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3rd Science Term 1 Unit 1: My Body | Personal Hygiene and Safety

3rd Science : Term 1 Unit 1 : My Body

Learning Objectives

After learning this lesson, students will be able to:

* know the benefits of hand washing

* understand self-hygiene

* acquire the habit of using toilets and avoiding open defecation

* know the precautions to protect the sense organs

* recognize the differences and similarities in people‛s appearance and abilities

* understand good touch and bad touch

* grasp the importance of physical exercises

Tick (✓) the items used for personal hygiene

I. Cleanliness

Mithra is watching TV with her parents. An advertisement in the TV shows a toilet with germs. Mithra asks her dad what they are. He explains to her about germs.

Do you know about germs?

Germs are microorganisms that affects our body. They are found in all the places. If we do not maintain cleanliness, the germs will infect us and cause many diseases to us.

1. Hand Washing

Dialogue between Shruthi and her mother.

(After playing outside, Shruthi returns home)

Shruthi: Mom, I am so hungry. Give me something to eat.

Mother: Shruthi, go and wash your hands.

Shruthi: No Mom, I want to eat first. Then I wash my hands.

Mother: No, you must wash your hands.

Mother and daughter dialogue illustration

Shruthi: Mom, is it necessary to wash hands before eating?

Mother: Yes, look at your hands. Is it clean or dirty?

Shruthi: My hands are too dirty, Mom.

Mother: Where does the dirt stick in your hands?

Shruthi: Under the nails, in the ridges…

Mother: Yes, these are the places where the germs are hiding.

Shruthi: Is it so?

Mother: Yes, it is important to wash your hands. It is a simple habit that keeps you healthy.

(Mom starts explaining the importance of simple hand washing technique)

Steps of Hand Washing

* Wet your hands

* Apply enough soap (coin size)

* Rub the palms

* Rub the back of each hand together

* Rub both hands holding all fingers (interlocked)

* Rub the back of the fingers

* Rub the tip of the fingers

* Rub the thumbs and the ends of the wrists

* Rinse both hands properly with enough water

Benefits of Hand Washing

* Kills or removes germs

* Lowers the risk of diseases like Diarrhoea

* Prevents eye infections

* Reduces the risk of respiratory infections

Let us draw – Germy Hand

1. Trace your hand in a white sheet using pencil.

2. Draw the pictures of small germs in the hand.

3. Colour the germs.

Let us Write - Complete the Worksheet

Name : Mahendran

Do germs make people sick? Yes No

Can you see germs? Yes No

Where are the germs found?

1. Germs are found in air and water.

2. Germs are found in our hands, nails and in the ridges.

What can I do to avoid the spread of germs?

1. We should wash our hands properly.

2. When we cough or sneeze, we should use handkerchief.

Think Zone

Preethi often bites her nails. Is it a good habit? Give reason.

Biting nails is not a good habit.

Because germs are found under the nails, in the ridges.

Do you know?

Global Hand Washing Day is observed on October 15.

2. Using Toilets

In the world, nearly one billion people have no access to toilet at all and they are forced to do open defecation. Open defecation spreads diseases such as cholera and diarrhoea. Groundwater is polluted by toilets and it also causes diseases. Children also get affected by intestinal worms which causes anaemia. So, it is necessary to use toilets.

Do you Know

World Toilet Day is observed on November 19.

Why is it important to have a toilet?

Having a toilet:

* Prevents diseases like cholera.

* offers privacy.

* is convenient.

* is safe.

Effects of open defecation

* Water borne diseases

* Diseases caused by insect carrier

* Pollution

What should we do to wash out the intestinal worms?

We should use toilet and avoid open defecation. We should wash our hands after using the toilet. These will help us to have a healthy life.

Do you know

National De-worming Day is observed on February 10.

More to know

Sulabh international Museum of Toilets is in Delhi, India. It exhibits different toilet models from 50 countries across the world – spanning from 3000 BC till the 20th century.

3. Bathing

Discuss with group:

* Why do we take bath?

* Why is it important to take bath daily?

* Why do we use soap to clean our body?

Importance of bathing

* Cleanses the body

* Removes dirt and odour

* Protects oneself from infection

* Improves blood circulation

Do you know

World Health Day is observed on April 7.

Safety measures in Bathroom

* Do not play with water or run around in the bathroom. You may get hurt.

* Do not touch the sharp objects such as blades, razors and scissors kept in the bathroom.

* Use hot water under the supervision of an adult only.

* Do not leave soap bar on the floor. Someone may slip due to it.

* Do not leave the bathroom floor wet. Dry the floor using a wiper before you leave.

* Do not touch electric switches with wet hands. You may get hurt with an electric shock.

Do you know

The fruits of Soapberry tree are Saponin-rich and used as a soap substitute.

Put a tick (✓) for correct action and cross (×) for wrong action.

Let us Answer

Read the statements given below. Identify whether it is Right (✓) or Wrong (×) and put the suitable symbol.

Let us Find

Varun is in the bathroom. He is searching the things to take bath. Help him to find the things by colouring them.

II. Protecting Sense organs

Sense organs helps us to observe and understand the world around us. There are five main ways we can do this: through sight (with our eyes), touch (with our skin), smell (with our nose), taste (with our tongue) and hearing (with our ears).

In this section we will discuss the ways by which we can take care of our sense organs.

By nature, all the sense organs have some ways to protect themselves. For example, the eyelids and lashes protect the eyes from dust and other foreign particles.

Think Zone

Anu plays with her baby brother. As he cries, she gives him a pencil. On seeing this, Anu‛s mom was shocked. She quickly takes the pencil back. Do you know why?

Answer: Pencil has a sharp end. It may hurt the baby's eye or ear.

a. EYES

Do‛s

* Read with proper light.

* Watch television from a minimum distance of 6 feet.

* If you feel your eyes are itchy, wash them with clean and cold water.

Don‛ts

* Read in too dim or too bright light.

* Play video games or watch TV for long hours.

* Rub your eyes with your fist.

If you find it difficult to read the classroom board from your place, tell your teacher/parents and consult a doctor. You must follow the suggestions of the ophthalmologist or eye doctor.

Do you know

A condition known as digital eye strain is common among people working at computers all day. This condition leads to dry eyes, eye strain, blurred vision and headache.

b. EARS

* Avoid loud noise.

* Do not listen to music at higher volume when using Earphones or Headphones.

* Do not clean ears with ear buds.

* Dry your ears after taking bath.

* Cover your ears while swimming and bathing.

* Use earplugs or ear muffs (when you are exposed to loud sound).

* Consult a doctor if you have pain in ear.

Hearing loss can not be prevented always. But hearing loss due to exposure to loud noises is can be avoided.

Do you know

Noise levels are measured in decibels (dB). Any sound over 85dB can be harmful to us.

c. NOSE

* Do not clean your nose by inserting any object into it

* If the nose is blocked due to cold it is better to use steam to clear it.

* Do not pick your nose.

d. TONGUE

* We should clean our tongue daily with the tongue cleaner while brushing our teeth.

* A dirty tongue may cause disease and bad breath.

e. SKIN

* Always use a mild soap.

* Keep your skin dry and clean.

* Dry your skin by rubbing gently with a clean cloth.

* Consult a doctor when you feel any itching, skin injury or infection.

Let us Do - True or False

Read the given statements carefully. Write ‘T‛ if it is True and ‘F‛ if it is False.

