(\frac{16 x+18(100-x)}{100}\)
or 16.2 = \(\frac{16 x+18(100-x)}{100}\) or 2x = 180 or x = 90%
Therefore, percentage of ₈¹⁶X = 90% and percentage of ₈¹⁸X = 100 – 90 = 10%

Question 12.
If Z = 3, what would be the valency of the element? Also, name the element.
Answer:
If the atomic number is 3, then the arrangement of electrons in an atom of the element will be:
K L
2 1
As the valence shell contains one electron only, so valency will be one. The element with atomic number 3 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?
Answer:
Since Mass number = Number of protons + Number of neutrons,
Therefore, the Mass numbers of X and Y are 12 and 14 respectively.
Again, Atomic number = Number of protons = Number of electrons.
Therefore, the atomic numbers of both X and Y are 6 each.
As X and Y have the same atomic number but different mass numbers, therefore, these are isotopes of each other and are represented as ₆¹²X and ₆¹⁴Y.

Question 14.
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 \(\frac {1}{2000}\) times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer:
(a) F (b) F (c) T (d) F.
Put a tick (√) against the correct choice and cross (✗) against the wrong choice in questions 15, 16 and 17.

Question 15.
Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic Nucleus
(b) Electron
(c) Proton
(d) Neutron.
Answer:
(a) √ (b) ✗ (c) ✗ (d) ✗.

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.
Answer:
(a) ✗ (b) ✗ (c) √ (d) ✗

Question 17.
Number of valence electrons in Cf ion are :
(a) 16
(b) 8
(c) 17
(d) 18
Answer:
(a) ✗ (b) √ (c) ✗ (d) ✗

Question 18.
Which one of the following is a correct electronic configuration of sodium ?
(«) 2, 8,
(b) 8, 2,1
(c) 2,1, 8
(d) 2, 8,1
Answer:
(a) ✗ (b) ✗ (c) ✗ (d) √

Question 19.
Complete the following table.

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Naine of the Atomic Species Sulphur
9	–	10	–	–	–
16	32	–	–	–	sulphur
–	24	–	12	–	–
–	2	–	1	–	–
–	32	1	1	0	–

Answer:

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Naine of the Atomic Species Sulphur
9	19	10	9	9	Fluorine
16	32	16	16	16	sulphur
12	24	12	12	12	Magnesium
1	2	1	1	1	Deuterium
1	32	1	1	0	Hydrogen

HBSE 9th Class Science Solutions Chapter 4 Structure of the Atom Read More »

HBSE 9th Class Science Solutions Chapter 3 Atoms and Molecules

September 19, 2022 / Class 9 / Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 3 Atoms and Molecules Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 3 Atoms and Molecules

HBSE 9th Class Science Atoms and Molecules Intext Questions and Answers

Questions from Sub-section 3.1

Question 1.
In a reaction, 5.3 g sodium carbonate reacted with 6.0 g of ethanoic acid. The products were 2.2 g of carbon dioxide. 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Solution:
Here,
Total mass of reactants = Mass of (Sodium carbonate + Ethanoic acid) = 5.3g + 6.0g = 11.3 g
Total mass of products = Mass of(sodium ethanoate + carbon dioxide + water) = 2.2g + 8.2g + 0.9 g = 11.3g
Since, the total mass of reactants is equal to that of the total mass of products, hence these observations are 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:
Here,
Mass of hydrogen:
mass of oxygen = 1:8
Therefore, the required mass of oxygen to combine it completely with 3g of hydrogen = 3g x 8 = 24 g

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
The postulate of Dalton’s atomic theory is that atoms are the smallest and indivisible particles which can neither be created nor destroyed in chemical reactions. It 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?
Answer:
The postulate which states that atoms of the same or different elements combine in the ratio of small whole numbers to form compounds can explain the law of definite proportions.

Questions from Sub-section 3.2

Question 1.
Define the atomic mass unit.
Answer:
1/12th part of the mass of one atom of carbon-12 isotope is taken to be the standard atomic mass unit. With respect to the mass of one atom of carbon-12 isotope atomic masses of all the elements have been obtained.

Question 2.
Why is it not possible to see an atom with naked eyes?
Answer:
Being very small in size, atom cannot be seen with naked eye. Its size is so small that its radius is measured in nanometre (nm). Where, nm = 10 9m. Now, with the help of modem technique, the magnified images of surfaces of elements can be displayed, in which the existing atoms are clearly visible.

Questions from Sub-section 3.4

Question 1.
Write down the formulae of:
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide.
Answer:
(i) Formula of sodium oxide Symbol Na₂O

∴ The formula of sodium oxide is Na₂O.

