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NCERT Solutions for Class 9 Science Chapter 3 - Atoms and Molecules
- NCERT Solutions
- Chapter 3 Atoms And Molecules
NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules - FREE PDF Download
The updated NCERT Solutions for Chapter 3 atoms and Molecules Class 9 is now available on Vedantu. Our subject experts prepare these solutions with close reference to the latest NCERT Class 9 Science textbook edition.
Students can download the PDF and refer to these solutions for free from our website. All the important topics and sub-topics covered in Class 9 science chapter 3 have been included in these solutions according to the latest Class 9 CBSE Science Syllabus . Therefore, students can rely upon NCERT Solutions for Class 9 Science for their exam preparation.
Glance on NCERT Solutions for Class 9 Science Chapter 3
Atoms and molecules class 9 NCERT will embark on a fascinating journey to understand fundamental concepts such as mass, atoms, and molecules. The total mass of the reactants and products remains constant during a chemical reaction. This is referred to as the Mass Conservation Law.
Class 9 science chapter 3 question answers deal with all the solutions related to the concepts such as elements, and compounds. Elements are always present in a defined proportion by mass in pure chemical composition. It is called the Law of Definite Proportions.
A molecule is the tiniest particle of an element or compound that may exist on its own under normal conditions. It displays all the qualities or properties of the compound.
A compound's chemical formula lists its constituent elements, as well as the number of atoms in each combining element.
Atoms and Molecules class 9 questions and answers can help students analyse their level of preparation and understanding of concepts.
There are two laws of chemical combination- the law of conservation of mass, and the law of constant proportion. Both these laws are discussed in detail in class 9 science chapter 3 PDF.
There is one important question related to the chemical reaction. Is there any change in the total mass of the system when a chemical reaction takes place? According to the law of conservation of mass, mass can neither be created nor be destroyed. The same logic stands true for chemical reactions as well. Several experiments carried out by Lavoisier standardized this law. This law is discussed in NCERT Solutions for class 9 Science chapter 3.
Access NCERT Solutions for Science Class 9 Chapter 3 – Atoms and Molecules
Intext Exercise 1
1. In a reaction, 5.3g of sodium carbonate reacted with 6g of ethanoic acid. The products were 2.2g of carbon dioxide, 0.9g water and 8.2g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Ans: Given,
Mass of sodium carbonate = 5.3 g
Mass of ethanoic acid = 6 g
Mass of sodium ethanoate = 8.2 g
Mass of carbon dioxide = 2.2 g
Mass of water = 0.9 g
Now, total mass before the reaction \[ = {\text{ }}\left( {5.3{\text{ }} + {\text{ }}6} \right){\text{ }}g{\text{ }} = {\text{ }}11.3{\text{ }}g\]
And, total mass after the reaction =\[\;\left( {8.2{\text{ }} + {\text{ }}2.2{\text{ }} + {\text{ }}0.9} \right){\text{ }}g{\text{ }} = {\text{ }}11.3{\text{ }}g\]
∴Total mass before the reaction = Total mass after the reaction
Hence, this is in agreement with the law of conservation of mass.
2. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Ans: In water, H:O (by mass) =$1\;:\;8$
The mass of oxygen gas required to react completely with 1 g of hydrogen gas = 8 g.
the mass of oxygen gas required to react completely with 3 g of hydrogen gas = \[(8\; \times \;3)\;g\; = \;24\;g\]
3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Ans: The postulate of Dalton’s atomic theory which is based on the law of conservation of mass is: “Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.”
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Ans: “The elements consist of atoms having fixed mass and that the number and kind of atoms of each element in a given compound is fixed.” This explains the law of definite proportion.
Intext Exercise 2
1. Define atomic mass unit.
Ans: Mass unit equal to exactly one-twelfth the $\left( {\dfrac{1}{{{{12}^{th}}}}} \right)$mass of one atom of carbon-12 is called one atomic mass unit. It is represented by as ‘a.m.u.’ or ‘u’.