1. Avoid playing video games or watching TV for a long time. (T)

2. Avoid loud noise. (T)

3. Do not clean your nose by inserting any object into it. (T)

4. A dirty tongue may cause disease and bad breath. (T)

5. Dry your skin by rubbing gently with a dirty cloth. (F)

Let us Practice - Make a figure ‘8‛ with your eyes

This is a good exercise to practice controlling the physical movement of your eyes.

• Draw a large figure of ‘8‛ on the floor, about 10 feet in front of you.

• Trace the figure ‘8‛ with your eyes, slowly, without shaking head.

• Do it clockwise for few minutes and anti-clockwise for few minutes.

III. Good Touch, Bad Touch and Don‛t Touch

(In the park, Janani plays with the puppy. She runs after a butterfly to catch it)

Mother: Janani, did you notice what happens when you touch the butterfly?

Janani: Yes mom, it flew away.

Mother: Now, tell me, why do you think it flew?

Janani: I think it did not like my touch.

Mother: Maybe, it did not. But, it wanted to be safe. Now, I am going to tell you three rules to keep yourself safe.

Janani: Okay, Mom.

Mother: There are body parts that you and others can see and touch like our arms and hands. But, there are also some body parts that no one should see or touch like our chest, between our legs and our bottom. These parts are called private parts. What are they called?

Janani: ___________________________.

Mother: So now, let me tell you rule number 1. It is never right for anyone to look at, touch or talk about your private parts. And it is never right for someone older to you to ask you to look, touch or talk about their private parts.

Janani: But mom, don‛t you give me bath daily?

Mother: Very good question! Who else knows that I give you a bath?

Janani: Dad, Grandpa, Grandma and Jagan know that you give me bath.

Mother: Correct, when we are small our mom or dad might touch us to help us to be clean. It is never a secret and it is okay to tell someone about it.

Janani: I got it mom. It is never right for anyone to _________, _________ or __________ about my private parts. But, it is never a _____________.

Mother: Very good! Some touches are good or safe, like when we shake hands. But, some touches are bad or unsafe, like when we hit someone. Can we hit someone?

Janani: _________________________.

Mother: Bad touches might make you feel sad, angry, scared, or confused. If you do not like any touch or if it makes you sad, angry or scared, tell them to STOP! Scream “STOP!” and run away. This is rule number 2.

Janani: Okay mom. If I do not like any touch, I scream __________ and ____________.

Mother: Very good! Now rule number 3. Tell an adult you trust, about the bad touch. Keep telling until you get the help you need.

Janani: I should ________________ adults I trust till I get _________.

Mother: Janani, remember, no matter who tells you. It is never your fault.

Janani: But mom, would people I know also do ‘bad touch‛?

Mother: Anyone would do. So it is important that you keep all the three rules in mind even if you know the person. Can you say the rules?

SAFE CIRCLE MEMBERS

Do you know

If someone abuses you or hurts you, you can call 1098 and get help.

IV. Physical or Sensory challenge

Ramu went for shopping with his father. On the way they saw a visually challenged person who tries to cross the road. Ramu‛s father went to him and asked, “May I help you sir?”, And the differently abled person replied, “Yes, I want to cross the road”.

Ramu‛s father joyfully helped him to cross the road. By seeing this, Ramu felt proud of his father, and he also wanted to help the needy. So, Ramu asked his father to teach him how to help differently abled persons.

1. Differently-abled persons

Not everybody can use all five senses. Some lack the ability to use any one or more of them, such people are called ‘Differently-abled’.

Do you know

Braille is a reading and writing method for visually challenged person. It has a raised pattern that can be read with the fingers.

2. Let us know how to help them

Helping people is very honourable, and there are plenty of ways by which we can help differently-abled people.

* Ask first if they need any help from you and follow their lead.

* Speak clearly and listen to their words well.

* Use direct words.

* Never tease them by calling with specific names.

* Be aware of their personal aids, do not cause any damage to the aids.

Your simple acts can be helpful to them.

* opening doors for the disabled.

* making way for them.

* bring them to cross the road.

* treat them as normal people.

By doing these you can exhibit that you want to help the differently-abled.

Do you know

We do not use the word ‘handicapped or disabled‛. We use ‘differently-abled'.

Match

A core tenet of a pedagogical approach is to present the problem before the solution. Match the items in the first column with the correct items in the second column to test your understanding.

V. Importance of Physical Exercise

Parents - teachers meet is going on in Sheela‛s school. As her parents are out of the town, her grandfather attends the meeting. He is a retired army person and even at the age of 90, he does all his work by himself. Everyone in the meeting wonders “What is the secret behind his good health?” and they asked him for his secret. Grandpa replied that he avoids junk food and takes healthy and fresh food.He does physical exercises like walking, running, jogging, and yoga. He advises all parents and children to do exercises regularly.

Physical Exercise:

Protects us from diseases

Balances our body weight

Strengthens our muscles and bones

Improves digestion

Improves sleep quality

Increases our energy level

Improves skin health

Boosts brain functions and memory

Find and circle the words related to fitness.

(SLEEP, ENERGY, HEALTHY, RUN, JUMP, PLAY, YOGA)

Expected Physics Problems with Solutions | TN 10th Science Public Exam 2026 எதிர்பார்க்கப்படும் இயற்பியல் கணக்கீடுகள் அரசுப் பொதுத் தேர்வு- 2026 - அறிவியல்

எதிர்பார்க்கப்படும் இயற்பியல் கணக்கீடுகள்

அரசுப் பொதுத் தேர்வு- 2026 - அறிவியல்

1. 5 கிகி நிறையுள்ள பொருளொன்றின் நேர்க் கோட்டு உந்தம் 2.5 கிகி மீ/வி எனில் அதன் திசைவேகத்தை கணக்கிடுக.

தீர்வு:

கொடுக்கப்பட்டவை:

நிறை (m) = 5 கிகி

நேர்க்கோட்டு உந்தம் (P) = 2.5 கிகி மீ/வி

நேர்க்கோட்டு உந்தம், $P = mv$

திசைவேகம் (v) = $\frac{P}{m}$

$V = \frac{2.5}{5} = 0.5$

திசைவேகம் V = 0.5 மீ/வி

2. கீழ் முனையில் இருந்து 90 செ.மீ தூரத்தில் கைப்பிடி கொண்ட கதவொன்று 40 N விசை கொண்டு திறக்கப்படுகிறது. கதவின் கீல் முனைப்பகுதியில் ஏற்படும் திருப்புத்திறன் மதிப்பினை கணக்கிடு?

தீர்வு:

கொடுக்கப்பட்டவை:

விசை F = 40 N

தொலைவு d = 90 செ.மீ = $90 \times 10^{-2}$ மீ

திருப்புத்திறன் $M = F \times d$

$M = 40 \times 90 \times 10^{-2}$

$M = 3600 \times 10^{-2}$

M = 36 Nm

3. 5N மற்றும் 15N விசை மதிப்புடைய இரு விசைகள் எதிரெதிர் திசையில் ஒரே நேரத்தில் பொருள் மீது செயல்படுகின்றன. இவைகளின் தொகுபயன் விசை யாது? எத்திசையில் அது செயல்படும்?

தீர்வு:

கொடுக்கப்பட்டவை:

$F_1 = 5N$

$F_2 = 15N$

தொகுபயன் விசை $F_{தொகு} = F_2 - F_1$ (விசைகள் எதிரெதிர் திசையில் உள்ளதால்)

$F_{தொகு} = 15 - 5 = 10N$

தொகுபயன் விசை 10N. இது 15N விசையின் திசையில் செயல்படும்.

