Class 9 – Page 34 – Haryana Board Solutions

HBSE 9th Class Science Notes in Hindi Medium & English Medium Haryana Board

September 16, 2024 / Class 9 / By Prasanna

Haryana Board HBSE 9th Class Science Notes

HBSE 9th Class Science Notes in English Medium

HBSE 9th Class Science Notes in Hindi Medium

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HBSE 9th Class Science Important Questions and Answers

September 16, 2024 / Class 9 / By Prasanna

Haryana Board HBSE 9th Class Science Important Questions and Answers

HBSE 9th Class Science Important Questions in English Medium

HBSE 9th Class Science Important Questions in Hindi Medium

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HBSE 9th Class Science Solutions Haryana Board

September 5, 2024 / Class 9 / By Prasanna

Haryana Board HBSE 9th Class Science Solutions

HBSE 9th Class Science Solutions in English Medium

HBSE 9th Class Science Solutions in Hindi Medium

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HBSE 9th Class Maths Solutions Haryana Board

September 5, 2024 / Class 9 / By Prasanna

Haryana Board HBSE 9th Class Maths Solutions

HBSE 9th Class Maths Solutions in English Medium

HBSE 9th Class Maths Chapter 1 Number Systems

HBSE 9th Class Maths Chapter 2 Polynomials

HBSE 9th Class Maths Chapter 3 Coordinate Geometry

HBSE 9th Class Maths Chapter 4 Linear Equations in Two Variables

HBSE 9th Class Maths Chapter 5 Introduction to Euclid’s Geometry

HBSE 9th Class Maths Chapter 6 Lines and Angles

HBSE 9th Class Maths Chapter 7 Triangles

HBSE 9th Class Maths Chapter 8 Quadrilaterals

HBSE 9th Class Maths Chapter 9 Areas of Parallelograms and Triangles

HBSE 9th Class Maths Chapter 10 Circles

HBSE 9th Class Maths Chapter 11 Constructions

HBSE 9th Class Maths Chapter 12 Heron’s Formula

HBSE 9th Class Maths Chapter 13 Surface Areas and Volumes

HBSE 9th Class Maths Chapter 14 Statistics

HBSE 9th Class Maths Chapter 15 Probability

Chapter 15 Probability Ex 15.1

HBSE 9th Class Maths Solutions in Hindi Medium

HBSE 9th Class Maths Chapter 1 संख्या पद्धति

HBSE 9th Class Maths Chapter 2 बहुपद

HBSE 9th Class Maths Chapter 3 निर्देशांक ज्यामिति

HBSE 9th Class Maths Chapter 4 दो चरों वाले रैखिक समीकरण

HBSE 9th Class Maths Chapter 5 युक्लिड के ज्यामिति का परिचय

HBSE 9th Class Maths Chapter 6 रेखाएँ और कोण

HBSE 9th Class Maths Chapter 7 त्रिभुज

HBSE 9th Class Maths Chapter 8 चतुर्भुज

HBSE 9th Class Maths Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल

HBSE 9th Class Maths Chapter 10 वृत्त

HBSE 9th Class Maths Chapter 11 रचनाएँ

HBSE 9th Class Maths Chapter 12 हीरोन का सूत्र

HBSE 9th Class Maths Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

HBSE 9th Class Maths Chapter 14 सांख्यिकी

HBSE 9th Class Maths Chapter 15 प्रायिकता

Chapter 15 प्रायिकता Ex 15.1

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HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.3

September 1, 2024 / Class 9 / By Prasanna

Haryana State Board HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 2 Polynomials Exercise 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by:
(i) x + 1
(ii) x – \(\frac {1}{2}\)
(iii) x
(iv) x + π
(v) 5 + 2x.
Solution:
Let p(x) = x³ + 3x² + 3x + 1
(i) We know that when p(x) is divided by (x + 1), the remainder is p(- 1).
(By Remainder Theorem)
So, p(-1) = (-1)³ + 3 x (-1)2 + 3 x (-1) + 1
= – 1 + 3 – 3 + 1 = 0
Hence, required remainder = 0.

(ii) We know that when p(x) is divided by (x – \(\frac {1}{2}\)), the remainder is p(\(\frac {1}{2}\))
(By Remainder Theorem)

(iii) We know that when p(x) is divided by x, the remainder is p(0).
[By Remainder Theorem]
So, p(0) = (0)³ + 3 × (0)² + 3 × 0 + 1
= 0 + 0 + 0 + 1 = 1
Hence,required remainder = 1.

(iv) We know that when p(x) is divided by (x + π), the remainder is p(-π).
(By Remainder Theorem)
So, p(-π) = (-π)³ + 3 × (-π)² + 3 × (-π) + 1
= – π³ + 3π² – 3π + 1
Hence,required remainder = – π³ + 3π² – 3π + 1.

(v) We know that when p(x) is divided by (5 + 2x), the remainder is p(\(\frac {-5}{2}\))
[By Remainder Theorem]

Question 2.
Find the remainder when x³ – ax³ + 6x – a is divided by x – a.
Solution :
Let p(x) = x³ – ax² + 6x – a
We know that when p(x) is divided by (x – a), the remainder is p(a).
[By Remainder Theorem]
So, p(a)=(a)³ – a × (a)² + 6 × a – a
= a³ – a³ + 6a – a = 5a
Hence required remainder = 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Solution :
Letp(x) = 3x³ + 7x
We know that, if 7 + 3x is a factor of p(x), then it is divisible by (7 + 3x) with leaving no
remainder i.e., remainder [p(\(\frac {-7}{3}\))] will be zero. (By Factor Theorem)

Since, remainder p(\(\frac {-7}{3}\)) ≠ 0. Therefore, (7 + 3x) is not a factor of p(x)

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HBSE 9th Class Maths Notes Chapter 15 Probability

August 28, 2024 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 15 Probability Notes.

