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Kerala SCERT Class 9 Chemistry Chapter 4 Solutions Redox Reactions

Kerala Syllabus Std 9 Chemistry Chapter 4 Redox Reactions Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 4 Let Us Assess Answers Redox Reactions

Question 1.
The unbalanced chemical equation regarding the formation of ammonia from nitrogen and hydrogen is given below.
N₂ + H₂ → NH₃
a) Balance the chemical equation.
b) Find out the total number of atoms of the same type in both the reactants and the products.
c) If 28 g of nitrogen combines with 6 g of hydrogen, find out the mass of ammonia formed. (Hint: Atomic mass H = 1 u N = 14 u)
Answer:
a) N₂ + 3H₂ → NH₃

b)

Elements	Nitrogen	Hydrogen
Reactants	2	6
Products	2	6

c) The balanced chemical equation for the formation of Ammonia
N₂ + 3H₂ → 2NH₃
28g + 6g → 2 × 17 (34g)
Mass of ammonia formed = 34g

Question 2.
C + 4HNO₃ → 2H₂O + CO₂ + 4NO₂
a) Find out and mark the oxidation number of carbon in this reaction.
b) What happens to the oxidation number of carbon in this reaction?
c) What happens to carbon – oxidation or reduction?
d) What are the oxidising and reducing agents in this reaction?
Answer:
a) C + 4HNO₃ → 2H₂O + CO₂ + 4NO₂
In reactants carbon is in the elemental state, So the oxidation number is 0.
In products, let the oxidation number of carbon be ‘x’
x + (2x – 2) = 0 1
x + – 4 = 0
x = +4

b) Increases from zero to +4.

c) Oxidation

d) Oxidising agent – HNO₃
Reducing agent – C

Question 3.
Find out the oxidation number of sulphur in the following compounds.
(Hint: Oxidation number H = +1, O = -2)
a) SO₂
b) SO₃
c) H₂SO₃
d) H₂SO₄
Answer:
a) SO₂
S + (2 × -2) = 0
S + -4 = 0
S = +4

b) SO₃
S + (-2 × 3) = 0
S + -6 = 0
S = + 6

c) H₂SO₃
(+1 × 2) + S + (-2 × 3) = 0
+ 2 + S + -6 = 0
S + -4 = 0
S = + 4

d) H₂SO₄
(+1 × 2) + S + (-2 × 4) = 0
+ 2 + S + -8 = 0
S + -6 = 0
S = +6

Question 4.
Certain statements are given below. Write whether they are true or false.
a) The process involving an increase in oxidation number is oxidation.
b) The process involving a decrease ini oxidation number is oxidation.
c) In a chemical reaction, oxidising agent undergoes reduction.
d) In a chemical reaction, oxidising agent undergoes oxidation.
Answer:
a) True
b) False
c) True
d) False

Question 5.
Balance the chemical equations given below.
a) SO₂ + O₂ → SO₃
b) H₂O₂ → H₂O + O₂
c) CH₄ + O₂ → H₂O + CO₂
d) Fe + HCl → FeCl₂ + H₂
Answer:
a) 2SO₂ + O₂ → 2SO₃
b) 2H₂O₂ → 2H₂O + O₂
c) CH₄ + 2O₂ → 2H₂O + CO₂
d) Fe + 2HCl → FeCl₂ + H₂

Question 6.
Two chemical reactions are given below. Find out the oxidation number of atoms and check whether these reactions are redox reactions.
a) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
b) Zn + 2HCl → ZnCl₂ + H₂
Answer:
a)
Not a redox reaction. Oxidation and reduction do not take place.

b)
It is a redox reaction, because oxidation and reduction take place simultaneously-.

Question 7.
The gaseous fuel carbon monoxide burns in oxygen to form carbon dioxide.
a) Write the balanced equation of this chemical reaction.
b) Is this a redox reaction? Why?
c) What is the oxidising agent in this reaction? What is the reducing agent?
Answer:
a) 2CO + O₂ → 2CO₂

b)
Yes, this is a redox reaction and reduction takes place simultaneously.

c) Oxidising agent – Oxygen (O)
Reducing agent – Carbon monoxide (CO)

Question 8.
Analyse the chemical equation given below.
Zn + 2HCl → ZnCl₂ + H₂
a) Mark the oxidation number of atoms before and after the chemical reaction.
b) Which atom undergoes oxidation?
c) Which atom undergoes reduction?
d) What are the oxidising and reducing agents?
Answer:
a)
b) Zn
c) H
d) Oxidising agent – HCl
Reducing agent – Zn

Question 9.
Analyse the chemical equations given below and find out whether they are redox reactions.
a) NaOH + HCl → NaCl + H₂O
b) H₂S + Cl₂ → 2HCl + S
Answer:
a)
It is not a redox reaction. No oxidation and reduction reactions.

b)
It is a redox reaction, because oxidation and reduction reactions takes place simultaneously.

Question 10.
A chemical reaction is given in the concept map below. Find out the oxidation number of each atom. On the basis of this, fill up the blanks.
(Hint: Valency S = 2 Fe = 2)

Answer:

Extended Activities

Question 1.
Mix iron powder and sulphur in the mass ratio 7 : 4 in a china dish. Heat the mixture well. After some time cool the china dish. Check whether iron can be separated using magnet.
Examine whether the product dissolves in carbon disulphide.
What is your inference?
Write down the equation of the chemical reaction. Check whether it is a redox reaction.
Answer:
Iron and sulphur on heating combines to gether to form Iron sulphide (ferrous sulphide). Thereafter iron cannot be separated using a magnet. The product obtained does not dissolve in carbon disulphide.

The constituents of a compound never retain their fundamental properties. (Iron is attracted by a magnet, sulphur dissolves in carbon disulphide).

This is a redox reaction because oxidation and reduction take place simultaneously

Question 2.
Take some sand in a tray. Place calcium carbide (CaC₂) on it. Place some more sand on top of it. Place some ice cubes on the sand. Ignite the ice cubes carefully. What do you see?
Calcium carbide reacts with water and forms acetylene (C₂H₂) gas. Acetylene is an inflammable gas.
Write the chemical equation of the combustion.
Check whether it is a redox reaction.
Answer:

It is a redox reaction.
Because oxidation and reduction take place simultaneously.

Question 3.
Make a mixture of aluminium powder and powdered iodine crystals in the mass ratio 1 : 2. Make a heap of it in a china dish. Make a small hole at the top of the heap. Add one or two drops of water into the hole. What do you see?
Here aluminium and iodine combine to form aluminium triiodide.
The valency of Al= 3 I = 1
a) Write the equation of the chemical reaction.
b) Find out the oxidation number of each atom. Check whether it is a redox reaction.
Answer:
a) 2Al + 3I₂ → 2AlI₃
b)
It is a redox reaction.
Because oxidation and reduction take place simultaneously.

Question 4.
Conduct a study tour to understand the importance of redox reactions in industry.
Answer:
You can visit one among the following to understand the importance of Redox reactions:

• Visit a metal refinery (e.g., aluminum or copper). Observe the smelting process, where a reducing agent (like carbon) extracts the desired metal from its oxide ore through a redox reaction. Learn about the role of electrolysis in purifying metals, another application of redox reactions.
• Visit a factory producing a metal product (e.g., car parts or appliances). Discuss how corrosion protection techniques, like galvanization (coating with zinc), utilize redox reactions to prevent rust.
• Tour a power plant (e.g., coal or natural gas). Understand the combustion process, a redox reaction releasing energy to generate electricity. Discuss the environmental implications of these reactions and potential cleaner alternatives.
• Visit a battery manufacturing facility. Learn about the different types of batteries (e.g., lithium – ion) and how they rely on redox reactions to store and release electrical energy. Explore advancements in battery technology.
• Tour a chemical plant producing a product that utilizes redox reactions (e.g., chlorine bleach or fertilizers). Understand the specific redox chemistry involved in the manufacturing process. Discuss safety protocols for handling these chemicals.
• Visit a water treatment facility. Leam about water disinfection methods like ozonation, which uses redox reactions to eliminate bacteria and contaminants. Explore wastewater treatment processes that employ redox reactions to break down pollutants.

Tips and Tricks for Your Redox Reactions Study Tour:
Preparation:

• Pre-tour Research: For each industry visit, delve deeper. Research the specific redox reactions . involved and the role of different chemicals. This will allow you to ask insightful questions during the tours.
• Focus and Note Taking: With so much information, prioritize! Identify key aspects of each visit that align with your learning goals. Take clear notes with diagrams or sketches for better recall.
• Bring Curiosity. Don’t be afraid to ask questions. The best way to learn is through active engagement with industry professionals.

During the Tour:

• Observe Keenly: Pay close attention to the processes and equipment used. Look for evidence of redox reactions happening (e.g., color changes, gas evolution).
• Engage with the Guides: Ask clarifying questions when something.is unclear. Don’t hesitate to request additional information or demonstrations.
• Network with Professionals: Introduce yourself and exchange contact information with industry experts. This can be valuable for future research or career opportunities.

Post-Tour:

• Consolidate Your Learnings: Review your notes and revisit the research you conducted. Organize the information to identify patterns and connections across different industries,
• Group Discussions: Share your experiences and insights with fellow participants. Discuss the broader implications of redox reactions in industry and brainstorm potential future applications.
• Project or Presentation: Consider creating a project or presentation summarizing your learnings. This will help solidify your understanding and allow you to share your knowledge with others.

Redox Reactions Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
What changes are generally observed during chemical reactions?
Answer:

• Different types of energy are absorbed or released.
• Occur change in the number of substances
• Occur colour change.
• Occur change in physical state.
• New substances are formed.

Question 2.
Let us do an experiment.
Take a trough and fill three-fourths of it with water. Add two drops of phenolphthalein to it and stir well. Cut a small piece of sodium and put it into the trough carefully.

a) What changes can be observed?
Answer:

• Sodium reacts with water vigorously.
• Sodium moves faster on the surface of the water and catches fire.
• A gas is evolved.
• Water in the trough turns pink.

b) What is the reason? Analyse the chemical equation and find out.
2Na + 2H₂O → 2NaOH + H₂↑
Answer:
Reason – Sodium reacts with water, forming sodium hydroxide alkali and hydrogen. In the presence of alkali, phenolphthalein turns pink. During this reaction, heat is liberated. Because of this heat, hydrogen gas bums.

Question 3.
Is there any change in the total mass of the substances during chemical reactions?
Answer:
No, there are no changes happening in the total mass of the substances during chemical reactions.

Question 4.
Take 20 mL barium chloride (BaCl₂) solution in a beaker. Take 20 mL sodium sulphate (Na₂SO₄) solution in another beaker. Place both the beakers together on an electronic balance and note the reading.
Now, pour the solution from one beaker to the other.

a) What do you observe?
Answer:
A chemical reaction takes place, and a white precipitate is formed.

b) After some time, note the reading of the electronic balance again. Compare this with the previous reading. What is your inference?

Answer:
We can see that the reading is not having any changes from the previous reading. Both are the same. From this, we can infer that the total mass of the reactants and the total mass of the products in the chemical reaction is constant.

c) Is there any change in the total mass as a result of this chemical reaction?
Answer:
No

d) Let us write the equation of this chemical reaction.
Answer:
BaCl₂ + Na₂SO₄ → BaSO₄ ↓ + 2NaCl
In this chemical reaction, barium chloride reacts with sodium sulphate to form barium sulphate and sodium chloride.

Question 5.
Take 20 mL of dilute hydrochloric acid (HCl) in a conical flask. Drop some zinc (Zn) granules in a balloon. Fix the balloon firmly to the mouth of the conical flask, as shown in the figure. Place the conical flask on an electronic balance and note the mass. Then, carefully raise the balloon and drop the zinc granules into the acid in the flask.

a) What do you see?
Answer:
Zinc reacts vigorously with hydrochloric acid. A gas evolves. The balloon gets inflated as the gas is collected in it.

b) Note the reading of the electronic balance. Compare this reading with the previous one. What do you understand?
Answer:
There is no change in both readings.

c) Which gas is collected in the balloon?
Answer:
Hydrogen gas (H₂)

d) Let us write the equation of the chemical reaction,
Answer:
Zn + 2HCl → ZnCl₂ + H₂ ↑

Question 6.
What can be inferred from these experiments?
Answer:
The total mass of the reactants and products does not change due to chemical reactions.

Question 7.
Does the total mass change during chemical reactions?
Answer:
No.
During chemical reactions, there is no change in the total mass of substances.
During the combustion of fuels and the burning paper, the total mass of the reactants and the products is constant. The major products formed during these reactions are carbon dioxide and water vapour. They are lost in the atmosphere. But if we collect these products without any loss and weighed, we can see that There will be no change in total mass.

Based on experiments and observations, the French scientist Antoine Lavoisier stated the law of conservation of mass,

LAW OF CONSERVATION OF MASS In a chemical reaction, the total mass of the reactants will be equal to the total mass of the products.

Question 8.
Carbon and oxygen combine to form carbon dioxide. Analyse the symbolic representation of this chemical reaction.

a) Complete table given below.

Total mass of the reactants	……………………….
Total mass of the products	……………………….

Answer:

Total mass of the reactants	12 u + 32 u = 44 u
Total mass of the products	16u + 12u + 16u = 44u

b) Record your inference.
Answer:
The total mass of the reactants is equal to the total mass of the products in a chemical reaction.

Question 9.
Methane (CH₄) burns in air to form carbon dioxide and water vapour. The symbolic representation of this chemical reaction is given below.

a) Write down the equation of the chemical reaction.
Answer:
CH₄ + 2O₂ → CO₂ + 2H₂O

b) Check whether this chemical reaction obeys the law of conservation of mass.
(Hint: H = 1 u, C = 12 u, O = 16 u.)
Answer:
Total mass of the reactants = (1 × 4) + 12 + (16 × 4)
= 4 + 12 + 64
= 80
Total mass of the products = (16 × 2) + 12 + (18 × 2)
= 32 + 12 + 36
= 80
The total mass of the reactants is equal to the total mass of the products. So, this chemical reaction obeys the law of conservation of mass.

Question 10.
You know that oxygen and hydrogen are diatomic molecules.
a) How are these molecules represented using symbols?
Oxygen ………………….., Hydrogen …………………
Answer:
Oxygen – O₂
Hydrogen – H₂

b) What is the total number of atoms in water (H₂O) molecule?
Answer:
3 (Three)

c) Calculate the number of molecules and the total number of atoms present in 5H₂O.
Total number of molecules …………….. Total number of atoms ………………..
Answer:
Total number of molecules – 5
Total number of atoms – 5 × 3 = 15
Balancing the chemical equation of the formation of water from hydrogen and oxygen:
Step 1 :

Number of atoms in the reactants	Hydrogen = 2	Oxygen = 2
Number of atoms in the products	Hydrogen = 2	Oxygen = 1

According to the law of conservation of mass, the total number of atoms should be the same.
So, the number of oxygen atoms in the products and reactants should be the same. We must make the number of oxygen atom must be 2. For this make the number of water molecules 2.

Step 2: H₂ + O₂ → 2H₂O
Number of atoms in the reactants
Number of atoms in the products

Number of atoms in the reactants	Hydrogen = 2	Oxygen = 2
Number of atoms in the products	Hydrogen = 4	Oxygen = 2

Now, the number of Hydrogen atoms is not the same. So we have to make it as four.

Step 3: 2H₂ + O₂ → 2H₂O

Number of atoms in the reactants	Hydrogen = 4	Oxygen = 2
Number of atoms in the products	Hydrogen = 4	Oxygen = 2

Now the number of atoms in reactants and the products are the same. Now we can say that the reaction is balanced.
The balanced chemical equation for this reaction is:
2H₂ + O₂ → 2H₂O

Balancing a chemical equation is the method of equalising the number of the same type of atoms in both the reactants and the products. The equation thus obtained is known as a balanced chemical equation.

More examples:
• Magnesium + Oxygen → Magnesium Oxide
Step 1: Mg + O₂ → MgO
Reactants
Mg – 1
O – 2
Products:
Mg – 1
O – 1
Number of oxygen atoms is not the same. Making it equal.

Step 2: Mg + O₂ → 2MgO
Reactants
Mg – 1
O – 2
Products:
Mg – 2
O – 2
Number of Magnesium atoms is not equal. Making it equal.

Step 3: 2Mg + O₂ → 2MgO
Reactants
Mg – 2
O – 2
Products:
Mg – 1
O – 1
Now, the number of both atoms is equal on both sides.

Balanced Chemical Equation: 2Mg + O₂ → 2MgO

• Hydrogen + Chlorine → Hydrogen chloride
Step 1: H₂ + Cl₂ → HCl
Reactants
H – 2
Cl – 2
Products:
H – 1
Cl – 1
Number of both the atoms are not same. Making it equal.

Step 2: H₂ + Cl₂ → 2HCl
Reactants
H – 2
Cl – 2
Products:
H – 2
Cl – 2
Now, the number of both atoms is equal on both sides. The equation is now balanced.

Balanced Chemical Equation: H₂ + Cl₂ → 2HCl
• Aluminium + Oxygen → Aluminium oxide
Step 1: Al + O₂ → Al₂O₃
Reactants
Al – 1
O – 2
Products:
Al – 2
O – 3
(Number of oxygen atoms is not the equal. Making it equal)

Step 2: Al + 3O₂ → Al₂O₃
Reactants
Al – 1
O – 6
Products:
Al – 2
O – 3
(Again, number of oxygen atoms is not the same. Making it equal))

Step 3: Al + 3O₂ → 2Al₂O₃
Reactants
Al – 1
O – 6
Products:
Al – 4
O – 6
Now, the number of oxygen atoms is the same. But the number of Aluminium atoms is not the same. Making it equal.

Step 4: 4Al + 3O₂ → 2Al₂O₃
Reactants
Al – 4
O – 6
Products:
Al- 4
O – 6
Now, the number of both atoms is equal on the reactant and products. The equation is balanced.

Balanced Chemical Equation: 4Al + 3O₂ → 2Al₂O₃
• Nitrogen + Hydrogen → Ammonia
Step 1: N₂ + H₂ → NH₃
Reactants
N – 2
H – 2
Products:
N – 1
H – 3
Making the number of Nitrogen atoms equal.

Step 2: N₂ + H₂ → 2NH₃
Reactants
N – 2
H – 2
Products:
N – 2
H – 6
Now, the number of hydrogen atoms is not the same. Making it equal.

Step 3: N₂ + 3H₂ → 2NH₃
Reactants
N – 2
H – 6
Products:
N – 2
H – 6
Now, the number of both atoms is equal on the reactant and products. The equation is balanced.
Balanced Chemical Equation:N₂ + 3H₂ → 2NH₃

Question 11.
Balance the chemical equations given below and record them in science diary.
a) H₂ + I₂ → HI
b) Na + H₂O → NaOH + H₂
c) Mg + HCl → MgCl₂ + H₂
Answer:
a) H₂ + I₂ → HI
Step 1: H₂ + I₂ → HI
Step 2: H₂ + I₂ → 2HI
Balanced Chemical Equation: H₂ + I₂ → 2HI

b) Na + H₂O → NaOH + H₂
Step 1: Na + 2H₂O → NaOH + H₂
Step 2: Na + 2H₂O → 2NaOH + H₂
Step 3: 2Na + 2H₂O → 2NaOH + H₂
Balanced Chemical Equation: 2Na + 2H₂O → 2NaOH + H₂

c) Mg + HCl → MgCl₂ + H₂
Step 1: Mg + HCl → MgCl₂ + H₂
Step 2: Mg + 2HCl → MgCl₂ + H₂
Balanced Chemical Equation: Mg + 2HCl → MgCl₂ + H₂

Question 12.
Look at the chemical equation regarding the formation of sodium chloride.
2Na + Cl₂ → 2NaCl
a) Which atom undergoes oxidation?
Answer:
Sodium (Loses electron)

b) Which atom supports oxidation? (sodium/chlorine)
Answer:
Chlorine (gains electron)

c) Which atom gets reduced? olojDceadm nilauxnxBac&jcm and)Qo
Answer:
Chlorine (gains electron)

d) Which atom helps reduction?
Answer:
Sodium (loses electron)

The species that helps oxidation in a chemical reaction is the Oxidising agent. The oxidising agent gets reduced in a chemical reaction.
The species that helps reduction is the reducing agent. The reducing agent gets oxidised in a chemical reaction.

Question 13.
Analyse the chemical reactions given below and complete the table.
1) Mg + F₂ → MgF₂
2) Ca + Cl₂ → CaCl₂
3) 4Fe + 3O₂ → 2Fe₂O₃

Answer:

Question 14.
The oxidation number of magnesium is +2 and that of oxygen is -2 in magnesium oxide (MgO). What do you understand from this?
Answer:
MgO (Magnesium oxide) is composed of Magnesium ion (Mg²⁺) and Oxide ion (O^2-).
When an electron is lost, a positive ion is formed and when an electron is gained, a negative ion is formed. Magnesium loses two electrons and become a positively charged ion, oxide gains two electrons and becomes negatively charged electron. The positively charged Magnesium ion (Mg²⁺) and the negatively charged oxide ion (O^2-) combine together to form Magnesium oxide (MgO). Covalent compounds are formed by the sharing of electrons. In such compounds, the oxidation number is assigned assuming that the shared electrons are shifted to the more electronegative element.

Example – In the covalent compound HF, it is considered that the more electronegative fluorine (F) attracts the electron pair and attains -1 oxidation number. Hydrogen is assumed to lose one electron and it attains +1 oxidation number.

The sum of the oxidation numbers of all atoms in a compound is zero.
In element molecules, electrons are equally shared by the atoms. So, at elemental state the oxidation number is considered to be zero.