4. கிட்டப்பார்வை குறைபாடு உடைய ஒரு மனிதரால் 4மீ தொலைவில் உள்ளப் பொருள்களை மட்டுமே காண இயலும். அவர் 20மீ தொலைவில் உள்ளப் பொருளை அவர் காண விரும்பினால் பயன்படுத்தப்பட வேண்டிய குழி லென்சின் குவியத் தொலைவு என்ன?

தீர்வு:

கொடுக்கப்பட்டவை:

x = 4 மீ (காணக்கூடிய தொலைவு)

y = 20 மீ (காண விரும்பும் தொலைவு)

குழி லென்சின் குவியத்தொலைவுக்கான சமன்பாடு:

$f = \frac{xy}{x-y}$

$f = \frac{4 \times 20}{4 - 20} = \frac{80}{-16}$

f = -5 மீ

லென்சின் திறன் $P = \frac{1}{f}$

$P = \frac{1}{-5} = -0.2 D$

திறன் P = -0.2D

5. தூரப்பார்வை குறைபாட்டால் பாதிக்கப்பட்ட மனிதர் ஒருவரின் அண்மைப் புள்ளியானது 1.5 மீ தொலைவில் உள்ளது. அவரது பார்வைக் குறைபாட்டை சரிசெய்ய பயன்படுத்தப்பட வேண்டிய குவிலென்சின் குவியத் தொலைவைக் கணக்கிடுக.

தீர்வு:

கொடுக்கப்பட்டவை:

அண்மைப் புள்ளி (d) = 1.5 மீ

தெளிவூறு காட்சியின் மீச்சிறு தொலைவு (D) = 25 செ.மீ = 0.25 மீ

குவி லென்சின் குவியத்தொலைவுக்கான சமன்பாடு:

$f = \frac{dD}{d-D}$

$f = \frac{1.5 \times 0.25}{1.5 - 0.25} = \frac{0.375}{1.25}$

f = 0.3 மீ

6. 12 கூலும் மின்னூட்டம் 5 வினாடி நேரம் ஒரு மின்விளக்கின் வழியாக பாய்கிறது எனில் அதன் வழியே செல்லும் மின்னோட்டத்தின் அளவு என்ன?

தீர்வு:

கொடுக்கப்பட்டவை:

மின்னூட்டம் (Q) = 12 கூலும்

நேரம் (t) = 5 வினாடி

மின்னோட்டம் (I) = $\frac{Q}{t}$

$I = \frac{12}{5} = 2.4 A$

மின்னோட்டம் I = 2.4A

7. 10 கூலும் மின்னூட்டத்தை ஒரு மின்சுற்றிலுள்ள இரண்டு புள்ளிகளுக்கிடையே நகர்த்த செய்யப்படும் வேலை 100 J எனில் அப்புள்ளிகளுக்கிடையே உள்ள மின்னழுத்த வேறுபாடு என்ன?

தீர்வு:

கொடுக்கப்பட்டவை:

மின்னூட்டம் (Q) = 10 கூலும்

செய்யப்பட்ட வேலை (W) = 100 J

மின்னழுத்த வேறுபாடு (V) = $\frac{W}{Q}$

$V = \frac{100}{10} = 10 V$

மின்னழுத்த வேறுபாடு V = 10 வோல்ட்

8. 30 வோல்ட் மின்னழுத்த வேறுபாடு கொண்ட ஒரு கடத்தியின் முனைகளுக்கு இடையே 2 ஆம்பியர் மின்னோட்டம் செல்கிறது எனில் அதன் மின்தடையை காண்க.

தீர்வு:

கொடுக்கப்பட்டவை:

மின்னோட்டம் (I) = 2 A

மின்னழுத்த வேறுபாடு (V) = 30 V

ஓம் விதிப்படி, $V = IR$

மின்தடை (R) = $\frac{V}{I}$

$R = \frac{30}{2} = 15 \Omega$

மின்தடை R = 15Ω

9. 5Ω, 3Ω மற்றும் 2Ω மின்தடை மதிப்புகள் கொண்ட மூன்று மின்தடையாக்கிகள் 10 வோல்ட் மின்கலத்துடன் தொடரிணைப்பில் இணைக்கப்பட்டுள்ளன. தொகுபயன் மின்தடை மற்றும் மின்சுற்றில் பாயும் மின்னோட்டத்தையும் காண்க.

தீர்வு:

கொடுக்கப்பட்டவை:

$R_1 = 5 \Omega, R_2 = 3 \Omega, R_3 = 2 \Omega, V = 10 V$

தொடரிணைப்பில் தொகுபயன் மின்தடை,

$R_s = R_1 + R_2 + R_3$

$R_s = 5 + 3 + 2 = 10 \Omega$

தொகுபயன் மின்தடை = 10Ω

மின்னோட்டம் $I = \frac{V}{R_s}$

$I = \frac{10}{10} = 1 A$

மின்னோட்டம் I = 1A

10. 5Ω மின்தடை கொண்ட மின் சூடேற்றி ஒரு மின் மூலத்துடன் இணைக்கப்படுகிறது. 6A மின்னோட்டமானது இந்த சூடேற்றி வழியாக பாய்கிறது எனில் 5 நிமிடங்களில் உருவாகும் வெப்பத்தின் அளவை காண்க.

தீர்வு:

கொடுக்கப்பட்டவை:

மின்தடை (R) = 5Ω

மின்னோட்டம் (I) = 6A

காலம் (t) = 5 நிமிடம் = $5 \times 60 = 300$ வினாடி

வெப்பத்தின் அளவு (H) = $I^2 Rt$

$H = 6^2 \times 5 \times 300$

$H = 36 \times 5 \times 300$

H = 54000 J

11. இராணுவத் தளவாடங்கள் தொழிற்சாலைகளில் காற்றின் வெப்பநிலை 46°C ஐ அடைய இயலும். அந்த வெப்ப நிலையில் காற்றில் ஒலியின் திசைவேகம் என்ன? ($V_0 = 331$ மீ/வி)

தீர்வு:

கொடுக்கப்பட்டவை:

வெப்பநிலை (T) = 46°C

$0°C$ இல் திசைவேகம் ($V_0$) = 331 மீ/வி

திசைவேகத்திற்கான சமன்பாடு:

$V = V_0 + 0.61 T$

$V = 331 + 0.61 \times 46$

$V = 331 + 28.06$

V = 359.06 மீ/வி

12. ஒரு ஊடகத்தில் 200 Hz அதிர்வெண் உடைய ஒலியானது 400 மீ/வி வேகத்தில் பரவுகிறது. ஒலி அலையின் அலைநீளம் காண்க.

தீர்வு:

கொடுக்கப்பட்டவை:

அதிர்வெண் (n) = 200 Hz

திசைவேகம் (c) = 400 மீ/வி

அலைநீளம் ($\lambda$), $c = n\lambda$

$\lambda = \frac{c}{n}$

$\lambda = \frac{400}{200}$

அலைநீளம் $\lambda$ = 2 மீ

13. 2 கிகி நிறையுடைய ஒரு கதிரியக்கப் பொருளானது அணுக்கரு இணைவின்போது வெளியாகும் மொத்த ஆற்றலைக் கணக்கிடுக.

தீர்வு:

கொடுக்கப்பட்டவை:

நிறை (m) = 2 கிகி

ஒளியின் திசைவேகம் (c) = $3 \times 10^8$ மீ/வி

ஐன்ஸ்டீன் நிறை ஆற்றல் சமன்பாடு, $E = mc^2$

$E = 2 \times (3 \times 10^8)^2$

$E = 2 \times 9 \times 10^{16}$

E = $18 \times 10^{16}$ J

14. $_{92}U^{238}$ என்ற தனிமம் ஆல்பா சிதைவிற்கு உட்படுகிறது எனில் சேய் தனிமத்தில் உள்ள நியூட்ரான்களின் எண்ணிக்கையை கணக்கிடுக.

