CBSE Academic

Motion Class 9 NCERT Solutions – The NCERT Solutions for Class 9 Science help students prepare effectively for their examinations by providing them with a comprehensive set of solutions. These solutions serve as a vital resource for students who may find it difficult to locate the answers to the questions on their own. Students can use these solutions to build a solid foundation in Science and achieve academic success.

Motion Class 9 NCERT Solutions

Q. A distance has been traversed by an object. Can it have zero displacement? If yes, give an example to prove up your assertion.

Answer – Yes, an object can have zero displacement even after moving through a distance.

Displacement is defined as the shortest distance between the initial and final positions of an object. It takes into account both the distance traveled by the object as well as the direction of the motion.

Consider an example where a person walks from point A to point B, which are 5 meters apart, and then returns to point A along the same path. In this case, the person has moved through a distance of 10 meters (5 meters from A to B and 5 meters from B to A), but the displacement is zero because the person has ended up at the same position as the starting position.

Q. A farmer walks 40 seconds along the perimeter of a square field with a side of 10 metres. What will the farmer’s displacement from his starting location be after 2 minutes and 20 seconds?

Answer –

Side square field = 10m
perimeter of square = 40m
Time taken to cover the boundary of 40m = 40s
Thus in 1s, the farmer covers a distance of 1m
Now, Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140m
The total number of rotations taken to cover distance

meters = total distance/perimeter
= 3.5

At this point, The farmer is at point B from the origin O

Therefore, the displacement theorem s = √(102+102)
s = 10√2
s = 14.14 m

Q. Which of the following statement is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is larger than the object’s travel distance.

Answer – Neither (a) nor (b) is true for displacement.
(a) is false because an object’s displacement can be zero if it returns to its initial position.
(b) is false because the magnitude of an object’s displacement can be equal to or less than the distance traveled, but never greater than it.

Q. Distinguish between speed and velocity.

Answer –

Q. What circumstance(s) must exist for an object’s average velocity to be equal to its average speed?

Answer –

Average speed = Total distance covered / Total time taken
Average velocity = Displacement / Total time taken

The magnitude of average velocity of an object is equal to its average speed only when the object moves in a straight line without changing its direction.

Q. What is measured by an automobile’s odometer?

Answer – An automobile’s odometer is a tool that calculates the overall distance the car has covered since it was first used. It usually sits on the dashboard and shows the distance travelled in miles or kilometres. Odometers are crucial instruments for keeping track of a car’s maintenance and figuring out how much it is worth to sell it. To intentionally lower a vehicle’s mileage reading by tampering with the odometer is prohibited in many nations.

Q. What does an object’s route look like when it is moving consistently?

Answer – The term “uniform motion” refers to a straight line motion in which the direction or speed of the item does not change.

Q. A signal from a spacecraft reached the ground station in an experiment in five minutes. How far away from the ground station was the spacecraft? The signal moves at the speed of light, or 3 × 10⁸ metres per second.

Answer –

distance = speed x time

In this case, the speed is the speed of light, which is 3 x 10⁸ m/s, and the time is 5 minutes, or 5 x 60 = 300 seconds.

So, the distance between the spacecraft and the ground station is:

distance = speed x time
distance = 3 x 10⁸ m/s x 300 s
distance = 9 x 10¹⁰ meters or 90,000,000,000 meters (in scientific notation)

Therefore, the spaceship was 90,000,000,000 meters or 90 billion meters away from the ground station when the signal was received.

Q. When can you state that a body is

(i) uniform acceleration?
(ii) non-uniform acceleration?

Answer –

Uniform Acceleration : An object is considered to be experiencing uniform acceleration when it is travelling straight and growing in velocity by the same amount at regular intervals. The motion’s acceleration stays constant the entire time. An object falling free is a good illustration of uniform acceleration.

Non – Uniform Acceleration : When an object’s velocity changes over time at different rates, it is said to be suffering non-uniform acceleration. The acceleration is therefore not constant during the motion, as a result. A bus departing a bus stop exhibits non-uniform acceleration since its velocity is varying, but not continuously.

Q. In 5 seconds, a bus may go from 80 km h^–1 to 60 km h^–1. Determine the bus’s acceleration.

Answer –

To find the acceleration of the bus, we can use the formula:

acceleration = (final velocity – initial velocity) / time

Here, the initial velocity (u) of the bus is 80 km/h, the final velocity (v) is 60 km/h, and the time taken (t) for the bus to change its velocity from u to v is 5 seconds.

First, we need to convert the velocities from km/h to m/s, since acceleration is measured in meters per second squared (m/s²).

So, initial velocity (u) = 80 km/h = 80 x (1000/3600) m/s = 22.22 m/s (rounded to two decimal places)

Final velocity (v) = 60 km/h = 60 x (1000/3600) m/s = 16.67 m/s (rounded to two decimal places)

We can now enter the values into the formula as follows:

acceleration = (final velocity – initial velocity) / time
acceleration = (16.67 m/s – 22.22 m/s) / 5 s
acceleration = -1.112 m/s² (rounded to two decimal places)

Therefore, the acceleration of the bus is -1.112 m/s², which means that it is decelerating or slowing down. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity of the bus.

Q. A train can reach a speed of 40 km h^–1 in 10 minutes when it departs from a railway station and moves with uniform acceleration. the acceleration of it.

Answer –

First, we need to convert the final velocity from km/h to m/s, since acceleration is measured in meters per second squared (m/s²).

Final velocity (v) = 40 km/h = 40 x (1000/3600) m/s = 11.11 m/s (rounded to two decimal places)

The time taken (t) for the train to reach this velocity is 10 minutes, which is 10 x 60 = 600 seconds.

Now, we can use the formula:

acceleration = (final velocity – initial velocity) / time

Here, the initial velocity (u) of the train is 0 m/s, since it starts from rest.

Substituting the values, we get:

acceleration = (11.11 m/s – 0 m/s) / 600 s

acceleration = 0.0185 m/s² (rounded to four decimal places)

Therefore, the acceleration of the train is 0.0185 m/s².

Q. What kind of graphs do uniform and non-uniform motion of an item have in terms of distance and time?

Answer – Non-uniform motion has a curving distance-time graph with a fluctuating slope, whereas uniform motion has a straight line with a constant slope. The object’s acceleration can be calculated from the graph’s curvature. The object is at rest when the graph shows a horizontal line.

Q. What may be said about an object’s motion when the distance-time graph is a straight line that runs parallel to the time axis?

Answer – The distance travelled by an item does not change over time if the object’s distance-time graph is a straight line parallel to the time axis. In other words, the object is not moving, and it is at rest.

Q. If an object’s speed-time graph is a straight line parallel to the time axis, what can you infer about its motion?

Answer – A straight line parallel to the time axis on an object’s speed-time graph indicates that the object is travelling at a constant speed. This is the case because the acceleration of the item is represented by the slope of the speed-time graph, and if the slope is zero (i.e., a horizontal line), then the acceleration is also zero.

Q. What is the value that the space below the velocity-time graph represents?

Answer –

Using the equation Area under the velocity-time graph = velocity*time, where velocity is the speed of the object and time is the duration of its travel, we can calculate the area under the velocity-time graph.

However, since velocity is the rate of change of displacement, we can also express velocity as displacement/time. Substituting this value into the previous equation, we get:

Area under the velocity-time graph = displacement

This means that the area under the velocity-time graph represents the total displacement or distance traveled by the object. Therefore, if we know the velocity-time graph of an object, we can find its displacement by calculating the area under the graph.

Q. A bus that starts off at rest travels for two minutes at a constant acceleration of 0.1 m s^-2. Find (a) the speed acquired, (b) the distance travelled.

Answer –

Initial velocity, u = 0 (as the bus is starting from rest)
Acceleration, a = 0.1 m/s²
Time, t = 2 minutes = 120 seconds

We need to find:
(a) The speed acquired by the bus.
(b) The distance traveled by the bus.

Solution:
(a) The final velocity (v) acquired by the bus can be calculated using the formula:

v = u + at

where v is the final velocity, u denotes the starting velocity, a denotes the acceleration, and t denotes the time required.
Substituting the given values, we get:

v = 0 + (0.1 m/s²) x (120 s)
v = 12 m/s

Therefore, the speed acquired by the bus is 12 m/s.
(b) The distance traveled (s) by the bus can be calculated using the formula:

s = ut + (1/2)at²

where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.
Substituting the given values, we get:

s = 0 x 120 + (1/2) x (0.1 m/s²) x (120 s)²
s = 720 m

Therefore, the distance traveled by the bus is 720 meters.

Q. 90 kilometres per hour is the speed of a train. The use of brakes results in a uniform acceleration of -–0.5 m s^-2. Find out how far the train will go before coming to a stop.

Answer –

Initial velocity, u = 90 km/h = 25 m/s (as 1 km/h = 1000 m/3600 s = 5/18 m/s) Acceleration, a = -0.5 m/s² (negative sign indicates deceleration) We need to find the distance traveled (s) by the train before it comes to rest.

We can use the following formula to find the distance traveled by the train:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

As the train is brought to rest, the final velocity v = 0.

Substituting the given values, we get:

0 = (25 m/s)² + 2(-0.5 m/s²) s

Simplifying the equation, we get:

s = (25 m/s)² / (2 x 0.5 m/s²) s = 625 m

Therefore, the train will travel a distance of 625 meters before it is brought to rest.

Q. A trolley accelerates by 2 cm s^-2 when moving down an inclined plane. What will its speed be three seconds after the start?

Answer –

Acceleration, a = 2 cm/s² Time, t = 3 s Initial velocity, u = 0 (as the trolley starts from rest)

We need to find the final velocity (v) of the trolley after 3 seconds.

We can use the following formula to find the final velocity of the trolley:

v = u + at

where u is the starting speed, v is the end speed, an is the acceleration, and t is the total duration.

Substituting the given values, we get:

v = 0 + (2 cm/s²) x (3 s) v = 6 cm/s

Therefore, the velocity of the trolley after 3 seconds is 6 cm/s.

Q. A race car accelerates uniformly at 4 m s^-2. How far will it travel in 10 seconds after starting?

Answer –

Acceleration, a = 4 m/s² Time, t = 10 s Initial velocity, u = 0 (as the racing car starts from rest)

We need to find the distance traveled (s) by the racing car in 10 seconds.

We can use the following formula to find the distance traveled by the racing car:

s = ut + (1/2)at²

where s denotes the distance covered, u the speed at which the journey began, a the acceleration, and t the amount of time required.

Substituting the given values, we get:

s = 0 x 10 + (1/2) x (4 m/s²) x (10 s)² s = 200 m

Therefore, after starting, the racing car will travel 200 metres in 10 seconds.

 Q. A stone is thrown with a 5 m s^-1 velocity in a vertical upward direction. What height will the stone reach and how long will it take to get there if the acceleration of the stone during its motion is 10 m s^–2 in the downward direction?

Answer –

Initial velocity, u = 5 m/s (upward) Acceleration, a = -10 m/s² (downward) We need to find the maximum height attained by the stone (h) and the time taken to reach there (t).

When the stone reaches its maximum height, its final velocity (v) becomes zero. We can use the following equation to find the maximum height reached by the stone:

v² = u² + 2as

where s is the distance travelled, an is the acceleration, v is the final velocity, and u is the initial velocity.

At the maximum height, v = 0, so we can write:

0² = (5 m/s)² + 2(-10 m/s²) s

Simplifying the equation, we get:

s = (5 m/s)² / (2 x 10 m/s²) s = 1.25 m

Therefore, the maximum height reached by the stone is 1.25 meters.

We can use the following equation to find the time taken to reach the maximum height:

v = u + at

where v is the final velocity, u denotes the starting velocity, a denotes the acceleration, and t denotes the time required.

Substituting the given values, we get:

0 = 5 m/s + (-10 m/s²) t

Simplifying the equation, we get:

t = 5/10 s

t = 0.5 s

Therefore, the stone takes 0.5 seconds to reach the maximum height.

Q. A 200 m-diameter circular track is traversed by an athlete in 40 seconds. What will the displacement and distance be after two minutes and twenty seconds?

Answer –

Diameter of circular track, d = 200 m
Radius of circular track, r = d/2 = 100 m
Time taken to complete one round, t = 40 s
We need to find the distance covered and the displacement of the athlete at the end of 2 minutes 20 seconds.
First, let’s calculate the distance covered by the athlete in one round of the circular track. Distance covered in one round = Circumference of the circle

Circumference of the circular track = 2πr
= 2 x 3.14 x 100 m
= 628 m

Therefore, the distance covered by the athlete in one round of the circular track is 628 meters.
Now, let’s calculate the total number of rounds completed by the athlete in 2 minutes 20 seconds.