2. Why is it not possible to see an atom with naked eyes?
Ans: Due to small size of an atom we cannot see them with naked eyes.
Intext Exercise 3
1. Write down the formulae of
(i) Sodium oxide
Ans: Sodium Oxide: \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}\]
(ii) Aluminium chloride
Ans: Aluminium chloride: \[{\text{AlC}}{{\text{l}}_{\text{3}}}\]
(iii) sodium sulphide
Ans: Sodium sulphide: \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}\]
(iv) Magnesium hydroxide
Ans: Magnesium hydroxide: ${\text{Mg(OH}}{{\text{)}}_{\text{2}}}$
2. Write down the names of compounds represented by the following formulae:
i) $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right) \mathrm{_3}$
Ans: Aluminium sulphate
ii) ${\text{CaC}}{{\text{l}}_{\text{2}}}$
Ans: Calcium chloride
iii) ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
Ans: :Potassium sulphate
iv) ${\text{KN}}{{\text{O}}_3}$
Ans: Potassium nitrate
v) \[{\text{CaC}}{{\text{O}}_{\text{3}}}\]
Ans: : Calcium carbonate
3. What is meant by the term chemical formula?
Ans: The symbolic representation of composition of a compound is known as chemical formula. Chemical formula gives us the idea of number of atoms present.
Example: from the chemical formula ${\text{C}}{{\text{O}}_{\text{2}}}$ of Carbon Dioxide, we come to know that one carbon atom and two oxygens atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.
4. How many atoms are present in a:
i) $\mathrm{H}_{2}$ S molecule
Ans: There are total 3 atoms present in ${{\text{H}}_{\text{2}}}{\text{S}}$ molecule, two hydrogen atoms and one Sulphur atom.
ii) \[{\text{PO}}_4^{3 - }\] ion
Ans: There are total 5 atoms in \[{\text{PO}}_4^{3 - }\] ion, one phosphorus atom and 4 oxygen atoms.
Intext Exercise 4
1. Calculate the molecular masses of \[{{\text{H}}_2},{{\text{O}}_2},{\text{C }}{{\text{l}}_2},{\text{C}}{{\text{O}}_2},{\text{C}}{{\text{H}}_4},{{\text{C}}_2}{{\text{H}}_6},{{\text{C}}_2}{{\text{H}}_4},{\text{N}}{{\text{H}}_3},{\text{C}}{{\text{H}}_3}{\text{OH}}.\]
Ans: Molecular mass of ${{\text{H}}_{\text{2}}}$ = $2\; \times $ Atomic mass of H
$ = \;2\; \times \;1\;{\text{u = }}\;{\text{2 u}}$
Molecular mass of ${{\text{O}}_{\text{2}}}$ = $2\; \times $ Atomic mass of O
$ = \;2\; \times \;16\;{\text{u = }}\;3{\text{2 u}}$
Molecular mass of ${\text{C }}{{\text{l}}_{\text{2}}}$ = $2\; \times $ Atomic mass of Cl
$ = \;2\; \times \;35.5\;{\text{u = }}\;71{\text{ u}}$
Molecular mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ = Atomic mass of C $ + $ $2\; \times $Atomic mass of O
$ = \;(12 + 2\; \times \;16)\;{\text{u = }}\;44{\text{ u}}$
Molecular mass of ${\text{C}}{{\text{H}}_4}$ = Atomic mass of C $ + $ $4\; \times $Atomic mass of H
$ = \;(12 + 4\; \times \;1)\;{\text{u = }}\;16{\text{ u}}$
Molecular mass of ${{\text{C}}_2}{{\text{H}}_6}$ = $2\; \times $Atomic mass of C $ + $ $6\; \times $Atomic mass of H
$ = \;(2 \times 12 + 6\; \times \;1)\;{\text{u = }}\;30{\text{ u}}$
Molecular mass of ${{\text{C}}_2}{{\text{H}}_4}$ = $2\; \times $Atomic mass of C $ + $ $4\; \times $Atomic mass of H
Molecular mass of ${\text{N}}{{\text{H}}_3}$ = Atomic mass of N $ + $ $3\; \times $Atomic mass of H