தீர்வு:

ஆல்பா சிதைவின் போது,

நிறை எண் (A) 4 குறையும் $\rightarrow A' = A - 4$

அணு எண் (Z) 2 குறையும் $\rightarrow Z' = Z - 2$

எனவே, $_{92}U^{238} \rightarrow _{90}Th^{234} + _2He^4$

சேய் தனிமத்தின் நிறை எண் = $238 - 4 = 234$

சேய் தனிமத்தின் அணு எண் = $92 - 2 = 90$

நியூட்ரான்களின் எண்ணிக்கை = நிறை எண் - அணு எண்

நியூட்ரான்களின் எண்ணிக்கை = $234 - 90 = 144$

நியூட்ரான்களின் எண்ணிக்கை = 144

15. ஒரு பொருளிலிருந்து செல்லும் ஒளிக்கற்றையானது 0.3 மீ குவியத் தொலைவு கொண்ட விரிக்கும் லென்சால் குவிக்கப்பட்டு 0.2 மீ என்ற தொலைவில் பிம்பத்தை ஏற்படுத்துகிறது எனில் பொருளின் தொலைவைக் கணக்கிடுக.

தீர்வு:

கொடுக்கப்பட்டவை:

குவியத் தொலைவு (f) = -0.3 மீ (விரிக்கும் லென்ஸ் என்பதால்)

பிம்பத்தின் தொலைவு (v) = -0.2 மீ (மெய் பிம்பம் இல்லை)

லென்ஸ் சமன்பாடு: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$

$\frac{1}{u} = \frac{1}{-0.2} - \frac{1}{-0.3} = -\frac{1}{0.2} + \frac{1}{0.3}$

$\frac{1}{u} = \frac{-0.3 + 0.2}{0.2 \times 0.3} = \frac{-0.1}{0.06}$

$u = \frac{-0.06}{0.1} = -0.6$ மீ

பொருளின் தொலைவு u = -0.6 மீ

Types of Chemical Reactions: Class 10 Science Book Back Questions and Answers

Book Back Questions with Answers - Types of Chemical Reactions

I. Choose the correct answer.

1. $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$ is a

a. Decomposition Reaction
b. Combination Reaction
c. Single Displacement Reaction
d. Double Displacement Reaction

2. Photolysis is a decomposition reaction caused by ___________

a. heat
b. electricity
c. light
d. mechanical energy

3. A reaction between carbon and oxygen is represented by $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} + \text{Heat}$. In which of the type(s), the above reaction can be classified?

(i) Combination Reaction
(ii) Combustion Reaction
(iii) Decomposition Reaction
(iv) Irreversible Reaction

a. i and ii
b. i and iv
c. i, ii and iii
d. i, ii and iv

4. The chemical equation $Na_2SO_{4(aq)} + BaCl_{2(aq)} \rightarrow BaSO_{4(s)}\downarrow + 2NaCl_{(aq)}$ represents which of the following types of reaction?

a. Neutralisation
b. Combustion
c. Precipitation
d. Single displacement

5. Which of the following statements are correct about a chemical equilibrium?

(i) It is dynamic in nature
(ii) The rate of the forward and backward reactions are equal at equilibrium
(iii) Irreversible reactions do not attain chemical equilibrium
(iv) The concentration of reactants and products may be different

a. i, ii and iii
b. i, ii and iv
c. ii, iii and iv
d. i, iii and iv

6. A single displacement reaction is represented by $X_{(s)} + 2HCl_{(aq)} \rightarrow XCl_{2(aq)} + H_{2(g)}$. Which of the following(s) could be X.

(i) Zn (ii) Ag (iii) Cu (iv) Mg.

Choose the best pair.

a. i and ii
b. ii and iii
c. iii and iv
d. i and iv

7. Which of the following is not an “element + element → compound” type reaction?

a. $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$
b. $2K_{(s)} + Br_{2(l)} \rightarrow 2KBr_{(s)}$
c. $2CO_{(g)} + O_{2(g)} \rightarrow 2CO_{2(g)}$
d. $4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)}$

8. Which of the following represents a precipitation reaction?

a. $A_{(s)} + B_{(s)} \rightarrow C_{(s)} + D_{(s)}$
b. $A_{(s)} + B_{(aq)} \rightarrow C_{(aq)} + D_{(l)}$
c. $A_{(aq)} + B_{(aq)} \rightarrow C_{(s)} + D_{(aq)}$
d. $A_{(aq)} + B_{(s)} \rightarrow C_{(aq)} + D_{(l)}$

9. The pH of a solution is 3. Its $[OH^-]$ concentration is

a. $1 \times 10^{-3} M$
b. $3 M$
c. $1 \times 10^{-11} M$
d. $11 M$

10. Powdered $CaCO_3$ reacts more rapidly than flaky $CaCO_3$ because of ___________.

a. large surface area
b. high pressure
c. high concentration
d. high temperature

II. Fill in the blanks

1. A reaction between an acid and a base is called neutralization.

2. When lithium metal is placed in hydrochloric acid, hydrogen gas is evolved.

3. The equilibrium attained during the melting of ice is known as physical equilibrium.

4. The pH of a fruit juice is 5.6. If you add slaked lime to this juice, its pH Increases (increse/decrese)

5. The value of ionic product of water at 25ºC is $1.00 \times 10^{-14}$.

6. The normal pH of human blood is 7.35 - 7.45.

7. Electrolysis is type of decomposition reaction.

8. The number of products formed in a synthesis reaction is one.

9. Chemical volcano is an example for decomposition type of reaction.

10. The ion formed by dissolution of $H^+$ in water is called Hydronium ion.

III. Match the following

Identify the types of reaction

Match the chemical reactions with their types.

Answer:

1. Single Displacement

2. Combination

3. Neutralization

4. Thermal decomposition

IV. True or False: (If false give the correct statement)

1. Silver metal can displace hydrogen gas from nitric acid.

False

Correct Statement: Silver metal will not displace hydrogen gas from nitric acid.

2. The pH of rain water containing dissolved gases like $SO_3, CO_2, NO_2$ will be less than 7.

True

3. At the equilibrium of a reversible reaction, the concentration of the reactants and the products will be equal.

False

Correct Statement: At the equilibrium of a reversible reaction, there is no change in the concentration of the reactants and the products.

4. Periodical removal of one of the products of a reversible reaction increases the yield.

True

5. On dipping a pH paper in a solution, it turns into yellow. Then the solution is basic.

False

Correct Statement: On dipping a pH paper in a solution, it turns into yellow. Then the solution is neutral.

V. Short answer questions:

1. When an aqueous solution of potassium chloride is added to an aqueous solution of silver nitrate, a white precipitate is formed. Give the chemical equation of this reaction.

$KCl + AgNO_3 \rightarrow KNO_3 + AgCl \downarrow$

2. Why does the reaction rate of a reaction increase on raising the temperature?

Increase in temperature provides energy to break more bonds and thus speeds up the reaction.

3. Define combination reaction. Give one example for an exothermic combination reaction.

A combination reaction is a reaction in which two or more reactants combine to form a compound. It is otherwise called synthesis reaction or composition reaction.