Haryana Board 9th Class Maths Notes Chapter 15 Probability

Introduction
In every life, we come across statements such as:
(i) Most probably it will rain today.
(ii) I doubt that he will pass in the examination.
(iii) Chances are high that price of gold will go up.
(iv) Probably Mumbai Indian team wins the IPL cricket tournament 2013.
In the above statements the words ‘most probably’, ‘doubt’, ‘chances’, ‘probably’ etc. shows uncertainty or probability of occurrence of some particular event. Through probability started with gambling, it has been used extensively in Science, Biological Science, Commerce, Weather Forecasting etc.

Key Words
→ Probability: Probability is a numerical measure of the likelihood that a specific event will occur.

→ Empirical probability: Probability that is based on some event occuring, which is calculated using collected empirical evidence.

→ Experiment: An experiment is a process which ends in some well defined result.

→ Outcome: A particular result of an experiment is called an outcome.

→ Random Experiment: An experiment in which all possible outcomesare known and the exact outcome cannot be predicted in advance, is called a random experiment.

→ Trial: Performing an experiment once is called a trial.

→ Events: The possible outcomes of an experiment are called events.

→ Impossible event: An event that has no chance of occuring is called an impossible event. The probability of impossible event is zero.

→ Sure event: An event that has a 100% probability of occuring is called sure event. The probability of sure event is 1.

→ Elementary Event: An event having only one outcome is called an elementary event. The sum of all the elementary events of an experiment is 1.

→ Equally likely events: Equally likely events are such events that have the equal chance to happen to an experiment.

→ Complementary events: Let E denotes the happening of an event and not E(or E) denotes its not happening, then E and not E are called complementary events.
Note: P(E) + P(not E) = 1.

Probability-an Experimental Approach:
(a) History of Probability: In 1654, a gambler Chevalier de Mere asked French philosopher and mathematician Blaise Pascal to solve certain dice problems. Pascal became interested to solve this problem and discussed them with French mathematician, Pierre de Fermat. Both solved the problems independently. This work was beginning of Probability Theory. The Italian mathematician, J. Cardan (1501-1576) wrote a book “Book on Games of Chance”. It was published by Liber de Ludo Aleae in 1663.

In developing the theory of probability notable contributions were also made by following mathematicians:
1. J. Bernoulli (1654 – 1705)
2. P. Laplace (1749 – 1827)
3. A. A. Markov (1856 – 1922)
4. A. N. Kolmogorov (born 1903)

b. Different Approaches to Probaility: There are three approaches to theory of probability:
(i) Experiment or empirical or observed frequency approach.
(ii) Classical approach.
(iii) Axiomatic approach.
In this chapter, we will study only experimental or empirical probability. The remaining two approaches would have been studying in higher classes.

Experimental or Empirical Probability: Let n be total number of possible outcomes. The empirical probability P(E) of an event E happening, is given by
P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

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HBSE 9th Class Science Solutions Chapter 10 Gravitation

March 21, 2024 / Class 9 / By Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 10 Gravitation Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 10 Gravitation

HBSE 9th Class Science Gravitation Intext Questions and Answers

Questions from Sub-section 10.1

Question 1.
State the universal law of gravitation.
Answer:
Universal Law of Gravitation:
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the

distance between them. The force is along the line joining the centres of two objects. Let two objects A and B of masses M and ‘m lies at a distance ‘d’ from each other as shown in figure. Let the force of attraction between two objects be F. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses, that is,
F ∝ M × m ……….(i)
And the force between two objects is inversely proportional to the square of the distance between them,
that is,
F ∝ \(\frac{1}{d^2}\) ………..(ii)
Combining eqs. (i) and (ii), we get
F ∝ \(\frac{M \times m}{d^2}\) or F = G \(\frac{M \times m}{d^2}\)
Where G is the constant of proportionality and is called universal gravitational constant.

Question 2.
Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:
Suppose the mass of earth = M_e
Mass of the object on the surface of earth = m
The radius of earth = R
∴ The magnitude of the gravitational force between the earth and an object on the surface of the earth (F) = G \(\frac{M_e \times m}{R^2}\) (where G is gravitational constant)

Questions from Sub-section 10.2

Question 1.
What do you mean by free fall?
Answer:
When the objects fall towards earth only because of gravitation, then it is said that the objects are in free faIl.

Question 2.
What do you mean by acceleration due to gravity?
Answer:
When a body falls towards earth, acceleration works. This acceleration is due to the force of gravity of earth. Therefore, this acceleration is called the acceleration due to the force of gravity of earth or acceleration due to gravity. it is denoted by ‘g’ and its unit is ms².

Questions from Sub-sections 10.3, 10.4

Question 1.
What are the differences between the mass of an object and its weight?
Answer:
The differences between the mass and weight of an object are as follows:

Mass:
1. Mass of a body is the quantity of matter contained in it.
2. The mass of an object cannot be zero.
3. Mass of an object is constant.
4. The mass of an object is found by physical balance.
5. Mass is measured ¡n kilogram (kg).
6. Mass is a scalar quantity.

Weight:
The force by which the earth attracts an object
towards itself is called the weight of that object on earth.
2. The weight of an object is zero on the centre of earth.
3. Weight is not constant. The weight of an object is more on poles as compared to equator.
4. The weight of an object is found by spring balance.
5. Weight is measured in newton.
6. Weight is a vector quantity.

Question 2.
Why is the weight of an object on the moon th its weight on the earth?
Answer:
We know that the value of acceleration due to gravity (g) on earth is 9.8 ms^-2. The value of g on moon is \(\frac {1}{6}\) th time of that on earth. Therefore, the weight of objects on moon is less than that on earth. That is g \(\frac {1}{6}\)th times.