$ = \;(14 + 3\; \times \;1)\;{\text{u = }}\;17{\text{ u}}$
Molecular mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ = Atomic mass of C $ + $ $3\; \times $Atomic mass of H $ + $ Atomic mass of O
$ + $ Atomic mass of $ = \;(12 + 4\; \times \;1\; + \;16)\;{\text{u = }}\;32{\text{ u}}$
2. Calculate the formula unit masses of \[{\text{ZnO}},\;{\text{N}}{{\text{a}}_2}{\text{O}},\;{{\text{K}}_2}{\text{C}}{{\text{O}}_3}\]given atomic masses of \[{\text{Z}} = 65{\text{u}},{\text{Na}} = 23\;{\text{u}},\;{\text{K}} = 39\;{\text{u}},\;{\text{C}} = 12{\text{u}},\;{\text{and O}} = 16{\text{u}}{\text{.}}\]
Ans: Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
\[ = \;(65 + 16)\;{\text{u = 81}}\;{\text{u}}\]
Formula unit mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}$ = $2\; \times $ Atomic mass of Na + Atomic mass of O
= \[(2\; \times \;23\; + \;16)\;{\text{u}}\; = \;62\;{\text{u}}\]
Formula unit mass of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ = $2\; \times $ Atomic mass of K + Atomic mass of C + $3\; \times $ Atomic mass O=
$=(2 \times 39+12+3 \times 16) \mathrm{u} $
$=138 \mathrm{u}$
Refer to page 42.
1. If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Ans: Given mass of One mole of carbon atoms = 12 g
Therefore , Mass of \[6.022 \times {10^{23}}\] number of carbon atoms = 12 g
Mass of 1 atom of carbon will be:
$ = \dfrac{{12}}{{6.022 \times {{10}^{23}}}}g$
$ = 1.9927 \times {{10}^{ - 23}}{\text{g}}$
2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?
Ans: Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = \[6.022 \times {10^{23}}\] number of Na atoms
Thus, 100 g of Na contains = \[\dfrac{{6.022 \times {{10}^{23}}}}{{23}} \times 100\] number of Na atoms
\[ = 2.6182 \times {10^{24}}{\text{number of}}\;{\text{Na}}\;{\text{atoms}}\]
Atomic mass of Fe = 56 u (Given)
Then, gram atomic mass of Fe = 56 g
Now, 56 g of Fe contains = \[6.022 \times {10^{23}}\] number of Fe atoms
Thus, 100 g of Fe contains = \[\dfrac{{6.022 \times {{10}^{23}}}}{{56}} \times 100\] number of Fe atoms
\[ = 1.0753 \times {10^{24}}{\text{number of Fe atoms}}\]
\[2.6182 \times {10^{24}} > 1.0753 \times {10^{24}}\]
Therefore, 100 grams of sodium contain a greater number of atoms than 100 grams of iron.
NCERT QUESTIONS:
1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Given,
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Mass of sample = 0.24 g
The percentage of boron by weight in the compound \[ = \dfrac{{0.096}}{{0.24}} \times 100\% \; = \;40\;\% \]
And, percentage of oxygen by weight in the compound \[ = \dfrac{{0.144}}{{0.24}} \times 100\% \; = \;60\;\% \]
2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?
Ans: Carbon + Oxygen ⎯⎯→ Carbon dioxide
3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.
If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen to form11 g of carbon dioxide.
The remaining (50 –8) = 42 g of oxygen will be left unreacted.
The above answer is governed by the law of constant proportions.
3. What are polyatomic ions? Give examples?
Ans: A polyatomic ion is a group of atoms carrying a charge either positive or negative.