Question 15.
Find out the oxidation number of nitrogen in HNO₂ and NO₂.
Answer:
HNO₂
+ 1 + N + (-2 × 2) = 0
+ 1 + N + – 4 = 0
N + -3 = 0
N = +3
Oxidation number of nitrogen in HNO₃ is +3.

NO₂
N + (-2 × 2) = 0
N + -4 = 0
N = +4
Oxidation number of nitrogen in NO₂ is +4.

Finding out the oxidation number of chromium (Cr) in potassium dichromate (K₂Cr₂O₇).
If the oxidation number of chromium is considered to be ‘x’.

Question 16.
Find out the oxidation number of chromium in Cr₂O₃.
Answer:
2Cr + (-2 × 3) = 0
2Cr + -6 = 0
2Cr = +6
Cr = \(\frac{+6}{2}\) = +3
Oxidation number of Chromium in Cr₂O₃ is + 3.

Question 17.
Find out the oxidation number of manganese (Mn) in the following compounds and record it in your science diary. (Hint: Oxidation number of O = -2, K = +1.)
a) MnO₂
b) Mn₂O₇
c) KMnO₄
Answer:
a) MnO₂
Mn + (-2 × 2) = 0
Mn + -4 = 0
Mn = +4
Oxidation number of Manganese in MnO₂ is +4.

b) Mn₂O₇
2Mn + (-2 × 2) = 0
2Mn + -14 = 0
2Mn = +14
Mn = \(\frac{+14}{2}\) = +7
Oxidation number of Manganese in Mn₂O₇ is +7.

c) KMnO₄
+ 1 + Mn + (-2 × 4) = 0
+ 1 + Mn + -8 = 0
Mn + -7 -0
Mn = +7
Oxidation number of manganese in KMnO₄ is +7.

Question 18.
Analysing the chemical equation of the formation of sodium chloride (NaCl).
2Na + Cl₂ → 2NaCl
a) What is the oxidation number of sodium and chlorine in their elemental state?
Answer:
Zero

b) Write the the chemical equation including their oxidation numbers.
Answer:

c) What happened to the oxidation number of sodium as a result of this reaction (increased/decreased)?
Answer:
Increased (Increased from 0 to +1)

d) What happened to the oxidation number of chlorine?
Answer:
Decreased (Decreased from 0 to -1)
Oxidation number increases during oxidation reactions. Reduction reactions involve a decrease in oxidation number.

e) Which atom undergoes oxidation during the formation of sodium chloride?
Answer:
Sodium

f) What is the oxidising agent in this reaction? Why?
Answer:
Chlorine. Because Chlorine helps oxidation by gaining electron.

g) Which atom undergoes reduction in this reaction? Why?
Answer:
Chlorine. Because the oxidation number of chlorine decreased from zero to -1.

g) What is the reducing agent in this case?
Answer:
Sodium. Because podium loses electron.

Question 19.
Analyse the chemical equation given below. Find out the oxidation number of atoms and complete the table given below.
H₂ + Cl₂ → 2HCl

Answer:

Question 20.
Let us analyse another chemical equation.
Mg + 2HCl → MgCl₂ + H₂

a) The oxidation number of magnesium changes from ………………. to ……………….
Answer:
Zero to +2

b) The change that happened to magnesium, (oxidation/ reduction).
Answer:
Oxidation

c) What is the oxidising agent in this case? ………………… (Mg/HCl)
Answer:
HCl

d) What is the reducing agent? (Mg/HCl)
Answer:
Mg

Question 21.
Analyse the chemical reaction given below and complete the table given below.

Answer:

Question 22.
The equation for the chemical reaction between hydrogen and chlorine to form hydrogen chloride is given below.

a) Which atom has undergone oxidation in this reaction?
Answer:
Hydrogen (H₂)

b) Which atom has undergone reduction?
Answer:
Chlorine (Cl)

A chemical reaction in which oxidation and reduction take place simultaneously is a redox reaction. In a redox reaction oxidising agent gets reduced and the reducing agent gets oxidised.

Familiar Redox reactions:

• Glucose molecules decompose and release energy during cellular respiration.
• Formation of oxide coating on the surface of metals.
• Combustion of fuels.
• Decomposition of organic substances in the presence of oxygen.
• Production of electricity in electrochemical cells.

Question 23.
Analyse the above redox reactions and present a seminar on the importance of redox reactions in daily life.
Answer:
Hints that can be used for seminar preparation

Redox Reactions

• Briefly explain chemical reactions as the rearranging of atoms to form new substances.
• Introduce the concept of redox reactions (reduction-oxidation) involving electron transfer.
• Mention that one element loses electrons (oxidation) while another gains them (reduction).

Redox Reactions in Action
• Energy for Life:

• Explain cellular respiration, where glucose (food) breaks down using oxygen, releasing energy to power our cells.
• Highlight that this is a redox reaction with glucose losing electrons (oxidized) and oxygen gaining them (reduced).
• Briefly mention ATP as the energy currency in cells produced during this process.

• Fire and Fuel:

• Explain combustion of fuels like wood or gasoline as a redox reaction.
• Show how the fuel loses electrons (oxidized) while reacting with oxygen (reduced), releasing heat and light energy.
• Briefly mention the importance of combustion for cooking, heating, and transportation.

• Metal Mysteries:

• Explain rusting of iron as a redox reaction.
• Show how iron atoms lose electrons (oxidized) to oxygen and water Vapour in the environment, forming the reddish-brown rust.
• Briefly mention that painting or using galvanized steel can prevent rust (oxidation).

Amazing Applications
• Briefly discuss how redox reactions are used in:

• Batteries: Electrochemical cells where reactions convert chemical energy to electrical energy (e.g., AA batteries).
• Food Preservation: Adding antioxidants (reducing agents) to food slows down spoilage by preventing oxidation.

Real-world Examples and Activities

• Show a simple demonstration (depending on safety guidelines):
  Observe the browning of an apple slice (oxidation) or the burning of a steel wool pad – (oxidation).
• Encourage class mates to brainstorm other examples of redox reactions in daily life (e.g., food
  browning, bleaching clothes).

Conclusion

• Summarize how redox reactions are crucial for our energy needs, material protection, and technological advancements.
• Briefly discuss the negative aspects of some redox reactions, like corrosion, and ways to mitigate them.

Students rely on Kerala Syllabus 9th Standard Chemistry Notes Pdf Download Chapter 3 Chemical Bonding Extra Questions and Answers to help self-study at home.

Kerala Syllabus Std 9 Chemistry Chapter 3 Chemical Bonding Extra Questions and Answers

Question 1.
What peculiarity do you see in the electronic configuration of noble elements except Helium?
Answer:
Except Helium all other elements have 8 electrons in the outermost shell, hence shall be considered to be chemically stable.

Question 2.
What is called an octet electron configuration?
Answer:
The arrangement of eight electrons in the outermost shell of an atom is called octet electron configuration.

Question 3.
Why helium is stable?
Answer:
In the Helium atom, there is only one shell. The maximum number of electrons in the first shell is 2. Hence, the electron pattern system of Helium is also stable.

Question 4.
The electronic configuration of some elements is given below.

Element	Atomic mass	Electronic configuration
Magnesium	12	2, 8, 2
Oxygen	8	2, 6
Sodium	11	2, 8, 1
Chlorine	17	2, 8, 7

(i) You are familiar with the compounds of these elements. Write the names of some compounds.
(ii) How are atoms in these compounds held together?
Answer:
(i) Magnesium chloride, Sodium oxide, Sodium chloride, Magnesium oxide.
(ii) Strong, attractive force holds these atoms together in a compound.

Question 5.
What is meant by Chemical Bonding?
Answer:
The attractive force that holds the atoms together in the formation of a molecule is called chemical bonding.

Question 6.
How many electrons are there in the outermost shell of sodium atom?
Answer:
Only one electron.

Question 7.
How the ionic bond formation of sodium oxide is represented? [Hint: Atomic No. of sodium 11, oxygen 8]
Answer:
Sodium oxide is Na₂O.

Question 8.
In the formation of sodium chloride which atoms are combining?
Answer:
Sodium, chlorine

Question 9.
How many electrons are there in outermost shell of chlorine?
Answer:
7

Question 10.
How do chlorine and sodium attain stability?
Answer:
Sodium donates one electron to chlorine to become a sodium ion [Na⁺], and chlorine gains one electron to become a chloride ion [Cl^–].

Question 11.
Draw the electron dot diagram of the transference of electrons of a sodium atom and chlorine atom in the formation of a sodium chloride molecule.
Answer:

Question 12.
Explain about the electron transfer during the formation of sodium chloride.
Answer:
During the formation of sodium chloride sodium atom donates an electron and gets converted to a sodium ion (Na⁺). Chlorine accepts an electron to form a chloride ion (Cl^–). Through this, sodium and chlorine atoms complete an octet in their outermost shell to attain stability. The electrostatic force of attraction holds together the oppositely charged ions thus formed. This attractive force is called the Ionic Bond.

Question 13.
Define Ionic Bond?
Answer:
An ionic bond is a chemical bond formed by electron transfer. In an ionic bond, the ions are held together by the electrostatic force of attraction between the oppositely charged ions.

Question 14.
Draw the electron dot diagram of the following compounds.
A) Sodium Fluoride and
B) Magnesium Fluoride [Hint: Atomic No. Na = 11, F = 9, Mg = 12].
Answer:

The distribution of electrons of fluorine is given.

Question 15.
What happens during the formation of fluorine molecule, electron transfer or electron sharing?
Answer:
Electron sharing

Question 16.
How covalent bonds are formed?
Answer:
The chemical bond formed as a result of the sharing of electrons between the combining atoms is called a covalent bond.

Question 17.
How single bonds are formed?
Answer:
Single bonds are formed by sharing one pair of electrons. A small line between the symbols of the combining elements represents it. For example, the Fluorine molecule can be represented as F – F.

Question 18.
Draw the electron dot diagram of the formation of a chlorine, molecule by combining two chlorine atoms.
Answer:

Chlorine atoms share one pair of electrons, forming a single covalent bond.

Question 19.
Complete the table given below related to covalent bonding.

Element molecule	Shared electron pairs	Chemical bond
F2	One pair	Single bond
Cl2	—	—
O2	—	—
N2	—	—

Answer:

Element molecule	Shared electron pairs	Chemical bond
F2	One pair	Single bond
Cl2	One pair	Single bond
O2	Two pair	Double bond
N2	Three pair	Triple bond

Question 20.
Examples of some covalent compounds are given. Draw the chemical bonds of the compounds using an electron dot diagram.
a) CH₄
b) HF
C) H₂O
Answer:

Question 21.
Who proposed the electronegativity scale?
Answer:
Linus Pauling

Question 22.
Explain the properties of ionic compounds and covalent compounds.
Answer:

Ionic compounds	Covalent compounds
Generally solids.	Either solid, liquid or gas
Soluble in water	Insoluble in water. But soluble in organic solvents like kerosene, benzene etc.
Conduct electricity in a fused or solution state.	Do not conduct electricity.
High melting and boiling point	Low melting and boiling point.

Question 23.
In the formation of magnesium oxide, how many electrons are donated by magnesium?
Answer:
Two electrons

Question 24.
What is the valency of oxygen? ood&cnJlsaanjj cuaejorSmfl o^)t«nauo6nf?
Answer:
Two

Question 25.
Write down the formulae of
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide
Answer:

Question 26.
Write down the names of compounds represented by the following formulae:
(i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃
Answer:
(i) Aluminium sulphate [Al₂(SO₄)₃]
(ii) Calcium chloride [CaCl₂]
(iii) Potassium sulphate [K₂SO₄]
(iv) Potassium nitrate [KNO₃]
(v) Calcium carbonate [CaCO₃]

Question 27.
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 is calcium oxide (CaO) – Ca and O
(b) Hydrogen bromide (HBr) – H and Br
(c) Baking powder (NaHCO₃) – Na, H, C and O
(d) Potassium sulphate (K₂SO₄) – K, S and O

Question 28.
Which of the following represents a correct chemical formula? Name it.
a) CaCl
b) BiPO₄
c) NaSO
d) NaS
Answer:
b) BiPO₄ – Bismuth phosphate
The correct formulas for other compounds are,
Calcium chloride – CaCl₂, Sodium sulphate – Na₂SO₄, Sodium sulphide – Na₂S.

Question 29.
Which type of force holds the atoms together in an ionic compound?
Answer:
Electrostatic force of attraction of oppositely charged ions.

Question 30.
The formation of sodium fluoride molecule is represented below.

Identify the type of bonding in this molecule.
Answer:
Ionic bonding

Question 31.
Electronic configuration of magnesium is 2, 8, 2. What is the valency of magnesium?
Answer:
Two.

Question 32.
Which type of force holds the atoms together in an ionic compound?
Answer:
Electrostatic force of attraction of oppositely charged ions.

Question 33.
Find the odd one out. [Ar, Ne, Kr, He]
Answer:
He (Duplet configuration)

Question 34.
Draw the electron dot structure of H₂S.
Answer:

Question 35.
Which is the anion of sodium sulphate?
Answer:
Sulphate (SO₄^2-) is the anion.

Question 36.
Which noble gas has no octet arrangement in the outermost shell of the atom?
Answer:
Helium (He)

Question 37.
The force that binds together constituent particles of a molecule is ………………….
Answer:
Chemical bonding

Question 38.
What change in energy takes place when a molecule is formed from its atoms?
Answer:
Energy decreases

Question 39.
Some elements and their electronic configuration are given in the table. (Hint: Symbols are not real).

Element	Electronic configuration
A	2, 8, 1
B	2, 8
C	2, 8, 7

a) Among these elements, which has the highest stability? Give reason.
Answer:
a) B
Reason: Element B has 8 electrons in the outermost shell. (Octet rule)

Question 40.
From the following statements, write those applicable to ionic compounds.
a) Usually do not dissolve in water.
b) Conduct electricity in the molten state and aqueous solution.
c) Generally not a conductor of electricity.
d) Exists in the solid state.
Answer:
b) Conduct electricity in a molten state and aqueous solution
d) Exists in a solid state.

Question 41.
Identify the relation and fill up the blank
(i) Electron dot diagram: Gilbert Newton Lewis
Electronegativity scale: ……………………..
(ii) Elements : Symbols
Compounds : ………………….
Answer:
(i) Electronegativity scale – Linus Pauling
(ii) Compounds: Chemical formula

Question 42.
Analyse the electronic configurations of the elements given below and answer the following questions, (symbols are not real).
p – 2, 7
Q – 2, 8, 2
R – 2, 8, 8
a. R is the most stable element. Why?
b. Which one of these elements donates its electrons in chemical reactions?
Answer:
a) Because of the octet electron configuration. (The arrangement of 8 electrons in the outermost shell).
b) Q

Question 43.
How many atoms are present in a (i) H₂S molecule and (ii) PO₄^3- ion?
Answer:
(i) H₂S
2 Hydrogen atoms + 1 sulphur atom = 3 atoms
(ii) PO₄^3-
1 phosphorous atoms + 4 oxygen atoms = 5 atoms.

Question 44.
In NaCl, sodium and chlorine are held together by the attraction between anions and cations in it. Identify the anion and cation in the NaCl molecule.
Answer:
Anion Cl^–
Cation Na⁺

Question 45.
Choose the compounds containing ionic bonds and covalent bonds out of the following.
HCl, NaCl, MgO, CH₄, CCl₄, MgCl₂
Answer:
Ionic bond – NaCl, MgO, MgCl₂ Covalent bond – HCl, ₄, CCl₄

Question 46.
The electronic configuration of oxygen is 2, 6.
a) Oxygen usually shows valency 2. What is the reason?
b) Illustrate the formation of sodium oxide using an electron-dot diagram.
(Hint: Electronic configuration of Na = 2, 8, 1)
Answer:
a) Oxygen has 6 electrons in the outermost shell. Thus, it tends to accept 2 electrons to attain stability. This usually shows valency 2.
b)

Question 47.
MgF₂ is an ionic compound. [Hint: Atomic number of Mg = 12, F = 9]
a) Which is the cation present in MgF₂?
b) Write the electronic configuration of the anion in this compound.
c) Write one property of ionic compounds.
Answer:
a) Magnesium ion – Mg²⁺
b) Anion is F^–.
Electronic configuration – 2, 8.
c) They are good conductors of electricity in a molten and aqueous state.

Question 48.
What are ionic and molecular compounds? Give examples.
Answer:

• Ionic compounds are made up of ions.
• Ionic compounds have strong electrostatic forces of attraction, called ionic bonds or electrovalent bonds, that hold cations and anions together.
• Examples are Sodium chloride, calcium oxide, etc.
• In molecular compounds, element atoms share electrons via covalent connections.
• Examples include methane (CH₄) and water (H₂O).

Question 49.
The electron dot diagram of the formation of sodium oxide is given below

(Hint: Atomic number Na = 11 and O = 8)
a) Which atom donates an electron in this reaction?
b) Write the electronic configuration of the oxide ion (O2 ).
c) Write the name of the cation.
d) Which type of chemical bonding is present in sodium oxide?
Answer:
a) Sodium (Na)
b) 2, 8
b) Sodium ion
c) Ionic bonding

Question 50.
The electron dot diagram of magnesium oxide is given below.
a) Which atom accepts electrons?

b) Write the equation showing the formation of Mg²⁺ ions.
c) Name the anion present in it.
d) Write the electronic configuration of the Mg²⁺ ion.
Answer:
a) Oxygen (O)
b) Mg → Mg²⁺ + 2e^–
c) Oxide ion (O^2-)
d) 2, 8

Question 51.
a) Why noble gases do not take part in chemical reactions?
b) Noble gas helium doesn’t have an octet electron configuration in the outermost shell. Yet, it is stable. Why?
Answer:
a) Noble gases do not take part in chemical reactions because they have attained stable octet electron configuration (except helium) in the outermost shell. So, they are chemically stable, b) Helium atom has only one shell (K) in it. The maximum number of electrons that can be accommodated in the first K shell is 2. Helium has 2 electrons in it. So it is stable.

Question 52.
Which among the following is a molecule showing a polar nature?
(H₂, HBr, N₂, Cl₂)
Answer:
HBr

Question 53.
Nitrogen (N₂) molecule doesn’t show polar nature. Why?
Answer:
It is a homodiatomic molecule in which the two atoms have the same electronegativity values.

Question 54.
Write the chemical formula of sodium sulphide.
Answer:
Sodium – 1
Sulphide – 2
Formula – Na₂S

Question 55.
How many pairs of electrons are mutually shared in the formation of a nitrogen molecule?
Answer:
3 pairs

Question 56.
Write the chemical formula of the salt formed by the reaction of ammonium ion and sulphate ion.
Answer:
(NH₄)₂SO₄ (Ammonium sulphate)

Question 57.
Write down the name of the compound represented by the formula KNO₃.
Answer:
Potassium nitrate

Question 58.
What is meant by the term chemical formula?
Answer:
A chemical formula is the most concise method of representing a compound using symbols and the valency of elements.

Question 59.
The chemical formula of calcium carbonate is CaCO₃ then what is the valency of Ca?
Answer:
The valency of calcium is 2.

Question 60.
The formula of a compound is X₃Y. The valencies of elements X and Y will be, _________ respectively
Answer:
Valency of X – 1
Valency of Y – 3

Question 61.
Magnesium ion (Mg²⁺) and Phosphate ion (PO₄^3-) combine to form magnesium phosphate.
a) Identify the cation in magnesium phosphate.
b) Write the chemical formula of magnesium phosphate.
Answer:
a) Magnesium ion (Mg²⁺)
b) Mg₃(PO₄₂)

Question 62.
Write the chemical formula of baking soda and caustic soda.
Answer:
Baking soda is the common name used for sodium hydrogen carbonate, and caustic soda is the name for sodium hydroxide.
Baking soda – NaHCO₃
Caustic soda – NaOH

Question 63.
Some elements and their valencies are given.

Element	Valency
Ba	2
Cl	1
O	2

a) Write the chemical formula of Barium chloride.
b) The chemical formula of Calcium oxide is CaO. What is the valency of Ca?
Answer:
a)BaCl₂
b) The valency of calcium is 2.

Question 64.
Identify the Ionic and Covalent compounds from the following.
a) CaO
b) CO₂
[Hint: Electronegativity values of O = 3.44, C = 2.5, Ca = 1.0]
Answer:
a) CaO is an ionic compound (3.44 – 1.0 = 2.44)
2.44 >1.7; hence, ionic bonding is present in it.

b) CO₂ is a covalent compound (3.44 – 2 .5 = 0.94)
0.94 < 1.7; hence, covalent bonding is present in it.

Question 65.
The element ‘X’ forms a diatomic molecule with a double bond.
a) How many electrons are there in the outermost shell of ‘X’?
b) How many pairs of electrons are shared for this bond formation?
Answer:
a) a) 6
b) Two pair

Question 66.
Some elements and their electronic configuration are given below. (Symbols are not real)

Element	Electronic configuration
A	2, 8, 7
B	2, 6
C	2, 8, 2
D	2, 5

a) Which among the elements shows the same valency?
b) Write the chemical formula of the compound formed by the combination of C and A.
Answer:
a) B and C (Valency 2)
b) CA₂

Question 67.
Electronic configuration of Nitrogen is 2, 5.
a) How many electrons are required by nitrogen to attain octet electronic configuration?
b) Draw the electron dot diagram of the formation of nitrogen (N₂) molecules.
Answer:
a) 3
b)

Question 68.
The atomic numbers of the elements X and Y are 13 and 16, respectively.
a) Write the electronic configuration of X and Y
b) Write the valency of X and Y
c) Write the molecular formula of the compound formed by the combination of X and Y.
Answer:
a) ₁₃X = 2, 8, 3
Y = 2, 8, 6
b) Valency of X = 3
Valency of Y = 2
c) X₂Y₃

Question 69.
Electronegativity values of some elements are given.
H = 2.20
O = 3.44
Mg = 1.31
On the basis of electronegativity values, identify the type of chemical bond in H₂O and MgO. Give reasons.
Answer:
If the electronegativity difference is greater than 1.7, the compound is ionic, and if it is less than 1.7, the compound is covalent.