Questions from Sub-section 10.5

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a strap made of a thin and strong string because its thrust affects a smaller area (in the center of the string).

Question 2.
What do you mean by buoyancy?
Answer:
When an object is immersed in a fluid, the fluid exerts a force on the body in an upward direction which is called buoyant force. The result of buoyant force depends upon the density of the fluid. This is known as buoyancy.

Question 3.
Why does an object float or sink when placed on the surface of the water?
Answer:
Objects of density less than that of a liquid float on the liquid. The objects of density greater than that of a liquid sink in the liquid. For example, when a nail and a cork of the same mass are put in water, the nail will sink because its density is more than water, whereas the cork will float because its density is less than that of water.

Questions from Sub-section 10.6

Question 1.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
if the mass is 42 kg on a weighing machine, then our mass will be 42 kg because mass remains constant.

Question 2.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one ¡s heavier and why?
Answer:
Iron bar will be heavier than a bag of cotton because the density of iron is more than cotton.

HBSE 9th Class Science Gravitation Textbook Questions and Answers

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
If the distance between two objects is reduced to half, then the gravitational force between them will increase 4 times, because gravitational force is inversely proportional to the square of the distance between them.

Question 2.
Gravitational force acts on all objects is proportional to their masses. Why then, a heavy object does not fall faster than a light object?
Answer:
Gravitational force acts on all objects is proportional to their masses, yet a heavy object will not fall faster than a light object because the mass of objects is negligible in comparison to that of earth.

Question 3.
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 x 10²⁴ kg and radius of the earth is 6.4 x 10⁶m.)
Solution:
Here, Mass of earth (Me) = 6 × 10²⁴kg
Mass of object (m) = 1 kg
Radius of earth (R) = 6.4 × 10⁶m
Gravitational constant (G) = 6.673 × 101 Nm² kg²
Force of gravity (g) = ?
We know that,
Force of gravity (g) = \(\frac{G M_e \times m}{R^2}\) = \(\frac{6.673 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^6\right)^2}\) = 9.8N

Question 4.
The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Answer:
The earth attracts the moon with greater force because the mass of earth is more than that of moon.

Question 5.
If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
The moon attracts the earth, but the earth does not move towards the moon because the mass of the moon is much less than that of earth.

Question 6.
What happens to the force between two objects, if:
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?
Answer:
(i) If the mass of one object is doubled, then the force between two objects will be doubled because gravitational force is proportional to mass.
(ii) If the distance between two objects is doubled, the gravitational force will become times and if the distance is tripled, the gravitational force will become times, because gravitational force is inversely proportional to the square of the distance.
(iii) As the force between two bodies is proportional to the product of masses of the bodies, so if the masses of both the bodies are doubled, then force between them will become four times.

Question 7.
What i.s the importance of universal law of gravitation?
Answer:
The importance of universal law of gravitation is as follows:
(i) This force binds us to the earth.
(ii) The motion of the moon around the earth is due to this force.
(iii) The motion of planets around the sun is due to this force.
(iv) The tides due to the moon and the sun are due to this force.

Question 8.
What is the acceleration of free fall?
Answer:
When an object falls towards the earth, then acceleration works. Acceleration of free fall is, acceleration due to the gravitational force of earth and is equal to g. Its value is 9.8 ms².

Question 9.
What do we cafl the gravitational force between the earth and an object?
Answer:
The gravitational force between the earth and an object is weight.

Question 10.
Amit buys few grams of gold at the poles as per the Instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why ? (Hint: The value of g is greater at the poles than at the equator.)
Answer:
No, the friend will not agree with the weight of gold bought, because the weight of gold is less on the equator than on the poles.

Question 11.
Why will a sheet of paper fall slower than one of that ¡s crumpled into a ball?
Answer:
The sheet of paper will fall slower than one of that is crumpled into a ball because the area of sheet is more than the area of ball, due to this, the pressure decreases.

Question 12.
The gravitational force on the surface of the moon is only as strong as gravitational force on the earth. What is the weight In newtons of a 10 kg object on the moon and on the earth?
Solution:
Here,
Mass of the object on earth (m) = 10kg
Acceleration due to gravity on earth (g) = 9.8 m/s²
Weight of the object on earth (w) m x g = 10 x 9.8 N = 98 N
Mass of object on moon (m) = 10 kg
Acceleration due to gravity on moon (g₁) = \(\frac {9.8}{6}\)m/s²
∴ Weight of the object on moon (w) = mg₁ = 10 x \(\frac {9.8}{6}\) N = 16.3 N

Question 13.
A ball ¡s thrown vertically upwards with a velocity 49 m’s. Calculate:
(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.
Solution:
Here,
Initial velocity of ball (u) = 49 m/s
Final velocity of ball (v) = 0 m/s
Acceleration due to gravity (g) = -9.8 m/s²
(i) We know that,
v² – u² = 2gs
s = \(\frac{v^2-u^2}{2 g}=\frac{(0)^2-(49)^2}{2(-9.8)}\) = \(\frac{-49 \times 49}{-19.6}\) = 122.5m
∴ The maximum height to which ball rises is 122.5 m

(ii) Time taken (i) = \(\frac{v-u}{g}\) = \(\frac{0-49}{-9.8}\) = 5s
∴ The total time it takes to return to the surface of earth 5s + 5s = 10

Question 14.
A stone is released from the top of a tower of a height of 19.6 m. Calculate Its final velocity just before touching the ground.
Solution:
Here,
Initial velocity of stone (u) = 0 m/s
The final velocity of stone (v) =?
Height of the top of the tower (s) = 19.6 m
Acceleration due to gravity (g) = 9.8 ms^-2
We know that,
v² – u²= 2gs
v² – (0)² = 2 x 9.8 x 19.6
or v² = 19.6 x 19.6
or v = \(\sqrt{19.6 \times 19.6}\)= 19.6 ms^-1
∴ Thus velocity of stone will be 19.6 ms^-1.