Total time taken = 2 minutes 20 seconds = 140 seconds
Number of rounds completed = Total time taken / Time taken to complete one round
= 140 seconds / 40 seconds
= 3.5 rounds (rounded to one decimal place)

Therefore, the athlete completes 3.5 rounds of the circular track in 2 minutes 20 seconds.
Now, let’s calculate the distance covered by the athlete in 3.5 rounds of the circular track:

Distance covered = Distance covered in one round x Number of rounds completed
= 628 m x 3.5
= 2198 m

Therefore, the athlete has travelled 2198 metres after 2 minutes and 20 seconds.
Finally, let’s calculate the displacement of the athlete at the end of 2 minutes 20 seconds. Since the athlete starts and ends at the same point, the displacement is zero.
Therefore, the displacement of the athlete at the end of 2 minutes 20 seconds is zero.

Q. In 2 minutes and 30 seconds, Joseph runs a straight 300-meter route from point A to point B. He then turns around and runs the final 100 metres to point C. What are Joseph’s typical running distances and velocities (a) from A to B and (b) from A to C?

Answer –

We can start by using the formulas for speed and velocity:

Speed = distance / time
Velocity = displacement / time

where displacement refers to a change in an object’s position.

(a) From A to B:
Distance = 300 m
Time = 2 minutes 30 seconds = 150 seconds

Speed = Distance / Time
Speed = 300 m / 150 s
Speed = 2 m/s

Note that since Joseph returns to point A, his displacement is zero. Therefore, his velocity from A to B is also zero.

(b) From A to C:
Distance = 300 m + 100 m = 400 m
Time = 2 minutes 30 seconds + 1 minute = 210 seconds

Speed = Distance / Time
Speed = 400 m / 210 s
Speed = 1.9 m/s

Joseph’s displacement from A to C is 100 m (he ends up at point C). Therefore, his velocity is:
Velocity = Displacement / Time
Velocity = 100 m / 210 s
Velocity = 0.48 m/s

So Joseph’s average speed from A to B is 2 m/s and his average velocity is zero. His average speed from A to C is 1.9 m/s and his average velocity is 0.48 m/s.

Q. Abdul calculates his average journey speed while driving to school to be 20 km.h^–1. The amount of traffic is lower and the average speed is 30 km.h^–1 on his return trip over the same route. What is Abdul’s journey’s average speed?

Answer –

The entire distance travelled divided by the total time taken is the formula we may use to determine the average speed:
Average speed = Total distance / Total time
Let’s assume the distance from Abdul’s home to school is d km. Then, the time taken for Abdul to travel from home to school at an average speed of 20 km/h would be:
Time taken = distance / speed
Time taken = d / 20

Similarly, the time taken for Abdul to travel from school to home at an average speed of 30 km/h would be:
Time taken = distance / speed
Time taken = d / 30

The total distance covered by Abdul is 2d (since he travels from home to school and then back from school to home). The total time taken is the sum of the time taken for the forward and return journeys:
Total time taken = d / 20 + d / 30
We can simplify this expression by finding a common denominator:

Total time taken = (3d / 60) + (2d / 60)
Total time taken = 5d / 60
Total time taken = d / 12

Now we can substitute the values for total distance and total time taken into the formula for average speed:
Average speed = Total distance / Total time taken
Average speed = 2d / (d / 12)
Average speed = 24 km/h

Abdul’s journey will therefore have an average speed of 24 km/h.

Q. A speedboat on a lake begins at rest and accelerates in a straight line for 8.0 seconds while maintaining a constant speed of 3.0 m s^–2. During this time, how far does the boat travel?

Answer –

The distance traveled with constant acceleration is:

Distance = initial velocity * time + (1/2) * acceleration * time²

where the initial velocity is the velocity of the object before acceleration begins.

In this case, the boat starts from rest, so its initial velocity is 0 m/s. The boat accelerates for 8.0 s at a steady speed of 3.0 m/s2. Hence, we may add these values to the formula as follows:

Distance = 0 m/s * 8.0 s + (1/2) * 3.0 m/s² * (8.0 s)²
Distance = 0 + 96.0 m
Distance = 96.0 meters

Therefore, the boat travels 96.0 meters during this time.

Q. A 52 km h^–1 vehicle’s driver evenly decelerates in one direction before accelerating in the opposite direction. In 5 seconds, the automobile stops. Another driver in another automobile slows down and stops in 10 seconds while travelling at a speed of 3 km h^–1. Plot the speed versus time graphs for the two cars on the same sheet of graph paper. Which of the two cars made a longer distance after using the brakes?

Answer – The equations for the two cars are:

First car: 0 m/s = 14.44 m/s + acceleration x 5 s
Second car: 0 m/s = 0.83 m/s + acceleration x 10 s

The speed versus time graphs for the two cars are:

To determine the distance traveled by each car, use the formula:
distance = (initial speed x time) + (0.5 x acceleration x time²)

For the first car:
distance = (14.44 m/s x 5 s) + (0.5 x -2.888 m/s² x (5 s)²) = 36.1 m

For the second car:
distance = (0.83 m/s x 10 s) + (0.5 x -0.083 m/s² x (10 s)²) = 4.17 m

Therefore, the first car traveled farther after the brakes were applied.

Q. From a distance of 20 metres, a ball is gently dropped. What speed will it hit the ground at if its velocity rises uniformly at a rate of 10 m s^-2? How long before it hits the ground?

Answer –

Using the kinematic equation:

v² = u² + 2as

where:
v = final velocity of the ball just before it strikes the ground
u = initial velocity of the ball (which is 0 m/s since it is dropped gently)
a = acceleration of the ball due to gravity (which is -9.8 m/s² since it is directed downwards)
s = distance travelled by the ball (which is 20 m since it is dropped from a height of 20 m)

We can rearrange this equation to solve for v:

v² = u² + 2as
v² = 0 + 2(-9.8 m/s²)(20 m)
v² = -392 m²/s²

Since we can’t have a negative velocity, we need to take the square root of both sides of the equation and include a negative sign to indicate the direction of the velocity (downwards):
v = – √392 m/s ≈ -19.8 m/s

Therefore, the ball will strike the ground with a velocity of approximately 19.8 m/s downwards.
To find the time it takes for the ball to strike the ground, we can use the kinematic equation:
s = ut + (1/2)at²

where:
t = time taken for the ball to strike the ground (which is what we’re solving for)

Rearranging this equation, we get:
t = √(2s/a)

Substituting the values, we get:
t = √(2 x 20 m / 9.8 m/s²) ≈ 2.02 s

Therefore, the ball will take approximately 2.02 seconds to strike the ground.

Q. The speed-time graph for a car is shown below

(A) Measure the distance the car covers in the first four seconds. Cover the portion of the graph that corresponds to the distance the car travelled during the time period. (b) Which area of the graph best captures the car’s constant motion?

Answer –

(a) The shaded area on the speed-time graph represents the displacement of the car over a time period of 4 seconds, and the value of the area is 12 meters. Therefore, the car travels a total distance of 12 meters in the first four seconds.

(b) From the 6th to the 10th second, the speed of the car remains constant, which means the car is said to be in uniform motion during that time interval. This uniform motion is represented by the horizontal section of the speed-time graph between x=6 and x=10.

Q. Provide examples for each of the following circumstances and indicate which are possible:
(A) a stationary object with continuous acceleration but no motion
(b) A moving item that exhibits an acceleration yet maintains a constant speed.
(c) A thing that is travelling in one direction while accelerating in the opposite direction.

Answer –

a) It is possible for an object to have constant acceleration but zero velocity, such as a ball thrown straight up into the air at its highest point.

b) It is not possible for an object to have an acceleration and also move with uniform speed.

c) It is possible for an object to have an acceleration perpendicular to its direction of motion, such as a car turning a corner, which experiences sideways acceleration directed towards the centre of the turning circle.

Q. An artificial satellite is travelling in a circular orbit with a 42250 km radius. If it takes 24 hours to complete one orbit of the earth, then determine its speed.

Answer –

We can use the formula for the speed of an object in circular motion:

v = 2πr/T

where v is the speed of the object, r is the radius of the circular orbit, and T is the time taken to complete one revolution.

In this case, the radius of the orbit is 42,250 km, and the time taken for one revolution is 24 hours or 86,400 seconds (since there are 24 hours in a day and 60 seconds in a minute).

Plugging these values into the formula, we get:

v = 2πr/T = 2π(42,250 km) = 265571.42 km

Therefore, the speed of the satellite in its circular orbit is approximately 265571.42 km.

CBSE Skill Education

Tissue Class 9 NCERT Solutions – The NCERT Solutions for Class 9 Science help students prepare effectively for their examinations by providing them with a comprehensive set of solutions. These solutions serve as a vital resource for students who may find it difficult to locate the answers to the questions on their own. Students can use these solutions to build a solid foundation in Science and achieve academic success.

Tissue Class 9 NCERT Solutions

Q. What is a tissue?

Answer – A tissue is a group of similar cells that work together to perform a specific function in a living organism.

Q. How are tissues used by multicellular organisms?

Answer – Tissues in multicellular organisms serve two main purposes:

Firstly, they provide structural and mechanical support to the organism. Different types of tissues such as connective tissues, bone tissues and vascular tissues, help maintain the shape, form and integrity of the organism.

Secondly, tissues enable division of labour among different cells in the organism, by allowing specialized functions to be carried out. For example, muscle tissue allows for movement, while nervous tissue allows for communication and coordination of different bodily functions. This specialization of cells allows for more complex and efficient functioning of the organism.

Q. Name the types of simple tissues.

Answer – Plant: Parenchyma, Collenchyma, Sclerenchyma

Q. Where is apical meristem found?

Answer – The apical meristem is located at the tips of the roots and shoots in plants, and it is responsible for the growth of the plant in length and height, respectively.

Q. Which tissue makes up a coconut’s husk?

Answer – The husk of a coconut is made up of a strong and tough tissue called sclerenchyma. It gives support and protection to the fruit, and makes it stiff and hard. Sclerenchyma contains dead cells with thick, durable walls that have lignin to make them strong.

Q. What are the constituents of phloem?

Answer – Phloem is a complex tissue found in vascular plants that is responsible for the transport of food materials, such as sugars and amino acids, from the leaves and other sources of the plant to the different parts of the plant where they are needed. There are four primary cell types found in phloem tissue:

1. Sieve tube elements: These are long, slender cells that form a continuous tube through which food materials are transported.
2. Companion cells: These are smaller cells that are closely associated with sieve tube elements and play a role in controlling their activities.
3. Phloem fibers: These are long, narrow cells that provide mechanical support to the phloem tissue.
4. Phloem parenchyma: These are relatively unspecialized cells that are involved in storage and other metabolic functions within the phloem tissue.

Q. Which body part’s movement is caused by which tissue?

Answer – Movement in our body is achieved through the coordination of two types of tissues: muscular tissue and nervous tissue.

Muscular tissue, consisting of skeletal, smooth, and cardiac muscle, is responsible for generating force and producing movement in response to nervous system signals.

Nervous tissue, on the other hand, is responsible for detecting, processing, and transmitting information in the form of electrical impulses throughout the body. This tissue coordinates and controls the activity of the muscular tissue by initiating and modulating the strength, duration, and timing of muscle contractions to produce smooth and coordinated movement.

Q. What does a neuron look like?

Answer – A neuron is a specialized type of nerve cell that plays a critical role in transmitting information throughout the nervous system. It has a unique shape and structure, consisting of a cell body that contains a nucleus and cytoplasm, and long, thin extensions called axons and shorter, branched extensions called dendrites.

Q. Name three characteristics of the cardiac muscle.

Answer – The heart’s muscles or cardiac muscles are highly specialised organs that circulate blood throughout the body.

• They are shaped like cylinders have striated muscle fibres.
• Cardiac muscles are uninucleated and branched, in contrast to skeletal muscles, which have numerous nuclei.
• These muscles cannot be actively controlled by the person because they are involuntary in nature. Instead, the cardiac conduction system—a network of specialised cells that manages the heart’s contractions—controls them.

Q. What purposes does areolar tissue serve?

Answer – Areolar tissues, which are a type of loose connective tissue, are commonly found in animals and serve a variety of functions. They can be found in the spaces between the skin and muscles, as well as around blood vessels and nerves throughout the body. They are also present in bone marrow and can be found inside organs, where they provide support and assist in tissue repair. Areolar tissues play a crucial role in maintaining the overall structure and integrity of the body, helping to connect and bind different tissues and organs together. They also play a role in immune defense, nutrient and waste exchange, and edema control.

Q. Define the term ’tissue’.

Answer – A collection of cells that share a common structure and cooperate to carry out a particular job in the body is referred to as a tissue. Cells that are designed to carry out a specific function make up tissues, and these cells cooperate to complete tasks.