(ii) Formula of aluminium chloride

∴ The formula of aluminium chloride is AlCl₃

(iii) Formula of sodium sulphide

∴ The formula of sodium sulphide is Na₂S.

(iv) Formula of magnesium hydroxide Symbol

∴ The formula of magnesium hydroxide is Mg (OH)₂.

Question 2.
Write down the names of compounds represented by the following formulae :
(i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃

Answer:

Chemical formula	Name of compound
(i) Al₂ (SO₄)₃	Aluminium Sulphate
(ii) CaCl₂	Calcium Chloride
(iii) K₂ SO₄	Potassium Sulphate
(iv) KNO₃	Potassium Nitrate
(v) CaCO₃	Calcium Carbonate

Question 3.
What is meant by the term chemical formula?
Answer:
The chemical formula of a compound is the symbolic representation of its composition,

Question 4.
How many atoms are present in a
(i) H₂S molecule
(ii) PO₄^3-ion?
Answer:
(i) Number of atoms in H₂S = 2 + 1 = 3
(ii) Number of atoms in PO^3-₄ ion = 1 + 4 = 5

Questions from Sub-section 3.5

Question 1.
Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, and CH₃OH.
Solution:
(1) Atomic mass of hydrogen = lu
H₂ contains two atoms of hydrogen.
Molecular mass of H₂ = 2 × 1 = 2u

(2) Atomic mass of oxygen = 16u
O₂ consists of two atoms of oxygen.
The molecular mass of O₂

(3) Atomic mass of chlorine Cl₂ consists of two atoms of chlorine.
The molecular mass of Cl₂

(4) Atomic mass of carbon = 12u
The atomic mass of oxygen = 16u
.’. CO₂ in which there is one atom of carbon and two atoms of oxygen.
Molecular mass of CO₂ =1 × 12 + 2 × 16= 12 + 32 = 44u

(5) Atomic mass of carbon = 12u
The atomic mass of hydrogen = lu
CH₄ where one atom of carbon and four atoms of hydrogen are there.
Molecular mass of CH₄ = 1 × 12 + 4 × 1 = 12 + 4 = 16u

(6) Atomic mass of carbon = 12u
The atomic mass of hydrogen = lu
C₂H₆ in which two atoms of carbon and six atoms of hydrogen are there.
Molecular mass of C₂H₆ = 2 × 12 + 6 × 1 = 24 + 6 = 30u

(7) Atomic mass of carbon = 12u
The atomic mass of hydrogen = 1u
C₂H₄ in which two atoms of carbon and four atoms of hydrogen are there.
Molecular mass of C₂H₄ = 2 × 12 + 4 × 1 = 24 + 4 = 28u

(8) Atomic mass of nitrogen = 14u.
The atomic mass of hydrogen = 1u
.’. NH₃ in which one atom of nitrogen and three atoms of hydrogen are there.
Molecular mass of NH₃ =1 × 14 + 3 × 1 = 14 + 3 = 17u

(9) Atomic mass of carbon = 12u
The atomic mass of hydrogen = 1u
The atomic mass of oxygen = 16u
CH₃OH, in which one atom of carbon, four atoms of hydrogen and one atom of oxygen are there.
Molecular mass of CH₃OH = 1 × 12 + 4 × 1 + 16 × 1 = 12 + 4 + 16 = 32u

Question 2.
Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
(1) Formula unit mass of ZnO = 1 × 65u + 1 × 16u = 65u + 16u = 81u
(2) Formula unit mass of Na₂O = 2 × 23u + 1 × 16u = 46u + 16u = 62u
(3) Formula unit mass of K₂CO₃ = 2 × 39u + 1 x 12u + 3 × 16u = 78u + 12u + 48u = 138u

Questions from Sub-section 3.5.3

Question 1.
If one mole of carbon atoms weighs 12 grams, what is the mass (in gram) of 1 atom of carbon 7
Solution:
Here,
Molar mass of carbon = 12g
1 mole = 6.022 x 10²³atom
That is, mass of 6.022 x 10²³ carbon atoms = 12 g
Mass of 1 carbon atom = \(\frac{12}{6.022 \times 10^{23}}\) g
= 1.993 x 10^-23g

Question 2.
Which has more atoms, 100 grams of sodium or 100 grams of iron (given, the atomic mass of Na = 23 u, Fe = 56u) ?
Solution:
The molar mass of sodium = 23g
1 mole = 6.022 x 10²³ atoms
Numbers of atoms in 23g of Na = 6.022 x 10²³

Molar mass of iron = 56g
1 mole = 6.022 x 10²³Numbers of atoms in 56g of Na = 6.022 x 10²³
= 10.75 × 10²³
Thus, there will be more atoms in 100 g of sodium as compared to 100 g of iron.