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 ms. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Solution:
Here,
Initial velocity of stone (u) = 40 m/s
Final velocity of stone (v) = 0 m/s
Acceleration due to gravity (s) = -10 m/s²

We know that,
v² – u² = 2gs
s = \(\frac{v^2-u^2}{2 g}=\frac{(0)^2-(40)^2}{2(-10)}\) = \(\frac{-40 \times 40}{-2 \times 10}\) = 80m
∴ The maximum height reached by the stone = 80 m
Total distance covered by the stone = 80 m + 80 m = 160m
TotaÌ displacement of the stone = Zero

Question 16.
Calculate the foíce of gravitation between the earth and the Sun, given that the mass of the earth = 6 x 10²⁴ kg and of the Sun 2 x 10³⁰ kg. The average distance between the two Is 1.5 x 10¹¹ m.
Solution:
Here, The mass of the earth (M_e) = 6 x 10²⁴ kg
Mass of the sun (M_e) = 2 x 10³⁰ kg
The average distance between the two (R_s) = 1.5 x 10¹¹ m
Gravitational constant (G) = 6.673 x 10¹¹ Nm²/kg²
∴ Force of gravitation between the earth and sun (F_s) = \(\frac{G M_e M_s}{\left(R_s\right)^2}\)
= \(\frac{6.673 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{\left(1.5 \times 10^{11}\right)^2}\) N
= \(\frac{6.673 \times 12 \times 10^{43}}{2.25 \times 10^{22}}\) N = 35.59 x 10²¹ N

Question 17.
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Solution:
Suppose the two stones will meet at a distance of m.
For the stone tell from the top;
initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Distance covered (s) = x m
We know that
S = ut + \(\frac {1}{2}\) gt²
x = (0) (t) + \(\frac {1}{2}\) (9.8)t²
or x = 4.9 t² ………….(i)

For the stone projected upwards;
Initial velocity (u) = 25 m/s
Acceleration due to gravity (g) = -9.8 ms²
Distance covered (s) = (100 – x)m
We know that,
(s) = ut+ \(\frac {1}{2}\) gt²
100 – x 251 + \(\frac {1}{2}\) (-9.8) t²
or 100 – 4.9t² = 25 t – 4.9 t² [From (i)]
or 100 = 25 t
or t = \(\frac {100}{25}\) = 4s

Both stones will meet after 4 seconds Ans
Height from top (x) = 4.9 (4)²
4.9 x 16 = 78.4 m
∴ Distance from down = 100 – 78.4 = 21.6 m

Question 18.
A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up.
(b) the maximum height It reaches and
(c) its position after 4s.
Solution:
Here,
Initial velocity (u) = ?
Final velocity (v) = 0 m/s
Time taken by ball to cover maximum height (t) = \(\frac {6s}{2}\) = 3s
Upward acceleration due to gravity (g) = – 9.8 m/s²
(a) We know that
v = u + gt
u = v – gt = 0 – (-.9.8)(3) = 29.4 m/s
Therefore, initial velocity of ball (u) = 29.4 m/s

(b) We know that
s = ut + \(\frac {1}{2}\) gt²
= (29.4) (3) + \(\frac {1}{2}\) (-9.8) (3)²
88.2 – 44.1 = 44.1 m
∴ Maximum height it reaches = 44.1 m

(c) For this state, distance will be found from up to down
u = 0
t = 4s – 3s = 1s
g = 9.8 m/s²
s = ut + \(\frac {1}{2}\) gt²
0(1) + \(\frac {1}{2}\) (9.8)(1)²
= 4.9m
∴ The ball will be at a height of 4.9 m from up.
Height of ball from down = 44.1 – 4.9 = 39.2 m

Question 19.
In what direction does the buoyant force on an object Immersed in a liquid act?
Answer:
The buoyant force on an object immersed in a liquid acts in the vertically upward direction.

Question 20.
Why does a block of plastic released under water come up to the surface of the water?
Answer:
The block of plastic when left under water comes to the surface because the thrust due to water is greater than the weight of the block itself.

Question 21.
The volume of 50 g of. the substance is 20. If the density of water is 1 gcm⁴, will the substance float or sink?
Solution:
Here,
Mass of the substance = 50g
Volume of the substance = 20 cm³
∴Density of substance = = \(\frac {50}{20}\) g cm^-3
= 2.5g cm^-3
Since the density of the substance is greater than that of water, hence, the substance will sink in water.

Question 22.
The volume of. 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water Is 1g cm³? What will be the mass of the water displaced by this packet?
Solution:
Here,
Mass of the packet = 500g
Volume of the packet = 350 cm³
Density of packet =
= \(\frac {500}{350}\) g cm^-3 = 1.43 gcm^-3
Since the density of packet is greater than that of water, hence, it will sink in water.
Mass of water displaced = volume x density = 350 x 1 g = 350 g.