For example,\[\;{\text{ammonium ion}}\;\left( {{\text{NH}}_4^ + } \right),\;{\text{hydroxide ion}}\;\left( {{\text{O}}{{\text{H}}^ - }} \right),\;{\text{carbonate ion}},\;\left( {{\text{CO}}_3^{2 - }} \right){\text{,}}\;\;{\text{sulphate ion}}\;\left( {{\text{SO}}_4^{2 - }} \right)\]
4. Write the chemical formulae of the following:
(a) Magnesium chloride
Ans: \[{\text{MgC}}{{\text{l}}_{\text{2}}}\]
(b) Calcium oxide
(c) Copper nitrate
Ans: \[{\text{Cu}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}\]
(d) Aluminium chloride
Ans: \[{\text{AlC}}{{\text{l}}_{\text{3}}}\]
(e) Calcium carbonate
Ans: \[{\text{CaC}}{{\text{O}}_{\text{3}}}\]
5. Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
6. Calculate the molar mass of the following substances:
a) Ethyne ${{\text{C}}_{_{\text{2}}}}{{\text{H}}_{\text{2}}}$
Ans: Molar mass of \[{{\text{C}}_2}{{\text{H}}_2} = 2 \times 12 + 2 \times 1 = 28{\text{g}}/{\text{mol}}\]
b) Sulphur molecule, ${{\text{S}}_{\text{8}}}$
Ans: Molar mass of \[{{\text{S}}_8} = 8 \times 32 = 256{\text{g}}/{\text{mol}}\]
c) Phosphorus molecule \[{{\text{P}}_{\text{4}}}\] (atomic mass of phosphorus = 31)
Ans: Molar mass of \[{{\text{P}}_4} = 4 \times 31 = 124{\text{g}}/{\text{mol}}\]
c) Hydrochloric acid, HCl
Ans: Molar mass of \[{\text{HCl}} = 1 + 35.5 = 36.5{\text{g}}/{\text{mol}}\]
d) Nitric acid, \[{\text{HN}}{{\text{O}}_{\text{3}}}\]
Ans: Molar mass of \[{\text{HN}}{{\text{O}}_3} = 1 + 14 + 3 \times 16 = 63{\text{g}}/{\text{mol}}\]
7. What is the mass of:
(a) 1 mole of nitrogen atoms?
Ans: The mass of 1 mole of N- atoms = 14 g
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
Ans: \[[{\text{Atomic mass of}}\;{\text{Al}} = 27{\text{u}}]\]
The mass of 4 moles of Al-atoms \[ = (4 \times 27){\text{g}}\;{\text{ = }}\;\;{\text{108}}\;{\text{g}}\]
(c) 10 moles of sodium sulphite \[\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}} \right)\]?
Ans: Atomic mass of Na = 23 u, Atomic mass of S = 32 u, Atomic mass of O = 16 u
The mass of 10 moles of sodium sulphite \[\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}} \right)\]\[ = 10 \times [2 \times 23 + 32 + 3 \times 16]{\text{g}}\]
\[ = 10 \times 126{\text{g}} = 1260{\text{g}}\]
Topics Covered in Class 9 Science Chapter 3 Atoms and Molecules
Deleted topics in class 9 science chapter 3 atoms and molecules.
Mole Concept
Benefits of NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules
NCERT Solutions for Class 9 Science Chapter 3 has the following features.
Chapter 3 Science Class 9, is presented comprehensively.
All the concepts covered in the Atoms and molecules class 9 are explained with the help of relevant examples in these NCERT Solutions.
The Class 9 science chapter 3 question answers are solved as per the CBSE guidelines so that students can score good marks in the examination.
The pointwise approach of atoms and molecules class 9 questions and answers help students revise the chapter before the examination.
The topics covered in toms and molecules class 9 PDF become easily understandable when students go through the atoms and molecules class 9 NCERT solutions.