HBSE 9th Class Science Atoms and Molecules Textbook Questions and Answers

Question 1.
A 0.24 g sample of a 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:
Quantity of boron in 0.24g compound = 0.096 g
Quantity of boron in 1g compound = \(\frac{0.096}{0.24}\)
Quantity of boron in 100g compound = \(\frac{0.096}{0.24} \times 100\) = 40g
Therefore, the quantity of boron in the compound = 40%
Quantity of oxygen in 0.24 g compound = 0.144g
Quantity of oxygen in 0.24 g compound = \(\frac{0.144}{0.24} \times 100\) = 60g
Therefore, the quantity of oxygen in the compound = 60%

Question 2.
When 3.0 g 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 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Answer:
3.0g carbon when burnt in 8.00g oxygen produces 11.00 g carbon dioxide, then on burning 3.00 g carbon in 50.00 g oxygen, 53.00 g carbon dioxide will be formed which is based on the law of conservation of mass of chemical combination.

Question 3.
What are polyatomic ions? Give examples.
Answer:
A group of atoms that acts as ions are called as polyatomic ions. For example:

Polyatomic ions	Symbol
ammonium	NH⁺₄
hydroxide	OH^–
nitrate	NO^–₃
hydrogen carbonate	HCO^–₃
carbonate	CO^2-₃
sulphate	SO^2-₄
phosphate	PO^3-₄

Question 4.
Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer:
(a) Formula of magnesium chloride

∴ Formula of magnesium chloride = MgCl₂

(b) Formula of calcium chloride

∴ The formula of calcium chloride = CaCl₂

(d) Formula of aluminium chloride

∴ Formula of aluminium chloride = AlCl₃

(e) Formula of Calcium Carbonate chloride

∴ The formula of calcium carbonate = CaCO₃

Question 5.
Give the names of the elements present in the following compounds :
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:
(a) Quick lime = Ca (OH)₂
Thus, in quick lime, the present elements are calcium (Ca), oxygen (O) and hydrogen (H).

(b) Hydrogen hromide = HBr
Thus, in hydrogen bromide, the present elements are hydrogen (H) and bromine (Br).

(c) Baking powder = NaHCO₃
Thus, in baking powder, the present elements are sodium (Na), hydrogen (H), carbon (C) and oxygen (O).

(d) Potassium sulphate = K₂SO₄
Thus, in potassium sulphate, the present elements are potassium (K), sulphur (S) and oxygen (O).

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, HCl
(e) Nitric acid, HNO₃
Solution:
We know that C = 12, H = 1, S = 32, P = 31, Cl = 35.5, N = 14, O = 16
(a) Molar mass of ethyne (C₂H₂) = 2 × 12 + 2 × 1 = 24 + 2 = 26g
(b) Molar mass of sulphur molecule (S₈) = 8 × 32 = 256g
(c) Molar mass of phosphorus molecule (P₄) = 4 × 31 = 124g
(d) Hydrochloric acid, HCl = 1 × 1 + 1 × 35.5 = 1 + 35.5 = 36.5g
(e) Nitric acid, HNO₃ = 1 × 1 + 1 × 14 + 3 × 16= 1 + 14 + 48 = 63g

Question 7.
What Is the mass of
(a) 1 mole of nitrogen atoms?
(b) 4 moles aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles sodium suiphite (Na₂SO₃)?
Solution:
(a) Atomic mass of 1 mole of nitrogen = 14 g
(b) Atomic mass of 1 mole of aluminium = 27 g
Atomic mass of 4 moIes of aluminium = 4 × 27 = 108 g
(c) Atomic mass of 1 mole of sodium suiphite (Na₂SO₃) = 2 × 23 + 1 × 32 + 3 ×16 = 46 + 32 + 48 = 126 g
Atomic mass of 10 moles of sodium sulphite (Na₂SO₃) = 10 × 126 g = 1260 g

Question 8.
Convert Into mole.
(a) 12 g oxygen gas
(b) 20 g water
(c) 22 g carbon dioxide
Solution:
(a) We know that 1 mole oxygen(O₂) 2 × 16 = 32 g
Therefore, 32g oxygen = 1 mole
1g oxygen = \(\frac {1}{2}\) mole
12g oxygen = \(\frac {1}{2}\) × 12 = 0.375 mole

(b) We know that l mole water (H₂O) = (2 × 1 + 1 × 16)g = (2 +1 6)g = 18g
Therefore, 18g water = 1 mole
1g water = \(\frac {1}{18}\) mole
20 gwater = \(\frac {1}{18}\) × 20 = 1.11 mole