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HBSE 9th Class Science Solutions Chapter 8 Motion

March 19, 2024 / Class 9 / By Prasanna

Haryana State Board HBSE 9th Class Science Solutions Chapter 8 Motion Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 8 Motion

HBSE 9th Class Science Motion Intext Questions and Answers

Question 1.
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Yes, displacement can be zero, if an object covers some distance, e.g., let an object moves from origin point along a straight line to point ‘A’ by covering a distance of 55 km and covers the same distance of 55 km from ‘A’ to ‘O’ on return as shown in figure In this state-

Distance travelled by the object = 55 km + 55 km = 110 km
Displacement = Zero (0)

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Solution:
Side of the square field ABCD = 10 metre
The perimeter of the square field ABCD = 4 × side
= 4 × 10 = 40 metres

Time taken for one round around the boundary = 40 s
Total time = 2 minutes 20 s (2 × 60 + 20) s = 140 s
Now, the distance covered by the farmer in 40 s = 40 m
The distance covered by the farmer in 1s m = \(\frac {40}{40}\) = 1 m
The distance covered by the farmer in 140 s 1 × 140 = 140 metres
If, the farmer starts moving from origin point A, then he will be at point C after covering a distance of 140 metres. In this way, the displacement of the farmer will be AC (the diagonal of the square.).
∴ AC = \(\sqrt{(\mathrm{AB})^2+(\mathrm{BC})^2}\)
= \(\sqrt{(10)^2+(10)^2}\)
= \(\sqrt{100+100}\)
= \(\sqrt{200}\)
= \(\sqrt{2 \times 100}\)
= 10\(\sqrt{2}\) m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m

Question 3.
Which of the following is true for displacement ?
(a) It cannot he zero.
(b) Its magnitude is greater than the distance travelled by the object.
Answer:
Both (a) and (b) are wrong for displacement.

Questions from Sub-section 8.2

Question 1.
Distinguish between speed and velocity.
Answer:
The differences between speed and velocity are as follows:

Speed:
1. It is the distance travelled by the body in unit time interval in any direction.
2. It is a scalar quantity, which has only magnitude.
3. It is always positive.

Velocity
1. It is the rate of distance travelled by the body in unit time interval in specified direction.
2. It is a vector quantity, which has both magnitude as well as direction.
3. It can be positive and negative.

Question 2.
Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer:
If the object is moving with uniform motion in a definite direction, then the magnitude of the average velocity of that object will be equal to its average speed.

Question 3.
What does the odometer of an automobile measure ?
Answer:
The odometer of the automobile measures the distance covered by the automobile.

Question 4.
What does the path of an object look like when it is in uniform motion ?
Answer:
When the object is in uniform motion, then its path seems as a straight line.

Question 5.
During an experiment, a signal from a spaceship tTactretf the ground-statianju five minutes. What was the distance of the spaceship from the ground station ? The signal travels at the speed of light, that is, 3 × 10⁸ ms^-1.
Solution:
The time taken by the signal from spaceship to earth = 5 minutes = 5 x 60 = 300 s
Speed of signal = 3 × 10⁸ms^-1

∴ The distance of spaceship from ground station = Speed × Time = 3 × 10s × 300m = 9 × 1010m

Questions from Sub-section 8.3

Question 1.
When will you say a body is in (i) uniform acceleration (ii) non-uniform acceleration ?
Ans.
(i) Uniform acceleration : A body is said to be moving with uniform acceleration if there is equal change in the velocity of the body in equal interval of time.
(ii) Non-uniform acceleration : A body is said to be moving with non-uniform acceleration, if its velocity changes unequally.

Question 2.
A bus decreases its speed from 80 kmh^-1 to 60 kmh^-1 in 5 s. Find the acceleration of the bus.
Solution:
Initial speed of the bus (u) = 80 kmh^-1
= \(\frac{80 \times 1000}{3600}\)ms^-1 = \(\frac{200}{9}\)ms^-1

Final speed of the bus (v) = 60 kmh^-1
\(\frac{60 \times 1000}{3600}\)ms^-1
\(\frac{50}{3}\)ms^-1
Time (t) = Ss
Acceleration of the bus (a) = \(\frac{v-u}{t}\) = \(\frac{\left(\frac{50}{3}-\frac{200}{9}\right) \mathrm{ms}^{-1}}{5 \mathrm{~s}}\) = \(\frac{150-200}{9 \times 5}\)ms^-2
\(\frac{-50}{45}=\frac{-10}{9}\)ms^-2 = -1.1 ms^-2

Question 3.
A train starting from a railway station and moving with uniform acceleration attains a speed of 40 kmh1 in 10 minutes. Find its acceleration.
Solution:
The initial speed of the train (u) = 0
The final speed of the train (v) = 40 kmh^-1
= \(\frac{40 \times 1000}{3600}\) = \(\frac{40 \times 1000}{3600}\)
Time (t) = 10 minutes = 10 x 60s = 600s
∴ Acceleration (a) = \(\frac{v-u}{t}\) = \(\frac{\frac{100}{9}-0}{600}\)ms^-1
= \(\frac{100}{9 \times 600}=\frac{1}{54}\) ~ 0.02 ms^-2

Questions from Sub-section 8.4

Question 1.
What is the nature of the distance-time graps for uniform and non-uniform motion of an object?
Ans.
For a uniform motion, the graph of the distance covered with time is a straight line. In graph, part OB shows that the distance is increasing with uniform rate.

Distance-time graph for an object possessing non-uniform motion is a curved line as shown in the graph. .

Question 2.
What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:
If the distance-time graph of an object, parallel to time axis is a straight line, then the object will move with uniform speed.

Question 3.
What can you say about the motion of an object, if its speed-time graph is a straight line parallel to the time axis ?
Answer:
If the speed-time graph of an object, parallel to time axis is a straight line, then the object will move with uniform speed.

Question 4.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
The quantity measured by the area occupied below the velocity-time graph shows the distance covered by the body.