Example: On burning magnesium in air, it combines with oxygen to form magnesium oxide. $2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)}$

4. Differentiate reversible and irreversible reactions

Reversible Reaction	Irreversible reaction
i. It can be reversed under suitable conditions	i. It cannot be reversed
ii. Both forward and backward reactions take place simultaneously	ii. It is unidirectional. It proceeds only in forward direction
iii. It attains equilibrium	iii. Equilibrium is not attained
iv. The reactants cannot be converted completely into products	iv. The reactants can be completely converted into products
v. It is relatively slow	v. It is fast

VI. Answer in detail

1. What are called thermolysis reactions?

In this type of reaction, the reactant is decomposed by applying heat.

As the molecule is dissociated by absorption of heat, it is otherwise called ‘Thermolysis’.

$2HgO_{(s)} \xrightarrow{\text{Heat}} 2Hg_{(l)} + O_{2(g)}$

2. Explain the types of double displacement reactions with examples.

There are two major classes of double displacement reactions. They are

(i) Precipitation Reactions

(ii) Neutralization Reactions

(i) Precipitation Reactions: When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.

$Pb(NO_3)_{2(aq)} + 2KI_{(aq)} \rightarrow PbI_{2(s)} + 2KNO_{3(aq)}$

(ii) Neutralization Reactions: Acid reacts with the base to form a salt and water. It is called neutralization reaction as both acid and base neutralise each other.

Acid + Base → Salt + Water

$HCl_{(aq)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$

3. Explain the factors influencing the rate of a reaction.

Important factors that affect rate of a reaction are:

(i) Nature of the reactants
(ii) Concentration of the reactants
(iii) Temperature
(iv) Catalyst
(v) Pressure
(vi) Surface area of the reactants

(i) Nature of the reactants: The reaction of sodium with hydrochloric acid is faster than that with acetic acid. Hydrochloric acid is stronger acid than acetic acid and thus more reactive. So the nature of the reactants influence reaction rate.

$2Na_{(s)} + 2HCl_{(aq)} \rightarrow 2NaCl_{(aq)} + H_{2(g)}$ (fast)

(ii) Concentration of the reactants: Changing the amount of the reactants also increase the reaction rate. More the concentration, more particles per volume exist in it and hence faster the reaction.

(iii) Temperature: Increase in temperature provides energy to break more bonds and thus speed up the reaction.

(iv) Pressure: If the reactants are gases, increasing their pressure increases the reaction rate. Since on increasing pressure the reacting particles come closer and collide frequently.

(v) Catalyst: A catalyst is a substance which increase reaction rate without being consumed in the reaction. In certain reactions, adding a substance as catalyst speeds up the reaction.

(vi) Surface area of the reactants: When solid reactants are involved in a reaction, their powdered form reacts more readily. Because powdering of the reactants increases the surface area and more energy is available on collision of reactant particles. Thus the reaction rate is increased.

4. How does pH play an important role in everyday life?

(i) Living organisms can survive only in a narrow range of pH change. Different body fluids have different pH values.
(ii) Our stomach produces hydrochloric acid which helps in the digestion of food without harming the stomach. During indigestion, if the stomach produces too much acid and this causes pain and irritation. pH of the stomach fluid is approximately 2.0.
(iii) pH of the saliva normally ranges between 6.5 to 7.5.
(iv) When the pH of the mouth saliva falls below 5.5, the enamel gets weathered.
(v) Toothpastes, which are generally basic are used for cleaning the teeth that can neutralise the excess acid and prevent tooth decay.
(vi) In agriculture, the pH of the soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.
(vii) The pH of rain water is approximately 7, which means that it is neutral and also represents its high purity. If the atmospheric air is polluted with oxide gases of sulphur and nitrogen, they get dissolved in the rain water and make its pH less than 7. Thus, if the pH of rain water is less than 7, then it is called acid rain. When acid rain flows into the rivers it lowers the pH of the river water also. The survival of aquatic life in such rivers becomes difficult.

5. What is a chemical equilibrium? What are its characteristics?

Chemical Equilibrium: It is state of a reversible chemical reaction in which no change in the amount of reactants and products takes place. At equilibrium,

Rate of forward reaction = Rate of backward reaction

Characteristics of equilibrium :

(i) In chemical equilibrium, the rates of forward and backward reactions are equal.
(ii) The observable properties such as pressure, concentration, color, density, viscosity etc., of the system remain unchanged with time.
(iii) The chemical equilibrium is a dynamic equilibrium, because both the forward and backward reactions continue to occur even though it appears static externally.
(iv) In physical equilibrium, the volume of all the phases remain constant.

VII. HOT questions

1. A solid compound ‘A’ decomposes on heating into ‘B’ and a gas ‘C’. On passing the gas ‘C’ through water, it becomes acidic. Identify A, B and C.

$CaCO_3 \rightarrow CaO + CO_2$

$CO_2 + H_2O \rightarrow H_2CO_3$ (Carbonic acid)

$CaCO_{3[A:Calcium \ carbonate]} \rightarrow CaO_{[B:Calcium \ oxide]} + CO_{2[C:Carbon \ di \ oxide]}$

$CO_2 + H_2O \rightarrow H_2CO_3$ (Carbonic acid)

A: $CaCO_3$ : Calcium carbonate

B: $CaO$ : Calcium oxide

C: $CO_2$ : Carbon di oxide

2. Can a nickel spatula be used to stir copper sulphate solution? Justify your answer.

Nickel spatula cannot be used to stir the $CuSO_4$ solution. Since Ni will displace Copper from $CuSO_4$ solution and Cu will be deposited on the Nickel spatula.

VIII. Solve the following problems

1. Lemon juice has a pH 2, what is the concentration of $H^+$ ions?

Solution

pH of lemon juice = 2

$[H^+] = ?$

$pH = -\log_{10}[H^+]$

$\log_{10}[H^+] = -2$

$[H^+] = 10^{-2}$

= 0.01 mole $litre^{-1}$

2. Calculate the pH of $1.0 \times 10^{-4}$ molar solution of $HNO_3$.

Solution

$[H^+] = 1.0 \times 10^{-4}$

$pH = -\log_{10}[H^+]$

$ = -\log_{10}[1 \times 10^{-4}]$

$pH = -(\log_{10}1 - 4 \log_{10}10)$

$= 0 + 4 \times \log_{10}10$

$= 0 + 4 \times 1$

$= 4$

pH = 4

3. What is the pH of $1.0 \times 10^{-5}$ molar solution of KOH?

Solution:

$KOH_{(aq)} \leftrightarrow K^+_{(aq)} + OH^-_{(aq)}$

One mole of KOH would give one mole of $OH^-$ ions.

$[OH^-] = 1 \times 10^{-5} \text{ mol } litre^{-1}$

$pOH = -\log_{10}[OH^-] = -\log_{10}[10^{-5}]$

$= - (-5 \times \log_{10}10) = - (-5) = 5$

$pH + pOH = 14$

$pH = 14 - pOH = 14 - 5 = 9$

4. The hydroxide ion concentration of a solution is $1 \times 10^{-11} M$. What is the pH of the solution?

Solution:

$[OH^-] = 1 \times 10^{-11} M$

$pOH = -\log_{10}[OH^-] = -\log_{10}[10^{-11}]$

$= - (-11 \times \log_{10}10) = - (-11) = 11$

$pH + pOH = 14$

$pH = 14 - pOH = 14 - 11 = 3$

Concept map for types of chemical reactions

10th Science: Solutions - Book Back Questions and Answers

Book Back Questions with Answers - Solutions | Science

I. Choose the correct answer.

A solution is a __________ mixture.
- a. homogeneous
- b. heterogeneous
- c. homogeneous and heterogeneous
- d. non homogeneous
The number of components in a binary solution is __________
- a. 2
- b. 3
- c. 4
- d. 5
Which of the following is the universal solvent?
- a. Acetone
- b. Benzene
- c. Water
- d. Alcohol
A solution in which no more solute can be dissolved in a definite amount of solvent at a given temperature is called _______
- a. Saturated solution
- b. Un saturated solution
- c. Super saturated solution
- d. Dilute solution
Identify the non aqueous solution.
- a. sodium chloride in water
- b. glucose in water
- c. copper sulphate in water
- d. sulphur in carbon-di-sulphide
When pressure is increased at constant temperature the solubility of gases in liquid ___________.
- a. No change
- b. increases
- c. decreases
- d. no reaction
Solubility of NaCl in 100 ml water is 36 g. If 25 g of salt is dissolved in 100 ml of water how much more salt is required for saturation _____________.
- a. 12g
- b. 11g
- c. 16g
- d. 20g
A 25% alcohol solution means
- a. 25 ml alcohol in 100 ml of water
- b. 25 ml alcohol in 25 ml of water
- c. 25 ml alcohol in 75 ml of water
- d. 75 ml alcohol in 25 ml of water
Deliquescence is due to __________
- a. Strong affinity to water
- b. Less affinity to water
- c. Strong hatred to water
- d. Inertness to water
Which of the following is hygroscopic in nature?
- a. ferric chloride
- b. copper sulphate penta hydrate
- c. silica gel
- d. none of the above

II. Fill in the blanks

The component present in lesser amount, in a solution is called Solute.
Example for liquid in solid type solution is Mercury with sodium (amalgam).
Solubility is the amount of solute dissolved in 100 g of solvent.
Polar compounds are soluble in polar solvents.
Volume percentage decreases with increases in temperature because of expansion of liquids.

III. Match the following

Problem:

Blue vitriol – $ \text{CaSO}_4 \cdot 2\text{H}_2\text{O} $
Gypsum – $ \text{CaO} $
Deliquescence – $ \text{CuSO}_4 \cdot 5\text{H}_2\text{O} $
Hygroscopic – $ \text{NaOH} $

Answer:

Blue vitriol – $ \text{CuSO}_4 \cdot 5\text{H}_2\text{O} $
Gypsum – $ \text{CaSO}_4 \cdot 2\text{H}_2\text{O} $
Deliquescence – $ \text{NaOH} $
Hygroscopic – $ \text{CaO} $

IV. True or False: (If false give the correct statement)

Solutions which contain three components are called binary solution. - False

Solutions which contain three components are called Trinary solution.
In a solution the component which is present in lesser amount is called solvent. - False

In a solution, the component present in a lesser amount is called the solute. The component in the larger amount is the solvent.
Sodium chloride dissolved in water forms a non-aqueous solution. - False

Sodium chloride dissolved in water forms an aqueous solution because water is the solvent.
The molecular formula of green vitriol is $ \text{MgSO}_4 \cdot 7\text{H}_2\text{O} $. - False

The molecular formula of green vitriol is Ferrous Sulphate: $ \text{FeSO}_4 \cdot 7\text{H}_2\text{O} $. $ \text{MgSO}_4 \cdot 7\text{H}_2\text{O} $ is Epsom salt.
When Silica gel is kept open, it absorbs moisture from the air, because it is hygroscopic in nature. - True

V. Short answer

Define the term: Solution

Solution is a homogeneous mixture of one or more substances.
What is mean by binary solution

Solutions which are made of one solute and one solvent (two components) are called binary solutions.
Give an example each i) gas in liquid ii) solid in liquid iii) solid in solid iv) gas in gas

(i) Gas - Liquid - carbon-di-oxide dissolved in water (Soda water)

(ii) Solid - Liquid - Sodium chloride dissolved in water

(iii) Solid - Solid - Copper dissolved in gold (Alloys)

(iv) Gas - Gas - Mixture of Helium-Oxygen gases
What is aqueous and non-aqueous solution? Give an example.

(i) Aqueous solution: The solution in which water acts as a solvent is called aqueous solution. E.g. Common salt in water, Sugar in water, Copper sulphate in water etc.

(ii) Non - Aqueous solution: The solution in which any liquid other than water acts as a solvent is called non - aqueous solution. E.g: Sulphur dissolved in carbon disulphide.
Define Volume percentage

Volume percentage is defined as the percentage by volume of solute (in ml) present in the given volume of the solution.

$$ \text{Volume Percentage} = \frac{\text{Volume of the solute}}{\text{Volume of the solution}} \times 100 $$
The aquatic animals live more in cold region Why?

Aquatic animals live more in cold regions because, more amount of dissolved oxygen is present in the water of cold regions. This shows that the solubility of oxygen is more in water at low temperature.
Define Hydrated salt.

When ionic substances are dissolved in water to make their saturated aqueous solution, their ions attract water molecules which then attached chemically in certain ratio. This process is called hydration.
A hot saturated solution of copper sulphate forms crystals as it cools. Why?

As the solution cools, the water molecules move closer together again and there's less room for the solution to hold onto as much of the dissolved solid. So copper sulphate crystallises as the solid is cooled.
Classify the following substances into deliquescent, hygroscopic.

Conc. Sulphuric acid, Copper sulphate penta hydrate, Silica gel, Calcium chloride, and Gypsum salt.

Deliquescent: Calcium chloride

Hygroscopic: Conc. Sulphuric acid, Silica gel

(Note: Copper sulphate pentahydrate and Gypsum are hydrated salts that exhibit efflorescence, not typically deliquescence or hygroscopy.)

VI. Long answer:

Write notes on i) saturated solution ii) unsaturated solution

Answer:

(i) Saturated solution : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.

E.g. 36 g of sodium chloride in 100g of water at 25° C forms saturated solution.

(ii) Unsaturated solution : Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature.

E.g. 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25° C
Write notes on various factors affecting solubility.

Answer:

There are three main factors which govern the solubility of a solute. They are: Nature of the solute and solvent, Temperature, Pressure

Nature of the solute and solvent:

(i) The nature of the solute and solvent plays an important role in solubility.

(ii) Solubility is based on the phrase “like dissolves like”. The expression means that dissolving occurs when similarities exist between the solvent and the solute. For example: Common salt is a polar compound and dissolves readily in polar solvent like water.

(iii) Non-polar compounds are soluble in non-polar solvents. For example: Sulphur dissolves in carbon disulphide.

Effect of Temperature :

Solubility of Solids in Liquid:

(i) Generally, solubility of a solid solute in a liquid solvent increases with increase in temperature.

(ii) In endothermic process, solubility increases with increase in temperature.

(iii) In exothermic process, solubility decreases with increase in temperature.

Solubility of Gases in liquid : Solubility of gases in liquid decrease with increase in temperature.

Effect of Pressure : When the pressure is increased, the solubility of a gas in liquid increases. When pressure is increased, more gas molecules strike the surface of the liquid and mix into the solution.
a) What happens when $ \text{MgSO}_4 \cdot 7\text{H}_2\text{O} $ is heated? Write the appropriate equation

b) Define solubility

Answer:

a) When magnesium sulphate heptahydrate crystals are gently heated, it loses seven water molecules, and becomes anhydrous magnesium sulphate.

(Magnesium sulphate heptahydrate) → (Anhydrous Magnesium sulphate)

b) Solubility is defined as the number of grams of a solute that can be dissolved in 100g of a solvent to form its saturated solution at a given temperature and pressure.

$$ \text{Solubility} = \frac{\text{Mass of the solute}}{\text{Mass of the solvent}} \times 100 $$
In what way hygroscopic substances differ from deliquescent substances.