Question 70.
a) Draw the electron dot diagram of the formation of hydrogen chloride (HCl) molecule. [Hint: Electronic configuration H – 1, Cl -2, 8, 7]
b) The HCl molecule shows a polar nature. Why?
Answer:
a)

b) HCl is a polar molecule. This is because the Chlorine (Cl) atom in the HCl molecule is more electronegative and does not share the bonding electrons equally with Hydrogen (H).

Question 71.
Analyse the given electron dot diagram and answer the following questions.

a) Which type of chemical bond is represented here?
b) What is the valency of Chlorine in this compound?
c) Write the chemical formula of Aluminium chloride (Hint: Valency of Aluminium = 3)
Answer:
a) Covalent Bond
b) One
c) Valency of Al = 3 Valency of Cl = 1
Chemical formula = AlCl₃

Question 72.
Electronegativity values of some elements are given. Analyse these values and answer the following questions. [Electro negativity of S = 2.58,03.44, Ca 1.0, F = 3.98]
a) Complete the table

Compound	Type of bond
SO2	A
CaF2	B

b) Justify your answer.
Answer:
a)

Compound	Type of bond
SO2	Covalent
CaF2	Ionic

b) If the difference in electronegativity values of elements in a compound is 1.7 or more compound shows ionic character, and if it is less than 1.7 compound shows a covalent nature. For SO₂
Difference = 3.44 – 2.58 = 0.86
For CaF₂
Difference = 3.98 – 1.0 = 2.98

Question 73.
Carbon combines with chlorine to form a compound carbon tetrachloride. [Hint: Atomic Number: C = 6, Cl = 17]
a) How many electrons are required to complete the octet of a carbon atom?
b) How many atoms of chlorine should combine with a carbon atom to complete the octet?
c) Which type of covalent bond is present in carbon tetrachloride? (Single bond, Double bond, Triple bond)
Answer:
a) 4
b) 4
c) Single bond

Question 74.
The molecular formula of Carbon tetrachloride is CCl₄
[Hint: Electronegativity C = 2.55 and CI = 3.16; Atomic number C =6 and CI = 17]
a) How many electrons are present in the outermost shell of carbon?
b) Which type of chemical bonding is present in CCl₄?
c) Draw the electron dot diagram of the formation of CCl₄.
Answer:
a) 4
b) Covalent bonding (single bond)
C – 2.55
Cl – 3.16
Electronegativity difference = 3.16 – 2.55 = 0.61 <1.7

Thus, covalent bonding.
c)

Question 75.
Oxygen molecule (O₂) and Nitrogen molecule (N₂) are formed by covalent bonding.
a) Which type of covalent bond is formed in N₂ molecule.?
b) How many pairs of electrons are shared in the O₂ molecule?
c) Draw the electron dot diagram illustrating the formation of chemical bonds in the O₂ molecule [Hint: Atomic number of O = 8, N = 7]
Answer:
a) Triple bond
b) 2 pair
c)

Question 76.
The table lists the valencies of different elements.

Symbol of element/ion	Valency
O	2
Na	1
OH	1

a) What will be the chemical formula of sodium oxide?
b) How many oxygen atoms are there in 2Fe₂O₃?
c) How many atoms are there in H₂SO₄?
d) Write the chemical formula of sodium hydroxide,
Answer:
a) Na₂O
b) 6 oxygen atoms
c) 2H + 1 S + 40 = 7
d) NaOH is the chemical formula of sodium hydroxide.

Students can read Kerala SSLC Chemistry Question Paper March 2020 with Answers Malayalam Medium and Kerala SSLC Chemistry Previous Year Question Papers with Answers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Chemistry Question Paper March 2020 Malayalam Medium

Time: 1½ Hours
Total Score: 40

പൊതുനിർദ്ദേശങ്ങൾ :

• ആദ്യത്തെ 15 മിനിട്ട് സമാശ്വാസ സമയമാണ്. ഈ സമയം ചോദ്യങ്ങൾ വായിക്കുന്നതിനും ഉത്തരങ്ങൾ ആസൂത്രണം ചെയ്യുന്നതിനും ഉപയോഗിക്കാവുന്നതാണ്.
• ചോദ്യങ്ങളും നിർദ്ദേശങ്ങളും ശരിയായി വായിച്ചതിനുശേഷം മാത്രം ഉത്തരം എഴുതുക.
• ഓരോ വിഭാഗത്തിലും 5 ചോദ്യങ്ങൾ വീതം ഉണ്ട്. അവയിൽ ഏതെങ്കിലും 4 എണ്ണത്തിന് ഉത്തരം എഴുതുക.

വിഭാഗം – A

1 മുതൽ 5 വരെയുള്ള ചോദ്യങ്ങളിൽ ഏതെങ്കിലും 4 എണ്ണത്തിന് ഉത്തരമെഴുതിയാൽ മതി. (ഓരോ ചോദ്യത്തിനും 1 സ്കോർ വീതം)

Question 1.
‘f’ സബ്ഷെല്ലിൽ ഉൾക്കൊള്ളുന്ന പരമാവധി ഇലക്ട്രോണുക ളുടെ എണ്ണം എത്ര? (1)
(2, 6, 10, 14)
Answer:
14

Question 2.
ലോഹങ്ങൾ നേർപ്പിച്ച് ആസിഡുമായി പ്രവർത്തിച്ചാൽ ഉണ്ടാ കുന്ന വാതകമേത്? (1)
Answer:
ഹൈഡ്രജൻ (H₂)

Question 3.
ബന്ധം കണ്ടെത്തി പൂരിപ്പിക്കുക. (1)
ബോക്സൈറ്റ് : ലീച്ചിങ്ങ്
ടിൻസ്റ്റോൺ : ________________
Answer:
കാന്തിക വിഭജനം

Question 4.
പ്രകൃതിദത്ത റബ്ബറിന്റെ മോണോമർ __________________ ആണ്. (1)
(ഈതിൻ, വിനൈൽ ക്ലോറൈഡ്, ഐസോപ്രീൻ, ട്രാഫ്ളൂറോ ഈതിൻ)
Answer:
ഐസോപ്രീൻ

Question 5.
STP യിൽ സ്ഥിതി ചെയ്യുന്ന ഒരു മോൾ എത് വാതകത്തിന്റെയും വ്യാപ്തം _________________ L ആയിരിക്കും. (1)
Answer:
22.4 ലിറ്റർ

വിഭാഗം – B

(6 മുതൽ 10 വരെയുള്ള ചോദ്യങ്ങളിൽ ഏതെങ്കിലും 4 എണ്ണത്തിന് ഉത്തരമെ ഴുതിയാൽ മതി. ഓരോ ചോദ്യത്തിനും 2 സ്കോർ വീതം)

Question 6.
താഴെ തന്നിരിക്കുന്ന സന്ദർഭങ്ങളുമായി ബന്ധപ്പെട്ട വാതക നിയ മങ്ങൾ ബോക്സിൽ നിന്ന് തെരെഞ്ഞെടുത്തെഴുതുക.
(ബോയിൽ നിയമം, ചാൾസ് നിയമം, അവൊഗാഡ്രോ നിയമം)
(a) വായു നിറച്ച ഒരു ബലൂൺ വെയിലത്തു വച്ചാൽ അൽപസ മയത്തിനു ശേഷം പൊട്ടുന്നു.
(b) ഒരു അക്വേറിയത്തിന്റെ ചുവട്ടിൽ നിന്ന് ഉയരുന്ന വായുകു മിളയുടെ വലിപ്പം മുകളിലേക്ക് എത്തുംതോറും കൂടിവരുന്നു.
Answer:
(a) ചാൾസ് നിയും
(b) ബോയിൽ നിയമം

Question 7.
(a) താഴെ തന്നിരിക്കുന്നവയിൽ 24Cr ന്റെ ശരിയായ സബ്ഷെൽ ഇലക്ട്രോൺ വിന്യാസം കണ്ടെത്തുക.
(i) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(ii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
(b) ഇലക്ട്രോൺ വിന്യാസം തെരെഞ്ഞെടുക്കാൻ കാരണമെന്ത്?
Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p³ 3d⁵ 4s¹
(b) പകുതി നിറഞ്ഞ, d സബ്ഷെൽ (d6) ക്രമീകരണം സ്ഥിരത യുള്ളതാണ്.

Question 8.
(a) ചുവടെ നൽകിയിരിക്കുന്നവയിൽ ഉരുക്കിവേർതിരിക്കൽ മുഖേന ശുദ്ധീകരിക്കുന്ന ലോഹമേത്?
(സിങ്ക്, ഇരുമ്പ്, കോപ്പർ, ടിൻ)
(b) ലോഹത്തിന്റെ ഏത് പ്രത്യേകതയാണ് ഇവിടെ പ്രയോജനപ്പെ ടുത്തിയിരിക്കുന്നത്?
Answer:
(a) ടിൽ
(b) താഴ്ന്ന ദ്രവണാങ്കം

Question 9.
താഴെ കൊടുത്ത സംയുക്തങ്ങളുടെ ഘടനാ വാക്യങ്ങൾ ബോക്സിൽ നിന്നും തെരെഞ്ഞെടുത്തെഴുതുക.
(a) പ്രൊപ്പീൽ
(b) ബ്യൂട്ട്-1-ഐൻ

CH2 = CH2	CH3 – CH = CH2
CH3 – CH2 – C ≡ CH	CH3 – C ≡ C – CH3

Answer:
(a) പ്രോപ്പിൽ CH₃ – CH = CH₂
(b) ബ്യൂട്ട്-1-ഐൻ CH₃ – CH₂ – C ≡ CH

Question 10.
(a) വ്യാവസായികമായി എഥനോൾ നിർമ്മിക്കുന്നതെങ്ങനെ?
(b) വ്യാവസായികാവശങ്ങൾക്ക് ഡീനേച്ചേർഡ് സ്പിരിറ്റ് ഉപയോ ഗിക്കുന്നു. എന്താണ് ഡീനേച്ചേർഡ് സ്പിരിറ്റ്?
Answer:
(a) മൊളാസസിനെ (പഞ്ചസാര ലായനി) യീസ്റ്റ് ചേർത്ത് ഫെർമെന്റേഷൻ നടത്തിയാൻ എതനോൾ നിർമ്മിക്കുന്നത്.
(b) മദ്യപാനത്തിനുവേണ്ടി ദുരുപയോഗപ്പെടുത്തുന്നത് തടയാൻ വ്യാവസായിക ആവശ്വത്തിനുള്ള ഏതനോളിൽ വിഷപദാർഥ ങ്ങൾ ചേർക്കുന്നു. ഇങ്ങനെ കിട്ടുന്ന ഉൽപന്നത്തെ ഡിനേ ച്ചേർഡ് സ്പിരിറ്റ് എന്നു പറയുന്നു.

വിഭാഗം – C

(11 മുതൽ 15 വരെയുള്ള ചോരങ്ങളിൽ ഏതെങ്കിലും 4 എണ്ണത്തിന് ഉത്തരം ഴുതിയാൽ മതി. ഓരോ ചോളത്തിനും 3 സ്കോർ വീതം)

Question 11.
ഉരുകിയ സോഡിയം ക്ലോറൈഡിന്റെ (NaCl) വൈദ്യുത വിശ്ലേ ഷണം നടത്തുന്നു.
(a) ഉരുകിയ സോഡിയം ക്ലോറൈഡിൽ അടങ്ങിയിരിക്കുന്ന അയോണുകൾ ഏവ?
(b) വൈദ്യുത വിശ്ലേഷണം നടത്തുമ്പോൾ പോസിറ്റീവ് ഇലക്ട്രോ ഡിൽ സ്വതന്ത്രമാക്കപ്പെടുന്ന വാതകമേത്?
(c) കാഥോഡിൽ നടക്കുന്ന രാസപ്രവർത്തനത്തിന്റെ രാസസമ വാക്യം എഴുതുക.
Answer:
(a) സോഡിയം അയോൺ (Na⁺), ക്ലോറൈഡ് അയോൺ (Cl^–)
(b) ക്ലോറിൻ വാതകം (Cl₂)
(c) Na⁺ + 1e^– → Na (നിരോക്സീകരണം)

Question 12.
ബ്ലാസ്റ്റ് ഫർണസിൽ നടക്കുന്ന ചില പ്രധാന രാസപ്രർത്തന ങ്ങളുടെ സമവാക്യങ്ങൾ താഴെ കൊടുത്തിരിക്കുന്നു.
(i) Fe₂O₃ + 3CO → 2Fe + 3CO₂
(ii) CaCO₃ → CaO + CO₂
(iii) CaO + SiO₂ → CaSiO₃
(a) ബ്ലാസ്റ്റ് ഫർണസിൽ ഇരുമ്പയിരിനൊപ്പം ചേർക്കുന്ന പദാർത്ഥങ്ങൾ ഏതെല്ലാം?
(b) ഇവിടെ നിരോക്സീകാരിയായി പ്രവർത്തിക്കുന്ന സംയുക്തം എത്ര്?
(c) സ്റ്റാഗ് രൂപീകരണത്തിന്റെ സമവാക്യം തെരെഞ്ഞെടുത്ത് എഴുതുക.
Answer:
(a) ചുണ്ണാമ്പ് കല്ല് (CaCO₃), കോക്ക് (C)
(b) കാർബൺ മോണോക്സൈഡ് (CO)
(c) CaO + SiO₃ → CaSiO₃

Question 13.
മീനിന്റെ (CH₄) മോളിക്യുലാർ മാസ് 16 ആണ്.
(a) 1 GMM CH₄ ൻ്റെ മാസ് എത്ര?
(b) 160 g CH₄
(c) 5 × 6.022 × 10²³ CH₄ തന്മാത്രകളുടെ മാസ് എത്ര?
Answer:
(a) 1 GMM CH₄ = 16 ഗ്രാം
(b) 10 മോൾ (160/16 = 10)
(c) 80 ഗ്രാം
1 GMM CH₄ = 6.022 × 10²³ = 16 ഗ്രാം
5 GMM CH₄ = 16 × 5 = 80 ഗ്രാം

Question 14.
(a) ലാബോറട്ടറിയിൽ അമോണിയ നിർമ്മാണത്തിന് ഉപയോഗി ക്കുന്ന രാസവസ്തുക്കൾ ഏതെല്ലാം?
(b) അമോണിയ വാതകത്തിന് മീതെ ഒരു നനഞ്ഞ ചുവന്ന ലിറ്റ്മസ് പേപ്പർ കാണിച്ചാൽ എന്തു നിരീക്ഷിക്കാം?
(c) അമോണിയയുടെ ഏതു ഗുണമാണ് ഇവിടെ പ്രകടമാകു ന്നത്?
Answer:
(a) കാൽസ്യം ഹൈഡ്രോക്സൈഡ് (Ca(OH)₂) അമോണിയം ക്ലോറൈഡ് (NH₄Cl)
(b) ചുവന്ന ലിറ്റ്മസ് പേപ്പർ നീലയായി മാറുന്നു.
(c) ബേസിക ഗുണം.

Question 15.
ഒരു ഹൈഡ്രോ കാർബണിന്റെ ഘടന നൽകിയിരിക്കുന്നു.

(a) മുഖ്യ ചെയിനിലെ കാർബൺ ആറ്റങ്ങളുടെ എണ്ണമെത്ര?
(b) ഇതിലെ ശാഖകളുടെ സ്ഥാനസംഖ്യകൾ എഴുതുക.
(c) ഈ സംയുക്തത്തിന്റെ IUPAC നാമം എഴുതുക.
Answer:
(a) ആറ്-(6)
(b) 2, 4
(c) 2, 4-ഡൈമീതൈൽ ഹെയ്ൻ

വിഭാഗം – D

(16 മുതൽ 20 വരെയുള്ള ചോദ്യങ്ങളിൽ ഏതെങ്കിലും 4 എണ്ണത്തിന് ഉത്തരം ഴുതിയാൽ മതി). ഓരോ ചോദ്യത്തിനും 4 സ്കോർ വീതം.

Question 16.
ചില സാമഗ്രികൾ നൽകിയിരിക്കുന്നു.

(a) തന്നിരിക്കുന്ന സാമഗ്രികളിൽ നിന്ന് ഒരു ഗാൽവനിക് സെൽ നിർമ്മിക്കാനാവശ്യമായവ തെരെഞ്ഞെടുത്ത് സെൽ ചിത്രീക രിക്കുക.
(ക്രിയാശീലത്തിന്റെ ക്രമം: Mg > Fe > Cu > Ag)
(b) ഈ സെല്ലിലെ ആനോഡേത്?
(c) കാഥോഡിൽ നടക്കുന്ന രാസപ്രവർത്തനത്തിന്റെ രാസസമ വാക്വം എഴുതുക?
Answer:
(a)

(b) മഗ്നീഷ്യം (Mg)
(c) Cu²⁺ + 2e^– → Cu (നിരോക്സീകരണം)

Question 17.
N₂ + 3H₂ = 2NH₃ + താപം
എന്ന ഉഭയദിശാ പ്രവർത്തനത്തിന്റെ ഗ്രാഫ് നൽകിയിരിക്കുന്നു. (ഗ്രാഫ് വിശകലനം ചെയ്ത് താഴെ കൊടുത്തിരിക്കുന്ന ചോദ ങ്ങൾക്ക് ഉത്തരം എഴുതുക.

(a) പുരോ പ്രവർത്തനത്തെ സൂചിപ്പിക്കുന്ന ഗ്രാഫിലെ ഭാഗം ഏത്? [OA, BA, AC]
(b) സംതുലനാവസ്ഥയെ സൂചിപ്പിക്കുന്ന ഗ്രാഫിലെ ഭാഗമേത്?
(c) ചുവടെ തന്നിരിക്കുന്ന പ്രസ്താവനകളിൽ നിന്നും രാസസം തുലനത്തെ സംബന്ധിച്ച് ശരിയായവ തെരെഞ്ഞെടുത്തെഴു
(i) സംതുലനാവസ്ഥയിൽ വ്യൂഹം തന്മാത്രാ തലത്തിൽ നിശ്ച ലമാണ്.
(ii) അഭികാരകങ്ങളും ഉൽപ്പന്നങ്ങളും സഹവർത്തിക്കുന്നു.
(iii) പുരോ-പാശ്ചാത് പ്രവർത്തന നിരക്കുകൾ തുല്യമായി രിക്കും.
(iv) തുറന്ന വ്യൂഹത്തിലാണ് രാസസംതുലനം കൈവരിക്കു ന്നത്.
Answer:
(a) BA
(b) AC
(c) (i) അഭികാരകങ്ങളും ഉൽപ്പന്നങ്ങളും സഹവർത്തിക്കുന്നു.
(ii) പുരോ പശ്ചാത് പ്രവർത്തന നിരക്കുകൾ തുല്യമായി രിക്കും.

Question 18.
ഒരു മൂലകത്തിന്റെ ബാഹ്യതമ സബ്ജെൽ ഇലക്ട്രോൺ വിന്യാസം 3d5 4s2 ആണ്.
(a) ഈ മൂലകത്തിന്റെ പൂർണ്ണ സബ് ഷെൽ ഇലക്ട്രോൺ വിന്യാസം എഴുതുക.
(b) ഈ മൂലകത്തിന്റെ അറ്റോമിക നമ്പർ എഴുതുക.
(c) ഇതിന്റെ ബ്ലോക്കും പിരിയഡും കണ്ടെത്തുക.
(d) ഈ മൂലകം വ്യത്യസ്ത ഓക്സീകരണാവസ്ഥകൾ കണിക്കുന്നു. കാരണമെന്ത്?
Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
(b) 25
(c) ബ്ലോക്ക് – d
പിരിയഡ് – 4
(d) ബാഹ്യതമ ‘s’ സബ്ഷെല്ലിലെ ഇലക്ട്രോണുകളും തൊട്ടു മുന്നിലുള്ള ഷെല്ലിലെ ‘d’ സബ്ഷെൽ ഇലക്ട്രോണുകളും തമ്മിൽ ഊർജ്ജനിലയിൽ വലിയ വ്യത്യാസമില്ല. അതിനാൽ രാസപ്രവർത്തനങ്ങളിൽ ബാഹ്യതമ ‘s’ സബ്ഷെല്ലിലെ ഇല ഓണുകൾക്കു പുറമെ തൊട്ടുമുന്നിലെ ഷെല്ലിലെ ‘d’ സബ്ഷെൽ ഇലക്ട്രോണുകളും വിട്ടുകൊടുക്കുന്നു.

Question 19.
ഒരു ഓർഗാനിക് സംയുക്തത്തെ സംബന്ധിച്ച ചില വിവരങ്ങൾ നൽകിയിരിക്കുന്നു.
(i) അതിന്റെ മുഖ്യ ചെയിനിൽ 3 കാർബൺ ആറ്റങ്ങൾ ഉണ്ട്.
(ii) അതിന്റെ രണ്ടാമത്തെ കാർബൺ ആറ്റത്തിൽ ഒരു ഹൈഡ്രോക്സിൽ (-OH) ഗ്രൂപ്പ് ഉണ്ട്.
(a) ഈ ഓർഗാനിക് സംയുക്തത്തിന്റെ ഘടനാവാക്യം എഴു തുക.
(b) ഇതിന്റെ തന്മാത്രാ സൂത്രം എഴുതുക.
(c) ഈ സംയുക്തത്തിന്റെ ഒരു ഫങ്ഷണൽ ഐസോമെ റിന്റെ ഘടനാവാക്യവും IUPAC നാമവും എഴുതുക.
Answer:

Question 20.
A, B, C കോളങ്ങളിൽ നിന്നും അനുയോജ്യമായവ കണ്ടെത്തി ചേർത്തെഴുതുക.