Questions front Sub-section 8.5

Question 1.
Abus starting from rest moves with a uniform acceleration of 0.1ms-2for 2 minutes. Find (a) the speed acquired (b) the distance travelled.
Solution:
Here,
Initial speed of the bus (u) = 0
Final speed of the bus (v) = ?
Acceleration (a) = 0.1 ms
Time (t) = 2 minutes = 2 x 60s = 120 s
(a) The final speed (v) of the bus = u + at = 0 + (120) + \(\frac {1}{2}\) = 12 ms^-1
(b) The distance travelled by the bus, (s) = ut + \(\frac {1}{2}\) at²
= 0(120) + \(\frac {1}{2}\)(0.1)(120)²

Question 2.
A train is travelling at a speed of 90 km h^-1. Brakes are applied so as to produce a uniform acceleration of – 0.5 ms^-2 Find how far the train will go before it is brought to rest?
Solution:
Here,
Initial speed of the train (u) = 90 km h^-1
\(\frac{90 \times 1000}{3600}\) ms^-1= 25 ms^-1

Final speed of the train (v) = 0
Acceleration (a) = – 0.5 ms²
Distance (s) = ?
We know that, v² – u² = 2as
s = \(\frac{90 \times 1000}{3600}\)
= \(\frac{(0)^2-(25)^2}{2(-0.5)}=\frac{-625}{-1.0}\) = 625 metres
∴ The train will cover a distance of 625 metres, before it is brought to rest.

Question 3.
A trolley, while going down on inclined plane, has an acceleration of 2 ms^-2. What will be is velocity 3s after the start?
Solution:
Here,
The initial speed of the trolley (u) = 0
Final speed of the trolley (v) = ?
Acceleration (a) = 2 ms^-2
Time (t) = 3s
We know that, v = u + at
= 0 + 2(3) = 6 ms^-1
∴ The velocity of the trolley after 3 s will be 6 ms^-1.

Question 4.
A racing car has a uniform acceleration of 4ms2. What distance will it cover in 10 s after start?
Solution:
Here,
The initial speed of the car (u) = 0
Acceleration (a) = 4 ms
Time (t) = 10s
Distance (s) = ?

We know that,
s = ut + \(\frac {1}{2}\) at²
= 0 x (10) + \(\frac {1}{2}\) (4) (10)²
= \(\frac {1}{2}\) × 4 × 100 = 200m
So, the covered distance will be 200m

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 ms. If the acceleration of the stone during its motion is 10 ms2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
Here,
The initial velocity of the stone (u) = 5 ms^-1
Final velocity of the stone (v) = 0
Acceleration (a) = – 10 ms^-2
Maximum height (s) = ?
We know that,
v²– u² = 2as
⇒ S = \(\frac{v^2-u^2}{2 a}=\frac{(0)^2-(5)^2}{2(-10)}\) = \(\frac {-25}{20}\) = 1.25m
Applying, v = u + at
⇒ t = \(\frac{v-u}{a}=\frac{0-5}{-10}\) = 0.5 s

HBSE 9th Class Science Motion Textbook Questions and Answers

Question 1.
An athlete completes one round of a circular track of diameter 200 metre in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
Here,
The diameter of the circular track = 200 m
Radius (r) of the circular track = \(\frac {200}{2}\) = 1oo m
The circumference of the circular track = 2πr
= 2 x \(\frac {22}{7}\) × 1oo m
= \(\frac {4400}{7}\)m
Total time = 2 minutes 20 se
= (2 x 60 + 20)s (120 + 20)s = 140s

∴ The distance covered by the athlete in 40 s = \(\frac {4400}{7}\) m
The distance covered by the athlete in 1 s = \(\frac{4400}{7 \times 40}\) m
The distance covered by the athlete in 140 s = \(\frac{4400}{7 \times 40}\) x 140m = 2200 metres
Here, the total time is 140 s, in which the athlete will complete three complete rounds and one half round. If he will start from point A of the circumference and reach at point B.
In this way, displacement (AB) = The diameter of the circular track = 200 metres

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 s and then returns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speed and velocitie in jogging (a) from point A to B (b) from point A to C?
Solution:
(a) The total distance covered from point A to B = 300 m
The total time taken from point A to B = 2 min 30 s = (2 x 60 + 30)s = 150 s
Total displacement from point A to B = 300 – 0 = 300 m

= \(\frac {4400}{7}\) = 2 ms
(b) Total distance covered from point A to C = AB + BC = (300 + 100)m = 400 m
Total time taken fr om point A to C = 2 minutes 30 s + 1 minute
= (2 x 60 + 30)s + 60s = (150 + 60)s = 210s
Total displacement from point A to C = 300 – 100 = 200 m

= \(\frac {400}{210}\) = 1.90 ms^-1
= \(\frac {200}{210}\) = 0.952 ms^-1
= 0.952 ms^-1

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20km h^-1 on his return trip along the same route there is less traffic and the average speed is 30km h^-1 . What is the average speed for Abdul’s trip?
Solution:
Average speed during the school trip v₁ = 20 km h^-1
Average speed for return trip v₂ = 30 km h^-1
Let, the distance from home to school = x km
Time taken for going school Speed = =\(\frac {x}{20}\) h
Time taken to return from school = =\(\frac {x}{30}\) h
Total distance covered for both (going and returning) trips = x + x = 2x km
Total time taken = \(\left(\frac{x}{20}+\frac{x}{30}\right) h\) = \(\left(\frac{3 x+2 x}{60}\right) h=\frac{5 x}{60} h=\frac{x}{12} h\)
∴ Average speed of Abdul during whole trip = = \(\frac{\frac{2 x}{x}}{12}=\frac{2 x \times 12}{x}=24 \mathrm{kmh}^{-1}\)

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3 mr² for 8 seconds. How far does the boat travel during this time?
Solution:
Here,
Initial speed of motorboat on lake (u) = 0
Acceleration (a) = 3.0 m^-2
Time (t) = 8.0 s
Distance (s) = ?
We know that, s = ut + at²
= 0(8) + (3) (8)²
= \(\frac {1}{2}\) × 3 × 64 = 96m
∴ Motorboat will cover a distance of 96m on lake. Ans.