(c) We know that I mole carbon dioxide (CO₂) = (1 × 12 + 2 × 16)g= (12 + 32)g = 44g
Therefore, 44g carbon dioxide = 1 mole
1g carbon dioxide = \(\frac {1}{44}\) mole
22g carbon dioxide = \(\frac {1}{44}\) × 22 = 0.5 mole

Question 9.
What is the mass of
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Solution:
(a) I mole of oxygen atoms = 16g
0.2 mole of oxygen atoms 16 × 0.2g = 3.2g
(b) 1 mole of water molecules (H₂O) (2 × 1 + 1 × 16)g = (2 + 16) g = 18g
0.5 mole of water molecules = 0.5 × 18 = 9.0g

Question 10.
Calculate the number of molecules of sulphur (S₈) present in 16g of solid sulphur.
Solution:
Molecular mass of sulphur S₈ = 8 × 32 = 256g
1 mole = 6.022 x 10²³ molecule
Therefore, the no. of molecules in 256g sulphur – 6.022 × 10²³
The no. of molecules m 1 g sulphur = \(\frac{6.022 \times 10^{23}}{256}\)
The no. of molecules in 16 g sulphur = \(\frac{6.022 \times 10^{23}}{256}\) × 16 = 3.76 x 10²²
Therefore, in 16g of solid sulphur, there will be 3.76 × 10²² molecules.

Question 11.
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Al₂O₃).
(Hint: The mass of an ion is the same as that of an atom of the same element. The atomic mass of Al = 27u)
Solution:
Mass of I mole aluminium oxide (Al₂O₃) = (2 x 27 + 3 × 16)g = 102g
102g of aluminium oxide = 1 mole
∴ 0.051g of aluminium oxide = \(\frac {1}{102}\) × 0.05 1 = 5 × 10^-4 mole
Number of Aliens in 1 mole Al₂O₃ 2 × 6.022 × 10²³
Number of ions in 5 × 10^-4 mole Al₂O₃ 2 × 6.022 × 10²³ × 5 x 10^-4= 6.022 × 10²⁰

HBSE 9th Class Science Solutions Chapter 3 Atoms and Molecules Read More »

HBSE 9th Class Science Solutions Chapter 2 Is Matter Around Us Pure

September 19, 2022 / Class 9 / Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 2 Is Matter Around Us Pure Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 2 Is Matter Around Us Pure

HBSE 9th Class Science Is Matter Around Us Pure Intext Questions and Answers

Questions from Sub-section 2.1

Question 1.
What is meant by a substance?
Answer:
A substance is a pure single form of matter. It consists of a single type of particle i.e. all the constituent particles in the substance are identical in their chemical nature.

Question 2.
List the points of differences between homogeneous and heterogeneous mixtures.
Answer:
Homogeneous mixtures: Mixtures that have uniform composition are called homogeneous mixtures. Heterogeneous mixtures: Mixtures that have non-uniform composition are called heterogeneous mixtures.

Questions from Sub-section 2.2

Question 1.
Differentiate between homogeneous and heterogeneous mixtures with examples.
Answer:
Homogeneous mixtures: Those mixtures that have uniform composition throughout their masses are called homogeneous mixtures, for example, a mixture of sugar in water, a mixture of salt in water, a mixture of alcohol in water.
Heterogeneous mixtures: Those mixtures that do not have uniform composition throughout their masses are called as heterogeneous mixtures, for example, a mixture of sand and salt, a mixture of salt and sugar.

Question 2.
How are the solution, suspension and colloid (sol) different from each other?
Answer:
Following are the differences among solution, suspension, colloid (sol.):

Solution:
1. This solution is homogeneous and transparent, e.g. solution of salt in water.
2. Here, the size of the solute particle is 10^-9m.
3. Here, the particles of solute cannot be seen under the microscope.
4. Here the solute particles cannot be separated by filtration.
5. Due to the small size, the particles of solution cannot scatter the rays of light passing through it. So that in the solution the path of light is not visible.

Suspension:
1. This solution is heterogeneous and opaque, e.g., muddy water, paint.
2. Here, the size of solute parti-cles used to be 10⁷ m or more than of that.
3. Here, the solute particles can be seen with naked eyes too.
4. Here, the solute particles can be separated by the filtration method.
5. Suspended particles scattered the rays of light, by which its path become visible

Colloid (Sol.):
1. This, the solution is homogeneous but little transparent e.g., milk, blood, ink, tooth-paste.
2. Here, the size of solute particles is between 10⁹ to 10⁷ m, i.e., their size is bigger than the size of solution particles.
3. Here, the solute particles can only be seen through the powerful microscope.
4. Here, the solute particles also cannot be separated by filtration.
5. Colloidal particles are as big as these scattered the rays of light and make its path, visible.’