A അഭികാരകങ്ങൾ	B ഉൽപന്നങ്ങൾ	C രാസ പ്രവർത്ത നത്തിന്റെ പേര്
CH4 + Cl2	CO2 + 2H2O	അഡീഷൻ
CH4 + 2O2	CH2 = CH2	താപിയ വിഘടനം
CH3 – CH2 – CH3	CH3Cl + HCl	ജ്വലനം
CH ≡ CH + H2	CH2 = CH2 + CH4	ആദേശ പ്രവർത്തനം

Answer:

A അഭികാരകങ്ങൾ	B ഉൽപന്നങ്ങൾ	C രാസ പ്രവർത്ത നത്തിന്റെ പേര്
CH4 + Cl2	CH3Cl + HCl	ആദേശ പ്രവർത്തനം
CH4 + 2O2	CO2 + 2H2O	ജ്വലനം
CH3 – CH2 – CH3	CH2 = CH2 + CH4	താപിയ വിഘടനം
CH ≡ CH + H2	CH2 = CH2	അഡിഷൻ

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 14 Malayalam Medium അനുപാതം can save valuable time.

Kerala SCERT Class 9 Maths Chapter 14 Solutions Malayalam Medium അനുപാതം

Class 9 Maths Chapter 14 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 14 Malayalam Medium Textual Questions and Answers

Question 1.
ചുവടെ പറയുന്ന ഓരോ സന്ദർഭത്തിലും ആദ്യത്തെ അളവിന് ആനുപാതികമായാണ് രണ്ടാമത്തെ അളവ് മാറുന്നത് എന്നു തെളിയിക്കുക. ഓരോന്നിലും ആനുപാതികസ്ഥിരം

കണക്കാക്കുക:
i) വശങ്ങളുടെ നീളം പലതായി വരയ്ക്കുന്ന സമചതുരങ്ങളുടെ വശത്തിന്റെ നീളവും, ചുറ്റളവും.
ii) പല നീളങ്ങളുള്ള കമ്പികൾ വളച്ചുണ്ടാക്കുന്ന സമചതുരങ്ങളിൽ കമ്പിയുടെ നീളവും, വശ ത്തിന്റെ നീളവും.
iii) ഒരു വരയിലൂടെ ഉരുളുന്ന വൃത്തത്തിന്റെ കറക്കങ്ങളുടെ എണ്ണവും, വരയിലൂടെ സഞ്ചരിച്ച ദൂരവും.
Answer:
i) സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം 5 എന്നും, ചുറ്റളവിനെ P എന്നും എടുക്കാം.
സമചതുരത്തിന്റെ ചുറ്റളവ്, P = 4s
ഇനി, ചുറ്റളവും ഒരു വശത്തിന്റെ നീളവും തമ്മിലുള്ള അനുപാതം കണ്ടെത്താം:
\(\frac{P}{s}=\frac{4 s}{s}\) = 4
സമചതുരത്തിന്റെ ചുറ്റളവ് വശങ്ങളുടെ നീളത്തിനൊപ്പം ആനുപാതികമായി മാറുന്നുവെന്നു കാണാം. ഇതിന്റെ, ആനുപാതികസ്ഥിരം 4 ആണ്.

ii) കമ്പിയുടെ മുഴുവൻ നീളം L എന്നും, സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം 5 എന്നും എടുക്കാം. സമചതുരത്തിന്റെ മുഴുവൻ നീളം, L = 45
കമ്പിയുടെ മുഴുവൻ നീളവും സമചതുരത്തിന്റെ വശത്തിന്റെ നീളവും തമ്മിലുള്ള അനുപാതം
കണ്ടെത്താം:
\(\frac{L}{s}=\frac{4 s}{s}\) = 4
കമ്പിയുടെ മുഴുവൻ നീളം സമചതുരത്തിന്റെ വശത്തിന്റെ നീളത്തിനൊപ്പം ആനുപാതികമായി മാറുന്നു. ഇതിന്റെ, ആനുപാതികസ്ഥിരം 4 ആണ്.

iii) വൃത്തത്തിന്റെ വ്യാസം r എന്നും, ചുറ്റളവിനെ C എന്നും എടുക്കാം.
വ്യത്തത്തിന്റെ ചുറ്റളവ്, C = 2πr
ഒരു വരയിലൂടെ ഉരുളുന്ന വൃത്തത്തിന്റെ കറക്കങ്ങളുടെ എണ്ണം n എന്നും, വരയിലൂടെ സഞ്ചരിച്ച ദൂരം d എന്നും എടുത്താൽ d =n C = n2πr

സഞ്ചരിച്ച ദൂരവും കറക്കത്തിന്റെ എണ്ണവും തമ്മിലുള്ള അനുപാതം കണ്ടത്താം:
\(\frac{d}{n}\) = 2πг
സഞ്ചരിച്ച ദൂരം കറക്കത്തിന്റെ എണ്ണത്തിനൊപ്പം ആനുപാതകമായി മാറുന്നു. ഇതിന്റെ ആനുപാതികസ്ഥിരം 21 ആണ്.

Question 2.
ചിത്രത്തിലെ ചരിഞ്ഞ വരയിലെ ബിന്ദുക്കളെല്ലാമെടുത്താൽ, കോണിന്റെ മൂലയിൽ നിന്നുള്ള അകലത്തിന് ആനുപാതികമായാണ്, താഴത്തെ വരയിൽ നിന്നുള്ള ഉയരം മാറുന്നത് എന്നു കണ്ടല്ലോ 30, 45°, 60° കോണുകളിൽ ആനുപാതികസ്ഥിരം കണക്കാക്കുക.

Answer:
കോൺ 30° ആയാൽ

ത്രികോണം ABC, AC = 2 സെമീ, BC = 1 സെമീ, AB = √3 സെ.മീ

ഇവിടെ ABC യും APQ യും സദൃശത്രികോണങ്ങളാണ്.
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)
BC യും PO യും ഉയരമായിട്ടും AC യും AQ യും അകലമായിട്ടും കണക്കാക്കുന്നു.
\(\frac{A C}{A Q}=\frac{B C}{P Q}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{1}\)
AQ = 2 PQ
അതിനാൽ, ആനുപാതികസ്ഥിരം = 2
കോൺ 60° ആയാൽ
ത്രികോണം ABC, AC = 2 സെമീ, AB = 1 സെമീ, BC = √3 സെമീ

ഇവിടെ ABC യും APO യും സദൃശത്രികോണങ്ങളാണ്.
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)
BC യും PO യും ഉയരമായിട്ടും AC യും AQ യും അകലമായിട്ടും കണക്കാക്കുന്നു.
\(\frac{A C}{A Q}=\frac{B C}{P Q}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{\sqrt{3}}\)
AQ = \(\frac{2}{\sqrt{3}}\) PQ
അതിനാൽ, ആനുപാതികസ്ഥിരം = \(\frac{2}{\sqrt{3}}\)

കോൺ 45° ആയാൽ
ത്രികോണം ABC, AC = √2 സെമീ, BC = 1 സെമീ, AB = 1 സെമീ
ഇവിടെ ABC യും APO യും സദൃശത്രികോണങ്ങളാണ്.
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)
BC യും PQ യും ഉയരമായിട്ടും AC യും AQ യും അകലമായിട്ടും കണക്കാക്കുന്നു.
\(\frac{A C}{A Q}=\frac{B C}{P Q}\)
AQ = PQ × \(\frac{A C}{B C}\)
AQ = PQ × √2
AQ = √2 PQ
അതിനാൽ, ആനുപാതികസ്ഥിരം = √2

Question 3.
വശങ്ങളുടെ നീളം പലതായി വരയ്ക്കുന്ന സമഭുജത്രികോണങ്ങളുടെ ചുറ്റളവ് വശത്തിന് ആനുപാതികമാണ് എന്ന് സമർത്ഥിക്കുക. ആനുപാതികസ്ഥിരം കണക്കാക്കുക. മറ്റു സമബഹുഭുജങ്ങളിലോ?
Answer:
വശങ്ങളുടെ നീളം 5 ഉള്ള ഒരു സമഭുജത്രികോണം പരിഗണിക്കാം.
സമഭുജത്രികോണത്തിന്റെ ചുറ്റളവ്, P = 35
ഒരു സമഭുജത്രികോണത്തിന്റെ ചുറ്റളവ് അതിന്റെ വശങ്ങളുടെ നീളവുമായി ആനുപാതികമായി മാറുന്നു, ആനുപാതികസ്ഥിരം 3 ആണ്.

n വശങ്ങളുള്ളതും 5 നീളമുള്ളതുമായ ഒരു സമബഹുഭുജത്തെ നമുക്ക് പരിഗണിക്കാം. സമബഹുഭുജത്തിന്റെ ചുറ്റളവ്, P = ns
ഏതൊരു സമബഹുഭുജത്തിന്റെയും ചുറ്റളവ് അതിന്റെ വശങ്ങളുടെ നീളവുമായി
ആനുപാതികമായി മാറുന്നു, ആനുപാതികസ്ഥിരം വശങ്ങളുടെ എണ്ണത്തിന് തുല്യമാണ്.

Question 4.
ഒരു നിശ്ചിത വൃത്തത്തിലെ വ്യത്യസ്ത ചാപങ്ങളുടെ നീളം, അവയുടെ കേന്ദ്രകോണിന് ആനുപാതികമാണ് എന്നു തെളിയിക്കുക. ആനുപാതികസ്ഥിരം എന്താണ്? വ്യത്യസ്ത വൃത്താംശങ്ങളുടെ പരപ്പളവും കേന്ദ്രകോണും തമ്മിലുള്ള ബന്ധമോ ?
Answer:
വ്യത്തത്തിന്റെ വ്യാസം = r സെമീ
കേന്ദ്രകോൺ = θ
ചാപത്തിന്റെ നീളം, L = θ x r
ഒരു നിശ്ചിത വൃത്തത്തിന്റെ ചാപത്തിന്റെ നീളം അവയുടെ കേന്ദ്രകോണുമായി ആനുപാതികമായി മാറുന്നു, ആനുപാതികസ്ഥിരം r ആണ്.
വൃത്താംശത്തിന്റെ പരപ്പളവ് A = \(\frac{1}{2}\)r² θ
ഒരു നിശ്ചിത വൃത്തത്തിന്റെ വൃത്താംശത്തിന്റെ പരപ്പളവ് അതിന്റെ കേന്ദ്രകോണുമായി ആനുപാതികമായി മാറുന്നു, ആനുപാതികസ്ഥിരം = \(\frac{1}{2}\)r² ആണ്.

Question 5.
i) ഒരു ത്രികോണത്തിന്റെ കോണുകളുടെ തുക എത്രയാണ്? ഒരു ഷഡ്ഭുജത്തിന്റെയോ ?
iii) ബഹുഭുജങ്ങളുടെ വശങ്ങളുടെ എണ്ണം മാറുന്നതിന് ആനുപാതികമായാണോ, കോണു കളുടെ തുക മാറുന്നത് ? കാരണം വിശദീകരിക്കുക.
Answer:
i) ത്രികോണത്തിന്റെ കോണുകളുടെ ആകെത്തുക = (3 – 2) × 180° = 180°
ഷഡ്ഭുജത്തിന്റെ കോണുകളുടെ ആകെത്തുക = (6 – 2) × 180° = 720°

ii) n വശങ്ങളുള്ള ഒരു ബഹുഭുജത്തിന്റെ കോണുകളുടെ ആകെത്തുക എന്നു പറയുന്നത്, S = (n – 2) × 180°
ബഹുഭുജങ്ങളുടെ കോണുകളുടെ ആകെത്തുക അതിന്റെ വശങ്ങളുടെ എണ്ണത്തിനു ആനുപാതികമായി മാറുന്നു, ആനുപാതികസ്ഥിരം 180° ആണ്.

Question 6.
പാദം 6 സെന്റിമീറ്ററും, ഉയരം 3 സെന്റിമീറ്ററും ആയ ത്രികോണത്തിനുള്ളിൽ പാദത്തിനു സമാന്തരമായി വരകൾ വരയ്ക്കുന്നു ഈ വരകളുടെ നീളം മുകളിലെ മൂലയിൽനിന്നുള്ള അക . ലങ്ങൾക്ക് ആനുപാതികമാണെന്നു തെളിയിക്കുക ആനുപാതികസ്ഥിരം എന്താണ് ?

Answer:
പാദം = 6 സെന്റിമീറ്റർ
ഉയരം = 3 സെന്റിമീറ്റർ
ത്രികോണത്തിന്റെ മുകളിലെ മൂലയിൽ നിന്നും നീളം വരുന്ന ഒരു സമാന്തര വര വരക്കുക. മുകളിലെ മൂലയിൽ നിന്നും സമാന്തര വരയിലേക്കുള്ള അകലം y ആയിട്ടെടുക്കുക. സദൃശത്രികോണങ്ങൾ ഉപയോഗിക്കുമ്പോൾ,
\(\frac{y}{x}=\frac{6}{3}\)
y = 2x
ത്രികോണത്തിന്റെ പാദത്തിനു സമാന്തരമായി വരക്കുന്ന വരകളുടെ നീളം അതിന്റെ മുകളിലെ മൂലയിൽ നിന്നും സമാന്തര വരയിലേക്കുള്ള അക്കാലത്തിനു ആനുപാതികമായി മാറുന്നു . ഇതിന്റെ ആനുപാതികസ്ഥിരം 2 ആണ് .

Question 7.
വ്യാസം 10 സെന്റിമീറ്റർ ആയ അർധവൃത്തത്തിനുള്ളിൽ വ്യാസത്തിനു സമാന്തരമായി വരകൾ വരയ്ക്കുന്നു.

i) ചുവടെയുള്ള ഓരോ ചിത്രത്തിലും, വ്യാസത്തിനു സമാന്തരമായ വരയുടെ നീളം കണക്കാക്കുക:

ii) വരകളുടെ നീളം അർധവൃത്തത്തിന്റെ ഏറ്റവും മുകളിൽ നിന്നുള്ള അകലങ്ങൾക്ക് ആനുപാതികമാണോ? കാരണം വിശദീകരിക്കുക.
Answer:
i) Figure 1

സമാന്തരമായ വരയുടെ നീളം = 2\(\sqrt{(5)^2-(4)^2}\) = 2 × 3 = 6 സെമീ

Figure 2

സമാന്തരമായ വരയുടെ നീളം = 2\(\sqrt{(5)^2-(3)^2}\) = 2 × 4 = 8 സെ.മീ

ii) സമാന്തരമായ വരയുടെ നീളം അർധവൃത്തത്തിന്റെ ഏറ്റവും മുകളിൽ നിന്നുള്ള അകലവും ആനുപാതികമല്ല.

Question 8.
(i) സമഭുജത്രികോണങ്ങളുടെ പരപ്പളവ്, വശത്തിന്റെ വർഗത്തിന് ആനുപാതികമാണെന്നു തെളിയിക്കുക. ആനുപാതികസ്ഥിരം എന്താാണ്?
Answer:

സമഭുജത്രികോണങ്ങളുടെ പരപ്പളവ് വശത്തിന്റെ വർഗത്തിന് ആനുപാതികമായാണ് മാറുന്നത്. ആനുപാതികസ്ഥിരം = \(\frac{\sqrt{3}}{4}\)

(ii) സമചതുരങ്ങളുടെ പരപ്പളവ്, വശത്തിന്റെ വർഗത്തിന് ആനുപാതികമാണോ? ആണെങ്കിൽ, ആനുപാതികസ്ഥിരം എന്താണ്?
Answer:

സമചതുരങ്ങളുടെ പരപ്പളവ് വശത്തിന്റെ വർഗത്തിന് ആനുപാതികമായാണ് മാറുന്നത്. ആനുപാതികസ്ഥിരം = 1.

Question 9.
പരപ്പളവ് ഒരു ചതുരശ്രമീറ്ററായ ചതുരങ്ങളിൽ, ഒരു വശത്തിന്റെ നീളം മാറുന്നതിനനുസരിച്ച് മറ്റേ വശത്തിന്റെ നീളവും മാറണം. ഈ ബന്ധം ബീജഗണിതസമവാക്യമായി എഴുതുക. അനുപാതത്തിന്റെ ഭാഷയിൽ ഈ ബന്ധം എങ്ങനെ പറയാം ?
Answer:
പരപ്പളവ് 1 ചതുരശ്രമീറ്ററായ ചതുരങ്ങളിൽ നീളം × ഉം വീതി y യും ആയി
പരിഗണിച്ചാൽ,
പരപ്പളവ് = നീളം × വീതി
1 = x × y
1 = xy
y = \(\frac{1}{x}\)
y, x നു വിപരീതാനുപാതത്തിലാണ്.

Question 10.
ഒരേ പരപ്പളവുള്ള ത്രികോണങ്ങളിൽ, ഏറ്റവും വലിയ വശത്തിന്റെ നീളവും, എതിർ മൂലയിൽ നിന്ന് ആ വശത്തിലേക്കുള്ള ലംബത്തിന്റെ നീളവും തമ്മിലുള്ള ബന്ധം അനുപാതമായി എങ്ങനെ പറയാം ? ഏറ്റവും വലിയ വശത്തിനു പകരം, ഏറ്റവും ചെറിയ വശമെടുത്താലോ ?
Answer:
വലിയ വശം a എന്നും , എതിർ മൂലയിൽ നിന്നുള്ള ലംബം h എന്നും, പരപ്പളവ് A എന്നും എടുത്താൽ,
A = \(\frac{1}{2}\)a × h,
a = \(\frac{2A}{h}\)
വലിയ വംശത്തിന്റെ നീളം എതിർമൂലയിൽ നിന്നുള്ള ലംബവുമായി
വിപരീതാനുപാതത്തിലാണ്.
അതുപോലെതന്നെ, ചതുരത്തിന്റെ ചെറിയവശത്തിന്റെ നീളം ആ വശത്തിന്റെ
എതിർമൂലയിൽ നിന്നുള്ള ലംബത്തിന് വിപരീതാനുപാതത്തിലാണ്.

Question 11.
സമബഹുഭുജങ്ങളിൽ, വശങ്ങളുടെ എണ്ണവും, ഒരു പുറംകോണിന്റെ അളവും തമ്മിലുള്ള ബന്ധത്തിന്റെ സമവാക്യമെന്താണ് ? ഈ ബന്ധം അനുപാതമായി പറയാൻ കഴിയുമോ ? ആനുപാതികസ്ഥിരം എന്താണ് ?
Answer:
ഏതൊരു ബഹുഭുജത്തിന്റെയും പുറംകോണുകളുടെ തുക 360° ആണ്.
വശങ്ങളുടെ എണ്ണം n എന്നെടുത്താൽ,

ഒരു പുറംകോണിന്റെ അളവ് × എന്നെടുത്താൽ,
x = \(\frac{360^{\circ}}{n}\)
സമബഹുഭുജങ്ങളിൽ വശങ്ങളുടെ എണ്ണവും ഒരു പുറംകോണിന്റെ അളവും വിപരീതാനുപാതത്തിലാണ്.
ആനുപാതികസ്ഥിരം = 360°

Proportion Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ചതുരാകൃതിയിലുള്ള ടാങ്കിലേക്ക് ഒരു നിശ്ചിത അളവിലുള്ള വെള്ളം ഒഴുകുന്നു . ഒഴുക്കിന്റെ നിരക്ക് വ്യത്യസ്ത പൈപ്പുകൾ ഉപയോഗിച്ച് മാറ്റാം . ഇനിപ്പറയുന്ന അളവുകൾ തമ്മിലുള്ള ബന്ധം ഒരു ബീജഗണിത സമവാക്യമായും അനുപാതങ്ങളുടെ അടിസ്ഥാനത്തിലും എഴുതുക.
i) ജലപ്രവാഹത്തിന്റെ തോതും ജലനിരപ്പിന്റെ ഉയരവും.
ii) ജലപ്രവാഹത്തിന്റെ തോതും ടാങ്ക് നിറയാൻ എടുത്ത സമയവും.
Answer:
i) ജലപ്രവാഹത്തിന്റെ തോത് x എന്നും, ടാങ്കിലെ ജലനിരപ്പ് y എന്നും , ടാങ്കിന്റെ പാദപരപ്പളവ് A എന്നും എടുത്താൽ, x = A y
അതായത്, ടാങ്കിലെ ജലനിരപ്പ് ജലപ്രവാഹത്തിന്റെ തോതില് നേരനുപാതത്തിലാണ്.

ii) ടാങ്കിന്റെ വ്യാപ്തം C എന്നും, ഒരു സെക്കന്റിൽ ഒഴുകുന്ന ജലത്തിന്റെ വ്യാപ്തം V എന്നും എടുത്താൽ,
t സെക്കന്റിൽ ഒഴുകിയ ജലത്തിന്റെ അളവ്,
C = V × t
V = C × \(\frac{1}{t}\)
അതായത്, ജലപ്രവാഹത്തിന്റെ തോതും നിറയാൻ എടുക്കുന്ന സമയവും വിപരീതാനുപാതത്തിലാണ്. ആനുപാതികസ്ഥിരം = C

Question 2.
രഘു 60,000 രൂപയും നാസർ 1,00,000 രൂപയും നിക്ഷേപിച്ചു ഒരു ബിസിനസ് തുടങ്ങി. ഒരു മാസത്തിനകം 4800 രൂപ ലാഭം കിട്ടി. ലാഭത്തിൽ 1800 രൂപ രഘുവും 3000 രൂപ നാസറും എടുത്തു.നിക്ഷേപത്തിന്റെ അനുപാതം എന്താണ്? നിക്ഷേപവും ലാഭവും വിഭജിച്ചത് അനുപാതികമായാണോ?
Answer:
നിക്ഷേപിച്ച തുകയുടെ അനുപാതം = 60000: 100000 = 6: 10 = 3: 5
ലാഭത്തിന്റെ അനുപാതം = 1800: 3000 = 18: 30 = 3: 5
നിക്ഷേപിച്ച തുകയുടെ അനുപാതവും ലാഭത്തിന്റെ അനുപാതവും
തുല്യമാണ്.അതിനാൽ നിക്ഷേപവും ലാഭവും വിഭജിച്ചത് ആനുപാതികമാണ്.