Important Links for Science Class 9 Chapter 3 Atoms and Molecules
NCERT Solutions for Atoms and Molecules class 9 by Vedantu is comprehensive and covers all the important concepts in the CBSE syllabus . They are also aligned with the latest CBSE exam pattern and the types of questions that are asked in the exams. Therefore, students who study NCERT Solutions class 9 atoms and molecules PDF from Vedantu are more likely to do well in the CBSE Exam. Specifically, for Chapter 3, students should mandatorily study all the NCERT Solutions to get a good understanding of the concepts and the types of questions that are asked in the exam. Class 9 Science Chapter 3 PDF is very important, and it covers a lot of the material that is tested on the CBSE Exam. We encourage students to download NCERT Solutions for Chapter 3 and to use them to prepare for the CBSE Exam.
NCERT Solutions for Class 9 Science - Other Chapter-Wise Links
Given below are the links for the other chapter-wise NCERT Solutions for Class 9 Science. These solutions are provided by the Science experts at Vedantu in a detailed manner. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
Important Study Material Links for Class 9 Science
Faqs on ncert solutions for class 9 science chapter 3 - atoms and molecules.
1. What is an Atom from atoms and molecules class 9?
When you refer to class 9 science chapter 3 question answers, you will have a clear concept about atoms. Atoms are the smallest component of any matter that can exist independently in nature. However, some atoms cannot exist freely and form molecules with the same type of atoms or form compounds with different atoms. Go through the class 9th science chapter 3 for a detailed explanation of the concepts related to atoms and molecules.
2. What is a Molecule, as per the NCERT solutions from class 9 science chapter 3?
When atoms combine, they form a molecule. A minimum of two atoms can form a molecule. A molecule can be made of the same or different atoms. To completely understand the concept of molecules, refer to the atoms and molecules class 9 solutions.
3. Mention the topics included in Chapter 3 Atoms and Molecules of NCERT Solutions for Class 9 Science.
The topics that are included in Chapter 3 of NCERT Solutions for Class 9 Science are Law of Chemical Combination, What is an Atom, What is a Molecule, Writing Chemical Formulae, Molecular Mass and Mole Concept. These topics are in a strategic manner such that they best serve the NCERT and impart the right kind of knowledge at the same time. The questions from these topics are the most frequent in the examination, and they are expanded upon in higher classes, so it’s best if the students do them thoroughly. Students can refer to atoms and molecules class 9 questions and answers to get all the solved answers related to this chapter.
4. What are the key features of NCERT Solutions for Chapter 3 Atoms and Molecules of Class 9 Science?
There are several features in NCERT Solutions for Class 9 Science Chapter 3.
The information in Chapter 3 Science Class 9 is provided in detail. In these NCERT Solutions, all the topics discussed in the Atom and Molecules Class 9 chapter are explained with applicable examples. The NCERT Class 9 Science Chapter 3 practice assignment is answered according to CBSE rules so that students may do well in the test. Chemistry Chapter 3 Class 9 NCERT answers use a point-by-point method to assist students in reviewing the material before the test. When students work through the atoms and molecules class 9 NCERT answers, the concepts covered in Chapter 3 Science Class 9 become clear.
5. What is the chemical formula of compounds as per the NCERT solutions of Chapter 3 atoms and molecules class 9 PDF?
Chemical formulas are believed to be the symbolic representation of substances. The net charge of each ion in the formula is taken into account in the chemical formula. The radical's valency is determined by the charge of individual ions. The valency of individual ions affects how many of each type of ion will participate in the interaction. The valency of individual ions must be balanced in order to properly represent the chemical formula. Metal names should appear first, followed by non-metal names.
6. What is a molecule made up of as per the NCERT solutions of class 9 chapter 3?
A molecule is made up of two or more atoms linked together. In nature, a molecule can exist without restriction. Hydrogen, oxygen, and calcium molecules, for example, can exist freely in nature. Molecules can be made from the same atoms or from atoms that are different. When molecules are made up of the same type of atoms, they are referred to as element molecules. Elements such as sodium and potassium, for example. Molecules can also be formed from atoms with different properties. A compound is formed when various atoms join together.