Q. What combination of different elements makes up the xylem tissue? Name them.

Answer – Transporting water and nutrients from the roots to the rest of the plant is the job of the specialised plant tissue known as xylem. It is made up of four major components, including:

• Vessels: These are lengthy, tube-like structures that are produced by the fusion of separate cells and serve as a continuous conduit for the transportation of water.
• Tracheids: Water can pass through tiny openings in the cell walls of these long, slender cells called tracheids, which have tapered ends.
• Xylem fibres: These elongated, cells with thick walls sustain the plant mechanically.
• Xylem parenchyma: These cells perform a variety of support functions, including respiration, storage, and other functions.

Q. How you will differentiate simple tissues in plants from complex tissues?

Answer –

Q. Determine the differences in cell walls between parenchyma, collenchyma, and sclerenchyma.

Answer –

Q. What are the stomata’s purposes?

Answer – Stomata are small openings or pores found on the epidermis of plant leaves, stems, and other organs. They allow for the exchange of gases such as carbon dioxide and oxygen, which is essential for photosynthesis. Stomata also play a role in transpiration, which involves the loss of water vapor from the plant through the stomatal pores.

Q. Diagrammatically depict the variations between the three categories of muscle fibres.

Answer – There are three different types of muscle fibers –

1. Striated muscles Fibers:

• Appear striated or striped under a microscope
• Have multiple nuclei
• Are cylindrical in shape
• Can be voluntarily controlled
• Attached to bones by tendons
• Generate rapid and powerful contractions

2. Smooth Muscle Fibers:

• Appear smooth and non-striated under a microscope
• Have a single nucleus
• Are spindle-shaped
• Are not under voluntary control
• located in the blood vessels and stomach, among other internal organs, walls.
• Generate slow and sustained contractions

3. Cardiac Muscle Fibers:

• Appear striated under a microscope
• Have a single nucleus
• Are branched in shape
• Are not under voluntary control
• Found only in the heart
• Generate strong and rhythmic contractions

Q. What job does the cardiac muscle perform?

Answer – Cardiac muscles are specialized muscle tissues found only in the heart. They are characterized by their unique features, which include:

• Branched and cylindrical shape
• Uninucleated cells
• Involuntary control
• Striated appearance under a microscope
• Generate strong and rhythmic contractions
• Responsible for pumping blood throughout the body
• Have the ability to contract and relax rhythmically throughout one’s lifetime
• Capable of sustaining activity for long periods without fatigue

Q. Based on their structure and position in the body, differentiate between striated, un-striated, and cardiac muscles.

Answer –

Q. Draw a labelled diagram of a neuron.

Answer –

Q. Name the following.

(A) The tissue that lines the inside of our mouths.
(b) In humans, the connective tissue between muscle and bone.
(c) Plants’ food-transporting tissue.
(d) Body tissue that stores fat.
(e) Tissue that is connected by a fluid matrix.
(f) Brain tissue is found.

Answer –

(a) Stratified squamous epithelium
(b) Tendon
(c) Phloem
(d) Adipose tissue
(e) Blood
(f) Nervous tissue

Q. Identify the type of tissue in the following:
Skin, lining of kidney tubule, bark of tree, bone, vascular bundle.

Answer –

• Skin – Stratified squamous epithelial tissue
• Bark of tree – Cork cambium (a type of meristematic tissue)
• Bone – Osseous tissue (a type of connective tissue)
• Lining of kidney tubule – Simple cuboidal epithelial tissue
• Vascular bundle – Complex tissue (contains both xylem and phloem)

Q. Identify the areas that have parenchyma tissue.

Answer – Parenchyma tissue is present in various regions of plants including:

• Cortex and pith of stems and roots
• Mesophyll of leaves
• Fruits and seeds
• Endosperm of seeds
• Resin ducts
• Medulla of stems
• Pericycle and xylem of roots.

Q. What function does the plant epidermis serve?

Answer – The epidermis is the outermost layer of cells in plants, which covers and protects the underlying tissues from mechanical injury, water loss, and invasion by pests and pathogens. It also plays a crucial role in the exchange of gases, regulating water loss through stomata and facilitating the absorption of water and minerals from the soil. Additionally, the epidermis can secrete waxy substances to reduce water loss and to prevent harmful substances from entering the plant.

Q. How does the cork function act as a barrier?

Answer – Cork tissue forms a protective layer in plants due to its dense arrangement of dead cells with no intercellular spaces. The walls of the cork cells are impervious to water and gases due to the deposition of suberin.

Q. Complete the following chart.

Answer –

CBSE Skill Education

Fundamental Unit of Life Class 9 NCERT Solutions – The notes provided on our website for CBSE students are designed to align with CBSE and NCERT syllabus guidelines. By referring to these notes, students can save time and gain a comprehensive understanding of important topics and questions in each chapter.

Fundamental Unit of Life Class 9 NCERT Solutions

Q. Who discovered cells, and how?

Answer – In 1665, the English scientist Robert Hooke discovered cells by examining a thin slice of cork through a microscope that he designed himself. Upon observing the cork, he noticed that it had a structure that resembled a honeycomb, consisting of many small compartments. Hooke described these small compartments as “cells”, which is the term still used today to describe the basic unit of life.

Q. Why the cell is referred to as the structural and junctional unit of life?

Answer – Cells are considered the fundamental and structural units of life because they are responsible for forming the basic structure of living entities. Groups of cells come together to form tissues, which further combine to form organs, and ultimately organ systems. These structures perform essential life processes such as respiration, digestion, and excretion, among others.

Q. How does water and carbon dioxide move through and out of the cell?

Answer – C0₂ moves in and out of the cell by diffusion, from an area of higher concentration inside the cell to an area of lower concentration outside the cell.

Water moves into and out of the cell by osmosis, from an area of higher concentration to an area of lower concentration across a selectively permeable membrane until an equilibrium is reached.

Fundamental Unit of Life Class 9 NCERT Solutions

Q. Why plasma membrane is referred to as a selectively permeable membrane?

Answer – The plasma membrane is selectively permeable because it only allows certain molecules to pass through it while restricting others. The movement of molecules in and out of cells is regulated by the membrane’s composition and structure, which selectively permits or prevents the passage of molecules based on their size, charge, and polarity. This feature ensures that only the necessary molecules enter and exit the cell, maintaining the cell’s internal environment and enabling it to function correctly.

Q. Fill in the blanks in the following table that shows how prokaryotic and eukaryotic cells differ from one another.

Answer –

Fundamental Unit of Life Class 9 NCERT Solutions

Q. Which of the two organelles we’ve studied that have their own genetic material can you name?

Answer – The two organelles we have studied that contain their own genetic material are:

1. Mitochondria
2. Chloroplasts (in plant cells)

Q. What will happen if a cell’s organisation is damaged by a physical or chemical influence?

Answer – Lysosomes are known as “suicide bags” because in case of damage to cells where cell revival is not possible, lysosomes burst and release enzymes which lead to the digestion of the damaged cells. This process is essential for maintaining the health and proper functioning of the surrounding cells and tissues.

Fundamental Unit of Life Class 9 NCERT Solutions

Q. Why are lysosomes referred to as “suicide” hags?

Answer – Lysosomes are called suicide bags because they contain enzymes that can break down biomolecules, and in case of cell damage, they can burst and cause autolysis or self-destruction of the cell.

Q. Where do proteins get synthesized inside of cells?

Answer – Proteins are synthesized inside the cell in the ribosomes, which are present in both prokaryotic and eukaryotic cells. In eukaryotic cells, ribosomes can be found either free in the cytoplasm or attached to the endoplasmic reticulum. In prokaryotic cells, ribosomes are free in the cytoplasm. The process of protein synthesis is called translation and involves the assembly of amino acids into a polypeptide chain according to the instructions encoded in messenger RNA (mRNA).

Q. Compare the differences between plant and animal cells, then list the ways in which they differ from each other.

Answer –

Fundamental Unit of Life Class 9 NCERT Solutions

Q. How are eukaryotic and prokaryotic cells distinct from one another?

Answer –

Q. What would happen if the plasma membrane was to break or degrade?

Answer – A ruptured or broken-down plasma membrane can allow harmful substances to enter the cell, cause the loss of important cell components, disrupt the cell’s energy balance, and ultimately lead to cell death.

Q. What would happen to a cell’s life if the Golgi apparatus were absent?

Answer – In eukaryotic cells, the Golgi apparatus is essential for the storage, packing, and manufacturing of substances. It participates in cell development as well as the storing and transport of components made by cells. Cells would be unable to adequately store and transfer materials without the Golgi apparatus, which could result in malfunction or cell death.

Fundamental Unit of Life Class 9 NCERT Solutions

Q. Which organelle is considered as the cellular powerhouse? Why?

Answer – The organelle known as the powerhouse of the cell is the mitochondrion. Mitochondria are responsible for producing the majority of the cell’s energy in the form of ATP (adenosine triphosphate) through the process of cellular respiration. This is why mitochondria are often referred to as the cell’s “powerhouse”. Mitochondria are unique in that they have their own DNA and can self-replicate, which suggests that they may have originated as independent prokaryotic organisms that were engulfed by eukaryotic cells through endosymbiosis. The presence of mitochondria in eukaryotic cells greatly increased the energy production capacity of these cells, allowing them to develop more complex structures and functions.

Q. What is the location of cell membrane lipid and protein synthesis?

Answer – The lipids and proteins that make up the cell membrane are synthesized in the endoplasmic reticulum (ER) of the cell. Specifically, the rough endoplasmic reticulum (RER) synthesizes and packages membrane proteins, while the smooth endoplasmic reticulum (SER) synthesizes membrane lipids. Once synthesized, these components are transported to the Golgi apparatus for modification and packaging, before being transported to the cell membrane.

Fundamental Unit of Life Class 9 NCERT Solutions

Q. How does Amoeba obtain it’s food?

Answer – Amoeba is a single-celled organism that feeds on other smaller organisms, such as bacteria, algae, and other protozoans. The process by which Amoeba obtains its food is known as phagocytosis.

During phagocytosis, the Amoeba extends its pseudopodia (temporary projections of the cell membrane and cytoplasm) to surround and engulf its prey, forming a food vacuole. The food vacuole is then transported into the cell, where it fuses with lysosomes containing digestive enzymes. These enzymes break down the food into smaller molecules that can be absorbed by the cell and used for energy or other cellular processes. Any undigested material is eliminated from the cell through exocytosis.

Fundamental Unit of Life Class 9 NCERT Solutions

Q. What is osmosis?

Answer – Osmosis is the transfer of water molecules from an area with a higher concentration of water to an area with a lower concentration of water through a membrane that is selectively permeable. A concentration gradient of solutes that can’t pass the membrane triggers this movement of water. Osmosis aims to reach equilibrium, where water and solute concentrations are equal on both sides of the membrane. Osmosis is an essential process for many living things because it controls the flow of nutrients and water into and out of cells.

Q. Carry out the following osmosis experiment:
Four potato halves should be peeled, then each half should be scooped out to create a potato cup, one of which should be constructed from a boiling potato. Each potato cup should be placed in a water-filled trough. Now,

(a) Leave cup A empty
(b) fill cup B with one teaspoon of sugar
(c) Fill cup C with one teaspoon of salt.
(d) Add one teaspoon of sugar to the cup of cooked potatoes
For two hours, keep these. Next, take a look at the four potato cups and respond to the following:
(i) Describe the reason water collects in B and C’s hollow areas.
(ii) What purpose does potato A serve in this experiment?
(iii) Describe why the hollowed-out areas of A and D do not collect water.

Answer –

In this osmosis experiment, four potato cups are created by scooping out the center of each potato half. Each potato cup is then placed in a trough containing water, and various substances are added to each cup to observe the effects of osmosis. The boiled potato cup is included to serve as a control for comparison.

After two hours, the following observations can be made:

(i) Water gathers in the hollowed portion of B and C due to the process of osmosis, where water moves from a region of higher concentration to a region of lower concentration, towards the solution with a higher concentration of solute.
(ii) Potato A is necessary to serve as a control group or a standard for comparison, as it has no solute added to it.
(iii) Water does not gather in the hollowed-out portions of A and D because the concentration of solutes in these potato cups is the same as that in the surrounding water, causing no net movement of water.

CBSE SKill Education

Structure of Atom Class 9 NCERT Solutions – The notes provided on our website for CBSE students are designed to align with CBSE and NCERT syllabus guidelines. By referring to these notes, students can save time and gain a comprehensive understanding of important topics and questions in each chapter. This can help them better retain key concepts and be well-equipped to answer any questions that may appear on exams.