Question 3.
ഒരേ ചുറ്റളവുള്ള ഒരു സമചതുരത്തിന്റെ നീളവും വീതിയും വിപരീതാനുപാതത്തിലാണോ ?
Answer:
സമചതുരത്തിന്റെ ചുറ്റളവ് = 20 സെന്റിമീറ്റർ
2(നീളം +വീതി) = 10
നീളം +വീതി = 10
നീളം x എന്നും വീതി y എന്നുമെടുത്താൽ,
x + y = 10
x ന്റെയും y യുടെയും സാധ്യമായ വിലകൾ,
x = 9 ആയാൽ y = 1
x = 8 ആയാൽ y = 2
ഇവിടെ xy ഒരു സ്ഥിരസംഖ്യയല്ല.
അതിനാൽ, നീളവും വീതിയും വിപരീതാനുപാതത്തിലല്ല.

Question 4.
5 ലിറ്റർ പെട്രോൾ ഉള്ള ഒരു കാർ 75 കിലോമീറ്റർ ദൂരം സഞ്ചരിക്കും. സഞ്ചരിച്ച ദൂരത്തിന്റെയും പെട്രോളിന്റെ അളവിന്റെയും ഇടയിലുള്ള ആനുപാതികസ്ഥിരം എന്താണ്? 180 കിലോമീറ്റർ ദൂരം സഞ്ചരിക്കാൻ എത്ര പെട്രോൾ വേണം?
Answer:
സഞ്ചരിച്ച ദൂരം x എന്നും പെട്രോളിന്റെ അളവ് y എന്നും എടുത്താൽ,

k = \(\frac{x}{y}=\frac{75}{5}\) = 15
ആനുപാതികസ്ഥിരം : 15
സഞ്ചരിച്ച ദൂരം =180 കിലോമീറ്റർ
k = \(\frac{x}{y}\) = 15
\(\frac{180}{x}\) = 15
x = \(\frac{180}{15}\) = 12
180 കിലോമീറ്റർ ദൂരം സഞ്ചരിക്കാൻ 12 ലിറ്റർ പെട്രോൾ വേണം.

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 2 Periodic Table Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 2 Solutions Periodic Table

Kerala Syllabus Std 9 Chemistry Chapter 2 Periodic Table Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 2 Let Us Assess Answers Periodic Table

Question 1.
The symbols of a few elements are given. Write the electron configurations of these elements and find the period and group to which they belong.
a) \({ }_{11}^{23} \mathrm{Na}\)
b) \({ }_{13}^{27} \mathrm{Al}\)
c) \({ }_{17}^{35} \mathrm{Cl}\)
d) \({ }_{8}^{16} \mathrm{O}\)
e) \({ }_{10}^{20} \mathrm{Ne}\)
f) \({ }_{6}^{12} \mathrm{C}\)
Answer:

Element	Atomic Number	Electron configuration	Period number	Group number
a) \({ }_{11}^{23} \mathrm{Na}\)	11	2, 8, 1	3	1
b) \({ }_{13}^{27} \mathrm{Al}\)	13	2, 8, 3	3	13
c) \({ }_{17}^{35} \mathrm{Cl}\)	17	2, 8, 7	3	17
d) \({ }_{8}^{16} \mathrm{O}\)	8	2, 6	2	16
e) \({ }_{10}^{20} \mathrm{Ne}\)	10	2, 8	2	18
f) \({ }_{6}^{12} \mathrm{C}\)	6	2, 4	2	14

Question 2.
The electron configuration of element X is 2, 8, 8, 1. (Symbol is not real.)
a. Find the atomic number of X.
b. To which group does it belong?
c. What is its period number?
d. To which family does it belong?
e. Write the electron configuration of the noble gas which comes just before X.
Answer:
a. 19
b. Group 1
c. Period 4
d. Alkali metals
e. 2, 8, 8

Question 3.
There are 3 shells in an atom of element P. There are 7 electrons in its outermost shell. (Symbol is not real.)
a. Write the electron configuration of element P.
b. What is its atomic number?
c. To which period does it belong?
d. To which group does it belong?
e. Draw the model of this atom.
Answer:
a. 2, 8, 7
b. Atomic number – 17
c. Period – 3
d. Group – 17
e.

Question 4.
The element M belongs to the 3rd period and group 1. (Symbol is not real)
a. Write the electron configuration of this element.
b. Write its name and symbol.
c. To which family does this element belong?
d. Write the electron configuration of the element belonging to the Same period and group 13.
Answer:
a. 2, 8, 1
b. Sodium, Na
c. Alkali metals
d. 2, 8, 3

Question 5.
Electron configurations of elements P, Q, R and S are given. (Symbols are not real)
P – 2, 7
Q – 2, 8
R- 2, 8, 1
S – 2, 8, 7
a) Which of these elements belong to the same period?
b) Which of these elements belong to the same group?
c) Identify the noble gas among these.
d) Find the group number and period number of element S.
Answer:
a) P and Q, R and S
b) P and S
c) Q (electron configuration: 2, 8)
d) Group number – 17
Period number – 3

Question 6.
Electron configurations of a few elements are given.
A – 2, 1
B – 2, 8, 1
C – 2, 8, 7
(Symbols are not real)
a. Which of these elements has bigger atom, A or B?
b. Which atom is bigger, B or C?
Answer:
a. B
b. B

Question 7.
A portion of the modern periodic table is given. (Symbols are not real) Answer the following questions.

a. Which of these elements belong to the halogen family?
b. Which are the transition elements?
c. Write the elements of group 1 in the decreasing order of their atomic size.
d. Which element has a smaller atom, B or I?
e. Write the elements of period 3 in the increasing order of their atomic size.
f. Which of these are alkaline earth metals?
g. Which element has 8 electrons in its outermost shell?
h. Find the real symbols of the given elements with the help of the periodic table.
Answer:
a. M and N
b. G and H
c. D > C > B > A
d. I
e. N < J < F < C
f. E and F
g. O
h. A – H, B – Li, C – Na, D – K, E – Be, F – Mg, G – Cr, H – Fe
I – B, J – Al, K – N, L – O, M – F, N – Cl, O – Ne

Question 8.
An element belonging to the 2nd period has 2 electrons in the outermost shell of its atom.
a. Write the electron configuration of this element.
b. Write the electron configuration of the noble gas belonging to the same period.
c. What is its group number?
d. Write the electron configuration of an element in the same group and in the third period.
Answer:
a. 2, 2
b. 2, 8
c. 2
d. 2, 8, 2

Question 9.
Analyse the table and answer the following questions.

Element	Mass number	Number of neutrons
A	9	5
B	35	18
C	39	20
D	40	22

(Hint: Symbols are not real)
a. Find the atomic number of these elements.
b. Write their electron configurations.
c. Which among these is a noble gas?
d. To which family does the element B belong?
e. To which period and group does the element C belong?
f. Which of these elements belong to the same period?
Answer:
a. Atomic numbers: A – 4, B – 17, C – 19, D – 18.
b. Electron configurations: A – 2, 2, B – 2, 8, 7, C – 2, 8, 8, 1, D – 2, 8, 8
c. D
d. Halogens
e. Period – 4, Group – 1
f. B and D

Extended Activities

Question 1.
Two English alphabets have not been used as symbols of elements so far. Find them with the help of the periodic table.
Answer:
The letters “J” and “Q” are the only two letters not found on the periodic table.

Question 2.
Prepare the biography of scientists involved in the classification of elements and publish it in the science magazine.
Answer:
Hints
Title: Pioneers of Periodicity
1. Dmitri Mendeleev (1834 – 1907):

• Formulated Periodic Law and created the first periodic table.
• Predicted properties of undiscovered elements.

2. Henry Moseley (1887 – 1915):

• Determined atomic number’s significance in element arrangement.
• Resolved periodic table inconsistencies.

3. Glenn T. Seaborg (1912 – 1999):

• Discovered transuranium elements.
• Proposed modem periodic table layout, including actinides.

Question 3.
Draw a model of the modern periodic table and exhibit it in your classroom.
Answer:
Hints

1. Materials: Poster board, markers, index cards.
2. Layout: Draw a grid with rows (periods) and columns (groups).
3. Element Blocks: Write element symbols and numbers on index cards.
4. Colour Coding: Use different colours for metals, non-metals, and metalloids.
5. Periods and Groups: Label rows 1 – 7 as periods and columns 1 – 18 as groups.
6. Element Placement: Arrange cards by atomic number and properties.
7. Display: Hang the model in the classroom for easy reference.

Question 4.
Prepare a table including the symbol, the electron configuration, and the physical state of elements having atomic numbers 1 to 36, using Kalzium software.
Answer:

Element	Symbol	Electron configuration	Physical state
Hydrogen	H	1	Gas
Helium	He	2	Gas
Lithium	Li	2, 1	Solid

Similarly, complete the table.

Question 5.
Using cardboard pieces, design a periodic table as shown in the figure given in the first page of this unit.
Answer:

• Gather cardboard, scissors, markers, and glue.
• Create a base structure with rows and columns.
• Cut out element tiles and label them.
• Arrange tiles by periods and groups.
• Add transition metals and separate lanthanides/actinides if desired.
• Decorate and label for clarity.
• Glue tiles onto the base.
• Display your cardboard periodic table.

Periodic Table Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
List the merits of Mendeleev’s periodic table.
Answer:

• Elements are classified in such a way that elements of similar properties were placed in the same group. Thus, learning chemistry was made easier.
• Corrected the wrongly determined atomic masses and gave the correct position to elements.
• Columns were left vacant for elements that were not known at that time, and predicted their properties.

Question 2.
List the limitations of Mendeleev’s periodic table.
Answer:

• Elements with large differences in properties were included in the same group.
• Eg: Hard metals like copper (Cu) and silver (Ag), along with soft metals like Sodium (Na) and potassium (K).
• No proper position could be given to the element hydrogen. Non-metallic hydrogen was placed along with metals like lithium (Li), Sodium (Na) and potassium (K).
• The increasing order of atomic mass was not strictly followed throughout.
• If elements are arranged in the order of atomic mass, isotopes should also be given a position. That was not done.

Question 3.
How do isotopes of the same element differ from one another?
Answer:
Isotopes are the atoms of the same element with the same atomic number and different mass numbers.
Eg: \({ }_{1}^{1} \mathrm{H}\), \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\)
You know that elements are arranged on the basis of atomic mass in Mendeleev’s periodic table. Since isotopes have different atomic masses, it is necessary to assign different positions for them in the periodic table.

For example, \({ }_{1}^{1} \mathrm{H}\), \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\) are the isotopes of hydrogen. As per Mendeleev’s periodic table, it is not possible to assign a specific position to each of them on the basis of atomic mass.

Through his X-ray diffraction experiments, Henry Moseley proved that the properties of elements depend mainly on atomic number rather than atomic mass. He then revised Mendeleev’s periodic law. This is known as modem periodic law.

Question 4.
How many periods are there?
Answer:
There are seven periods in the modem periodic table.

Question 5.
Write the total number of groups.
Answer:
Eighteen

Question 6.
Which period has the least number of elements?
Answer:
Period 1. (Two elements only: Hydrogen and Helium)

Question 7.
Are the number of elements in periods 2 and 3 the same?
Answer:
Yes (eight elements each)

Question 8.
How many elements are included in the 4th period?
Answer:
Eighteen

Question 9.
What all information about an element can be obtained from the periodic table? Note down in the science diary.
Answer:
Name, Symbol, Atomic number, Mass number, Electronic configuration, Physical state, Group number, Period number.

Question 10.
Elements of group 1 are given in the table. Complete the table

Name of the element	Symbol	Atomic number	Electron configuration
Lithium	Li	3	–
Sodium	Na	11	–
Potassium	–	–	2, 8, 8, 1
Rubedium	Rb	–	2, 8, 18, 8, 1
Caesium	–	55	2, 8, 18, 18, 8, 1
Francium	Fr	–	2, 8, 18, 32, 18, 8, 1

Answer:

Name of the element	Symbol	Atomic number	Electron configuration
Lithium	Li	3	2,1
Sodium	Na	11	2, 8, 1
Potassium	K	19	2, 8, 8, 1
Rubedium	Rb	37	2, 8, 18, 8, 1
Caesium	Cs	55	2,8, 18, 18, 8, 1
Francium	Fr	87	2,8, 18, 32, 18, 8, 1

Question 11.
Have you noticed any peculiarity regarding the number of outermost electrons in the elements of group 1?
Answer:
All elements in group 1 have the same number of outermost electrons.

Question 12.
With the help of the periodic table, write the electron configuration of the elements in group 2.
Answer:

Name of the Element	Symbol	Atomic Number	Electron Configuration
Beryllium	Be	4	2,2
Magnesium	Mg	12	2, 8, 2
Calcium	Ca	20	2, 8, 8, 2
Strontium	Sr	38	2, 8, 18, 8, 2
Barium	Ba	56	2, 8, 18,18, 8, 2
Radium	Ra	58	2, 8, 18, 32, 18, 8, 2

It is clear that the number of outermost electrons of the elements in a given group is the same.The
chemical properties of elements are based on the number of outermost electrons in them. Usually,
these electrons take part in chemical reactions.

Based on the common characteristics of elements in each group, they can be considered as families.
A table enlisting the various families of elements is given below.

Group number	Name of family
1	Alkali metals
2	Alkaline earth metals
From 3 to 12	Transition elements
13	Boron family
14	Carbon family
15	Nitrogen family
16	Oxygen family
17	Halogens
18	Noble gases

Question 13.
Which of these elements are familiar to you?
Answer:
Hydrogen, Nitrogen, Oxygen, Chlorine, Aluminium, Sodium.

Question 14.
Write the examples of metals among these elements.
Answer:
Li, Na, K, Rb, Cs, Fr, Be, Mg, Ca, Sr, Ba, Ra, Al, Ga, Sn, Pb etc.

Question 15.
Do these elements include non-metals?
Answer:
Yes. B, C, Si, N, P, O, S, F, Br, He, Ne, Ar etc.

Question 16.
Do these groups include elements belonging to the solid state, liquid state and gaseous state?
Answer:
Solids: – Li, Na, K, Rb, Be, Mg, Ca, Sr, Ba, Ra, B, Al, In, Tl, C, Si, Ge, Sn, Pb, S, Se, Te, Po, I, At
Liquids:- Cs, Fr, Hg, Ga, Br
Gases:- H, N, O, F, Cl, He, Ne, Ar, Kr, Xe, Rn

Question 17.
How does electron filling take place in the outermost shell of these elements?
Answer:
From left to right, with an increase in atomic number, the number of electrons in the outermost shell increases by one.

Question 18.
What change do you observe in the number of outer electrons on moving from left to right along a period?
Answer:
On moving along a period from left to right, there is an increase of one electron in the outermost shell of the main group elements until eight electrons are gained.

Question 19.
Which are the families included in the main group elements?
Answer:
Alkali metals, Alkaline earth metals, Boron family, Carbon family, Nitrogen family, Oxygen family, halogens and noble gases.

Question 20.
In which groups are metalloids present?
Answer:
The metalloids are found in a zig-zag arrangement in the periodic table between group 13 and group 17.
They are found between the metals and non-metals in the periodic table.
Eg: silicon is in group 14 along with germanium, whereas arsenic belongs to group 15.

Question 21.
A few elements of groups 1 and 2 are given in the Table below.
Complete the table and record it in your science diary.

Answer:

Question 22.
What is the relation between the number of outermost electrons and the group number here?
Answer:
Both are same.
In the elements of groups 1 and 2, the number of outermost electrons represents the group number
Let us examine whether groups 13 to 18 follow the same relation.

Question 23.
Complete the Table on the basis of the periodic table.

Answer:

Question 24.
Find the number that is added to the number of outermost electrons to get the group number of elements in groups 13 to 18.
Answer:
10

Question 25.
Have you ever thought why the number 10 is added to the number of outermost electrons?
Answer:
Between 2 and 13 groups of elements, 10 groups of transition elements are included.

Question 26.
In how many groups are they distributed?
Answer:
From 3 to 12
The position of transition elements is after the second group elements in the periodic table. The elements from groups 13 to 18 are placed after these 10 groups of transition elements.
It is clear why the number 10 is added to the number of outermost electrons to get the group number of groups 13 to 18.

Question 27.
Complete the Table with the help of the periodic table.

Answer:

Question 28.
Can you find any relation between the period number and the number of shells of the given elements?
Answer:
In these elements, the number of shells is the period number.
The number of shells in the atoms of elements is their period number

Question 29.
Certain data regarding the main group elements are given in the following table. Complete the Table and record it in your science diary.

Answer:

Question 30.
You know that the elements given in the table are noble gases. To which group do they
belong?
Answer:
Group 18

Question 31.
What peculiarity do you notice in the number of the outermost electrons of elements except helium?
Answer:
All elements have 8 electrons in the outermost shell except helium.
If elements other than hydrogen and helium have 8 electrons in their outermost shell, they attain . stability. It is to attain this stability that atoms of all elements undergo chemical reactions. (You can leam more about this in the next unit.)

Usually, 18th-group elements do not take part in chemical reactions because of the stable arrangement of electrons.

Question 32.
Elements ₈P, ₁₀Q, ₁₂R, ₁₈S are given, (symbols are not real)
a) Write down the electron configuration of these elements.
b) Which among these are noble gases?
Answer:
a) P : 2, 6
Q : 2, 8
R : 2, 8, 2
S : 2, 8, 8

b) Q and S

Question 33.
Which transition elements are familiar to you? List them with the help of the periodic table.
tsonSmiloaand ntg^ejoanjjjdajgpem mlsBnc/dt©® torel^jlanaocojcoi? (dlcdHeayocuildBs
Answer:
Iron, copper, zinc, silver, mercury etc.

Question 34.
Are all of them metals?
Answer:
Yes. All of them are metals.

Question 35.
From which period onwards can you locate transition elements in the periodic table?
Answer:
From the fourth period.

The elements of groups 1 and 2 are generally more metallic in nature and are placed on the left side of the periodic table. Meanwhile, the elements from groups 13 to 18 are placed on the right side, of the periodic table and are generally less metallic in nature. Based on this, how will you indicate the position of the transition elements?

The transition elements lie in between the more metallic elements and the comparatively less metallic ones.

The elements from group 3 upto group 12 are known as the transition elements because they indicate a regular change or transition from more metallic elements of group 2 to less metallic elements of group 13
Let us consider another peculiarity of the transition elements. The electron configuration of a few elements in the 4th period is given in the Table below

It is evident from the table that in the elements of group 1 and 2, the electron is being added to the last shell.

However, in groups 3,4 and 5, electrons are being added to the penultimate shell.

In ten groups from group 3 to 12 (transition elements), electron filling takes place in the penultimate shell.

You have learnt that elements in the same group show similarity in properties.
Generally, transition metals also show such similarity in groups.
Let us examine whether they exhibit any peculiarity along a period.

Question 36.
Analyse the transition elements of the 4th period given in the Table above.
Do they have any peculiarity in the number of outermost electrons?
Answer:
They have the same number of outermost electrons.
Usually, transition elements in the same period have the same number of outermost electrons. Hence,they show similarity in properties along a period too.

Question 37.
You have seen coloured chemicals in your lab. Examine the chemicals given in the Table given below. Find their molecular formulae and identify their colours with the help of your teacher. Complete the table and record it in your science diary.

Answer:

Name of the chemical	Molecular formula	Colour
Nickel sulphate	NiSO4	Bluish green
Copper sulphate	CuSO4	Blue
Calcium carbonate	CaCO3	No Colour
Potassium permanganate	KMnO4	Pink
Cabolt nitrate	CO(NO3)2	Blue
Potassium diehromate	K2Cr2O7	Orange
Ferrous sulphate	FeSO4	light green

It is clear that transition elements are present in the coloured compounds given in the table.Usually,
transition elements form coloured compounds.

• Elements included in groups 3 to 12 are transition elements.
• The filling of electrons takes place in the penultimate shell.
• Generally, they exhibit similanty in chemical properties in groups as well as in periods.
• They are metals.
• They generally form coloured compounds.

Question 38.
Have you noticed the number of elements included in the 6th period of the periodic table?
Answer:
18

Question 39.
Identify the position of lanthanum (atomic number -57) and the 14 elements following it.
Answer:
They are placed just below the main body of the periodic table as the first row.

Question 40.
Similarly, find the position of actinium (atomic number-89) and the 14 elements following it in the 7th period.
Answer:
They are placed just below the main body of the periodic table as the second row.

In the 6th period, lanthanum and the 14 elements following it, have been arranged separately at the bottom of the periodic table. The elements from lanthanum, (La, atomic number – 57) to lutetium (Lu, atomic number – 71) are known as lanthanoids.

In the 7th period, actinium and the 14 elements following it have ’been given a separate position below lanthanoids. The elements from actinium (Ac, atomic number – 89) to lawrencium (Lr, atomic number – 103) are called actinoids.

Lanthanoids and actinoids are known as inner transition elements. Lanthanoids are also called rare earths. Actinoids coming after uranium (U) are man-made elements.

Question 41.
You are familiar with situations in which transition elements and their compounds are used in our daily life. What are they?
Answer:

• All transition elements are useful metals.
• They are used to make alloys.
• Compounds of transition elements are used for adding colour to glass, fireworks etc.
• Transition elements and their compounds are used as catalysts.