7. What is atomic mass as per the NCERT solutions from class 9 science chapter 3?
The atomic mass unit is the unit of mass for each atom. The atomic mass is an atom's property. The law of constant proportion and the law of mass conservation are both supported by the idea of atomic mass. The NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules provides a thorough examination of atomic mass. There are a lot of properties of atoms that are discussed in Chapter 3 of Class 9 Science NCERT.
8. What are some important questions from class 9 science chapter 3?
Students can expect questions from:
Law of Chemical Combustion
Molecular Mass
Chemical Formulae
Students can download and refer to the class 9 atoms and molecules PDF to learn and practice all the solved questions.
9. Do NCERT Solutions for Chapter 3 class 9 cover all question types?
Yes, NCERT Solutions typically covers all the question types covered in this chapter’s exercise section, including:
Multiple choice questions
Short answer questions
Balancing chemical equations
Calculation-based problems
10. What are some important topics covered in this chapter?
Here are some important topics:
Structure of the atom
Atomic mass unit
Distinguish between atoms and molecules
Types of molecules
Writing chemical formulae
Law of constant proportions
Students can access Atoms and Molecules class 9 PDF for their exam preparation.
NCERT Solutions for Class 9 Science
Ncert solutions for class 9.
Atoms and Molecules
Class 9 - ncert science solutions, intext questions 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + acetic acid ⟶ Sodium acetate + carbon dioxide + water
As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products.
5.3 g + 6 g ⟶ 8.2 g + 2.2 g + 0.9 g
⇒ 11.3 g ⟶ 11.3 g
As per the above reaction, L.H.S. = R.H.S. = 11.3 g
Hence, the observations are in agreement with the law of conservation of mass.
Hydrogen and oxygen combine in a ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
hydrogen : water = 1 : 8
So, for every 1 g of hydrogen, 8 g of oxygen is required
∴ For 3 g of hydrogen, oxygen required is 8 1 \dfrac{8}{1} 1 8 x 3 = 24 g
Hence, 24 g of oxygen would be required for the complete reaction with 3 g of hydrogen gas.
Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?
One of the postulates of Dalton's atomic theory states that : 'Atoms cannot be created nor be destroyed in a chemical reaction'. This law is the result of law of conservation of mass.
Which postulate of Dalton's atomic theory can explain the law of definite proportions?
The postulate of Dalton's atomic theory that can explain the law of definite proportions is : 'Relative number and kinds of atoms are equal in given compounds.'
Intext Questions 2
Define the atomic mass unit.
Atomic mass unit is defined as 1⁄12 th the mass of an carbon atom C-12.
Why is it not possible to see an atom with naked eyes?
It is not possible to see an atom with naked eyes because:
- Atoms are very small in size, measured in nanometres.
- Except for atoms of noble gases, they do not exist independently.