Structure of Atom Class 9 NCERT Solutions

Q. What are canal rays?

Answer – Canal rays are a type of radiation that is positively charged. The discovery of canal rays was important in the discovery of protons, which are also positively charged subatomic particles.

Q. Will an atom have any charge if it has one proton and one electron?

Answer – The proton and electron carry opposite charges, with the proton being positively charged and the electron being negatively charged. Due to this opposite charge, when an atom contains one electron and one proton, the net charge of the atom becomes neutral. In other words, the positive charge of the proton is exactly balanced by the negative charge of the electron, resulting in a neutral charge for the atom.

Q. Explain how the atom is neutral as a whole using Thomson’s model of an atom as your foundation.

Answer – According to Thomson’s model of an atom:

(i) An atom consists of a positively charged sphere, and the negatively charged electrons are embedded within it.

(ii) Electrons and protons have equal but opposite charges, and they are uniformly distributed within the atom. This results in an atom having no overall charge, making it electrically neutral.

Q. Which subatomic particle is present in an atom’s nucleus according to Rutherford’s model of an atom?

Answer – The majority of an atom’s mass is contained in the nucleus, which contains positively charged protons, according to Rutherford’s model of an atom. Similar to how planets circle around the sun, the electrons in the nucleus move in clearly defined orbits.

Q. Draw a rough sketch of Bohr’s three-shelled atom.

Answer –

Q. What do you think the results of the α-particle scattering experiment would be if a foil made of a different metal than gold was used?

Answer – Regardless of which metal foil is used in the α-particle scattering experiment, the observations remain the same because the fundamental structure of an atom, when examined on an individual level, remains constant across different elements. In other words, the behavior of α-particles is determined by the structure of individual atoms rather than the specific properties of the metal foil used as a target.

Q. Identify the three atom’s subatomic components.

Answer – The general pattern of scattering in the α-particle scattering experiment would still be observed if a metal foil other than gold is used, but the exact scattering angles and intensities of the scattered particles would be different due to variations in the atomic structure of the foil.

Q. The nucleus of a helium atom contains two protons and has an atomic mass of 4 u. What is the number of neutrons in it?

Answer – The equation to calculate the number of neutrons in an atom is:

Number of neutrons = Atomic mass – Number of protons

Using the given information, we can substitute the values as follows:

Number of neutrons = 4 u – 2

Simplifying, we get:

Number of neutrons = 2

Therefore, the helium atom with an atomic mass of 4 u and two protons in its nucleus has 2 neutrons.

Q. Describe how the electrons are distributed among the carbon and sodium atoms.

Answer –

Carbon
Atomic number = 6
No. of protons = 6
and Number of protons = Number of electrons
Distribution of electrons = KL
24

Sodium
Atomic number = 11
No. of protons = 11 = No. of electrons
Distribution of electrons = K L M
2 8 1

Q. What would be the total number of electrons in an atom if the K and L shells were both filled?

Answer – If the K and L shells of an atom are full, then the total number of electrons in the atom would be:

2 in the K shell + 8 in the L shell = 10 electrons

The K shell can hold a maximum of 2 electrons, and the L shell can hold a maximum of 8 electrons. Therefore, if both shells are full, the total number of electrons in the atom would be 10.

Q. How do you determine the valencies of magnesium, sulphur, and chlorine?

Answer – The valency of an element can be determined by subtracting the number of valence electrons in the outermost shell from 8 (for elements in groups 1, 2, 13) or from 18 (for elements in groups 3-18) since the maximum number of valence electrons an atom can have is 8 or 18.

The valency of chlorine is 7 because it has 7 valence electrons in the third energy level (3s and 3p).

The valency of sulphur can be 2, 4, or 6, depending on the chemical reaction, because it has 6 valence electrons in the third energy level (3s and 3p).

The valency of magnesium is 2 because it has 2 valence electrons in the third energy level (3s).

Q. If number of proton in atom is 8 and number of electorns is also 8, then

(i) what is the atomic number of the atom

(ii) what is the charge on the atom?

Answer – If the number of electrons in an atom is 8 and the number of protons is also 8, then:

(i) The atomic number of the atom is equal to the number of protons, which is 8. This is because the atomic number of an element is defined as the number of protons in the nucleus of an atom of that element. Therefore, if an atom has 8 protons, its atomic number is 8, which indicates that it is an oxygen atom.

(ii) The charge on the atom would be neutral or zero. This is because the charge on an atom is determined by the number of protons and electrons. In a neutral atom, the number of protons is equal to the number of electrons, and their charges cancel out, resulting in a net charge of zero. Therefore, if an atom has 8 protons and 8 electrons, it has a neutral charge.

Q. With the help of below table find out the mass number of sulphur and oxygen atom.

Answer – To determine the mass number of an atom, we add the number of protons and neutrons present in the nucleus of the atom.

(a) For oxygen:

• Number of protons = 8
• Number of neutrons = 8
• Mass number = number of protons + number of neutrons = 8 + 8 = 16

Therefore, the mass number of oxygen is 16.

(b) For sulphur:

• Number of protons = 16
• Number of neutrons = 16
• Mass number = number of protons + number of neutrons = 16 + 16 = 32

Therefore, the mass number of sulphur is 32.

Q. List the three subatomic particles that are present in each of the symbols H, D, and T.

Answer – The symbols H, D, and T represent different isotopes of the element hydrogen. Here is the table for the three subatomic particles found in each isotope:

Note:

• The number of protons determines the identity of the element, so all three isotopes are hydrogen and have one proton each.
• The number of neutrons can vary among isotopes of the same element, leading to different atomic masses.
• The number of electrons in a neutral atom is equal to the number of protons, so all three isotopes have one electron each.

Q. Any particular pair of isotopes and isobars’ electronic configuration should be written down.

Answer – Isotopes are atoms of the same element that have the same number of protons (atomic number), but different numbers of neutrons (and hence different atomic masses). Isobars are atoms that have the same mass number, but different atomic numbers.

Here are the electronic configurations of one pair of isotopes and one pair of isobars:

Pair of isotopes: Hydrogen-1 and Hydrogen-2 (also known as deuterium)

Hydrogen-1: 1s1

Hydrogen-2 (deuterium): 1s1 2s1

Both isotopes have one electron in the first shell, but deuterium has an additional neutron in the nucleus.

Pair of isobars: Carbon-12 and Nitrogen-12

Carbon-12: 1s2 2s2 2p2

Nitrogen-12: 1s2 2s2 2p3

Both isotopes have the same mass number of 12, but different atomic numbers (6 for carbon and 7 for nitrogen). They have the same electronic configuration up to the 2p subshell, but differ in the number of electrons in the 2p subshell.

Q. Write the difference between electrons, protons and neutrons.

Answer –

Note: u refers to atomic mass unit.

Q. What are the drawbacks of J.J. Thomson’s atomic model?

Answer – According to J.J. Thomson’s atomic model, electrons are evenly dispersed across a positively charged sphere. However, later research by other scientists showed that protons are found in the atom’s core nucleus and that electrons orbit the nucleus at distinct energy levels. The Bohr model, a redesigned representation of the atom, is the cornerstone of contemporary knowledge of atomic structure.

Q. What are the drawbacks of Rutherford’s atomic model?

Answer – According to J.J. Thomson’s atomic model, electrons are evenly dispersed across a positively charged sphere. However, later research by other scientists showed that protons are found in the atom’s core nucleus and that electrons orbit the nucleus at distinct energy levels. The Bohr model, a redesigned representation of the atom, is the cornerstone of contemporary knowledge of atomic structure.

Q. Describe Bohr’s atomic model.

Answer – Bohr’s model of the atom proposed that electrons exist in discrete energy levels or shells around the nucleus, and that they can move between these shells by absorbing or emitting energy in the form of photons. The model also explains the spectral lines observed in the emission and absorption spectra of atoms.

Q. Compare each of the atomic model proposals presented in this chapter.

Answer –

Q. Provide a brief summary of the guidelines for describing the distribution of electrons among the first 18 elements’ different shells.

Answer – The rules for writing the distribution of electrons in various shells for the first eighteen elements can be rephrased as follows:

The maximum number of electrons that can be present in a shell is determined by the formula 2 n², where n is the orbit number (1, 2, 3, etc.). For example, the maximum number of electrons that can be present in the K shell (n=1) is 2, in the L shell (n=2) is 8, in the M shell (n=3) is 18, and in the N shell (n=4) is 32.

The maximum number of electrons that can be present in the outermost orbit (valence shell) is 8.

Electrons are not accommodated in a given shell unless the inner shells are filled first. In other words, shells are filled step-wise, with the lowest energy level being filled first before moving to higher energy levels. This is known as the Aufbau principle.

Q. Using silicon and oxygen as examples, define valency.

Answer – Valency is defined as the combining capacity of an atom to form chemical bonds with other atoms. It is determined by the number of electrons in the outermost shell or valence shell of an atom. The electronic configurations of oxygen and silicon are 2-6 and 2-8-4, respectively.

In the case of oxygen, the atomic number is 8, which means it has 8 electrons in total. Out of these, 6 electrons are present in the outermost shell. To fill the outermost shell, 2 more electrons are required. Therefore, the valency of oxygen is 2.

In the case of silicon, the atomic number is 14, which means it has 14 electrons in total. Out of these, 4 electrons are present in the outermost shell. To fill the outermost shell, 4 more electrons are required. Therefore, the valency of silicon is 4.

Q. Explain with examples
(i) Atomic number,
(ii) Mass number,
(iii) Isotopes and
(iv) Isobars. Give any two uses of isotopes.

Answer –

(i) Atomic number: Number of protons in the nucleus of an atom, denoted by symbol Z. Example: The atomic number of oxygen is 8.

(ii) Mass number: Sum of protons and neutrons in the nucleus of an atom, denoted by symbol A. Example: The mass number of carbon-12 is 12.

(iii) Isotopes: Atoms of the same element with the same atomic number but different mass numbers. Example: Carbon-12 and carbon-14 are isotopes of carbon.

(iv) Isobars: Atoms of different elements with the same mass number but different atomic numbers. Example: Carbon-14 and nitrogen-14 are isobars.

Uses of isotopes:

1. Medical applications: Technetium-99m is used in medical imaging to diagnose various diseases, such as cancer and heart disease.
2. Industrial applications: Cobalt-60 is used to irradiate food to extend its shelf life, while iridium-192 is used to detect flaws in welds and other materials.

Q. K and L shells are entirely filled with Na+. Explain.

Answer – Sodium (Na) has an atomic number of 11, which means it has 11 protons in its nucleus and 11 electrons surrounding the nucleus. The electronic configuration of sodium is 2, 8, 1, where 2 electrons occupy the first energy level or K-shell, 8 electrons occupy the second energy level or L-shell, and 1 electron occupies the third energy level or M-shell.

When sodium loses an electron, it forms a sodium ion (Na+). This is because sodium has a tendency to lose an electron to achieve the stable configuration of the previous noble gas, neon (2, 8). Thus, Na+ has 10 electrons, with 2 electrons in the K-shell and 8 electrons in the L-shell. Since both the K and L shells are completely filled, Na+ has achieved a stable electronic configuration and is thus an inert gas-like cation.

Q. Calculate the average atomic mass of the bromine atom if it exists in the form of, say, two isotopes: 79.35 Br (49.7%) and 81.35 Br (50.3%).

Answer – To calculate the average atomic mass of bromine, we need to take into account the relative abundance and mass of each of its isotopes. Given that the atomic masses of the two isotopes, 79Br and 81Br, are 78.9183 u and 80.9163 u respectively, we can use the following formula to calculate the average atomic mass:

Average atomic mass = (relative abundance of isotope 1 × mass of isotope 1) + (relative abundance of isotope 2 × mass of isotope 2)

Given that the relative abundance of 79Br is 49.7% and that of 81Br is 50.3%, we can substitute these values into the formula to get:

Average atomic mass = (0.497 × 78.9183 u) + (0.503 × 80.9163 u)
= 39.228 u + 40.683 u
= 79.911 u

Therefore, the average atomic mass of bromine is approximately 79.911 u.

Q. A sample of element X has an average atomic mass of 16.2 u. What proportions of the isotopes 16 8 X and 18 8 X are present in the sample?

Answer – The average atomic mass of the sample is 16.2 u. Let the percentage of the isotope ₈X¹⁶ be ‘a’, and the percentage of the isotope ₈X¹⁸ be ‘100 – a’.

Using the given equation, we get:

(16a/100) + (18(100 – a)/100) = 16.2

Simplifying this equation, we get:

16a + 1800 – 18a = 1620

-2a = -180

a = 90

Therefore, the percentage of the isotope ₈X¹⁶ is 90% and the percentage of the isotope 8X18 is 10%.