Question 3.
To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Solution:
At temperature 293K –
Mass of solute substance (Sodium Chloride) = 36g
Mass of solvent (Water) = 100g
Mass of solution = Mass of solute substance + Mass of solvent = 36g + 100g = 136g

= \(\frac {36}{136}\) × 100 = 26.47

Questions from Sub-section 2.3

Question 1.
How will you separate a mixture containing kerosene and petrol, which are miscible with each other ? The difference in their boiling points is more than 25°C.
Answer:
Petrol and kerosene oil which are miscible to each other, their mixture is separated by the fractional distillation method. This method is based upon the fact that different components have different boiling points. Because the difference of boiling points of petrol and kerosene is more than 25°C, liquid with low boiling point will separate first and after some intervals liquid with high boiling point will be separated after becoming distilled.

Question 2.
Name the technique to separate:
(i) butter from curd
(ii) salt from sea-water
(iii) camphor from salt
Answer:
(i) Butter is separated from curd by centrifugation method.
(ii) Salt from sea water is separated by an evaporation method.
(iii) Camphor is separated from salt by the sublimation method.

Question 3.
What type of mixtures are separated by the technique of crystallization?
Answer:
Crystallization is a method by which pure solid is separated from a solution in the form of a pure crystal for example obtaining of salt from seawater.

Questions from Sub-section 2.4

Question 1.
Classify the following as chemical and physical 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 the water breaking down into hydrogen and oxygen gases,
dissolving common salt in water,
making a fruit salad with raw fruits, and
burning of paper and wood.

Answer:
Chemical change: Rusting of almirah; passing of electric current, through the water and the water breaking down into hydrogen and oxygen gases; burning of paper and wood.

Physical change: Cutting of trees, melting of butter in a pan; boiling of water to form steam, dissolving of common salt in water; making of fruit salad with raw fruits.

Question 2.
Try segregating the things around you as pure substances or mixtures.
Answer:
Pure Substances: Iron, gold, silver, copper, aluminium, sugar, salt etc.
Mixture: Sea-water, minerals, soil, air, beverages etc.

HBSE 9th Class Science Is Matter Around Us Pure Textbook Questions and Answers

Question 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) Butter from curd.
(e) Oil from water.
(f) Tea leaves from tea.
(g) Iron pins from sand.
(h) Wheat grains from husk.
(i) Fine mud particles suspended in water.
(j) hues from the essence of the crushed flower petals.
Answer:
(a) To separate sodium chloride from the solution of water, the evaporation method is adopted.
(b) To separate ammonium chloride from a mixture of sodium chloride and ammonium chloride, a sublimation method is adopted since ammonium chloride is a volatile substance.
(c) To separate a piece of metal from the engine oil of car, the filtration method is adopted.
(d) In order to separate butter from curd centrifugation method is adopted.
(e) To separate oil from water separating funnel is used, since water and oil both are immiscible liquids.
(f) To separate tea leaves from tea, the filtration method is brought in use. For filtration, tea strainer is used.
(g) To separate iron from sand, the magnetic separation method is adopted, because iron is attracted towards the magnet.
(h) To separate wheat grains from chaff, the threshing method is adopted, because with the threshing method chaff being lighter in weight, flies away with the wind and the wheat grains being heavier in weight, directly falls bn the ground.

(i) Tiny particles of soil floating in water can be separated by means of loading method for the particulates of soil get heavier by alum and thus, they settle down at the bottom.
(j)To separate various hues (dyes) from the essence of the crushed flower petals chromatography method is followed.

Question 2.
Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer:
We will prepare tea by using the given words in the following manner:
1. Solvent: Take water in the pan in the form of solvent and keep it on the burner.
2. Solute: Add sugar to water in the form of solute.
3. Solution: The mixture of water and sugar will become the solution.
4. Dissolve: Sugar will dissolve in water and make a solution.
5. Soluble: Sugar dissolves in water being miscible and after boiling, it is also a soluble substance in milk.
6. Insoluble: In the mixture of water and sugar, add tea leaves in the form of an insoluble substance and boil it.
7. Filtrate and residue: After boiling of tea leaves, filter the tea with a filtrate strainer. Use filtrate tea to drink arid and throw away the residue remaining in the strainer.

Question 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 substance dissolved in 100 grams of water to form a saturated solution:

Substance Dissolved	Temperature in Kelvin (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 SO grams of water at 313 K?
(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 a change of temperature on the solubility of a salt?