Question 5.
A driver of a car travelling at 52km h applies the brakes and accelerates uniformly in the opposite direction. The car stops in Ss. Another driver going at 30km h in another car applies his brakes slowly and stops in lOs. On the same graph paper, plot the speed versus time graph for two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
Speed-time graph for both drivers is shown in the figure. Suppose.
first driver starts from point A and second driver starts from point B.
Distance covered by first car before rest = area of ∆ AOC
= \(\frac {1}{2}\) × base × height
\(\frac {1}{2}\) × 5s × 52km/h
\(\frac {1}{2}\) × 3600 × 52 km
=\(\frac {1}{2}\) × 300 × 52 x 1000 m
= 36.11 metres

In same way, the distance covered by second car before
rest = area of ∆BOD
= \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 10 s × 30 km/h
= \(\frac {1}{2}\) × \(\frac {10}{3600}\) × 30km
= \(\frac {1}{2}\) × \(\frac {10}{3600}\) × 30 × 1000 m = 41.67m
It is clear from above solution that second car travels larger distance after applying brakes.

Question 6.
Figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b)Are all three ever at the same point on the road?
(e) How far has C travelled whenB passes A?
(d) How far has B travelled by the time it passes C?
Solution:

(a) B is travelling fastest as the slope in the graph of B is maximum as compared to A and C.
(b) Three can never be at the same point on the road because the three graphs do not meet at any single point.
(c) C has travelled the distance of 9 km, when B passes A.
(d) B has travelled a distance more than 4 km during the time he passed C.

Question 7.
A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10ms2, with what velocity will it strike the ground ? After what time will it strike the ground?
Solution:
Here,
Initial velocity of ball (u) = 0
Height (s) = 20 m
Acceleration (a) = 10 ms^-2
Final velocity of ball (v) = ?
Time (t) = ?
We know that,
v² – u² = 2as
v² = u² + 2as
= (0)² + 2(10)(20) = 400
or v = \(\sqrt{400}\) = 20 ms^-1
Now, acceleration, a = \(\frac{v-u}{t}\)
or t = \(\frac{v-u}{a}\) = \(\frac{20-0}{10}\) = 2s
∴The velocity of the ball will be 20 ms’ before it strike the ground and it will strike the ground in 2s.

Question 8.
The speed-time graph for a car is shown in figure.

(a) Find how far does the car travel in first 4 seconds? Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Solution:
(a) The area of the distance covered by the car in first 4 s is OAB, that is almost a right angled triangle.

∴The distance covered by the car in first 4s = \(\frac {1}{2}\) × OA × AB
The distance covered by the car in first 4s = \(\frac {1}{2}\) × 4 × 6 = 12m
The distance covered by the car OABin the figure.

(b) In the graph, the speed after 6 s, shows the uniform motion of the car.

Question 9.
State which of the following situations are possible and give an eamnle for eich of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Solution:
(a) Yes, this situation is possible. When a body is thrown up with some velocity, velocity is zero at the highest point but acceleration is non-zero and constant.
(b) Yes, this is possible. A body moving with uniform velocity on circular path is its example.

Question 10.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
Here,
Radius of circular orbit of artificial satellite (r) = 42250 km
‘Time taken by the satellite to revolve around the earth (t) = 24 hours
= 24 × 3600s = 86400s
Speed of artificial satellite (v) = \(\frac {2πr}{t}\) = \(\frac{2 \times 3.14 \times 42250}{86400}\) k ms^-1 = 3.07 k ms^-1

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

March 13, 2024 / Class 9 / By Prasanna

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

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

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

Questions from Sub-section 4.1

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

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

Questions from Sub-section 4.2

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

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

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

Question 4.
What do you think would be the observation if the a-particle scattering experiment is carried out using a foil of a metal other than gold ?
Answer:
Yes, the a-particie scattering experiment will be possible with any metal foil other than gold foil.

Questions from Subsection 4.2

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

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

Questions from Subsection 4.3

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

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

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

Questions from Sub-section 4.4

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

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

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

Questions from Subsection 4.5

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

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

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

Questions from Sub-section 4.6

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 12.
If Z = 3, what would be the valency of the element? Also, name the element.
Answer:
If the atomic number is 3, then the arrangement of electrons in an atom of the element will be:
K L
2 1
As the valence shell contains one electron only, so valency will be one. The element with atomic number 3 is Lithium.

Question 13.
Composition of the nuclei of two atomic species X and Y are given as under:
X Y
Protons 6 6
Neutrons 6 8
Give the mass numbers of X and Y. What is the relation between the two species?
Answer:
Since Mass number = Number of protons + Number of neutrons,
Therefore, the Mass numbers of X and Y are 12 and 14 respectively.
Again, Atomic number = Number of protons = Number of electrons.
Therefore, the atomic numbers of both X and Y are 6 each.
As X and Y have the same atomic number but different mass numbers, therefore, these are isotopes of each other and are represented as ₆¹²X and ₆¹⁴Y.

Question 14.
For the following statements write T for True and F for False.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about \(\frac {1}{2000}\) times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer:
(a) F (b) F (c) T (d) F.
Put a tick (√) against the correct choice and cross (✗) against the wrong choice in questions 15, 16 and 17.

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

Question 16.
Isotopes of an element have (a) the same physical properties (b) different chemical properties (c) different number of neutrons (d) different atomic numbers.
Answer:
(a) ✗ (b) ✗ (c) √ (d) ✗

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

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

Question 19.
Complete the following table.