Q. What would the element’s valency be if Z = 3? Name the element.

Answer – The given atomic number is 3, and the electronic configuration of the element is K-2 and L-1, which means that the first shell (K-shell) has 2 electrons, and the second shell (L-shell) has 1 electron. Since the outermost shell of the element has 1 electron, its valency is 1.

Therefore, the element with atomic number 3 and electronic configuration K-2; L-1, and a valency of 1 is Lithium.
The atomic number 3 is Lithium.

Q. The two atomic species X and Y’s nuclei’s composition is listed under X Y.
Protons = 6 6
Neutrons = 6 8
Please provide the masses of X and Y. What connection do the two species have to one another?

Answer – Two atomic species X and Y are given with the following composition of their nuclei:

• X has 6 protons and 6 neutrons
• Y has 6 protons and 8 neutrons

The mass numbers of X and Y can be calculated by adding the number of protons and neutrons in each species’ nucleus:

• Mass number of X = 6 protons + 6 neutrons = 12
• Mass number of Y = 6 protons + 8 neutrons = 14

X and Y are isotopes of different elements because they have the same number of protons but a different number of neutrons. Therefore, they have the same atomic number (6) but different mass numbers.

Q. Write T for True and F for False for the following statements.
(A) J.J. Thomson suggested that an atom’s nucleus is made up entirely of nucleons.
(b) The union of an electron and a proton results in the creation of a neutron. It is hence impartial.
(c) An electron has a mass that is around 1 2000 times more than a proton.
(d) Iodine tincture, a type of medicine, is created using an isotope of the element.
In questions, place a checkmark (ü) next to the correct answer and a cross (x) next to the incorrect answer.

Answer –

(a) The statement is False.
(b) The statement is False.
(c) The statement is True.
(d) The statement is False.

Q. The finding of was due to Rutherford’s alpha-particle scattering experiment.
(a) Atomic Nucleus
(b) Electron
(c) Proton
(d) Neutron

Answer – (a) Atomic Nucleus

Q. Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.

Answer – (c) different number of neutrons

Q. Number of valence electrons in Cl– ion are:
(a) 16 (b) 8 (c) 17 (d) 18

Answer – (b) 8

CBSE Skill Education

Atoms and Molecules Class 9 NCERT Solutions – The notes provided on our website for CBSE students are designed to align with CBSE and NCERT syllabus guidelines. By referring to these notes, students can save time and gain a comprehensive understanding of important topics and questions in each chapter. This can help them better retain key concepts and be well-equipped to answer any questions that may appear on exams.

Atoms and Molecules Class 9 NCERT Solutions

Q. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 28.2 g of sodium acetate, 2.2 g of carbon dioxide, and 0.9 g of water. Establish the consistency of these data with the law of mass conservation. sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water.

Answer – The law of conservation of mass states that in any chemical reaction, the total mass of the reactants must be equal to the total mass of the products.

Given:

Mass of sodium carbonate = 5.3 g
Mass of acetic acid = 6 g
Mass of carbon dioxide produced = 2.2 g
Mass of water produced = 0.9 g
Mass of sodium acetate produced = 8.2 g

Total mass of reactants = 5.3 g + 6 g = 11.3 g
Total mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g

Q. Water is created when the mass ratio of hydrogen to oxygen is 1:8. What mass of oxygen gas would be necessary for 3 g of hydrogen gas to totally react?

Answer – According to the given ratio, for every 1 part of hydrogen, 8 parts of oxygen are required to form water.

Therefore, the mass of oxygen required to react completely with 3 g of hydrogen can be calculated as follows:

Mass of oxygen = (8/1) x 3 g of hydrogen
Mass of oxygen = 24 g of oxygen

Therefore, 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Q. Which atomic theory postulate is a direct effect of the law of conservation of mass?

Answer – The concept of conservation of mass leads to the following assumption in Dalton’s atomic theory:

“A chemical process cannot create, destroy, or divide atoms into smaller particles, which are the building blocks of all elements.”

Q. What atomic theory postulate best explains the law of definite proportions?

Answer – The second postulate of Dalton’s atomic theory—according to which atoms of various elements combine in compounds in predetermined ratios—is the only one that can adequately explain the rule of definite proportions. This postulate states that the ratio of atoms from various elements that combine to create a compound is a constant for that particular compound, independent of the origin of the elements, and that this ratio is known as the atomic ratio. In other words, the rule of definite proportions results from the fact that the ratio of the masses of the constituent elements in a compound is always the same.

Q. Define the atomic mass unit.

Answer – The atomic mass unit (amu) is a unit of mass used for expressing atomic and molecular weights. It is defined as exactly one-twelfth of the mass of one atom of the carbon-12 isotope. The mass of all other elements is measured relative to this standard. Therefore, the relative atomic masses of all elements are determined with respect to an atom of carbon-12, which is assigned a mass of exactly 12 amu.

Q. Why can’t we see an atom with our naked eyes?

Answer – It is not possible to see an atom with naked eyes because atoms are extremely small, even compared to the smallest visible objects. The size of an atom is on the order of 10^-10 meters, which is much smaller than the wavelength of visible light.

Q. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide

Answer – The formulae are:

(i) Sodium oxide – Na₂O
(ii) Aluminium chloride – AlCl₃
(iii) Sodium sulphide – Na₂S
(iv) Magnesium hydroxide – Mg(OH)₂

Q. List the names of the compounds that the following equations represent:
(i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃

Answer –

Below are the names of the compounds for each of the following formulae –

(i) Al2(SO4)₃ – Aluminium sulphate
(ii) CaCl₂ – Calcium chloride
(iii) K2SO₄ – Potassium sulphate
(iv) KNO₃ – Potassium nitrate
(v) CaCO₃ – Calcium carbonate

Q. What does the phrase “chemical formula” mean?

Answer – Chemical formula is a symbolic representation of a chemical compound that consists of the chemical symbols of the elements present in the compound, along with their subscripts to indicate the relative proportions of each element. It provides information about the composition of the compound, such as the type and number of atoms of each element in the compound. The chemical formula helps in identifying the compound and predicting its chemical and physical properties.

Q. How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄ ^3– ion?

Answer –

(i) There are 3 atoms in an H2S molecule – 2 atoms of hydrogen and 1 atom of sulfur.
(ii) There are 5 atoms in a PO4 3- ion – 1 atom of phosphorus and 4 atoms of oxygen.

Q. Determine the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Answer – The molecular masses of the given compounds can be calculated using the following equations:

(i) Molecular mass of H₂ = 2 × Atomic mass of H
= 2 × 1
= 2 u

(ii) Molecular mass of O₂ = 2 × Atomic mass of O
= 2 × 16
= 32 u

(iii) Molecular mass of Cl₂ = 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u

(iv) Molecular mass of CO₂ = Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u

(v) Molecular mass of CH₄ = Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u

(vi) Molecular mass of C₂H₆ = 2 × Atomic mass of C + 6 × Atomic mass of H
= 2 × 12 + 6 × 1
= 30 u

(vii) Molecular mass of C₂H₄ = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28 u

(viii) Molecular mass of NH₃ = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3 × 1
= 17 u

(ix) Molecular mass of CH₃OH = Atomic mass of C + 4 × Atomic mass of H + Atomic mass of O + Atomic mass of H
= 12 + 4 × 1 + 16 + 1
= 32 u

Q. Determine the formula unit masses of ZnO, Na₂O and K₂CO₃, Using the atomic weights of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

Answer – The formula unit mass is the sum of the atomic masses of all the atoms in a formula unit of the compound.

The formula of ZnO is ZnO.
Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 u + 16 u
= 81 u

The formula of Na₂O is Na₂O.
Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 u + 16 u
= 62 u

The formula of K₂CO₃ is K₂CO₃.
Formula unit mass of K₂CO₃ = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 u + 12 u + 3 × 16 u
= 138 u

Therefore, the formula unit masses of ZnO, Na₂O, K₂CO₃ are 81 u, 62 u, and 138 u, respectively.

Q. What is the mass of one carbon atom in grammes if a mole of carbon atoms weighs 12 grammes?

Answer – The mass of one mole of carbon atoms is 12 grams, which is also known as the molar mass of carbon.

To find the mass of one atom of carbon, we can use Avogadro’s number, which states that there are 6.022 x 10²³ atoms in one mole of an element.

So, the mass of 1 atom of carbon is given by:

mass of 1 mole of carbon atoms / Avogadro’s number
= 12 g / 6.022 x 10²³ atoms
= 1.99 x 10^-23 g

Therefore, the mass of 1 atom of carbon is 1.99 x 10^-23 grams.

Q. Which contains more atoms, 100 grammes of sodium or 100 grammes of iron (assuming that Na atomic mass is 23 u and Fe is 56 u)?

Answer – To determine which substance has more number of atoms, we need to calculate the number of moles of each substance using their given masses and molar masses, and then compare the number of atoms in each substance.

The molar mass of sodium (Na) is 23 g/mol and the molar mass of iron (Fe) is 56 g/mol.

For 100 grams of sodium:
Number of moles of Na = Mass of Na / Molar mass of Na
Number of moles of Na = 100 g / 23 g/mol
Number of moles of Na = 4.35 moles

Number of atoms of Na = Number of moles of Na × Avogadro’s number
Number of atoms of Na = 4.35 moles × 6.022 × 10²³ atoms/mol
Number of atoms of Na = 2.6193 × 10²⁴ atoms

For 100 grams of iron:
Number of moles of Fe = Mass of Fe / Molar mass of Fe
Number of moles of Fe = 100 g / 56 g/mol
Number of moles of Fe = 1.79 moles

Number of atoms of Fe = Number of moles of Fe × Avogadro’s number
Number of atoms of Fe = 1.79 moles × 6.022 × 10²³ atoms/mol
Number of atoms of Fe = 1.07938 × 10²⁴ atoms

Therefore, 100 grams of iron has more number of atoms than 100 grams of sodium.

Q. Analysis revealed that a 0.24 g sample of an oxygen and boron combination had 0.144 g of oxygen and 0.096 g of boron. Determine the compound’s weight-based % composition.

Answer –

Mass of the compound = 0.24 g
Mass of boron in the compound = 0.096 g
Mass of oxygen in the compound = 0.144 g

To calculate the percentage composition by weight, we need to find the total mass of all the elements in the compound.

Total mass of the compound = Mass of boron + Mass of oxygen = 0.096 g + 0.144 g = 0.24 g

Now we can calculate the percentage composition of each element by weight:

Percentage composition of boron = (mass of boron in the compound / total mass of the compound) x 100%
= (0.096 g / 0.24 g) x 100%
= 40%

Percentage composition of oxygen = (mass of oxygen in the compound / total mass of the compound) x 100%
= (0.144 g / 0.24 g) x 100%
= 60%

As a result, the compound has a weight percentage of 40% boron and 60% oxygen.

Q. When 11.00 g of carbon dioxide is created when 3.0 g of carbon burn in 8.00 g of oxygen. When 3.00 g of carbon is burned in 50.00 g of oxygen, what mass of carbon dioxide results? Which chemical combination law will control your response?

Answer – The law of definite proportions will govern the answer to the question.

To calculate the mass of carbon dioxide formed when 3.00 g of carbon is burnt in 50.00 g of oxygen, we need to first calculate the amount of oxygen required to react with 3.00 g of carbon:

1 mol of carbon reacts with 1 mol of oxygen to form 1 mol of carbon dioxide.

Carbon dioxide has a molar mass of 44 g/mol.
Oxygen has a molar mass of 32 g/mol.

Therefore, 3.00 g of carbon requires (3.00/12) = 0.25 mol of carbon.

This requires 0.25 mol of oxygen.
0.25 mol of oxygen is equal to (0.25 x 32) = 8.0 g of oxygen.
Since we have 50.00 g of oxygen, it is in excess and all of the carbon will react with 8.0 g of the oxygen.
So, 1 mol of carbon reacts with 1 mol of oxygen to form 1 mol of carbon dioxide.

Therefore, 0.25 mol of carbon will react with 0.25 mol of oxygen to form 0.25 mol of carbon dioxide.

The mass of 0.25 mol of carbon dioxide is (0.25 x 44) = 11.0 g.
Therefore, when 3.00 g of carbon is burnt in 50.00 g of oxygen, 11.0 g of carbon dioxide will be formed.

Q. What are polyatomic ions? Provide examples.

Answer – Ions made up of two or more atoms that are covalently bound together and have a total electrical charge are referred to as polyatomic ions. These ions play a significant role in many biochemical processes and are frequently used in chemical reactions.