Solution:
(a) According to the Question:
The essential quantity of potassium nitrate for a saturated solution of potassium nitrate in 100 g of water at 313 K= 62 g
The required quantity of potassium nitrate for a saturated solution of potassium nitrate in 1 g of water at 313 K = \(\frac {62}{100}\)g
The required quantity of potassium nitrate for saturated solution in 50 g of water at 313 K = \(\frac {62}{100}\) × 50g = 31 g

(b) Pragya obtains a saturated solution of potassium chloride at 353 K and leaves the solution at room temperature (293 K) to cool down when the solution will cool down, then it will be a most saturated solution because at room temperature, it will have (54-35) 19 g more potassium chloride than saturation.

(c) At 293 K temperature in 100 g of water the solubility of potassium nitrate, sodium chloride, potassium chloride and ammonium chloride are 32 g, 36 g, 35 g, and 37 g, respectively. So, at this temperature, the solubility of ammonium chloride salt will be the most.

(d) On changing the temperature, the solubility of the salt change positively, i.e. the solubility of salt increases with the increase in temperature.

Question 4.
Explain the following with examples:
(a) saturated solution
(A) pure substance
(c) colloid
(d) suspension.
Answer:
(a) Saturated Solution:
If at the given fixed temperature, the solute does not dissolve in solution, it is called a saturated solution, i.e., at the given temperature when in a solution the solute dissolves more than the capacity of the solution, it is called saturated solution. For instance-take 50 ml of water in a beaker, now add little quantity of salt into it gradually and stir it, when the salt stops dissolving any more, then it will be called a saturated solution.

(b) Pure Substance:
Matter formed of molecules Of a similar type is called as a pure substance. Either element or compound is pure like iron, gold, silver, sugar, water etc.

(c) Colloid:
Colloid is a heterogeneous mixture whose molecules are of the size in between lnm to 100 nm. These molecules cannot be seen with naked eyes and they diverge the rays of light; like – milk, shaving cream, toothpaste, jelly, face cream etc.

(d) Suspension:
Suspension is a heterogeneous mixture in which a solute substance does not dissolve, rather than they remain suspended in the medium. Suspended molecules are bigger in size than 100 nm (10⁷m) and can be seen with naked eyes like contaminated water of a river, the mixture of thick lime mortar stones and water, etc.

Question 5.
Classify each of the following as a homogeneous and heterogeneous mixture: soda water, wood, ice, air, soil, vinegar, filtered tea.
Answer:
Homogeneous Mixture: Soda water, ice, vinegar, filtered tea.
Heterogeneous Mixture: Wood, air, soil.

Question 6.
How would you confirm that a colorless liquid given to you is pure water?
Answer:
We will find out the boiling point of the given colorless liquid. If that boiling point comes out to 373 K, we will approve that the given colorless liquid is pure water, but if it does not fulfill the condition then it is not pure water.

Question 7.
Which of the following materials fall in the category of a “pure substance”?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air.
Answer:
In the given substances following are the pure substances :
(a) ice
(c) iron
(d) hydrochloric acid
(e) calcium oxide
(f) Mercury.

Question 8.
Identify the solutions among the following mixtures :
(a) soil
(b) Seawater
(c) Air
(d) Coal
(e) Soda water.
Answer:
Soda water is a solution.

Question 9.
Which of the following will show the “Tyndall effect”?
(a) Salt solution
(A) Milk
(c) Copper sulfate solution
(d) Starch solution.
Answer:
Milk exhibits the Tyndall effect.

Question 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.
Answer:
Element:
(a) Sodium
(d) Silver
(f) Tin
(g) Silicon.

Compound:
(e) Calcium carbonate
(j) Soap
(A) Methane
(f) Carbon dioxide.

Mixture:
(A) Soil
(c) Sugar solution
(h) Coal
(i) Air
(m) Blood.

Question 11.
Which of the following are chemical changes?
(a) Growth of a plant
(h) 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 a candle.
Answer:
Chemical changes are as follows:
(a) growth of a plant
(h) rusting of iron
(d) cooking of food
(e) digestion of food
(g) burning of a candle.

HBSE 9th Class Science Solutions Chapter 2 Is Matter Around Us Pure Read More »

HBSE 9th Class Science Solutions Chapter 1 Matter in Our Surroundings

September 19, 2022 / Class 9 / Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 1 Matter in Our Surroundings Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 1 Matter in Our Surroundings

HBSE 9th Class Science Matter in Our Surroundings Intext Questions and Answers

Questions from Sub-sections 1.1,1.2

Question 1.
Which of the following matter? Chair, air, love, smell, hate, almonds, thought, cold, cold drink, the smell of perfume.
Answer:
Chair, air, almonds, and cold drink matter because they occupy space and mass.

Question 2.
Give reasons for the following observation:
The smell of hot sizzling food reaches you several metres away, but to get the smell of cold food you have to go close.
Answer:
Since, the diffusion becomes faster with the increase in temperature, therefore the smell of hot sizzling? food reaches us from several metres away, whereas to smell the flavour of cold food, we have to go close to them.