Transuranium Elements

All the 118 elements discovered till now are included in the modem periodic table. Among elements from atomic number 1 to 92, the elements except technitium (atomic number 43) and promethium (atomic number 61) are naturally occurring. Elements coming after atomic number 92 are made artificially. Artificial elements are less stable and exhibit radio activity.
The elements coming after uranium (atomic number – 92) are known as transuranium elements

Question 42.
What change do you observe in the number of shells, on moving down the group?
Answer:
The number of shells increases.

Question 43.
How does the increase in the number of shells influence the size of an atom?
Answer:
When the number of shells increases, the size of the atom also increases.The nuclear charge depends on the number of protons present in the nucleus.

Question 44.
What change do you observe in the number of protons with the increase in the atomic number?
Answer:
Increases

Question 45.
If so, what happens to the nuclear charge with the increase in the atomic number?
Answer:
Increases
With an increase in nuclear charge, the force of attraction between the nucleus and the outermost electron increases.

Question 46.
If so, what happens to the size of the atom? Why?
Answer:
Size of atom decreases.
Because when nuclear charge increases, the attraction of nucleus on electrons increases.
Though nuclear charge increases down a group, its effect is overcome by the increase in the number of shells and hence, the size of the atom increases.
Though nuclear charge increases down a group, its effect is overcome by the increase in the number of shells and hence, the size of the atom increases.
The electron configuration of the elements belonging to the 2nd period of the periodic table is given below.

Question 47.
Do you observe any change in the number of shells on moving along a period from left to right?
Answer:
No

Question 48.
Does the nuclear charge increase?
Answer:
Nuclear charge increases on moving along a period from left to right, but there is no change in the number of shells.

Question 49.
What happens to the attractive force of the nucleus towards the outermost electrons? (increases/ decreases)
Answer:
Increases

Question 50.
What change takes place in the size of the atom?
Answer:
The size of the atom will decrease.
Moving along a period from left to right, there is no change in the number of shells. But nuclear charge increases gradually. The attractive force of the nucleus on the outermost electron increases. Hence, the size of the atom gradually decreases.

Screening Effect (Shielding Effect)
The number of shells increases down a group. As a result, the outermost electrons move away from the nucleus. As the number of electrons in the inner shells increases, the attractive force of the nucleus on the outermost electrons decreases gradually. This is known as the screening effect.
You have seen the change in the size of the atom in group and period.

Question 51.
If so, where can you locate the comparatively bigger atoms in the periodic table?
Answer:
Left bottom area

Question 52.
Where are the smaller atoms located?
Answer:
Right top area
Moving down the group, the size of an atom increases. The size of an atom decreases on moving from left to right along a period.
You will learn about periodic trends, such as ionisation energy, electronegativity etc in the next unit.

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 3 Chemical Bonding Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 3 Solutions Chemical Bonding

Kerala Syllabus Std 9 Chemistry Chapter 3 Chemical Bonding Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 3 Let Us Assess Answers Chemical Bonding

Question 1.
Draw the electron dot diagram of hydrogen (H), helium (He), lithium (Li), beryllium (Be) and fluorine (F).
Answer:

Question 2.
Illustrate the formation of the chemical bond in chlorine (Cl₂) using an electron dot diagram as illustrated in the fluorine (F₂) molecule.
Answer:

Question 3.
Represent the covalent bond in chlorine molecules using symbols.
Answer:
Cl – Cl

Question 4.
Represent the formation of ionic bonds in the following ionic compounds using the electron dot diagram and orbit model.
a) Sodium fluoride (NaF)
b) Sodium oxide (Na₂O)
c) Magnesium fluoride (MgF₂)
d) Calcium oxide (CaO)
Answer:

Question 5.
Assume that calcium (Ca) and fluorine (F) combine.
a) Complete the following table accordingly.

Element	Atomic number	Electronic configuration	Number of electrons received or donated
Ca	20	……………………………	………………………………….
F	9	……………………………..	……………………………..

b) Write the chemical formula of calcium fluoride.
c) Similarly, write the chemical formula of magnesium chloride and aluminium chloride.
Answer:

Element	Atomic number	Electronic configuration	Number of electrons received or donated
Ca	20	2, 8, 8, 2	2
F	9	2,7	1

b) CaF₂

c)

Element	Atomic number	Electronic configuration	Number of electrons received or donated
Mg	12	2, 8, 2	2
Al	13	2, 8, 3	3
Cl	17	2, 8, 7	1

Magnesium chloride – MgCl₂
Aluminium chloride – AlCl₃

Question 6.
Some cations and anions are given in the table. Fill in the blanks.

Answer:

Question 7.
Complete the following chemical equations and answer the questions given below.
(Hint: Atomic number Mg – 12, Cl – 17)
Mg → Mg²⁺ + ________
Cl + 1 e^– → ________
________ + ________ → MgCl²
(a) Identify the cation and anion in these compounds.
(b) What is the nature of the chemical bond in MgCh?
Answer:
Mg → Mg²⁺ + 2e^–
Cl + l e^– → Cl^–
Mg²⁺ + 2Cl^– → MgCl₂

(a) Cation – Mg²⁺
Anion – Cl^–
(b) Ionic bonding

Question 8.
Complete the following table. (Hint: Atomic number F – 9, H – 1, O – 8, N – 7)

Answer:

Question 9.
Complete the following table. (Symbols are not real)

Element	Atomic number	Electron configuration
P	12	………………..
Q	………………..	2, 7
R	10	………………..
S	17	………………..

a) Which among these is the most stable element?
b) Which element donates electrons during chemical reactions?
c) Write the chemical formula of the compound formed when the elements P and S combine.
Answer:

Element	Atomic number	Electron configuration
P	12	2, 8, 2
Q	9	2, 7
R	10	2, 8
S	17	2, 8, 7

a) R is the most stable element.
b) P is the element that donates electrons.

Formula – PS₂

Question 10.
Atom models of two elements are represented below.

a) Draw the electron dot diagram of the formation of sodium fluoride.
b) What is the nature of the chemical bond in sodium fluoride?
c) Write any two characteristics of compounds having this type of bond.
Answer:
a)
b) Ionic bond
c)

• Dissolves in polar solvents like water.
• Exhibits high melting and boiling points.
• Conducts electricity in a molten state or solutions.

Question 11.
The electron configuration of the elements P, Q, and R are given below. (Symbols are not real)
P – 2, 8, 6
Q – 2, 8, 1
R – 2, 8, 8
a) Which is the most stable element among these? What is the reason?
b) What is the atomic number of Q?
c) Draw the atom model of Q.
d) What are the valencies of the elements P and Q?
e) Write the chemical formula of the compound formed when P and Q combine.
Answer:
a) R is the most stable element among these. Because it has a stable octet electron configuration in
the outermost shell.
b) Atomic number of Q = 11
c)
d) Valency of P = 2
Valency of Q = 2

e)
Formula – Q₂P

Question 12.
A, B, C and D are four elements (Symbols are not real). Information about them is given in the following table.

Element	Atomic number	Electronegativity
A	6	2.55
B	8	3.44
C	12	1.31
D	17	3.16

Based on these, find the type of bond in the compounds formed by the combination of the
following pairs of elements.
1. C, B
2. C, D
3. A, B
Answer:
1. C, B
Difference in electronegativity = 3.44 – 1.31 = 2 : 13
Greater than 1.7, so it is ionic bonding.

2. C, D
The difference in electronegativity = 3.16 – 1.31 = 1.85
Greater than 1.7, so it is ionic bonding.

3. A, B
Difference in electronegativity = 3.44 – 2.55 = 0.89
Less than 1.7, so it is covalent bonding.

Extended Activities

Question 1.
Magnesium nitride is obtained when nitrogen is passed over heated magnesium. Write the chemical equation of this reaction. Find out whether the formed compound is ionic or covalent using the electronegativity scale given in this unit. (Hint – Valency: Nitrogen – 3, Magnesium – 2)
Answer:
The chemical equation for the reaction is 3Mg + N₂ → Mg₃N₂
The difference in electronegativity = 3.04 – 1.31 = 1.73
Greater than 1.7, so it is an ionic compound.

Question 2.
Draw the electron dot diagram of the chemical bonds in ethane (C₂H₆), ethene (C₂H₄) and ethyne (C₂H₂). Find out whether these compounds are ionic or covalent. Calculate the total number of bonds in each compound.
Answer:
a) Ethane (C₂H₆)

Total number of bonds = 7

b) Ethane (C₂H₄)

Total number of bonds = 5

c) Ethyne (C₂H₂)

Total number of bonds = 3

Question 3.
Conduct the experiment arranging the apparatus as shown in the figure.

Record your observations and identify what types of compounds are sodium chloride and glucose.
Answer:
The galvanometer in the first experiment shows deflection, indicating that electricity passes through a common salt solution. This observation also notes that gases evolve from metal rods. Thus, common salt is confirmed as an ionic compound, as ionic compounds conduct electricity in their molten state or solutions.

No deflection is observed in the second experiment, signifying that electricity does not pass through the glucose solution. Therefore, glucose is identified as a covalent compound. Typically, covalent compounds do not conduct electricity.

Question 4.
Draw the chemical bonds in different compounds and exhibit them on the bulletin board.
Answer:

Cl – Cl


a)
b) Ionic bond
c)

• Dissolves in polar solvents like water.
• Exhibits high melting and boiling points.
• Conducts electricity in a molten state or solutions.

Chemical Bonding Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Some substances are given below. Differentiate them into elements and compounds and list them.
Potassium, oxygen, water, common salt, nitrogen, helium, hydrogen, and sugar.

Answer:

You know that there are two atoms in one molecule of hydrogen. If so, how many atoms are there in each substance given.

Question 2.
How many atoms are there in each substance given below?

Answer:

Molecule	Number of atoms
Oxygen (O2)	2
Water (H2O)	3
Nitrogen (N2)	2
Helium (He)	1
Methane (CH4)	5
Sugar (C12H22O11)	45

Some molecules have more than one atom.

Question 3.
Why do atoms in a molecule stay together?
Answer:
The atoms of a molecule stay together because of chemical bonds.

Question 4.
Why do atoms combine to form molecules?
Answer:
An atom combines in order to attain stability. Also, different compounds are formed by the combination of different atoms.

Question 5.
How do atoms combine?
Answer:
Atoms can combine either by sharing electrons or by completely transferring the electrons.

Question 6.
Do all atoms combine in the same way?
Answer:
No, not all atoms combine in the same way.

Question 7.
Do all atoms combine with other atoms?
Answer:
Not all atoms combine with other atoms.

Question 8.
How many atoms are there in a molecule of noble gases?
Answer:
One atom only.
Generally, noble gases do not combine with other atoms due to their stability. They are able to exist independently. Examples of noble gases, along with their atomic number and electronic configuration, are given in the table below.

Element (Symbol)	Atomic number	Electronic Configuration
Helium (He)	2	2
Neon (Ne)	10	2, 8
Argon (Ar)	18	2, 8, 8
Krypton (Kr)	36	2, 8, 18, 8
Xenon (Xe)	54	2, 8, 18, 18, 8
Radon (Rn)	86	2, 8, 18, 32, 18, 8

Question 9.
How many electrons are there in the outermost shell of noble gases except helium?
Answer:
Eight electrons.
The atomic number of helium is 2; hence, it contains a maximum of two electrons in its outermost shell. All the other noble gases have a total of eight electrons in their outermost shell.

The arrangement of eight electrons in the outermost shell is called octet configuration.

Atoms having octet configuration are more stable. Such atoms are generally reluctant to take part in chemical reactions. So noble gases are also called inert gases. In the case of helium, the configuration is called duplet configuration, which is stable like that of the other noble gases.

Question 10.
Look at the electron configuration of magnesium and oxygen given in the table.

Element	Atomic number	Electronic configuration
Magnesium	12	2, 8, 2
Oxygen	8	2, 6

(i) Are these atoms stable?
(ii) How can they attain stability?
(iii) What is the name of the compound formed when these atoms combine?
Answer:
(i) No
(ii) Stability can be attained by gaining octet electron configuration in the outermost shell through chemical bonding. (In here, magnesium must lose two electrons, and oxygen must gain two- electrons).
(iii) Magnesium oxide (MgO).
The force that binds together the component particles in a compound is called chemical bonding,

Question 11.
The chemical name of table salt is sodium chloride. Answer the following questions.
(i) What are the constituent elements of sodium chloride?
(ii) Write the electron configuration of the sodium atom (atomic number – 11).
(iii) How many electrons are there in the outermost shell of a sodium atom?
(iv) How does the sodium atom attain octet electron configuration?
Answer:
(i) Sodium and Chlorine
(ii) ₁₁Na – 2, 8, 1
(iii) One atom
(iv) By losing one electron in the outermost shell.
In the case of a sodium atom, it loses one electron and changes into a sodium ion.
Na → Na⁺ + le^–

The removal of the outermost electron from the sodium atom can only be achieved by overcoming the force of attraction exerted by the nucleus.

This energy required, or in other terms, the amount of energy required to remove the most loosely bound electron from the outermost shell of an isolated gaseous atom of an element is called its ionisation energy or ionisation enthalpy.

The amount of energy required to remove the most loosely bound electron from the outermost shell of an isolated gaseous atom of an element is called its ionisation energy.

Question 12.
Write the electron configuration of a chlorine atom (atomic number 17).
Answer:
₁₇Cl – 2, 8, 7

Question 13.
How many electrons are needed for the chlorine atom to attain octet electron configuration?
Answer:
One.
I Non-metals like chlorine accept an electron to become a chloride ion.

Energy is released when atoms become negative ions by accepting electrons. This energy released when an electron is added to a neutral gaseous atom to form a negative ion is called electron gain enthalpy.

Illustrate the electron exchange in the formation of sodium chloride through shell electron configuration.

Question 14.
Represent the electron dot diagram of a chlorine atom.
Answer:


The electron dot diagram of the formation of sodium chloride can be represented as

Question 15.
Observe the electron dot diagram and shell electron configuration diagram of sodium chloride formation and complete the table.

Answer:

The equations of the electron transfer during the formation of sodium-chloride are:
Na → Na⁺ + le^–
Cl + le- → Cl^–
In, the reaction between sodium and chlorine to form sodium chloride, the sodium atom loses one electron and gets converted into sodium ion (Na⁺). The chlorine atom accepts an electron to form a chloride ion (Cl^–).

The positive ions formed by losing electrons during chemical reactions are called cations, and the negative ions formed by accepting electrons are called anions.

The particles Na⁺ and Cl^–, which possess opposite charges, show mutual attraction and are bound together by an extremely strong electrostatic force of attraction, resulting in the formation of NaCl.

The electrostatic force of attraction that holds together the oppositely charged ions in an ionic compound is called an ionic bond. An ionic bond is also known as an electrovalent bond. An ionic bond is always formed between a metal and a non-

Question 16.
What is the compound formed when magnesium burns in the air?
Answer:
Magnesium oxide (MgO)
The equation for the chemical reaction between magnesium and oxygen can be represented as
2Mg + O₂ → 2MgO

Question 17.
The electron dot diagram of magnesium oxide formation is given. Analyse the diagram and complete the table.

Answer:

Question 18.
Which are the ions present in magnesium oxide?
Answer:
Magnesium ion (Mg²⁺) and oxygen ion (O^2-).

Question 19.
How many electrons are transferred from magnesium to oxygen during the formation of magnesium oxide?
Answer:
Two electrons.
By the transfer of two electrons from magnesium to oxygen, an ionic bond is formed between them. The
compounds that are formed by ionic bonding are known as ionic compounds or electrovalent compounds.
The distribution of electrons of fluorine is given.

Question 20.
How many electrons are there in the outermost shell of fluorine?

Answer:
Seven electrons.

Question 21.
How many more electrons are required for one fluorine atom to attain octet configuration? 63caj
Answer:
One electron.

Question 22.
Is it possible to transfer electrons from one fluorine atom to another? If so, what, type of arrangement might have taken place, between the atoms in order to attain octet configuration?
Answer:
No, it is not possible to move electrons from one fluorine atom to another. In order to attain stability, atoms will share electrons.
In the case of some molecules like fluorine, the octet configuration is attained by the sharing of electrons.

Question 23.
Analyse the electron dot diagram of fluorine molecule formation through chemical bonding and answer the following questions.

(i) How many electrons are donated by each fluorine atom for sharing?
(ii) How many pairs of electrons are shared in the chemical bonding of fluorine molecule?
Answer:
(i) One each.
(ii) One pair.
The chemical bond formed as a result of the sharing of electrons between the combining atoms is called a covalent bond. The covalent bond formed by the sharing of one pair of electrons is a single bond.
A single bond is represented by a small line (-) between the symbols of the combining elements in molecules. The single bond in fluorine molecule can be represented using symbols such as F – F.

Question 24.
Oxygen is a diatomic molecule. Now answer the following questions.
(i) What is the atomic number of oxygen?
(ii) Write the electronic configuration of oxygen.
(iii) How many more electrons are required for one oxygen atom to attain the octet configuration?
Answer:
(i) Atomic number of oxygen is eight.
(ii) ₈O – 2, 6.
(iii) 2 more electrons.

Question 25.
The illustration of a chemical bond in an oxygen molecule is given below

How many pairs of electrons are shared in the oxygen molecule?
Answer:
Two pairs
The covalent bond formed by the sharing of two electron pairs or four electrons is called a double covalent bond.
The double bond in oxygen molecules can be represented by using symbols such as 0 = 0.

Question 26.
The illustration of a chemical bond in a nitrogen molecule is given below

How many pairs of electrons are shared here to complete the octet configuration?
Answer:
3 pairs
The covalent bond formed by the sharing of three electron pairs or six electrons is called a triple covalent bond.
The triple bond in nitrogen molecules can be represented by using symbols such as N ≡ N

Question 27.
Illustrate the chemical bond in a hydrogen molecule using an electron dot diagram.
Answer:

In this case, two hydrogen atoms share one pair of electrons, which forms a single bond. Stability is achieved by adopting the electronic configuration of helium, which is the nearest noble gas.

Formation of hydrogen chloride molecule:
The atomic number of hydrogen is 1.
Electronic configuration = 1.
The atomic number of chlorine is 17.
Electronic configuration = 2, 8, 7.
Chlorine needs one more electron to complete the octet, which the hydrogen atom will share, thereby forming the hydrogen chloride molecule. Hence, a single bond is present in hydrogen chloride.

Question 28.
Represent the covalent bond in hydrogen chloride using symbols.
Answer:
H – Cl

Question 29.
Depict the chemical bonding in hydrogen fluoride molecules.
Answer:

A water molecule consists of two hydrogen atoms and one oxygen atom. The chemical bonding present in water molecules can be represented as:

Question 30.
How many covalent bonds are formed here?
Answer:
2 covalent bonds.
Compounds formed by covalent bonding are called covalent compounds. When non-metals combine, usually covalent compounds are formed.

Question 31.
Is the shared pair of electrons in the HF molecule attracted equally by both atoms?
Answer:
No, the electrons are more attracted by fluorine.

Question 32.
Find out the electronegativity difference of the constituent elements and complete the table.

Compounds	Difference in electronegativity of constituent elements	Nature of the compound
Sodium chloride (NaCl)	3.16 – 0.93 = …………………	Ionic
Hydrogen Chloride (HCl)	3.16 – 2.20 = …………………	Covalent
Sodium Oxide (Na2O)	………………..	…………………
Calcium Chloride (CaCl2)	…………………	…………………
Methane (CH4)	…………………	…………………
Magnesium Fluoride (MgF2)	…………………	…………………

Answer:

Compounds	Difference in electronegativity of constituent elements	Nature of the compound
Sodium chloride (NaCl)	3.16 – 0.93 = 2.23	Ionic
Hydrogen Chloride (HCl)	3.16 – 2.20 = 0.96	Covalent
Sodium Oxide (Na2O)	3.44 – 0.93 = 2.51	Ionic
Calcium Chloride (CaCl2)	3.16 – 1.00 = 2.16	Ionic
Methane (CH4)	2.55 – 2.20 = 0.35	Covalent
Magnesium Fluoride (MgF2)	3.98 – 1.31 = 2.67	Ionic

Question 33.
Answer the questions given below
(i) What is the electronegativity value of hydrogen?
(ii) What is the electronegativity value of chlorine?
(iii) The nucleus of which of these atoms has a greater tendency to attract the shared pair of electrons involved in covalent bonding?
Answer:
(i) 2.20
(ii) 3.16
(iii) Chlorine will attract the shared pair of electrons.

Question 34.
Analyse the change in the electron arrangement in atoms during the formation of each compound and
Complete the table given below.

Answer:

Question 35.
Complete the following table regarding the combination of magnesium (Mg) and fluorine (F).

Element	Atomic number	Electron configuration	Number of electrons donated or accepted
Mg	12	–	–
F	9	–	–

Answer:

Element	Atomic number	Electron configuration	Number of electrons donated or accepted
Mg	12	2, 8, 2	2
F	9	2, 7	1

Question 36.
How many fluorine atoms are required to receive the electrons donated by magnesium?
Answer:
2 fluorine atoms.
During the formation of magnesium fluoride, one magnesium atom combines with two fluorine atoms. Hence, the chemical formula of magnesium fluoride will be MgF₂.

Question 37.
What are the constituent elements of aluminium oxide?
Answer:
Aluminium and Oxygen

Question 38.
What is the valency of aluminium? (Atomic number – 13)
Answer:
Electron configuration – 2, 8, 3
Therefore, the valency of aluminium is three.

Question 39.
What is the valency of oxygen? (Atomic number – 8)
Answer:
Electron configuration – 2, 6
Therefore, the valency of oxygen is two.

Question 40.
What are the constituent elements of carbon dioxide?
Answer:
Carbon and oxygen

Question 41.
Write the symbols of elements together, considering their electronegativity.
Answer:
C and O.