Here are some instances of polyatomic ions:

Ammonium ion (NH4+)
Nitrate ion (NO3-)
Sulfate ion (SO42-)
Carbonate ion (CO32-)
Hydroxide ion (OH-)
Phosphate ion (PO43-)

Q. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

Answer –

(a) MgCl₂
(b) CaO
(c) Cu(NO₃)₂
(d) AlCl₃
(e) CaCO₃

Q. Name the elements that make up the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

Answer –

(a) Quick lime – Calcium and Oxygen
(b) Hydrogen bromide – Hydrogen and Bromine
(c) Baking powder – Sodium, Hydrogen, Carbon, and Oxygen
(d) Potassium sulphate – Potassium, Sulphur, and Oxygen

Q. The following compounds’ molar masses should be determined.
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃

Answer –

(a) Molar mass of C2H2 = 2(atomic mass of C) + 2(atomic mass of H) = 2(12.01) + 2(1.01) = 26.04 g/mol
(b) Molar mass of S8 = 8(atomic mass of S) = 8(32.06) = 256.48 g/mol
(c) Molar mass of P4 = 4(atomic mass of P) = 4(31.00) = 124.00 g/mol
(d) Molar mass of HCl = atomic mass of H + atomic mass of Cl = 1.01 + 35.45 = 36.46 g/mol
(e) Molar mass of HNO3 = atomic mass of H + atomic mass of N + 3(atomic mass of O) = 1.01 + 14.01 + 3(16.00) = 63.01 g/mol

Q. What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na₂SO₃)?

Answer –

(a) The atomic mass of nitrogen (N) is 14 u. Therefore, the mass of 1 mole of nitrogen atoms will be 14 g/mol.

(b) The atomic mass of aluminium (Al) is 27 u. Therefore, the mass of 1 mole of aluminium atoms will be 27 g/mol. So, the mass of 4 moles of aluminium atoms will be:

mass = 4 moles × 27 g/mol mass = 108 g

(c) The formula mass of sodium sulphite (Na2SO3) can be calculated as follows:

Na2SO3 = (2 × atomic mass of Na) + atomic mass of S + (3 × atomic mass of O) Na2SO3 = (2 × 23) + 32 + (3 × 16) Na2SO3 = 46 + 32 + 48 Na2SO3 = 126 u/mol

Therefore, the mass of 10 moles of sodium sulphite (Na2SO3) will be:

mass = 10 moles × 126 g/mol mass = 1260 g

Q. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.

Answer –

(a) The molar mass of oxygen is 32 g/mol.
Therefore, the number of moles in 12 g of oxygen gas can be calculated as follows:
Number of moles = Mass / Molar mass = 12 g / 32 g/mol = 0.375 mol
Therefore, 12 g of oxygen gas is equal to 0.375 mole.

(b) The molar mass of water is 18 g/mol.
Therefore, the number of moles in 20 g of water can be calculated as follows:
Number of moles = Mass / Molar mass = 20 g / 18 g/mol = 1.11 mol
Therefore, 20 g of water is equal to 1.11 moles.

(c) The molar mass of carbon dioxide is 44 g/mol.
Therefore, the number of moles in 22 g of carbon dioxide can be calculated as follows:
Number of moles = Mass / Molar mass = 22 g / 44 g/mol = 0.5 mol
Therefore, 22 g of carbon dioxide is equal to 0.5 mole.

Q. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Answer –

(a) The molar mass of oxygen is 16 g/mol. Therefore, the mass of 0.2 mole of oxygen atoms is:
0.2 mol x 16 g/mol = 3.2 g
Hence, the mass of 0.2 mole of oxygen atoms is 3.2 g.

(b) The molar mass of water (H2O) is 18 g/mol. Therefore, the mass of 0.5 mole of water molecules is:
0.5 mol x 18 g/mol = 9 g
Hence, the mass of 0.5 mole of water molecules is 9 g.

Q. Determine how many sulphur (S₈) molecules are contained in 16 g of solid sulphur.

Answer –

The molar mass of S8 is:

Molar mass of S8 = 8 x Atomic mass of S = 8 x 32 g/mol = 256 g/mol
Now, we can calculate the number of moles of S8 present in 16 g of solid sulphur using the formula:
Number of moles = Mass ÷ Molar mass
Number of moles of S8 = 16 g ÷ 256 g/mol = 0.0625 mol
One mole of S8 contains Avogadro’s number (NA) of molecules. Therefore, the number of molecules of S8 present in 0.0625 mol of S8 is:
Number of molecules of S8 = 0.0625 mol x NA

= 0.0625 mol x 6.022 x 10²³/mol
= 3.765 x 10²² molecules

Therefore, there are approximately 3.765 x 10²² molecules of S8 present in 16 g of solid sulphur.

Q. Determine how many aluminium ions are contained in 0.051 g of aluminium oxide.
(Hint: An ion’s mass is equal to an atom’s of the same element. Al has an atomic mass of 27 u.

Answer –

The formula of aluminium oxide is Al₂O₃.

The molar mass of Al₂O₃ = (2 × atomic mass of Al) + (3 × atomic mass of O)
= (2 × 27 u) + (3 × 16 u) = 102 u

So, 1 mole of Al₂O₃ has a mass of 102 g and contains 6.022 × 10²³ formula units (Avogadro’s number).

Therefore, the number of formula units in 0.051 g of Al₂O₃ = (0.051 g) / (102 g/mol) × (6.022 × 10²³ formula units/mol) = 3.01 × 10²⁰ formula units.

Since there are 2 Al atoms in each formula unit of Al₂O₃, the number of aluminium ions present in 0.051 g of Al₂O₃ = 2 × (3.01 × 10²⁰) = 6.02 × 10²⁰ aluminium ions.

CBSE Skill Education

For 9th grade students, Science Class 9 NCERT Solution play a significant role in their academic journey. These solutions contain complete answers to the in-text and chapter-end questions in the Science textbook, making it easier for students to comprehend the concepts and apply them to the questions. The Science textbook covers three primary branches of science: Physics, Chemistry, and Biology, and the NCERT Solutions cover all the chapters of these subjects.

The language used in the Science Class 9 NCERT Solution is simple and easy-to-understand, presented in a step-by-step manner to help students grasp the concepts logically. By referring to these solutions, students can identify their weak areas and revise the concepts regularly to perform better in their examinations.

The NCERT Solutions for Class 9 Science help students prepare effectively for their examinations by providing them with a comprehensive set of solutions. These solutions serve as a vital resource for students who may find it difficult to locate the answers to the questions on their own. Students can use these solutions to build a solid foundation in Science and achieve academic success.

Science Class 9 NCERT Solution

• Chapter 1 Matter in Our Surroundings
• Chapter 2 Is Matter Around Us Pure
• Chapter 3 Atoms and Molecules
• Chapter 4 Structure of the Atom
• Chapter 5 The Fundamental Unit of Life
• Chapter 6 Tissues
• Chapter 7 Diversity in Living Organisms
• Chapter 8 Motion
• Chapter 9 Force and Laws of Motion
• Chapter 10 Gravitation
• Chapter 11 Work, Power And Energy
• Chapter 12 Sound
• Chapter 13 Why Do we Fall Ill
• Chapter 14 Natural Resources
• Chapter 15 Improvement in Food Resources

Science Class 9 Important Points Chapter Wise

Chapter 1: Matter in Our Surrounding

The first chapter of the Class 9 Science textbook, entitled Matter in Our Surroundings, exposes students to the idea of matter, as well as to its various states and physical characteristics. These are some key topics this chapter covered:

1. Matter is anything that has mass and occupies space.
2. It is composed of atoms and molecules, which are little particles.
3. Solid, liquid, and gas are the three distinct states of matter. The arrangement of a substance’s particles determines its state.
4. Shape, volume, density, and compressibility are some of matter’s properties.
5. The variation in the intermolecular forces between the constituents of matter contributes to state changes. Melting, freezing, evaporation, condensation, and sublimation are a few of the several forms of transformations.
6. Surface area, temperature, humidity, and wind speed are just a few of the variables that affect how quickly water evaporates.
7. Air pressure affects a liquid’s boiling point. Because of this, water boils more slowly at high altitudes.
8. When a solid transforms into a gas without first becoming liquid, this is known as sublimation.
9. This chapter introduces the idea of the Kinetic Theory of Matter, which describes how matter behaves in terms of its constituent particles.

Chapter 2: Is Matter Around Us Pure?

1. A pure substance is made up of only one type of particles with definite properties.
2. A mixture contains two or more pure substances mixed in varying proportions.
3. Mixtures can be homogeneous or heterogeneous.
4. Homogeneous mixtures have uniform composition and appearance, while heterogeneous mixtures have non-uniform composition and appearance.
5. Homogeneous mixtures of two or more components are called solutions.
6. The concentration of a solution is measured by the amount of solute dissolved in a given amount of solvent.
7. Solids in a mixture can be separated from liquids using filtration or centrifugation.
8. Chromatography is a technique used to separate the components of a mixture based on their solubility in a particular solvent.
9. Distillation is a technique used to separate the components of a mixture based on their boiling points.
10. Crystallisation is a technique used to obtain pure solids from their impure solutions by cooling the solution slowly.
11. Impurities can be detected by using various methods like melting point determination, boiling point determination, and chromatography.
12. Different methods are used to purify water like boiling, filtration, chlorination, etc.
13. Common techniques used for purification of metals are liquation, distillation, electrolysis, and zone refining.

Chapter 3: Atoms and Molecules

1. Atoms and molecules that are too small for the human eye to see make up matter.
2. As they cannot be formed, destroyed, or further divided, atoms are the fundamental building units of matter.
3. According to Dalton’s atomic theory, molecules are created when atoms join in a particular ratio to form compounds.
4. An element’s molecules are identical, and they can be combined to create a compound’s molecules.
5. The total atomic masses of every atom in a molecule make up the substance’s molecular mass.
6. According to Avogadro’s law, equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
7. The quantity and types of atoms that make up a molecule are represented by the chemical formula of a compound.
8. An element’s valency is determined by how many electrons it must receive or lose in order to establish a stable configuration.
9. Whereas covalent compounds are created when atoms share electrons, ionic compounds are created when electrons are transferred from one atom to another.
10. Atoms of the same element that have variable numbers of neutrons but the same number of protons are called isotopes.
11. The mass number is the total amount of protons and neutrons in an atom, whereas an element’s atomic number is the quantity of protons in its nucleus.
12. The number of particles in 12 grammes of carbon-12, which is used as a unit of measurement for substance amounts, is equal to one mole.

Chapter 4: Structure of the Atom

1. The concept of an atom was first proposed by John Dalton in the early 19th century.
2. Protons, neutrons, and electrons are the three primary subatomic particles that make up an atom.
3. Protons are positively charged particles present in the nucleus of an atom, while neutrons are neutral particles also present in the nucleus.
4. Electrons are negatively charged particles that revolve around the nucleus in shells or energy levels.
5. The atomic number of an element is the number of protons present in its nucleus.
6. The mass number of an atom is the sum of the number of protons and neutrons present in its nucleus.
7. The electrons in an atom are arranged in different shells or energy levels, with each shell having a specific energy level.
8. The maximum number of electrons that can be accommodated in a shell is given by 2n^2, where n is the number of the shell.
9. The quantity of electrons in an atom’s outermost shell is known as an element’s valency.
10. The octet rule states that elements tend to gain, lose, or share electrons to attain a stable configuration of 8 electrons in their outermost shell.
11. Atoms of the same element that have variable numbers of neutrons but the same number of protons are called isotopes.
12. The discovery of isotopes led to the revision of the atomic mass of some elements, which is the weighted average of the masses of all the naturally occurring isotopes of an element.

Chapter 5: The Fundamental Unit of Life

1. The smallest structural and functional units of life, cells are the building blocks of all living things.
2. Robert Hooke used a simple microscope to make the first observation of cells in 1665. These reminded him of the little apartments in a monastery, therefore the term “cells,” he gave them.
3. The first person to view living cells with a microscope was a Dutch scientist named Anton van Leeuwenhoek.
4. Prokaryotic and eukaryotic cells can be divided into two groups. Eukaryotic cells are complex and have nuclei, whereas prokaryotic cells are simple and lack a nucleus.
5. The cell is surrounded by a thin, pliable membrane called the cell membrane, which isolates the interior of the cell from the external environment.
6. Several organelles are found in the cytoplasm, a jelly-like fluid that fills the cell.
7. The most significant organelle in eukaryotic cells is the nucleus. DNA, which carries genetic data and regulates a cell’s actions, is present in it.
8. The nucleolus, a tiny, compact structure within the nucleus, is responsible for producing ribosomes.
9. Protein synthesis is carried out by microscopic organelles called ribosomes.
10. Cells manufacture energy through a process called respiration, which is carried out by mitochondria, which are organelles.
11. The organelles known as chloroplasts are located in plant cells and are in charge of photosynthesis, which is how plants produce their own food.
12. Moreover, plant cells have a stiff cell wall formed of cellulose that serves as the cell’s structural support.
13. Both asexual and sexual reproduction are possible in cells. Sexual reproduction involves two parents and produces kids with a combination of genetic features from both parents as opposed to asexual reproduction, which only involves one parent and produces identical offspring.