Answer:

Deliquescence substances
- Certain water soluble substances, when exposed to the atmosphere at ordinary temperature, absorb moisture from the atmospheric air and become moist or pasty or watery.
- Deliquescent substances change its state on exposure to air.
- Deliquescent substances are crystalline solids.
Hygroscopic Substances
- Certain substances, when exposed to the atmosphere at ordinary temperature, absorb moisture without dissolving it.
- Hygroscopic substances do not change its state on exposure to air.
- Hygroscopic substances may be amorphous solids or liquids.
A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute.

Answer:

$$ \text{Mass percentage} = \frac{\text{Mass of the solute}}{\text{Mass of the solution}} \times 100 $$

$$ \text{Mass percentage} = \frac{\text{Mass of the solute}}{(\text{Mass of the solute} + \text{Mass of the solvent})} \times 100 $$

$$ \text{Mass percentage} = \frac{45}{(45+180)} \times 100 = \frac{45}{225} \times 100 $$

Mass percentage = 20%
3.5 litres of ethanol is present in 15 litres of aqueous solution of ethanol. Calculate volume percent of ethanol solution.

Answer:

$$ \text{Volume percentage} = \frac{\text{Volume of the solute}}{\text{Volume of the solution}} \times 100 $$

$$ \text{Volume percentage} = \frac{3.5}{15} \times 100 $$

Volume percentage = 23.33%

VII. HOT (Higher-Order Thinking)

Vinu dissolves 50 g of sugar in 250 ml of hot water, Sarath dissolves 50 g of same sugar in 250 ml of cold water. Who will get faster dissolution of sugar? and Why?

50 g of sugar in 250 ml of hot water will dissolve faster as solubility of solid solute in liquid solvent increases with increase in temperature.
'A' is a blue coloured crystaline salt. On heating it loses blue colour and to give 'B'. When water is added, 'B' gives back to 'A'. Identify A and B, write the equation.

$$ \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \rightleftharpoons \text{CuSO}_4 + 5\text{H}_2\text{O} $$

(A) : $ \text{CuSO}_4 \cdot 5\text{H}_2\text{O} $ : Copper sulphate Pentahydrate - Blue

(B) : $ \text{CuSO}_4 $ : Anhydrous Copper sulphate - White
Will the cool drinks give more fizz at top of the hills or at the foot? Explain

Cool drinks will fizz more at the top of the hill since when pressure decrease solubility of a gas in liquid decreases so $ \text{CO}_2 $ bubbles out as a gas in a soda can.

Atoms and Molecules: Class 10 Science Chapter 7 Book Back Questions and Answers

Atoms and Molecules - Book Back Questions

I. Choose the best answer.

Which of the following has the smallest mass?

a. 6.023 × 10²³ atoms of He

b. 1 atom of He

c. 2 g of He

d. 1 mole atoms of He
Which of the following is a triatomic molecule?

a. Glucose

b. Helium

c. Carbon dioxide

d. Hydrogen
The volume occupied by 4.4 g of CO₂ at S.T.P

a. 22.4 litre

b. 2.24 litre

c. 0.24 litre

d. 0.1 litre
Mass of 1 mole of Nitrogen atom is

a. 28 amu

b. 14 amu

c. 28 g

d. 14 g
Which of the following represents 1 amu?

a. Mass of a C – 12 atom

b. Mass of a hydrogen atom

c. 1/12^th of the mass of a C – 12 atom

d. Mass of O – 16 atom
Which of the following statement is incorrect?

a. One gram of C – 12 contains Avogadro’s number of atoms.

b. One mole of oxygen gas contains Avogadro’s number of molecules.

c. One mole of hydrogen gas contains Avogadro’s number of atoms.

d. One mole of electrons stands for 6.023 × 10²³ electrons.
The volume occupied by 1 mole of a diatomic gas at S.T.P is

a. 11.2 litre

b. 5.6 litre

c. 22.4 litre

d. 44.8 litre
In the nucleus of $^{40}_{20}\text{Ca}$, there are

a. 20 protons and 40 neutrons

b. 20 protons and 20 neutrons

c. 20 protons and 40 electrons

d. 40 protons and 20 electrons
The gram molecular mass of oxygen molecule is

a. 16 g

b. 18 g

c. 32 g

d. 17 g
1 mole of any substance contains ____ molecules.

a. 6.023 × 10²³

b. 6.023 × 10^-23

c. 3.0115 × 10²³

d. 12.046 × 10²³

II. Fill in the blanks

Atoms of different elements having same mass number, but different atomic numbers are called isobars.
Atoms of different elements having same number of neutrons are called isotones.
Atoms of one element can be transmuted into atoms of other element by artificial transmutation.
The sum of the numbers of protons and neutrons of an atom is called its mass number.
Relative atomic mass is otherwise known as standard atomic weight.
The average atomic mass of hydrogen is 1.0079 amu.
If a molecule is made of similar kind of atoms, then it is called Homo atomic molecule.
The number of atoms present in a molecule is called its atomicity.
One mole of any gas occupies 22400 ml at S.T.P.
Atomicity of phosphorous is 4.

III. Match the following

Column A	Column B (Correct Match)
1. 8 g of O₂	0.25 moles
2. 4 g of H₂	2 moles
3. 52 g of He	13 moles
4. 112 g of N₂	4 moles
5. 35.5 g of Cl₂	0.5 moles

IV. True or False: (If false give the correct statement)

Two elements sometimes can form more than one compound.

Answer: True
Noble gases are Diatomic.

Answer: False

Correct statement: Noble gases are monoatomic.
The gram atomic mass of an element has no unit.

Answer: False

Correct statement: The gram atomic mass of an element is expressed in grams.
1 mole of Gold and Silver contain same number of atoms.

Answer: True
Molar mass of CO₂ is 42g.

Answer: False

Correct statement: Molar mass of CO₂ is 44g.

V. Assertion and Reason

Answer the following questions using the data given below:

i) A and R are correct, R explains the A.
ii) A is correct, R is wrong.
iii) A is wrong, R is correct.
iv) A and R are correct, R doesn’t explains A.

Assertion: Atomic mass of aluminium is 27.

Reason: An atom of aluminium is 27 times heavier than 1/12th of the mass of the C – 12 atom.

Answer: i) A and R are correct, R explains the A.
Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.

Reason: The natural abundance of Chlorine isotopes are not equal.

Answer: i) A and R are correct, R explains the A.

VI. Short answer questions

Define: Relative atomic mass.

The Relative Molecular Mass of a molecule is the ratio between the mass of one molecule of the substance to 1/12th mass of an atom of Carbon-12.

Write the different types of isotopes of oxygen and its percentage abundance.

Oxygen exists as a mixture of three stable isotopes in nature.

Isotope	Mass (amu)	% abundance
$^{16}_8\text{O}$	15.9949	99.757
$^{17}_8\text{O}$	16.9991	0.038
$^{18}_8\text{O}$	17.9992	0.205

Define: Atomicity

The number of atoms present in the molecule is called as its atomicity.
Give any two examples for heterodiatomic molecules.

Examples of heterodiatomic molecules are HCl, CO.
What is Molar volume of a gas?

(i) The volume occupied by one mole of any gas at S.T.P is called molar volume.

(ii) One mole of any gas occupies = 22.4 litre or 22400 ml at S.T.P.
Find the percentage of nitrogen in ammonia.