Question 42.
The valency of carbon is 4, and that of oxygen is 2. Interchange the valencies and write them as base indices.
Answer:

Chemical formula – C₂O₄
Divide the base indices by common factor, C_2/2O_4/2 = C₁O₂.
If the base index is 1, then there is no need to write. Therefore, the chemical formula of carbon dioxide is CO₂.

Question 43.
The constituent elements of some compounds and the valencies of their constituent elements are given in the following table. Find out the chemical formula.

Answer:

Question 44.
Which are the ions derived from hydrochloric acid? Why is it a monobasic acid?
Answer:
Hydrochloric acid contains H⁺ and Cl^–. The ionisation of one molecule of HC1 releases one H⁺ ion; hence, it is a monobasic acid.
In the case of sulphuric acid, two H⁺ and one SO₄^2- ions are released. Therefore, it is a dibasic acid. Hence, the chemical formula of sulphuric acid is H₂SO₄.

Question 45.
The basicity and the negative ions of certain acids are given in the table. Find their chemical formula and complete the table.

Answer:

Bases that are soluble in water are called alkalies. The number of OH” ions in an alkali will be equal to the number of positive ions.

Question 46.
Which is the positive ion present in sodium hydroxide?
Answer:
Sodium ion (Na⁺)

Question 47.
How many OH” ions, equal to the positive charge on sodium ion, will be present in sodium hydroxide?
Answer:
One

Question 48.
If so, what is the chemical formula of sodium hydroxide?
Answer:
NaOH

Question 49.
The positive ions of some bases are given in the table below. Find the chemical formulae and complete the table.

Answer:

Question 50.
Which is the positive ion in magnesium hydroxide, Mg(OH)₂?
Answer:
Magnesium ion (Mg²⁺).

Question 51.
Which is the negative ion in phosphoric acid, H₃PO₄?
Answer:
Phosphate ion (PO₄^3-).
Let us write the chemical formula of magnesium phosphate, which is formed from magnesium hydroxide and phosphoric acid.
Step 1 – Write the symbols of the ions Mg²⁺ + PO₄^3-
Step 2 – Interchange the number indicating charge and write as base indices. Mg₄(PO₄)₂
Therefore, the chemical formula of magnesium phosphate is Mg₃(PO₄)₂.

Question 52.
The reaction between sulphuric acid and calcium hydroxide forms salt calcium sulphate. To find the chemical formula of calcium sulphate, answer the following questions.
(i) Which is the positive ion in calcium hydroxide, Ca(OH)₂?
(ii) Which is the negative ion in sulphuric acid, H₂SO₄?
(iii) Write the symbol of the positive ion and then the symbol of the negative ion.
(iv) Write the number indicating the charge of each ion/radical as the base index after interchanging them.
Answer:
(i) Calcium ion, Ca²⁺
(ii) Sulphate ion, SO₄^2-
(iii) Ca²⁺ SO₄^2-
(iv) Ca₂(SO₄)₂
Simplify the base indices into simple whole number ratio
Ca_2/2(SO₄)_2/2 = CaSO₄
Therefore, the chemical formula of calcium sulphate is CaSO₄.

Question 53.
Certain positive ions and negative ions are given in the following’ table. Complete the table by writing the chemical formula and the name of the salt formed from these ions.

Answer:

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 15 Malayalam Medium സ്ഥിതിവിവരക്കണക്ക് can save valuable time.

Kerala SCERT Class 9 Maths Chapter 15 Solutions Malayalam Medium സ്ഥിതിവിവരക്കണക്ക്

Class 9 Maths Chapter 15 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 15 Malayalam Medium Textual Questions and Answers

Question 1.
ട്വന്റി-ട്വന്റി ക്രിക്കറ്റിൽ ഒരു ടീം ആദ്യത്തെ 5 ഓവറിൽ 51 റൺ നേടി.
i) അപ്പോഴത്തെ റൺ നിരക്ക് എത്രയാണ്?
Answer:

ii) ഇതേ റൺ നിരക്ക് തുടരുകയാണെങ്കിൽ 20 ഓവറിൽ എത്ര റൺ പ്രതീക്ഷിക്കാം?
Answer:
ഇതേ റൺ നിരക്ക് തുടർന്നാൽ 20 ഓവറിൽ എടുക്കാവുന്ന റൺസ് = 10.2 × 20 = 204

Question 2.
ക്ലാസിൽ ഒരു കണക്കു പരീക്ഷ നടത്തി, മാർക്കിന്റെ അടിസ്ഥാനത്തിൽ കുട്ടികളെ തരംതിരിച്ച പട്ടികയാണ് ചുവടെ കാണിച്ചിരിക്കുന്നത്:

i) കുട്ടികൾക്ക് കിട്ടിയ മാർക്കുകളുടെ മാധ്യം കണക്കാക്കുക.
ii) മാധ്യത്തെക്കാൾ കുറവ് മാർക്ക് കിട്ടിയവർ എത്ര പേരാണ്?
iii) മാധ്യത്തെക്കാൾ കൂടുതൽ മാർക്ക് കിട്ടിയവർ എത്ര പേരാണ് ?
Answer:

മാർക്ക്	കുട്ടികൾ	ആകെ മാർക്ക്
2	1	2 × 1 = 2
3	2	3 × 2 = 6
4	5	4 × 5 = 20
5	4	5 × 4 = 20
6	6	6 × 6 = 36
7	11	7 × 11 = 77
8 ‘	10	8 × 10 = 80
9	4	9 × 4 = 36
10	2	10 × 2 = 20
ആകെ	45	297

കുട്ടികൾക്ക് കിട്ടിയ മാർക്കുകളുടെ മാധ്യം = \(\frac{297}{45}\) = 6.6
ii) മാധ്യത്തെക്കാൾ കുറവ് മാർക്ക് കിട്ടിയവർ = 1 + 2 + 5 + 4 + 6 = 18
iii) മാധ്യത്തെക്കാൾ കൂടുതൽ മാർക്ക് കിട്ടിയവർ = 11 + 10 + 4 + 2 = 27

Question 3.
ഒരു കർഷകന് ഒരു മാസം കിട്ടിയ റബ്ബർഷീറ്റിന്റെ വിവരങ്ങൾ ചുവടെയുള്ള പട്ടികയിലുണ്ട്:

i) ഈ മാസത്തിൽ കിട്ടിയ റബ്ബർഷീറ്റിന്റെ മാധ്യ ഭാരം എത്രയാണ്?
Answer:

റബ്ബർ (കിഗ്രാം)	ദിവസങ്ങൾ	ആകെ റബ്ബർ (കിഗ്രാം)
09	3	9 × 3 = 27
10	4	10 × 4 = 40
11	3	11 × 3 = 33
12	3	12 × 3 = 36
13	5	13 × 5 = 65
14	6	14 × 6 = 84
16	6	16 × 6 = 96
ആകെ	30	381

ഈ മാസത്തിൽ കിട്ടിയ റബ്ബർഷീറ്റിന്റെ മാധ്യ ഭാരം = \(\frac{381}{30}\) = 12.7 കിഗ്രാം

ii) റബ്ബറിന്റെ വില കിലോഗ്രാമിന് 175 രൂപയാണ്. ഈ മാസത്തിൽ റബ്ബറിൽ നിന്നു കിട്ടിയ മാധ്യ വരുമാനം എത്ര രൂപയാണ്?
Answer:
റബ്ബറിന്റെ വില കിലോഗ്രാമിന് 175 രൂപ ആയതിനാൽ ഈ മാസത്തിൽ റബ്ബറിൽ നിന്നു കിട്ടിയ മാധ്യ വരുമാനം = 12.7 × 175
= 2222.5 രൂപ

Question 4.
ഒരു പ്രദേശത്തു പെയ്ത മഴയുടെ അളവനുസരിച്ച് ഒരു മാസത്തിലെ ദിവസങ്ങളെ തരം തിരിച്ച പട്ടികയാണിത്.

ആ മാസം അവിടെ ഒരു ദിവസം പെയ്ത മഴയുടെ മാധ്യ അളവെന്താണ്?
Answer:

മഴ (മിമീ)	ദിവസങ്ങൾ	ആകെ മഴ (മിമീ)
54	3	54 × 3 = 162
56	5	56 × 5 = 280
58	6	58 × 6 = 348
55	3	55 × 3 = 165
50	2	50 × 2 = 100
47	4	47 × 4 = 188
44	5	44 × 5 = 220
41	2	41 × 2 = 82
ആകെ	30	1545

ആ മാസം അവിടെ ഒരു ദിവസം പെയ്ത മഴയുടെ മാധ്യം

= \(\frac{1545}{30}\)
= 51.5 മിമീ

Question 5.
ഒരു ക്ലാസിലെ കുട്ടികളെ ഉയരത്തിന്റെ അടിസ്ഥാനത്തിൽ തരംതിരിച്ച പട്ടികയാണ് ചുവടെ കാണുന്നത്.

ഈ ക്ലാസിലെ കുട്ടികളുടെ മാധ്യഉയരം എത്രയാണ് ?
Answer:

ഉയരം (സെമീ)	കുട്ടികളുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ ഉയരം
148- 152	8	150	1200
152- 156	10	154	1540
156 – 160	15	158	2370
160- 164	10	162	1620
164- 168	7	166	1162
ആകെ	50		7892

= \(\frac{7892}{50}\)
= 157.84 സെമീ

Question 6.
ഒരു സർവകലാശാലയിലെ അധ്യാപകരുടെ എണ്ണം പ്രായമനുസരിച്ച് തരംതിരിച്ചെഴുതിയ പട്ടികയാണ് ചുവടെയുള്ളത്.

അധ്യാപകരുടെ മാധ്യ പ്രായം കണക്കാക്കുക.
Answer:

പ്രായം	അധ്യാപകരുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ പ്രായം
25-30	06	27.5	165
30-35	14	32.5	455
35-40	18	37.5	675
40-45	20	42.5	850
45-50	05	47.5	237.5
50-55	04	52.5	210
55-60	03	57.5	172.5
ആകെ	70		2765

= \(\frac{2765}{70}\)
= 39.5

Question 7.
ഒരു ക്ലാസിലെ കുട്ടികളെ ഭാരമനുസരിച്ചു തരംതിരിച്ച പട്ടികയാണിത്.

മാധ്യ ഭാരം കണക്കാക്കുക.
Answer:

ഭാരം (കിഗ്രാം)	കുട്ടികളുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ ഭാരം
21-23	4	22	88
23-25	7	24	168
25-27	8	26	208
27-29	6	28	168
29-31	3	30	90
31-33	1	32	32
ആകെ	29		754

= \(\frac{754}{29}\)
= 26 കിഗ്രാം

Class 9 Maths Chapter 15 Malayalam Medium Intext Questions and Answers

Question 1.
ഏതെങ്കിലും കുറേ സംഖ്യകൾ എടുത്തു മാധ്യം കണക്കാക്കുക; ഓരോ സംഖ്യയും മാധ്യത്തേക്കാൾ എത്ര കൂടുതൽ, അല്ലെങ്കിൽ എത്ര കുറവ് എന്നു കണക്കാക്കി, കൂടുതലും കുറവും വെവ്വേറെ കൂട്ടി നോക്കുക. ഒരേ തുകയാണോ?
Answer:
ഇത് എന്തുകൊണ്ട് എന്നു വിശദീകരിക്കാമോ?
10, 15, 20, 25, 30 എന്നീ സംഖ്യകൾ പരിഗണിക്കുക.
മാധ്യം = \(\frac{10+15+20+25+30}{5}=\frac{100}{5}\) = 20
ഓരോ സംഖ്യയും മാധ്യത്തേക്കാൾ എത്ര കൂടുതൽ, അല്ലെങ്കിൽ എത്ര കുറവ് എന്ന് കണക്കാക്കാം.

സംഖ്യകൾ	മാധ്യത്തിൽ നിന്ന് എത്ര കൂടുതൽ	മാധ്യത്തിൽ നിന്ന് എത്ര കുറവ്
10		20 – 10 = 10
15		20 – 15 = 5
20
25	25 – 20 = 5
30	30 – 20 = 10

മാധ്യത്തിൽ നിന്ന് കൂടുതൽ വന്ന സംഖ്യകളുടെ തുക = 10 + 5 = 15
മാധ്യത്തിൽ നിന്ന് കുറവ് വന്ന സംഖ്യകളുടെ തുക = 10 + 5 = 15
ഇവിടെ, മാധ്യത്തിൽ നിന്ന് കൂടുതൽ വന്ന സംഖ്യകളുടെ തുകയും മാധ്യത്തിൽ നിന്ന് കുറവ് വന്ന സംഖ്യകളുടെ തുകയും തുല്യമാണ്. അതായത്, മാധ്യം അഥവാ ശരാശരി ഒരു സംഖ്യാസമൂഹത്തിന്റെ തുലനബിന്ദുവായാണ് പ്രവർത്തിക്കുന്നത്.

Statistics Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ഒരു തൊഴിൽ ശാലയിലെ തൊഴിലാളികളുടെ ദിവസക്കൂലി പട്ടികപ്പെടുത്തിയിരിക്കുന്നു. ശരാശരി കൂലി എത്ര?

കൂലി	എണ്ണം
500	3
600	7
700	10
900	8
1000	2

Answer:

കൂലി	എണ്ണം	ആകെ കൂലി
500	3	1500
600	7	4200
700	10	7000
900	8	7200
1000	2	2000
ആകെ	30	21900

ശരാശരി കൂലി = \(\frac{21900}{30}\) = 730

Question 2.
ഒരു ക്ലാസിലെ കുട്ടികളെ പരീക്ഷയ്ക്ക് കിട്ടിയ മാർക്കിന്റെ അടിസ്ഥാനത്തിൽ തരംതിരിച്ച പട്ടികയാണിത്

i) ശരാശരി മാർക്ക് 6 ആണ്. എത്ര കുട്ടികൾക്കാണ് 8 മാർക്ക് കിട്ടിയത്?
ii) ക്ലാസിൽ അകെ എത്ര കുട്ടികളുണ്ട്?
Answer:

മാർക്ക്	കുട്ടികൾ	ആകെ മാർക്ക്
3	2	6
4	4	16
5	5	25
6	6	36
7	7	49
8	x	8x
9	2	18
10	1	10
ആകെ	27 + x	160 + 8x

i) ശരാശരി മാർക്ക്

6 = \(\frac{160+8 x}{27+x}\)
6(27 + x) = 160 + 8x
162 + 6x = 160 + 8x
2x = 2
x = 1
8 മാർക്ക് കിട്ടിയ കുട്ടികളുടെ എണ്ണം = 1
ii) ക്ലാസ്സിലെ ആകെ കുട്ടികളുടെ എണ്ണം = 27 + x = 27 + 1 = 28

Question 3.
ഒരു സ്ഥാപനത്തിലെ തൊഴിലാളികളുടെ ദിവസവേതനം പട്ടികയിൽ ചുവടെ കൊടുത്തി രിക്കുന്നു. മാധ്യവേതനം കണക്കാക്കുക.

ദിവസ വേതനം (രൂപ)	തൊഴിലാളികളുടെ എണ്ണം
15000 – 18000	1
18000 – 21000	3
21000 – 24000	5
24000 – 27000	4
27000 – 30000	1
30000 – 33000	1

Answer:

ദിവസ വേതനം (രൂപ)	എണ്ണം തൊഴിലാളികളുടെ	വിഭാഗമാധ്യം	ആകെ വേതനം
15000 – 18000	1	16500	16500
18000 – 21000	3′	19500	58500
21000 – 24000	5	22500	112500
24000 – 27000	4	25500	102000
27000 – 30000	1	28500	28500
30000 – 33000	1	31500	31500
ആകെ	15		349500

= \(\frac{3,49,500}{15}\)
= 23,300 രൂപ

Question 4.
ഒരു സ്ഥാപനത്തിലെ തൊഴിലാളികളുടെ ദിവസവേതനം പട്ടികയിൽ ചുവടെ കൊടുത്തിരി ക്കുന്നു. മാധ്യവേതനം കണക്കാക്കുക.

ദിവസ വേതനം (രൂപയിൽ)	തൊഴിലാളികളുടെ എണ്ണം
450 – 550	7
550 – 650	8
650 – 750	10
750 – 850	10
850 – 950	9
950 – 1050	6

Answer:

ദിവസ വേതനം (രൂപയിൽ)	തൊഴിലാളികളുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ വേതനം
450 – 550	7	500	3500
550 – 650	8	600	4800
650 – 750	10	700	7000
750 – 850	10	800	8000
850 – 950	9	900	8100
950 – 1050	6	1000	6000
ആകെ	50		37,400

= \(\frac{37,400}{50}\)
= 748 രൂപ

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 15 Statistics Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 15 Solutions Statistics

Statistics Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 15 Statistics Solutions Questions and Answers

Class 9 Maths Chapter 15 Kerala Syllabus – Average

Intext Questions And Answers

Question 1.
Take some numbers and calculate the arithmetic means. Calculate the excess or deficit of each of the numbers from the mean and add them up separately. Are the sums equal? Can you explain the reason for this?
Answer:
Let’s consider some numbers 10, 15, 20, 25, 30
Arithmetic mean = \(\frac{10+15+20+25+30}{5}=\frac{100}{5}\) = 20
Now calculate the excess or deficit of each of the number from the mean.

Numbers	Excess of the number from 20	Deficit of the number from 20
10		20 – 10 = 10
15		20 – 15 = 5
20
25	25 – 20 = 5
30	30 – 20 = 10

Sum of excess of the number from 20 = 10 + 5 = 15
Sum of deficit of the number from 20 = 10 + 5 = 15
Here, the sum of excess of the numbers from the mean and deficit of the numbers from the mean are equal. Because the mean act as a balance between the values of the numbers.

Class 9 Maths Kerala Syllabus Chapter 15 Solutions – Tables

Textual Questions And Answers

Question 1.
In a T20 match, 51 runs were scored in the first 5 overs
i) What is the mean run rate then?
ii) If this run rate is maintained, what is the total they can expect?
Answer:
i) To find the mean run rate, we divide the total runs scored by the number of overs:
Mean Run Rate = \(\frac{\text { Total Runs Scored }}{\text { Overs Bowled }}\) = \(\frac{51}{5}\) = 10.2 runs per over

ii) Expected Total = Mean Run Rate × Total Overs
= 10.2 × 20
= 204 runs

Question 2.
The table below shows the children in a class grouped according to their marks in a math test:

i) What is the mean marks of the class?
ii) How many got less marks than the mean?
iii) How many got more marks than the mean?
Answer:

Marks	Children	Total Mark
2	1	2 × 1 = 2
3	2	3 × 2 = 6
4	5	4 × 5 = 20
5	4	5 × 4 = 20
6	6	6 × 6 = 36
7	11	7 × 11 = 77
8 ‘	10	8 × 10 = 80
9	4	9 × 4 = 36
10	2	10 × 2 = 20
Total	45	297

Mean mark = \(\frac{297}{45}\) = 6.6
i) Children who got mark less than mean is = 1 + 2 + 5 + 4 + 6 = 18
ii) Children who got mark more than mean is = 11 + 10 + 4 + 2 = 27

Question 3.
The details of rubber sheets a farmer got during a month are shown below:

i) How many kilograms of rubber did he get a day on average in this month?
Answer:

Rubber(kg)	Days	Total Rubber (kg)
09	3	9 × 3 = 27
10	4	10 × 4 = 40
11	3	11 × 3 = 33
12	3	12 × 3 = 36
13	5	13 × 5 = 65
14	6	14 × 6 = 84
16	6	16 × 6 = 96
Total	30	381

Average Quantity of rubber per day = \(\frac{381}{30}\) = 12.77 kg

ii) The price of a kilogram of rubber is 175 rupees. How much did he get a day on average this month from rubber?
Answer:
If the price is Rs. 175 per kg, then average income per day = 12.77 × 175
= Rs.2234.75

Question 4.
The table below shows the days in a month sorted according to the amount of rainfall in a locality:

What is the mean rainfall per day during this month?
Answer:

Rainfall (mm)	Days	Total Rainfall (mm)
54	3	54 × 3 = 162
56	5	56 × 5 = 280
58	6	58 × 6 = 348
55	3	55 × 3 = 165
50	2	50 × 2 = 100
47	4	47 × 4 = 188
44	5	44 × 5 = 220
41	2	41 × 2 = 82
Total	30	1545

The average of the rain fall per day during that month = \(\frac{\text { Total rain fall }}{\text { Number of days }}=\frac{1545}{30}\) = 51.5 m

Textual Questions And Answers

Question 1.
The table below shows the children in a class, grouped according to their heights:

What is the mean height?
Answer:

Height (cm)	Number of children	Class Mark	Total Height
148 – 152	8	150	1200
152 – 156	10	154	1540
156 – 160	15	158	2370
160 – 164	10	162	1620
164 – 168	7	166	1162
Total	50		7892

Mean height = \(frac{\text { Total height }}{\text { Number of children }}\)
= \(\frac{7892}{50}\)
= 157.84 cm

Question 2.
The table below shows the classification of teachers in a university based on their ages:

Calculate the mean age of the teachers
Answer:

Age	Number of teachers	Class Mark	Total Age
25-30	06	27.5	165
30-35	14	32.5	455
35-40	18	37.5	675
40-45	20	42.5	850
45-50	05	47.5	237.5
50-55	04	52.5	210
55-60	03	57.5	172.5
Total	70		2765

Mean age = \(\frac{\text { Total age }}{\text { Number of persons }}\)
= \(\frac{2765}{70}\)
= 39.5

Question 3.
The classification of a group of children according to their weights is given in the table below:

Calculate the mean weight.
Answer:

Weight (kg)	Number of children	Class Mark	Total Weight
21-23	4	22	88
23-25	7	24	168
25-27	8	26	208
27-29	6	28	168
29-31	3	30	90
31-33	1	32	32
Total	29		754

Mean Weight = \(=\frac{\text { Total Weight }}{\text { Number of children }}\)
= \(\frac{754}{29}\)
= 26

Statistics Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
The daily wages of workers of a factory are given below. Find the average daily wage.