Chapter 6: Tissues

1. Tissue is a group of similar cells that perform a particular function in an organism.
2. Tissues are of four types: epithelial, connective, muscular, and nervous.
3. Epithelial tissue covers the surface of the body and lines the internal organs and cavities.
4. Connective tissue provides support, protection, and framework to the organs and tissues.
5. Muscular tissue helps in movement and locomotion.
6. Nervous tissue is responsible for communication and coordination between various body parts.
7. Meristem is a type of tissue that helps in the growth of the plant.
8. Permanent tissues are the ones that have attained maturity and cannot divide further.
9. Parenchyma, collenchyma, and sclerenchyma are examples of plant tissues.
10. The study of tissues is called histology.

Chapter 7: Diversity in Living Organism

1. Diversity in Living Organisms is the study of the vast variety of living organisms that exist on our planet.
2. The classification of living organisms is based on their structural and functional characteristics.
3. The two major classification systems are the traditional classification system and the modern classification system.
4. Carolus Linnaeus is known as the father of taxonomy and is credited with developing the traditional classification system.
5. The modern classification system is based on the evolutionary relationships among living organisms and is called the phylogenetic classification system.
6. The five kingdoms of living organisms are Monera, Protista, Fungi, Plantae and Animalia.
7. Viruses are not considered living organisms as they cannot reproduce on their own and require a host cell to do so.
8. Bacteria are unicellular organisms and are classified under the kingdom Monera.
9. Protists are mostly unicellular eukaryotic organisms and are classified under the kingdom Protista.
10. Fungi are heterotrophic organisms that feed on dead and decaying matter and are classified under the kingdom Fungi.
11. Plants are multicellular eukaryotic organisms that have chlorophyll and perform photosynthesis. They are classified under the kingdom Plantae.
12. Animals are multicellular eukaryotic organisms that are heterotrophic and have specialized tissues and organs. They are classified under the kingdom Animalia.
13. The classification of organisms helps in the identification and study of different organisms and their relationships with each other.
14. The classification system is important in conservation efforts as it helps in identifying endangered species and their habitats.

Chapter 8: Motion

1. Motion refers to the change in position of an object with respect to its surroundings.
2. The distance travelled by an object in a given time is known as speed, while the rate of change of displacement is known as velocity.
3. The rate at which velocity changes in relation to time is referred to as acceleration.
4. The three equations of motion relate to the displacement, initial velocity, final velocity, time taken and acceleration of an object.
5. Uniform motion is when an object travels with a constant speed while non-uniform motion is when the object travels with a varying speed.
6. Graphical representation of motion can be made using distance-time and velocity-time graphs.
7. Galileo’s experiment demonstrated that objects of different masses fall at the same rate in the absence of air resistance, whereas air resistance can greatly affect the motion of objects.
8. Newton’s laws of motion state that an object at rest will remain at rest and an object in motion will continue to move with uniform velocity unless acted upon by an external force.
9. The second law of motion states that the force acting on an object is directly proportional to the rate of change of its momentum.
10. Every action has an equal and opposite reaction, according to the third rule of motion.
11. Friction is a force that opposes the relative motion between two surfaces in contact and can be both beneficial and harmful.
12. Air resistance is a form of friction that opposes the motion of objects through air.
13. The force of gravity is a universal force of attraction between any two objects with mass, and its strength is determined by the mass and distance between the objects.

Chapter 9: Force and Law of Motion

1. Force is a physical quantity that changes or tends to change the state of motion of an object.
2. The SI unit of force is Newton (N), and it is measured using a device called a spring balance.
3. Unless acted upon by an outside force, an object will continue to be in its condition of rest or uniform motion in a straight line, according to Newton’s first law of motion.
4. The first law is also called the law of inertia. An object’s ability to resist changes in its state of motion is known as inertia.
5. The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, according to Newton’s second rule of motion.
6. Mathematically, F = ma, where F is the net force, m is the mass of the object, and a is its acceleration.
7. The second law is also known as the law of acceleration.
8. The acceleration is moving in the same general direction as the net force.
9. With every action, there is an equal and opposite response, states Newton’s third rule of motion.
10. The third law explains why two objects in contact always experience equal and opposite forces.
11. The law of action and reaction is another name for the third law.
12. The force that prevents motion between two surfaces in contact is known as the force of friction.
13. Friction can be static or kinetic, depending on whether the surfaces are at rest or in motion.
14. Air resistance is a type of friction that opposes the motion of objects through the air.
15. The force of gravity is the force of attraction between two objects with mass.
16. Gravity is the reason why objects fall to the ground and why planets orbit around the sun.

Chapter 10: Gravitation

1. The pull between any two objects in the universe is known as gravity.
2. The force of gravity is inversely proportional to the square of the distance between objects and directly proportional to their mass.
3. Sir Issac Newton made the initial discovery of the gravitational law in 1687.
4. The gravitational force is measured with a tool called a spring balance and has the SI unit of Newton (N).
5. The acceleration a body experiences as a result of gravity is known as the acceleration due to gravity.
6. Near the surface of the Earth, the acceleration caused by gravity is 9.8 m/s2.
7. The gravitational pull on an object is measured by its weight.
8. An object’s weight varies depending on its location and gravitational field intensity.
9. The amount of matter that makes up an item is its mass, which is constant regardless of location or gravitational field intensity.
10. Due to the moon’s weaker gravitational field than the Earth’s, an object’s weight on the moon is one-sixth that of the object’s weight on the Earth.
11. Astronauts in space feel as though they are weightless since there is no external force supporting them.
12. The gravitational attraction between the planets and the sun causes their orbital motion.
13. On Earth, tides are caused by the gravitational pull of the earth.
14. The orbital motion of satellites around the Earth is also impacted by gravitational pull.
15. The minimal velocity necessary for an object to escape the gravitational field of a planet or other celestial body is known as the escape velocity.
16. The planet’s mass and radius affect the escape velocity.
17. By putting more space between two things, the gravitational attraction between them can be weakened.

Chapter 11: Work and Energy

1. The product of the force acting on an object and the distance it travels while being moved by the force is known as work.
2. Joule is the SI unit of work .
3. Work might be zero, negative, or in between.
4. When the force being applied and the motion are moving in the same direction, positive work is being done.
5. When the force being applied and the motion’s direction are in opposition, negative work is being done.
6. Zero work is done when the force applied and the direction of motion are perpendicular to each other.
7. Power is the rate at which work is done, and it is defined as the amount of work done per unit time.
8. The SI unit of power is Watt (W).
9. Energy is the ability of an object to do work.
10. Energy can be classified into two types: kinetic energy and potential energy.
11. The energy an object possesses as a result of its motion is known as kinetic energy.
12. The formula for kinetic energy is KE = 1/2mv², where m is the mass of the object and v is its velocity.
13. Potential energy is the energy possessed by an object due to its position or state.
14. The equation for potential energy is PE = mgh, where m is the object’s mass, g is its gravitational acceleration, and h is its height above the ground.
15. Energy cannot be created or destroyed; it can only be changed from one form to another, according to the rule of conservation of energy.
16. The total energy of a system remains constant unless an external force acts on it.
17. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
18. The principle of conservation of mechanical energy states that the total mechanical energy of a system remains constant if only conservative forces act on it.
19. Conservative forces are those forces that do not dissipate energy as heat.
20. Non-conservative forces are those forces that dissipate energy as heat, such as friction.
21. The efficiency of a machine is the ratio of output work to input work, and it is always less than 1.

Chapter 12: Sound

1. Energy in the form of sound causes our ears to experience hearing.
2. Sound waves travel in a straight line through a medium.
3. The speed of sound waves in air is approximately 340 m/s.
4. The frequency of a sound wave is the number of complete vibrations it makes per second, and it is measured in Hertz (Hz).
5. The amplitude of a sound wave is the maximum displacement of the particles of the medium from their equilibrium position.
6. A sound wave’s amplitude determines how loud it is.
7. A sound’s frequency and pitch are correlated.
8. The range of frequencies that the human ear can hear is from 20 Hz to 20,000 Hz.
9. Different animals have various audible frequency ranges.
10. A sound wave’s strength is measured in decibels and is the amount of energy it contains per unit of space (dB).

Chapter 13: Why Do We Fall Ill?

1. Health is a complete state of physical, mental, and social well-being.
2. Disease is a state of an organism’s reduced functioning for a variety of causes.
3. Pathogens are organisms that cause disease, including bacteria, viruses, fungus, and protozoa.
4. Infectious and non-infectious diseases can be categorised.
5. Pathogens that are contagious from person to person are what cause infectious diseases.
6. Non-infectious diseases are brought on by things like heredity, way of living, and environmental factors.
7. Air, water, food, direct contact, and indirect contact are all possible ways for infectious diseases to spread.
8. Immunity is an organism’s capacity to fend against illnesses and infections.
9. Natural or artificial means can be used to develop immunity.
10. Via pathogen exposure, immunisation, and antibody transfer from the mother, one can develop natural immunity.
11. Proteins called antibodies are created by the immune system to combat particular infections.
12. Disease prevention, diagnosis, and treatment are all included in health care.
13. The transmission of infections can be stopped by using good personal hygiene and sanitation techniques like handwashing, clean water, and proper waste disposal.
14. A strong immune system can be maintained with the right diet, exercise, and rest.
15. Mental health is an important aspect of overall health, and it can be affected by factors such as stress, trauma, and mental disorders.

Chapter 14: Natural Resources

1. Resources that can be used by humans are known as natural resources.
2. Renewable and non-renewable resources can be categorised into two groups.
3. Resources like water, wind, and solar energy are examples of renewable resources since they can be regenerated naturally.
4. Minerals and fossil fuels are examples of non-renewable resources because they are limited and cannot be regenerated after usage.
5. The creation of oxygen, the storage of carbon dioxide, the production of timber, and the provision of wildlife habitat are just a few of the numerous ecological, economic, and social advantages that forests offer.

Chapter 15: Improvement in Food Resources

1. Growing crops and raising animals for food and other things is called agriculture.
2. Crop production, which entails the cultivation of plants for food, fibre, and other goods, is impacted by elements like climate, soil, water availability, and pest control.
3. The technique of irrigation involves providing water to crops, and it can boost crop production in places with little rainfall.
4. Fertilizers are compounds that give plants vital nutrients and can boost agricultural yields and soil fertility.
5. Pesticides are chemicals that are used to manage pests including weeds, fungi, and insects. They can also shield crops from harm.
6. Crop rotation, integrated pest control, and organic farming are examples of activities that fall under the category of sustainable agriculture. These strategies strive to increase crop yields while reducing their detrimental effects on the environment.
7. Breeding, feeding, and disease control are all part of the practise of raising animals for food, fibre, and other goods, which is known as animal husbandry.
8. A significant source of nourishment and income for many people, dairy farming entails the production of milk and other dairy products.
9. Poultry farming is a significant source of protein and entails raising chickens, ducks, and other birds for their meat and eggs.
10. Fisheries are significant sources of protein and revenue because they entail the raising and harvesting of fish and other aquatic species.

Frequently Asked Questions

Q. What are NCERT Solutions for Class 9 Science?

NCERT Solutions for Class 9 Science are a set of comprehensive answers to the in-text and chapter-end questions in the Science textbook for students studying in the 9th grade.

Q. Why are NCERT Solutions for Class 9 Science important?

In order to help students understand the ideas and apply them to the questions, NCERT Answers for Class 9 Science are crucial since they offer comprehensive answers to the questions in the Science textbook. With the aid of these solutions, students can pinpoint their areas of weakness and successfully get ready for exams.

Q. What topics are covered in NCERT Solutions for Class 9 Science?

NCERT Solutions for Class 9 Science cover three primary branches of science: Physics, Chemistry, and Biology. The solutions cover all the chapters of these subjects in the Science textbook.

Q. Are NCERT Solutions for Class 9 Science written in an easy-to-understand language?

Truly, the language used in the NCERT Answers for Class 9 Science is straightforward and clear, making it simpler for pupils to understand the concepts and apply them to the questions.

Q. Can students use NCERT Solutions for Class 9 Science to revise for their examinations?

Yes, students can use NCERT Solutions for Class 9 Science to revise the concepts regularly and retain the information, which will help them perform better in their examinations.

Q. Are NCERT Solutions for Class 9 Science free?

Yes, you can easily find and get the NCERT Answers for Class 9 Science online for free.