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

Answer:

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

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HBSE 9th Class Science Solutions Chapter 3 Atoms and Molecules

March 12, 2024 / Class 9 / By Prasanna

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

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

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

Questions from Sub-section 3.1

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

Question 2.
Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Solution:
Here,
Mass of hydrogen:
mass of oxygen = 1:8
Therefore, the required mass of oxygen to combine it completely with 3g of hydrogen = 3g x 8 = 24 g

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
The postulate of Dalton’s atomic theory is that atoms are the smallest and indivisible particles which can neither be created nor destroyed in chemical reactions. It is the result of the law of conservation of mass.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
The postulate which states that atoms of the same or different elements combine in the ratio of small whole numbers to form compounds can explain the law of definite proportions.

Questions from Sub-section 3.2

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

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

Questions from Sub-section 3.4

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

∴ The formula of sodium oxide is Na₂O.

(ii) Formula of aluminium chloride

∴ The formula of aluminium chloride is AlCl₃

(iii) Formula of sodium sulphide

∴ The formula of sodium sulphide is Na₂S.

(iv) Formula of magnesium hydroxide Symbol

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

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

Answer:

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

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

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

Questions from Sub-section 3.5

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

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

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

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

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

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

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

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

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

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

Questions from Sub-section 3.5.3

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

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

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

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

Question 1.
A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Solution:
Quantity of boron in 0.24g compound = 0.096 g
Quantity of boron in 1g compound = \(\frac{0.096}{0.24}\)
Quantity of boron in 100g compound = \(\frac{0.096}{0.24} \times 100\) = 40g
Therefore, the quantity of boron in the compound = 40%
Quantity of oxygen in 0.24 g compound = 0.144g
Quantity of oxygen in 0.24 g compound = \(\frac{0.144}{0.24} \times 100\) = 60g
Therefore, the quantity of oxygen in the compound = 60%

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

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

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

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

∴ Formula of magnesium chloride = MgCl₂

(b) Formula of calcium chloride

∴ The formula of calcium chloride = CaCl₂

(d) Formula of aluminium chloride

∴ Formula of aluminium chloride = AlCl₃

(e) Formula of Calcium Carbonate chloride

∴ The formula of calcium carbonate = CaCO₃

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

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

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

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

Question 6.
Calculate the molar mass of the following substances:
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
Solution:
We know that C = 12, H = 1, S = 32, P = 31, Cl = 35.5, N = 14, O = 16
(a) Molar mass of ethyne (C₂H₂) = 2 × 12 + 2 × 1 = 24 + 2 = 26g
(b) Molar mass of sulphur molecule (S₈) = 8 × 32 = 256g
(c) Molar mass of phosphorus molecule (P₄) = 4 × 31 = 124g
(d) Hydrochloric acid, HCl = 1 × 1 + 1 × 35.5 = 1 + 35.5 = 36.5g
(e) Nitric acid, HNO₃ = 1 × 1 + 1 × 14 + 3 × 16= 1 + 14 + 48 = 63g

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

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

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

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

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

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

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

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

March 6, 2024 / Class 9 / By Prasanna

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

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

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

Questions from Sub-section 2.1

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

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

Questions from Sub-section 2.2

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

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

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

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

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

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

= \(\frac {36}{136}\) × 100 = 26.47

Questions from Sub-section 2.3

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

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

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

Questions from Sub-section 2.4

Question 1.
Classify the following as chemical and physical changes:

cutting of trees,
melting of butter in a pan,
rusting of almirah,
boiling of water- to form steam,
passing of electric current, through water and the water breaking down into hydrogen and oxygen gases,
dissolving common salt in water,
making a fruit salad with raw fruits, and
burning of paper and wood.

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

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

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

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

Question 1.
Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Butter from curd.
(e) Oil from water.
(f) Tea leaves from tea.
(g) Iron pins from sand.
(h) Wheat grains from husk.
(i) Fine mud particles suspended in water.
(j) hues from the essence of the crushed flower petals.
Answer:
(a) To separate sodium chloride from the solution of water, the evaporation method is adopted.
(b) To separate ammonium chloride from a mixture of sodium chloride and ammonium chloride, a sublimation method is adopted since ammonium chloride is a volatile substance.
(c) To separate a piece of metal from the engine oil of car, the filtration method is adopted.
(d) In order to separate butter from curd centrifugation method is adopted.
(e) To separate oil from water separating funnel is used, since water and oil both are immiscible liquids.
(f) To separate tea leaves from tea, the filtration method is brought in use. For filtration, tea strainer is used.
(g) To separate iron from sand, the magnetic separation method is adopted, because iron is attracted towards the magnet.
(h) To separate wheat grains from chaff, the threshing method is adopted, because with the threshing method chaff being lighter in weight, flies away with the wind and the wheat grains being heavier in weight, directly falls bn the ground.

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

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

Question 3.
Pragya tested the solubility of three different substances at different temperatures and collected the data as given below. Results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution:

Substance Dissolved	Temperature in Kelvin (K)
	283	293	313	333	353
	Solubility
Potassium nitrate	21	32	62	106	167
Sodium chloride	36	36	36	37	37
Potassium chloride	35	35	40	46	54
Ammonium chloride	24	37	41	55	66

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in SO grams of water at 313 K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
(d) What is the effect of a change of temperature on the solubility of a salt?

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 10.
Classify the following into elements, compounds and mixtures:
(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood.
Answer:
Element:
(a) Sodium
(d) Silver
(f) Tin
(g) Silicon.

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

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

Question 11.
Which of the following are chemical changes?
(a) Growth of a plant
(h) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food,
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle.
Answer:
Chemical changes are as follows:
(a) growth of a plant
(h) rusting of iron
(d) cooking of food
(e) digestion of food
(g) burning of a candle.

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