Question 3.
A diver is able to cut through water in a swimming pool? Which property of matter does this observation show?
Answer:
Due to comparatively more distance in between the particles of water, it has the property of compressibility in it, for this reason, a diver able to cut through water in the swimming pool.

Question 4.
What are the characteristics of the particles of matter?
Answer:
The characteristics of the particles of matter are as follows:

The particles of matter have empty spaces in between them.
The particles of matter constantly remain in motion i.e., they have kinetic energy in them.
The kinetic energy of particles of matter increases with an increase in temperature.
The particles of matter automatically get mixed finally.
The particles of matter attract one another.

Questions from Sub-section 1.3

Question 1.

The mass per unit volume of a substance is called density. (Density = mass/volume) Arrange the following in order of increasing density – air, exhaust from chimneys, honey, water, chalk, cotton and iron.
Answer:
On writing the given substances in order of increasing density, the following sequence will be obtained:
exhaust from chimneys, air, cotton, chalk, water, honey and iron.

Question 2.
(a) Tabulate the differences in the characteristics of states of matter.
(b) Comment upon the following: rigidity, compressibility, fluidity, Ailing a gas container, shape, kinetic energy and density.
Answer:
(a) Given differences are found in case of different properties of different states of matter:

Property	Solids	Liquids	Gases
Size	They have definite size.	They do not have definite size.	They do not have definite size.
Volume	They have definite volume.	They too have definite volume.	They do not have definite volume.
Hardness	They are hard.	They are not hard.	They do not have this property.
Piling up	They can be piled up.	They flow.	They also flow.
Capacity of compression	They cannot be compressed.	They cannot be compressed.	They can be compressed.
Position of Particles	Particles are very close with each other.	Particles are away from each other.	They have too much vacant space in between particles.

(b) (i) Rigidity: The force of cohesion working in between the particles of a substance determines the rigidity of the substance. Due to the maximum force of cohesion in solids, rigidity is there. Liquids have yet lesser and the gasses have the least rigidity.

(ii) Compressibility: Reducing of interstitial gaps in the particles of a substance with the application of external force is called compressibility. Gases have the property of compressibility.

(iii) Fluidity: Materials that have the property to flow are known as liquid materials. In liquids, the property of fluidity is there.

(iv) Filling of gas Container: Due to weak intermolecular force between the gas particles in a container, the gas occupies the whole of the provided space, which means gases do not have a definite volume. The gases are filled in the container at high pressure.

(v) Shape: Since the particles of solids are attached to more intermolecular force, hence they provide a definite shape to the solids. Whereas, in liquids the intermolecular force is comparatively low, due to this reason liquids do not have a definite shape. The same is the case with gases too.

(vi) Kinetic Energy: The energy produced due to the speed of particles is called kinetic energy. On increasing the temperature the kinetic energy of particles also increases. There is not much kinetic energy in solids, some kinetic energy in liquids whereas gases have high kinetic energy. Density of soilds is high, low in liquids whereas it is nil in gases.

(vii) Density: The mass per unit volume of a substance is called density, i.e.,

Question 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 air but to do the same through a solid block of wood we need a karate expert
Answer:
(a) As the volume of gas is not definite, the gas completely fills up die vessel in which it is kept.

(b) In the gaseous state the speed of particles is irregular and it is fairly random. Due to this irregular speed, the particles of gas collide with themselves and with the walls of the container. Due to the cohesive force exerted by die particles of gas per unit area; on the walls of the container, the pressure of the gas is maintained.

(c) As the wooden table constitutes a definite shape, also it has a definite volume and the property of incompressibility; therefore, it is called to be solid.

(d) Due to maximising space in the particles of air, we can move our hand quite easily in the air, whereas, due to less intermolecular space between the particles in the piece of wood, we have to be quite expert at Karate Technique.

Question 4.
Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.
Answer:
In comparison to solids, liquids have low density, but water has maximum density at 4°C. When it is cooled down below 4°C, the ice starts thawing at 0°C which has lower density than the density of water. Therefore ice cube floats on the surface of water.

Questions from Sub-section 1.4

Question 1.
Convert the following temperatures to Celsius scale:
(a) 300 K
(b) 573 K.
Answer:
(a) 300 K = (300 – 273)°C = 27°C
(b) 573 K = (573 -273)°C = 300°C

Question 2.
What is the physical state of water at:
(a) 250°C
(b) 100°C?
Answer:
(a) At 250°C water remains boiling.
(b) At 100°C water will start boiling, because this is the boiling point of water.