Matter In Our Surroundings NCERT Solutions – Here are simplified solutions for Class 9 Science Chapter 1, “Matter in Our Surroundings.” This chapter covers various topics related to matter, its properties, and changes it undergoes. The solutions provide easy-to-understand explanations, step-by-step guidance, and relevant images to help students comprehend the concepts better. These solutions are designed for students who use the NCERT textbook to study science.

Matter In Our Surroundings NCERT Solutions

Q. Which of the following are matter?
Chair, air, love, hate, smell, thought, almonds, cold-drink, cold, smell of perfume.

Answer – The matter is anything that has mass and takes up space. Based on this definition, the following are matters:

• Chair: It has mass and takes up space.
• Air: It has mass and takes up space.
• Almonds: They have mass and take up space.
• Cold-drink: It has mass and takes up space.
• Smell of perfume: It involves particles moving through the air, so it has mass and takes up space.

Q. Give reasons for the following observation:
The smell of hot, sizzling food may be several meters away, But, you must come close to smell cold food.

Answer – Hot food molecules have more kinetic energy than cold food molecules, so they evaporate faster and spread over a larger area, reaching you several meters away. Cold food molecules have less kinetic energy, so they evaporate slower and don’t spread as far, requiring you to be closer to smell them.

 Q. A diver can easily cut through the water in a pool. Which property of matter observation shows?

Answer – The ease with which the diver is able to cut through the water in the swimming pool is due to the low intermolecular forces of attraction between water molecules. This property of water allows it to flow easily and is responsible for the diver’s ability to move through it with minimal resistance.

Q. What distinguishing features do matter particles have?

Answer – Characteristics of the particles of matter are –

1. We refer to the space between matter particles as intermolecular space. Depending on the condition of the matter, this space can differ, with solids having the least and gases having the greatest.
2. Intermolecular forces of attraction also affect the behaviour and characteristics of matter particles. Depending on the type of particle and the distance between them, these forces might be weak or powerful.
3. Matter particles always move because they have kinetic energy. The temperature is one factor that affects this mobility randomly, with higher temperatures causing more movement of the particles.

Q. Density is the measure of a substance’s mass per unit volume.
(density = mass/volume).
Arrange the following in increasing order: air, honey, exhaust from chimneys, chalk, water, cotton and iron.

Answer – Increasing density

Air < Cotton < Exhaust from chimneys < Water < Honey < Chalk < Iron

Q.

(a) Tabulate the variations in the properties of states of matter.
(b) Comment upon the following: rigidity, compressibility, fluidity, , kinetic energy filling a gas container, shape and density.

Answer –

(a)

(b)

• Rigidity: The ability of a substance to resist deformation or movement. Solids have high rigidity.
• Compressibility: The ability of a substance to be compressed or squeezed. Gases have high compressibility.
• Fluidity: The ability of a substance to flow or take the shape of the container it is in. Liquids and gases have fluidity.
• Filling a gas container: Gases will completely fill any container they are placed in, regardless of the shape or size of the container.
• Shape: Solids have a definite shape, while liquids and gases take the shape of their container.
• Kinetic energy: The energy that particles possess due to their motion. The kinetic energy of particles increases as temperature increases.
• Density: The mass per unit volume of a substance. Solids are generally the densest, followed by liquids, and then gases.

Q. Give reasons
(a) A gas fills completely the container in which it is contained.
(b) A gas experience pressure on the walls of the container.
(c) A wooden table need to be called a solid.
(d) We can easily move our hand in air but a karate master is required to move a hand through a solid piece of wood.

Answer –

(a) A gas fills completely the vessel in which it is kept because gases have weak intermolecular forces and the particles move randomly in all directions. Therefore, they tend to fill the available space completely.

(b) A gas exerts pressure on the walls of the container because gas particles are in constant random motion, and they collide with the walls of the container, which results in pressure on the walls.

(c) A wooden table should be called a solid because it has a definite shape and volume due to strong intermolecular forces between its particles. The molecules of a solid are closely packed and have minimum intermolecular spaces.

(d) We can easily move our hand in air, but to do the same through a solid block of wood, we need a karate expert because solids have strong intermolecular forces, and the molecules are closely packed. It requires a significant amount of force to break the intermolecular forces and move the particles apart. A karate expert has trained their muscles to generate a large amount of force, which helps them to break the intermolecular forces and move their hand through the solid block of wood.

Q. Liquids generally have less density as compared to solid. But ice floats on water. Find out why.

Answer – Due of its lower density than liquid water, ice floats on water. This is because ice has a special structural makeup. When the molecules are closely clustered, a solid has a smaller volume and a higher density. Ice, on the other hand, has molecules that arrange themselves into a crystal lattice structure with empty spaces in between. As a result, the ice’s density decreases and its overall volume increases. Ice floats on liquid water because liquid water is denser than ice. This characteristic of ice is important because it enables the preservation of aquatic life over the winter.

Q. Convert temperature to Celsius scale:
(a) 300 K (b) 573 K

Answer –

(a) To convert 300 K to Celsius scale, we need to subtract 273.15 from it.

Therefore, 300 K – 273.15 = 26.85 °C

(b) To convert 573 K to Celsius scale, we need to subtract 273.15 from it.

Therefore, 573 K – 273.15 = 299.85 °C

Q. What physical state is water in at the following:
(a) 250°C (b) 100°C

Answer – (a) 250°C = gas (b) 100°C liquid as well as gas

Q. Why does the temperature stay the same during the change of state for any substance?

Answer – During a change of state, such as from a solid to a liquid or from a liquid to a gas, the temperature of a substance remains constant because the energy being added or removed is used to overcome the intermolecular forces that hold the substance in its current state, rather than increasing or decreasing the kinetic energy of the molecules.

Q. Suggest a technique for liquefying atmospheric gases?

Answer – One common method to liquefy atmospheric gases is through a process called cryogenic distillation, which involves cooling the gases to very low temperatures and then separating them based on their boiling points. The steps involved in this process are as follows:

1. Compress the atmospheric gases, such as nitrogen, oxygen, and argon, to increase their pressure and make them easier to liquefy.
2. Cool the compressed gases using a refrigeration system, such as a Joule-Thomson expansion, to reduce their temperature to below their boiling points.
3. Separate the cooled gases based on their boiling points using a distillation column. The column is typically filled with packing material to increase the surface area for heat exchange and promote separation of the different components. The lower boiling gases will evaporate first, followed by the higher boiling gases, with each component being collected separately as a liquid.
4. Further cool and purify the separated gases to remove any remaining impurities and improve their quality as commercial products.

Q. Why does a desert cooler cool more effectively in the desert on a hot, dry day?

Answer – A desert cooler works best in hot and dry climates because the rate of evaporation is directly proportional to the dryness of the air. In a hot and dry environment, the air has low humidity, which means that it has a high capacity to absorb moisture. When the dry air is passed through the wet cooling media of the cooler, the water molecules readily evaporate into the air, resulting in a greater cooling effect.

Q. How can water cool in a matka (earthen pot) throughout the summer?

Answer – Water kept in an earthen pot, also known as a matka, becomes cool during summer through a process called evaporative cooling.

The earthen pot is made of porous clay that allows air and water to pass through its walls. When water is stored in the pot, it seeps through the tiny pores and wets the surface of the pot. As the water evaporates from the surface of the pot due to the surrounding hot and dry air, it takes away heat from the water inside the pot, resulting in a cooling effect.

Q. Why does applying acetone, gasoline, or perfume to our palm make us feel cold?

Answer – Our palm feels cold when we put some acetone or petrol or perfume on it due to a process called evaporative cooling.

When a volatile liquid, such as acetone, petrol or perfume, is applied to the skin, it quickly evaporates into the surrounding air. During this process, the liquid takes away heat from the skin surface to overcome the intermolecular forces and enter the gaseous state. This transfer of heat from the skin surface to the evaporating liquid results in a cooling effect on the skin.

Q. Why can we drink hot tea or milk from a saucer more quickly than a cup?

Answer – Due to the increased surface area of the saucer, which speeds up the rate of heat transmission from the liquid to the surrounding air, it is possible to drink hot tea or milk more quickly from a saucer than from a cup.

Q. What kind of clothing is appropriate for summer?

Answer – In the summer, it’s crucial to dress in light, breathable clothing that will keep you cool and comfortable in the hot, muggy weather. Consider the following possibilities for clothing: garments with a loose fit, light-colored attire skirts and shorts, as well as breathable materials.

Q. The following temperatures should be converted to Celsius.
(a) 293 K (b) 470 K.

Answer – To convert temperatures from the Kelvin (K) scale to the Celsius (°C) scale, we need to subtract 273.15 from the given temperature.

(a) 293 K = 293 – 273.15 = 19.85 °C

Therefore, 293 K is equivalent to 19.85 °C.

(b) 470 K = 470 – 273.15 = 196.85 °C

Therefore, 470 K is equivalent to 196.85 °C.

Q. The following temperatures should be converted to Kelvin scale.
(a) 25°C (b) 373°C.

Answer – To convert temperatures from the Celsius (°C) scale to the Kelvin (K) scale, we need to add 273.15 to the given temperature.

(a) 25°C = 25 + 273.15 = 298.15 K

Therefore, 25°C is equivalent to 298.15 K.

(b) 373°C = 373 + 273.15 = 646.15 K

Therefore, 373°C is equivalent to 646.15 K.

Q. Give reason for the following observations.
(a) Naphthalene balls vanish over time without leaving any solid.
(b) We can get the smell of perfume from several metres away.

Answer –

(a) Naphthalene balls disappear with time without leaving any solid because of a process called sublimation. Naphthalene balls are made of a solid chemical compound called naphthalene, which can change directly from a solid to a gas without going through the liquid phase. This means that when the naphthalene balls are exposed to air, they slowly undergo sublimation, turning from a solid directly into a gas without leaving any solid residue behind.

(b) We can get the smell of perfume sitting several metres away because perfume molecules are highly volatile, meaning they evaporate easily into the air. When we apply perfume to our skin or clothing, the perfume molecules begin to evaporate and spread out into the surrounding air. As these molecules travel through the air, they can be carried by air currents over a distance, which is why we can often smell perfume even if the person wearing it is several metres away. Additionally, our sense of smell is highly sensitive, so even a small amount of perfume molecules in the air can be detected by our noses.

Q. Arrange the following substances in ascending order of the forces of attraction between their particles—water, sugar, oxygen.

Answer – Oxygen —> water —> sugar.

Q. How does water physically behave at—
(a) 25°C (bj 0°C (cj 100°C

Answer – The physical state of water at different temperatures is as follows:

(a) At 25°C, water is in a liquid state.

(b) At 0°C, water freezes and turns into a solid state, i.e., ice.

(c) At 100°C, water boils and turns into a gaseous state, i.e., steam or water vapor.

Q. Give two reasons to justify
(a) water at room temperature is a liquid.
(b) at room temperature, an iron almirah is solid.

Answer –

(a) Two reasons to justify why water at room temperature is a liquid are:

1. At room temperature, the average kinetic energy of water molecules is not enough to break the intermolecular bonds between them completely, and they remain closely packed but in a mobile state. As a result, water molecules can flow and take the shape of the container they are placed in, making it a liquid.
2. The boiling point of water is 100°C and the freezing point of water is 0°C at standard atmospheric pressure. Since room temperature is typically around 25°C, it is well within the liquid phase range of water.

(b) Two reasons to justify why an iron almirah is a solid at room temperature are:

1. Iron atoms in the almirah are closely packed together due to strong interatomic forces, making it difficult for them to move freely. As a result, the almirah maintains its shape and does not flow or take the shape of the container it is placed in.
2. Iron has a high melting point of 1535°C and a high boiling point of 2750°C. At room temperature, which is typically around 25°C, the temperature is much lower than the melting point of iron, so the iron almirah remains in a solid state.

Q. Why does 273 K ice cool more efficiently than water does at the same temperature?

Answer – Ice at 273 K is more effective in cooling than water at the same temperature because of the difference in the heat capacity of ice and water. Heat capacity is defined as the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius.

Q. What causes more severe boiling water, bums or steam?

Answer – More severe burns can result from steam than from boiling water.

This is due to the fact that steam contains a significant quantity of latent heat, or the energy needed to transform a gas into a liquid. Steam quickly loses heat and condenses into water droplets when it comes into contact with the skin. Large amounts of latent heat are released during this process, and this heat can transfer a lot of energy to the skin, leading to serious burns.

Q. Names A, B, C, D, E, and F in the picture below that depicts a change in status

Answer – A > Liquefication/melting/fusion B > Vapourisation/evaporation C >Condensation D > Solidification E > Sublimation F > Sublimation

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