The percentage of nitrogen in ammonia ($\text{NH}_3$) is calculated as:
$$ \% \text{ of Nitrogen in NH}_3 = \frac{\text{Mass of Nitrogen}}{\text{Molar Mass of NH}_3} \times 100 $$ $$ = \frac{14}{17} \times 100 \approx 82\% $$

VII. Long answer questions

Calculate the number of water molecule present in one drop of water which weighs 0.18 g.

Answer:

Mass of water = 0.18 g

Molecular Mass of water ($\text{H}_2\text{O}$) = 18 g/mol
$$ \text{No of moles} = \frac{\text{Given Mass}}{\text{Molecular Mass}} = \frac{0.18}{18} = 0.01 \text{ mole} $$ $$ \text{Number of molecules} = \text{No of moles} \times \text{Avogadro number} $$ $$ = 0.01 \times 6.023 \times 10^{23} $$ $$ = 6.023 \times 10^{21} \text{ molecules} $$
N₂ + 3 H₂ → 2 NH₃

(The atomic mass of nitrogen is 14, and that of hydrogen is 1)

1 mole of nitrogen (____ g) + 3 moles of hydrogen (____ g) → 2 moles of ammonia (____ g)

Answer: 1 mole of nitrogen (28 g) + 3 moles of hydrogen (6 g) → 2 moles of ammonia (34 g)
Calculate the number of moles in

i) 27g of Al ii) 1.51 × 10²³ molecules of NH₄Cl

Answer:

(i) 27 g of Al
$$ \text{No of moles} = \frac{\text{Mass}}{\text{Atomic Mass}} = \frac{27}{27} = 1 \text{ mole} $$
(ii) 1.51 x 10²³ molecules of NH₄Cl
$$ \text{No of moles} = \frac{\text{Number of molecules}}{\text{Avogadro number}} $$ $$ = \frac{1.51 \times 10^{23}}{6.023 \times 10^{23}} = 0.25 \text{ mole} $$
Give the salient features of “Modern atomic theory”.

Answer:

The main postulates of modern atomic theory are as follows:
- (i) An atom is no longer indivisible.
- (ii) Atoms of the same element may have different atomic mass. (e.g., Isotopes $^{35}_{17}\text{Cl}$, $^{37}_{17}\text{Cl}$).
- (iii) Atoms of different elements may have same atomic masses. (e.g., Isobars $^{40}_{18}\text{Ar}$, $^{40}_{20}\text{Ca}$).
- (iv) Atoms of one element can be transmuted into atoms of other elements.
- (v) Atoms may not always combine in a simple whole number ratio. (e.g., Glucose C₆H₁₂O₆).
- (vi) Atom is the smallest particle that take part in a chemical reaction.
- (vii) Mass of an atom can be converted into energy, according to the equation $E = mc^2$.
Derive the relationship between Relative molecular mass and Vapour density.

Answer:

Relative molecular mass (Hydrogen scale): The Relative Molecular Mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of Hydrogen.

Vapour Density: Vapour density is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
$$ \text{Vapour Density (VD)} = \frac{\text{Mass of a given volume of gas or vapour at S.T.P}}{\text{Mass of the same volume of hydrogen}} $$
According to Avogadro's law, equal volumes of all gases contain an equal number of molecules. Let this number be 'n'.
$$ \text{VD} = \frac{\text{Mass of 'n' molecules of gas or vapour}}{\text{Mass of 'n' molecules of hydrogen}} $$
Cancelling 'n', we get:
$$ \text{VD} = \frac{\text{Mass of 1 molecule of gas or vapour}}{\text{Mass of 1 molecule of hydrogen}} $$
Since hydrogen is diatomic (H₂), one molecule of hydrogen contains 2 atoms.
$$ \text{VD} = \frac{\text{Mass of 1 molecule of gas or vapour}}{2 \times \text{Mass of 1 atom of hydrogen}} $$
From the definition of relative molecular mass:
$$ \text{Relative Molecular Mass} = \frac{\text{Mass of 1 molecule of gas or vapour}}{\text{Mass of 1 atom of hydrogen}} $$
Substituting this into the V.D. equation:
$$ \text{VD} = \frac{\text{Relative Molecular Mass}}{2} $$
Therefore, the relationship is:
$$ \text{Relative Molecular Mass} = 2 \times \text{Vapour Density} $$

VIII. HOT question

Calcium carbonate is decomposed on heating in the following reaction:

$$ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 $$

i. How many moles of Calcium carbonate are involved in this reaction?

ii. Calculate the gram molecular mass of calcium carbonate involved in this reaction.

iii. How many moles of CO₂ are there in this equation?

Answer:

(i) The number of moles of CaCO₃ involved is 1 mole.

(ii) Gram molecular mass of CaCO₃ = (Mass of Ca) + (Mass of C) + (3 × Mass of O)

= 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100 g/mol.

(iii) The number of moles of CO₂ in the reaction is 1 mole.

IX. Solve the following problems

How many grams are there in the following?

i. 2 moles of hydrogen molecule, H₂

ii. 3 moles of chlorine molecule, Cl₂

iii. 5 moles of sulphur molecule, S₈

iv. 4 moles of phosphorous molecule, P₄

Answer:

(i) 2 moles of hydrogen molecule (H₂)

Mass = No. of moles × Molecular mass = 2 × 2 = 4 g

(ii) 3 moles of Chlorine molecule (Cl₂)

Mass = No. of moles × Molecular mass = 3 × 70.9 = 212.7 g

(iii) 5 moles of sulphur molecule (S₈)

Mass = No. of moles × Molecular mass = 5 × (32 × 8) = 5 × 256 = 1280 g

(iv) 4 moles of phosphorous molecule (P₄)

Mass = No. of moles × Molecular mass = 4 × (31 × 4) = 4 × 124 = 496 g
Calculate the % of each element in calcium carbonate. (Atomic mass: C-12, O-16, Ca-40)

Answer: Molar mass of CaCO₃ = 100 g/mol.
$$ \% \text{ of Ca} = \frac{40}{100} \times 100 = 40\% $$ $$ \% \text{ of C} = \frac{12}{100} \times 100 = 12\% $$ $$ \% \text{ of O} = \frac{3 \times 16}{100} \times 100 = \frac{48}{100} \times 100 = 48\% $$
Calculate the % of oxygen in Al₂(SO₄)₃. (Atomic mass: Al-27, O-16, S-32)

Answer:

Molar mass of Al₂(SO₄)₃ = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342 g/mol.
$$ \% \text{ of Oxygen} = \frac{\text{Mass of Oxygen in compound}}{\text{Total Molar Mass}} \times 100 $$ $$ = \frac{192}{342} \times 100 \approx 56.1\% $$
Calculate the % relative abundance of B-10 and B-11, if its average atomic mass is 10.804 amu.

Answer:

Let the % abundance of B-10 be $x$ and B-11 be $y$.

$ x + y = 100 \implies y = 100 - x $

Atomic mass of B-10 = 10.01294 amu

Atomic mass of B-11 = 11.00931 amu (using a more standard value)
$$ \text{Average Atomic Mass} = \frac{( \text{Mass}_1 \times x ) + ( \text{Mass}_2 \times y )}{100} $$ $$ 10.804 = \frac{(10.01294 \times x) + (11.00931 \times (100 - x))}{100} $$ $$ 1080.4 = 10.01294x + 1100.931 - 11.00931x $$ $$ 1080.4 = 1100.931 - 0.99637x $$ $$ 0.99637x = 1100.931 - 1080.4 $$ $$ 0.99637x = 20.531 $$ $$ x = \frac{20.531}{0.99637} \approx 20.6\% $$
So, $ y = 100 - 20.6 = 79.4\% $

The relative abundance of B-10 is approximately 20.6% and B-11 is 79.4%.