Question 3.
For any substance, why does the temperature remain constant during the change of state?
Answer:
During the change of state in any material the temperature remains constant because during the change of state the entire heat energy being yielded to the material brings under control intercohesive force of the particles, which is utilised to change the state of a material.

Question 4.
Suggest a method to liquefy atmospheric gases.
Answer:
The atmospheric gases present in the air zone are confined to a utensil and by increasing its pressure and reducing the temperature, they can be converted further into liquid.

Questions from Sub-section 1.5

Question 1.
Why does a desert cooler cool better on a hot dry day?
Answer:
On a hot and dry day, due to lack of humidity, the air is generally dry, when this dry air is sucked inwards by the exhaust fan of the desert cooler, then the water trickling on the mats of the desert cooler turns into vapours and thereby sucks the heat present in the dry air, as a result, the air around cools down. This cool air cools down the room from the inside.

Question 2.
How does the water kept in an earthen pot (matka) become cool during summer?
Answer:
The earthen pot consists of very small pores in its walls through which the water leaks out from inside the pot with capillarity process. These small drops of water get evaporated and it receives heat energy from water itself for evaporation. Consequently, the water kept in the earthen pot remains cold.

Question 3.
Why does our palm feel cold when we put some acetone or petrol or perfume on it?
Answer:
By pouring down acetone/petrol or perfume on the palm, its particles eliminate heat energy from the palm and further, evaporate it into air due to which the palm feels cold.

Question 4.
Why are we able to sip hot tea or milk faster from a saucer rather than a cup?
Answer:
The surface area of the saucer is far more than that of a cup due to which, the evaporation in case of saucer is more rapid from its surface and with that the hot milk or tea comparatively cools down quicker and due to their cooling down, they can be drunk quickly.

Question 5.
What type of clothes should we wear in summer?
Answer:
In summer, we should wear cotton clothes, it is because of the reason that in summer, due to physical activity we perspire more, with a result we feel cool. As we know during evaporation the particles of the liquid’s surface receive energy from our body and further convert it into vapours.

The exposed dormant heat energy of evaporation combines with the equal amount of heat energy absorbed from our body, which lets our body feel cool. By wearing cotton garments the absorption of water gets more, consequently the sweat after absorption into it gets easily evaporated into the atmosphere.

HBSE 9th Class Science Matter in Our Surroundings Textbook Questions and Answers

Question 1.
Convert the following temperatures to the Celsius scale:
(a) 293 K
(b) 4701L
Answer:
(a) 293K = (293 – 273)°C = 20°C
(b) 470K = (470-273)°C = 197°C

Question 2.
Convert the following temperatures to the Kelvin scale:
(a) 25°C
(b) 373°C.
Answer:
(a) 25°C = (25 ÷ 273)K = 298 K
(b) 373°C = (373 + 273)K = 646 K

Question 3.
Give the reason for the following observations.
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.
Answer:
(a)Naphthalene ball is after some time without losing its solid state disappear, it is because of the reason that naphthalene is a volatile substance which directly converts from a solid state into a gaseous state.

(b) We can feel the fragrance of perfume sitting several metres away because perfume has the property to diffuse into air.

Question 4.
Arrange the following substances in increasing order of forces of attraction between the particles: water, sugar, and oxygen.
Answer:
In increasing order the applicable intermolecular force among the molecules will be as follows:
Oxygen < Water < Sugar.

Question 5.
What is the physical state of water:
(a) 25°C,
(b) 0°C, (c) 100°C?
Answer:
(a) At 25°C water will be in a liquid state.
(b) At 0°C water will be in a solid state (ice).
(c) At 100°C water will be in a boiling state (steam).

Question 6.
Give two reasons to justify:
(a) water at room temperature is a liquid.
(b) an iron almirah is solid at room temperature.
Answer:

(a) Water at room temperature is in a liquid state, because

At this temperature the applied force of cohesion between its molecules used to be normal.
Its molecules have normal elliptical speed.

(b) The iron almirah at room temperature is in solid state, because

The applicable force of cohesion in between the molecules of iron is fairly excessive.
The gap distance between the molecules of iron is almost negligible due to that reason they can move around at a fixed distance.

Question 7.
Why is ice at 273 K more effective in cooling than water at the same temperature?
Answer:
At 273 K temperature, in comparison to water, ice releases more coolness, because in ice inexposed melting heat energy is more.

Question 8.
What produces more severe burns, boiling water or steam?
Answer:
Rather, steam will give severe burning sensation it is because of the reasons that steam does have extra heat energy in it which is called in exposed heat energy of evaporation.

Question 9.
Name A, B, C, D, E and F in the following diagram showing changes in its state:

Answer:
(A) Melting
(B) Vaporisation
(C) Condensation
(D) Freezing
(E) Sublimation
(F) Sublima