Daily wages	Number
500	3
600	7
700	10
900	8
1000	2

Answer:

Daily wages	Number	Total Wages
500	3	1500
600	7	4200
700	10	7000
900	8	7200
1000	2	2000
Total	30	21900

Average daily wage = \(\frac{21900}{30}\) = 730

Question 2.
The table shows the students in a class sorted according to their marks in an exam

i) The average marks is 6 .How many students got 8 marks?
ii) How many students are there in the class?
Answer:

Mark	Students	Total mark
3	2	6
4	4	16
5	5	25
6	6	36
7	7	49
8	X	8x
9	2	18
10	1	10
Total	27 + x	160 + 8x

i) Average mark = \(\frac{\text { Total Mark }}{\text { Number of Students }}\)
6 = \(\frac{160+8 x}{27+x}\)
6(27 + x)= 160 + 8x
162 + 6x = 160 + 8x
2x = 2
x = 1
Therefore only one student got 8 marks ii) Total Number of students in the class = 27 + x = 27 + 1 = 28

Question 3.
A table categorizing the workers in an office on the basis of their salary is given below.

Salary (Rs)	Number of workers
15000 -18000	1
18000 – 21000	3
21000 – 24000	5
24000 – 27000	4
27000 – 30000	1
30000-33000	1

Find the mean of salary.
Answer:

Salary (Rs)	Number of workers	Class interval	Total salary
15000- 18000	1	16500	16500
18000-21000	3	19500	58500
21000-24000	5	22500	112500
24000 – 27000	4	25500	102000
27000 – 30000	1	28500	28500
30000 – 33000	1	31500	31500
Total	15		349500

Mean income = \(=\frac{\text { Total salary }}{\text { Number of workers }}\)
= \(\frac{349500}{15}\)
= 23,300
Therefore the mean salary of the workers = 23,300

Question 4.
The table below shows the daily wages of the workers in a firm. Calculate the mean daily wage.

Daily wage (Rs)	Number of workers
450 – 550	7
550 – 650	8
650 – 750	10
750 – 850	10
850 – 950	9
950 -1050	6

Answer:

Daily wages (Rs)	Number of workers	Class interval	Total salary
450-550	7	500	3500
550-650	8	600	4800
650 – 750	10	700	7000
750-850	10	800	8000
850-950	9	900	8100
950- 1050	6	1000	6000
Total	50		37,400

Mean value = \(\frac{\text { Total salary }}{\text { Number of workers }}\)
= \(\frac{37,400}{50}\)
= 748
Therefore the mean daily wage = 748

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 13 Malayalam Medium ബഹുപദചിത്രങ്ങൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 13 Solutions Malayalam Medium ബഹുപദചിത്രങ്ങൾ

Class 9 Maths Chapter 13 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 13 Malayalam Medium Textual Questions and Answers

Question 1.
ചുവടെയുള്ള ബഹുപദങ്ങളുടെ ചിത്രരൂപം വരയ്ക്കുക
i) p(x) = 2x – 1
ii) p(x) = x – 1
iii) p(x) = 1 – x
iv) p(x) = x
v) p(x) = −x
Answer:
i) p(x) = 2x – 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-1	1	3

ii) p(x) = x – 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-1	0	1

iii) p(x) = 1 – x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	1	0	-1

iv) p(x) = x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	1	2

v) p(x) = -x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	-1	-2

Question 2.
ചുവടെയുള്ള വരകളുടെ ബഹുപദരൂപം കണ്ടുപിടിക്കുക

Answer:
i) ചിത്രത്തിൽ നിന്നും
p(0) = 1
p(\(\frac{-1}{2}\)) = 0 എന്ന് കിട്ടും
ഇവിടെ P(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = 1
b = 1
p(\(\frac{-1}{2}\)) = a × (\(\frac{-1}{2}\)) + b = 0
\(\frac{-a}{2}\) + 1 = 0
a = 2
ബഹുപദ രൂപം = 2x + 1

ii) ചിത്രത്തിൽ നിന്നും
p(0) = 0
p(\(\frac{1}{2}\)) = 1 എന്ന് കിട്ടും
ഇവിടെ P(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = 0
b = 0
p(\(\frac{1}{2}\)) = a × (\(\frac{1}{2}\)) + b = 1
\(\frac{a}{2}\) = 1
a = 2
ബഹുപദ രൂപം = 2x

iii) ചിത്രത്തിൽ നിന്നും
p(0) = -2
p(2) = 0 എന്ന് കിട്ടും
ഇവിടെ P(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = -2
b = -2
p(2) = a × (2) + b = 0
2a – 2 = 0
a = 1
ബഹുപദ രൂപം = x – 2

Question 3.
ചില രണ്ടാംകൃതി ബഹുപദങ്ങളുടെ ചിത്രങ്ങളാണ് ചുവടെ കൊടുത്തിരിക്കുന്നത്.

ഓരോന്നിന്റെയും ബഹുപദരൂപം കണക്കാക്കുക.
Answer:
i) ചിത്രത്തിൽ നിന്നും
P(0) = 3
p(1) = 0
p(3) = 0 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = 3
c = 3
p(1) = a × 1² + b × 1 + c = 0
= a + b + 3 = 0 … (1)
p(3) = a × 3² + b × 3 + c = 0
= 9a + 3b + 3 = 0
= 3a + b + 1 = 0 … (2)

സമവാക്യം (2) ൽ നിന്നും സമവാക്യം (1) കുറച്ചാൽ
2a + (1 – 3) = 0
2a = 2
a = 1

a = 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
1 + b + 3 = 0
b + 4 = 0
b = -4
ആയതിനാൽ, ബഹുപദരൂപം= p(x) = x² – 4x + 3.

ii) ചിത്രത്തിൽ നിന്നും
2a + 2 = 0
a = -1
a = – 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
-1 + b – 3 = 0
b – 4 = 0
b = 4
ആയതിനാൽ, ബഹുപദരൂപം = p(x) = -x² + 4x – 3.

p(0) = 4
P(1) = 0
p(2) = 0 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b x 0 + c = 4.
c = 4
p(1) = a x 1² + b x 1 + c = 0
= a + b + 4 = 0… (1)

p(2) = a x 2² + b x 2 + c = 0
= 4a + 2b + 4 = 0
= 2a + b + 2 = 0… (2)

സമവാക്യം (2) ൽ നിന്നും സമവാക്യം (1) കുറച്ചാൽ
a + (2 – 4) = 0
a = 2

2 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
2 + b + 4 = 0
b + 6 = 0
b = -6

ആയതിനാൽ, ബഹുപദരൂപം= p(x) = 2x² – 6x + 4.

iii) ചിത്രത്തിൽ നിന്നും
p(0) = – 3
p(1) = 0
p(3) = 0 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = -3
c = -3
p(1) = a × 1² + b × 1 + c = 0
= a + b + -3 = 0… (1)

p(3) = a × 3² + b × 3 + c = 0
= 9a + 3b – 3 = 0
= 3a + b – 10… (2)

സമവാക്യം (2) ൽ നിന്നും സമവാക്യം (1) കുറച്ചാൽ
2a + (-1 – -3) = 0

Class 9 Maths Chapter 13 Malayalam Medium Intext Questions and Answers

Question 1.
ചുവടെപ്പറയുന്ന ബഹുപദങ്ങളുടെ ചിത്രം വരയ്ക്കുക:
i) p(x) = x
ii) p(x) = 2x
iii) p(x) = x
iv) p(x) = -x 1
v) p(x) = -2x
ഈ വരകൾക്കെല്ലാം പൊതുവായ എന്തെങ്കിലും സവിശേഷതയുണ്ടോ?
Answer:
i) p(x) = x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	1	2

ii) p(x) = 2x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	2	4

iii) p(x) = \(\frac{1}{2}\)x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	2	4
p(x)	0	1	2

iv) p(x) = – x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	-1	-2

v) p(x) = −2x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	l	2
P(x)	0	-2	-4

ഈ വരകൾ എല്ലാം ആധാര ബിന്ദുവിലൂടെ കടന്നു പോകുന്നു.

Question 2.
ഇതുപോലെ ചുവടെയുള്ള ബഹുപദങ്ങളുടെയും ചിത്രം വരയ്ക്കുക:
i) p(x) = x + 1
ii) p(x) = x + 2
iii) p(x) = x + \(\frac{1}{2}\)
iv) p(x) = x – 1
v) p(x) = x – 2
Answer:
i) p(x) = x + 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	1	2	3

ii) p(x) = x + 2 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	2	3	4

iii) p(x) = x + \(\frac{1}{2}\) ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	\(\frac{1}{2}\)	1
p(x)	\(\frac{1}{2}\)	1	1\(\frac{1}{2}\)

iv) p(x) = x – 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-1	0	1

v) p(x) = x – 2 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-2	-1	0

ഈ ചിത്രങ്ങളുടെ എല്ലാം ചരിവ് തുല്യമാണ്

Polynomial Pictures Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ചുവടെയുള്ള ബഹുപദങ്ങളുടെ ചിത്രരൂപം വരയ്ക്കുക
i) p(x) = 3-x
ii) p(x) = 3x + 1
iii) p(x) = 2x – 3
Answer:
i) p(x) = 3 – x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
P(x)	3	2	1

ii) p(x) = 3x + 1 ൽ x ആയി -1, 0, 1 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	-1	0	1
p(x)	-2	1	4

iii) p(x) = p(x) = 2x – 3 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-3	-1	1

Question 2.
ചുവടെയുള്ള വരകളുടെ ബഹുപദരൂപം കണ്ടുപിടിക്കുക

Answer:
ചിത്രത്തിൽ നിന്നും
p(0) = 8
p(-4) = 0 എന്ന് കിട്ടും
ഇവിടെ p(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = 8
b = 8
p(-4) = a × (-4)+ b = 0
= 4a + 8 = 0
a = 2
ബഹുപദ രൂപം = 2x + 8

Answer:
ചിത്രത്തിൽ നിന്നും
p(0) = -4
p(4) = 0 എന്ന് കിട്ടും
ഇവിടെ p(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = -4
b = -4
p(4) = a × (4) + b = 0
= 4a – 4 = 0
a = 1
ബഹുപദ രൂപം = x – 4

Question 3.
ചില രണ്ടാംകൃതി ബഹുപദങ്ങളുടെ ചിത്രങ്ങളാണ് ചുവടെ കൊടുത്തിരിക്കുന്നത്.

Answer:
i) ചിത്രത്തിൽ നിന്നും
P(0) = 1
p(-2) = 5
p(2) = 5 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = 1
c = 1

p(-2) = a × (-2)² + b × -2 + c = 5
= 4a – 2b + 1 = 5
= 4a- 2b = 4
= 2a – b = 2…(1)

p(2) = a × 2² + b × 2 + c = 0
= 4a + 2b + 1 = 5
= 4a + 2b = 4
= 2a + b = 2… (2)

സമവാക്യം (1) ഉം സമവാക്യം (2) ഉം കൂട്ടിയാൽ
4a= 4
a = 1

a = 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
2 × 1 – b = 2
b = 0
ആയതിനാൽ, ബഹുപദരൂപം= p(x) = x² + 1

Answer:
ചിത്രത്തിൽ നിന്നും
p(0) = -2
p(-2) = 2
p(2) = 2 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = -2
c = -2

p(-2) = a × (-2)² + b x -2 + c = 2
= 4a – 2b – 2 = 2
= 4a- 2b = 4
= 2a – b = 2…(i)

p(2) = a × 2² + bx² + c = 0
= 4a + 2b – 2 = 2
= 4a + 2b = 4
= 2a + b = 2… (2)

സമവാക്യം (1) ഉം സമവാക്യം (2) ഉം കൂട്ടിയാൽ
4a = 4
a = 1

a = 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
2 × 1 – b = 2
b = 0
ആയതിനാൽ, ബഹുപദരൂപം= p(x) = x – 2

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 14 Proportion Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 14 Solutions Proportion

Proportion Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 14 Proportion Solutions Questions and Answers

Class 9 Maths Chapter 14 Kerala Syllabus – Proportional Changes

Textual Questions And Answers

Question 1.
In each of the instances below, show that the second quantity changes proportionally with respect to the first. Also find the proportionally constant in each:
i) The length of sides of squares and their perimeters.
ii) The lengths of wires bent into squares and the length of the sides of the squares.
iii) The number of rotations of a circle rolling along a line, and the distance travelled along the line.
Answer:
i) Let s be the length of a side of a square. The perimeter P of a square is given by:
P = 4s
Now, we can express the ratio of the perimeter to the side length:
\(\frac{\mathrm{P}}{\mathrm{~s}}=\frac{4 \mathrm{~s}}{\mathrm{~s}}\) = 4
Since this ratio is constant (4), we conclude that the perimeter changes proportionally with respect to the length of the sides of the squares. The proportionality constant is 4.

ii) Let L be the total length of the wire, and let s be the length of a side of the square formed by bending the wire. The total length of the wire is given by:
L = 4s
Now, we express the ratio of the total length of the wire to the side length:
\(\frac{L}{s}=\frac{4 s}{s}\) = 4
Since the ratio is constant (4), we find that the lengths of wires bent into squares change proportionally with respect to the length of the sides of the squares. The proportionality constant is 4.

iii) Let r be the radius of the circle. The circumference C of the circle is given by:
C = 2πr
When a circle rolls along a line, the distance d travelled along the line is equal to the number of rotations n times the circumference of the circle:
d = nC
Substituting the circumference:
d = n (2πr)
Now, we express the ratio of the ratio of the distance travelled to the number of rotations:
\(\frac{\mathrm{d}}{\mathrm{n}}\) = 2πr
Since this ratio is constant (specifically, 27rr), we find that the distance travelled along the line changes proportionally with respect to the number of rotations of the circle. The proportionality constant is 2πr.

Question 2.
We have seen that in the picture, the height of a point on the slanted line from the horizontal line changes proportionally with respect to its distance from the corner. Calculate the proportionality constants 30°, 45° and 60°.

Answer:

When the angle is 30°
A base triangle ΔABC is drawn such that AC = 2cm, BC = 1cm and AB = √3 cm
Here triangle ABC and APQ are similar, we get
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)

BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{1}\)
AQ = 2 PQ
Therefore, the proportionality constant is 2
Similarly, when the angle is 60°

A base triangle ΔABC is drawn such that AC = 2cm, BC = √3 cm and AB = 1cm
Here triangle ABC and APQ are similar, we get
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)

BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{\sqrt{3}}\)
AQ = \(\frac{2}{\sqrt{3}}\)PQ

Therefore, the proportionality constant is \(\frac{2}{\sqrt{3}}\)
Similarly, when the angle is 45°
A base triangle A ABC is drawn such that AC = √2 cm, BC = 1 cm and AB = 1 cm
Here triangle ABC and APQ are similar, we get
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{AB}}{\mathrm{AP}}\)
BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{\sqrt{2}}{1}\)
AQ = \(\frac{\sqrt{2}}{1}\) PQ
Therefore, the proportionality constant is √2.

Question 3.
Prove that in equilateral triangles, the perimeter changes proportionally with respect to the length of the sides. What is the proportionality constant? What can we say about other regular polygons?
Answer:
Let’s consider an equilateral triangle with side length’s’.
Perimeter of an equilateral triangle, P = 3s
The perimeter of an equilateral triangle changes proportionally with respect to the length of its sides, with a proportionality constant of 3.

Let’s consider a regular polygon with ‘n’ sides, each of length’s’.
Perimeter of a regular polygon, P = ns
The perimeter of any regular polygon changes proportionally with respect to the length of its sides,
with a proportionality constant equal to the number of sides.

Question 4.
Prove that the lengths of arcs of a fixed circle change proportionally with respect to their central angles. What is the proportionality constant? What about the relation between the area of a sector and its central angle?
Answer:
Let’s consider a circle with radius ‘r’ and central angle ‘θ’
Arc Length (L) = θ × r
The lengths of arcs of a fixed circle change proportionally with respect to their central angles, with a proportionality constant of r.
Area of a sector (A) = \(\frac{1}{2}\) r² θ
The area of a sector of a fixed circle changes proportionally with respect to its central angle with a proportionality constant of \(\frac{1}{2}\) r².

Class 9 Maths Kerala Syllabus Chapter 14 Solutions – Scale And Proportion

Textual Questions And Answers

Question 1.
i) What is the sum of the angles of a triangle? And the sum of the angles of a hexagon?
ii) Does the sum of the angles of polygons change proportionally with respect to the number of sides? Explain the reason.
Answer:
i) Sum of the interior angles of a triangle = (3 – 2) × 180° = 180°
Sum of the interior angles of a hexagon = (6 – 2) × 180° = 720°

ii) Yes, the sum of the angles of polygons changes proportionally with respect to the number of sides.
The sum of interior angles (S) of a polygon with ‘n’ sides is given by:
S = (n- 2) × 180°
The sum of the angles of polygons changes proportionally with respect to the number of sides with proportionality constant 180°

Question 2.
Inside a triangle of base 6 centimeters and height 3 centimeters, lines are drawn parallel to the base. Prove that the lengths of these lines change proportionally with respect to the distance from the top vertex. Find the proportionality constant.

Answer:
Given, Base = 6 cm
Height = 3 cm
Consider parallel lines drawn from the top vertex.
Distance from top vertex = x
Length of parallel line = y
Using similar triangle property,
\(\frac{y}{x}=\frac{6}{3}\)
y = 2x
Thus we can say that lengths of these lines change proportionally with respect to the distance from the top vertex with proportionality constant 2

Question 3.
Within a semicircle of diameter 10 centimeters, lines are drawn parallel to the diameter:

(i) In each of the pictures below, calculate the length of the line parallel to the diameter:

ii) Does the length of the parallel line change proportionally with respect to the distance from the top of the semicircle. Explain the reason.
Answer:
i) Figure 1

Length of the parallel line = 2\(\sqrt{(5)^2-(4)^2}\) = 2 × 3 = 6 cm

Figure 2

Length of the parallel line = 2\(\sqrt{(5)^2-(3)^2}\) = 2 × 4 = 8 cm
ii) The length of the parallel line does not change proportionally with respect to the distance from the top of the semicircle.

SCERT Class 9 Maths Chapter 14 Solutions – Different Proportions

Textual Questions And Answers

Question 1.
i) Prove that the areas of equilateral triangles change proportionally with respect to the squares of the lengths of sides. What is the proportionality constant?
Answer:

Areas of equilateral triangles change proportionally with respect to the squares of the lengths of sides with proportionality constant of \(\frac{\sqrt{3}}{4}\).

ii) Are the areas of squares proportional to the squares of the lengths of sides? If so, what is the proportionality constant?
Answer:

Area of the squares are proportional to the squares of the lengths of side with proportionality constant of 1.

Question 2.
Consider all rectangles of area 1 square meter. The length of one side of such a rectangle depends on the length of the other side. Write this relation as an algebraic equation. How do we state it in terms of proportion?
Answer:
In rectangles of area one square meter
Let x be the length and y be the breadth of the rectangle
Then, area = length × breadth
1 = x × y
1 = x y
y = \(\frac{1}{x}\)
y is inversely proportional to x

Question 3.
Consider all triangles of a fixed area. How do we state in terms of proportion, the relation between the length of the longest side and the length of the perpendicular to it from the opposite vertex? What if we use the shortest side instead of the longest?
Answer:
Let ‘a’ be the longest side, ‘h’ be the length of perpendicular from opposite vertices, ‘A’ be the area then
A = \(\frac{1}{2}\)ah,
a = \(\frac{2A}{h}\)
Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.
That is, the length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

Question 4.
In regular polygons, can we say the relation between the number of sides and the measure of an outer angle, in terms of proportion? What is the proportionality constant?
Answer:
The sum of the exterior angles of all polygon is 360°.
If ‘n’ is the number of sides.
Measure of an exterior angle = \(\frac{\text { sum of exterior angle }}{\text { number of sides }}\)
If the measure of an outer angle is ‘x’
x = \(\frac{360^{\circ}}{n}\)
One outer angle and number of sides are inversely proportional.
The constant of proportionality is \(\frac{1}{n}\)

Proportion Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.
i) The rate of water flow and the height of the water level.
ii) The rate of water flow and the time taken to fill the tank.
Answer:
i) Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then x = Ay
Height of the water level in the tank is proportional to the rate of the water flowing.

ii) If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in’t’ second is given by C = V t
V = C × \(\frac{1}{t}\)
That is the rate of water flow and the time taken for filling the tank are inversely proportional.
C is the constant of proportionality.

Question 2.
Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Answer:
Ratio of investments = 60000: 100000 = 6:10 = 3:5
Ratio of profit divided 1800: 3000 = 18:30 = 3:5
Ratio of investments and Ratio of profit divided are equal.
Hence they are proportional.

Question 3.
Are the length and breadth of a square having same perimeter inversely proportional?
Answer:
For a square the perimeter is 20 cm. so, length + breadth = 10.
Let x be the length and y be the breadth then possible values of x and y are
When x = 9 then y = 1
When x = 8 then y = 2
Here xy is not a constant term i.e. the changes in the x and y is not in the firm of xy = kept
Hence length and breadth are not in inversely proportional.

Question 4.
A car with 5L of petrol travels a distance of 75 km. What is the proportionality constant between the distance travelled and the quantity of petrol? How much petrol is needed for travelling 180 km?
Answer:
Taking the distance travelled as x and quantity of petrol as y, then the constant of proportionality 15
k = \(\frac{y}{x}=\frac{75}{5}\) = 15
k = \(\frac{y}{x}\) = 15 = \(\frac{100}{x}\)
x = \(\frac{180}{15}\) = 12
The quantity of petrol needed to travel 180 km = 12 litre