Intext Questions 3
Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
(i) Sodium oxide:
Step 1 — Write each symbol with its valency
Na 1 + ↗ O 2 − \text{Na}^{1+} \phantom{\nearrow} \text{O}^{2-} Na 1 + ↗ O 2 −
Step 2 — Interchange the valencies
Na 2 1 ↘ ↙ O 2 ⇒ Na 1 2 ↘ ↙ O 1 \overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{O}} \\[0.5em] Na 2 1 ↘ ↙ O 2 ⇒ 1 2 Na ↘ ↙ 1 O
Step 3 — Write the interchanged number, ignore equal numbers & hence the formula
Therefore, we get
Formula of Sodium oxide : N a 2 O \bold{Na_2O} N a 2 O
(ii) Aluminium chloride
Al 3 + ↗ Cl 1 − \text{Al}^{3+} \phantom{\nearrow} \text{Cl}^{1-} Al 3 + ↗ Cl 1 −
Al 2 3 ↘ ↙ Cl 1 ⇒ Al 1 1 ↘ ↙ Cl 3 \overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \\[0.5em] Al 2 3 ↘ ↙ Cl 1 ⇒ 1 1 Al ↘ ↙ 3 Cl
Formula of Aluminium chloride : A l C l 3 \bold{AlCl_3} AlC l 3
Na 1 + ↗ S 2 − \text{Na}^{1+} \phantom{\nearrow} \text{S}^{2-} Na 1 + ↗ S 2 −
Na 2 1 ↘ ↙ S 2 ⇒ Na 1 2 ↘ ↙ S 1 \overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{S}} \\[0.5em] Na 2 1 ↘ ↙ S 2 ⇒ 1 2 Na ↘ ↙ 1 S
Formula of Sodium sulphide : N a 2 S \bold{Na_2S} N a 2 S
(iv) Magnesium hydroxide
Mg 2 + ↗ OH 1 − \text{Mg}^{2+} \phantom{\nearrow} \text{OH}^{1-} Mg 2 + ↗ OH 1 −
Mg 2 2 ↘ ↙ OH 1 ⇒ Mg 1 1 ↘ ↙ OH 2 \overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \\[0.5em] Mg 2 2 ↘ ↙ OH 1 ⇒ 1 1 Mg ↘ ↙ 2 OH
Formula of Magnesium hydroxide : M g ( O H ) 2 \bold{Mg(OH)_2} Mg ( OH ) 2
Write down the names of compounds represented by the following formulae:
(i) Al 2 (SO 4 ) 3
(ii) CaCl 2
(iii) K 2 SO 4
(i) Al 2 (SO 4 ) 3 — Aluminium sulphate
(ii) CaCl 2 — Calcium chloride
(iii) K 2 SO 4 — Potassium sulphate
(iv) KNO 3 — Potassium nitrate
(v) CaCO 3 — Calcium carbonate
What is meant by the term chemical formula?
The chemical formula of a compound is a symbolic representation of its composition. It denotes in a compound, the number of atoms of each element present.
How many atoms are present in a
(i) H 2 S molecule and
(ii) PO 4 3- ion ?
(i) In one molecule of H 2 S, 2 atoms of hydrogen and 1 atom of sulphur are present. Hence, 3 atoms in total are present.
(ii) In one ion [PO 4 3- ], 1 atom of phosphorus and 4 atoms of oxygen are present. Hence, 5 atoms in total are present.
Intext Questions 4
Calculate the molecular masses of
The molecular mass of H 2 = 2 x atoms atomic mass of H = 2 x 1u = 2u
The molecular mass of O 2 = 2 x atoms atomic mass of O = 2 x 16u = 32u
The molecular mass of Cl 2 = 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u
The molecular mass of CO 2 = atomic mass of C + (2 x atomic mass of O) = 12 + (2 × 16)u = 12 + 32 = 44u
The molecular mass of CH 4 = atomic mass of C + (4 x atomic mass of H) = 12 + (4 x 1) = 16u
The molecular mass of C 2 H 6 = (2 x atomic mass of C) + (6 x atomic mass of H) = (2 x 12) + (6 x 1) = 24 + 6 = 30u
The molecular mass of C 2 H 4 = (2 x atomic mass of C) + (4 x atomic mass of H) = (2 x 12) + (4 x 1) = 24 + 4 = 28u
The molecular mass of NH 3 = atomic mass of N + (3 x atomic mass of H) = 14 + (3 x 1) = 17u
The molecular mass of CH 3 OH = atomic mass of C + (3 x atomic mass of H) + atomic mass of O + atomic mass of H = [12 + (3 × 1) + 16 + 1] = (12 + 3 + 17) = 32u
Calculate the formula unit masses of ZnO, Na 2 O, K 2 CO 3 , given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and O = 16u.
Atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and O = 16u
The formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u
The formula unit mass of Na 2 O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u
The formula unit mass of K 2 CO 3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u
A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Mass of the sample compound = 0.24 g,
mass of boron = 0.096 g,
mass of oxygen = 0.144 g
To calculate the percentage composition of the compound,
Percentage of boron :
= mass of boron mass of the compound \dfrac{\text{mass of boron}}{\text{mass of the compound}} mass of the compound mass of boron x 100
= 0.096 0.24 \dfrac{0.096}{0.24} 0.24 0.096 x 100
Percentage of oxygen
= mass of oxygen mass of the compound \dfrac{\text{mass of oxygen}}{\text{mass of the compound}} mass of the compound mass of oxygen x 100
= 0.144 0.24 \dfrac{0.144}{0.24} 0.24 0.144 x 100
Hence, percentage of boron = 40% and percentage of oxygen = 60% in the compound.
When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
C 3 g + O 2 8 g ⟶ CO 2 11 g \underset{\text{ 3 g}}{\text{C}} + \underset{\text{ 8 g}}{\text{O}_2} \longrightarrow \underset{\text{11 g}}{\text{CO}_2} 3 g C + 8 g O 2 ⟶ 11 g CO 2
When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
C 3 g + O 2 50 g ⟶ CO 2 ? \underset{\text{ 3 g}}{\text{C}} + \underset{\text{ 50 g}}{\text{O}_2} \longrightarrow \underset{\text{?}}{\text{CO}_2} 3 g C + 50 g O 2 ⟶ ? CO 2
Carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8. Hence, 3 g of carbon will react with 50 g of oxygen but only 8 g of it will be used in producing 11 g of CO 2 and 42 g [i.e., 50 g - 8 g] of oxygen will be left unused.
The above answer is governed by the law of constant proportions.
What are polyatomic ions? Give examples.
Cluster of atoms that act as an ion are called polyatomic ions. They carry a fixed charge on them.
Example: Hydroxide [OH - ], Cyanide [CN - ]
Write the chemical formula of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Mg 2 + ↗ Cl 1 − \text{Mg}^{2+} \phantom{\nearrow} \text{Cl}^{1-} Mg 2 + ↗ Cl 1 −
Mg 2 2 ↘ ↙ Cl 1 ⇒ Mg 1 1 ↘ ↙ Cl 2 \overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{Cl}} \\[0.5em] Mg 2 2 ↘ ↙ Cl 1 ⇒ 1 1 Mg ↘ ↙ 2 Cl
Formula of Magnesium chloride : M g C l 2 \bold{MgCl_2} MgC l 2
Ca 2 + ↗ O 2 − \text{Ca}^{2+} \phantom{\nearrow} \text{O}^{2-} Ca 2 + ↗ O 2 −
Ca 2 2 ↘ ↙ O 2 ⇒ Ca 1 2 ↘ ↙ O 2 \overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}} \\[0.5em] Ca 2 2 ↘ ↙ O 2 ⇒ 1 2 Ca ↘ ↙ 2 O
Formula of Calcium oxide : C a O \bold{CaO} CaO
Cu 2 + ↗ NO 3 1 − \text{Cu}^{2+} \phantom{\nearrow} \text{NO}_3^{1-} Cu 2 + ↗ NO 3 1 −
Cu 2 2 ↘ ↙ NO 3 1 ⇒ Cu 1 1 ↘ ↙ NO 3 2 \overset{\phantom{2}{2}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{NO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{NO}_3} \\[0.5em] Cu 2 2 ↘ ↙ NO 3 1 ⇒ 1 1 Cu ↘ ↙ 2 NO 3
Formula of copper nitrate : C u ( N O 3 ) 2 \bold{Cu(NO_3)_2} Cu ( N O 3 ) 2
(e) Calcium carbonate –
Ca 2 + ↗ CO 3 1 − \text{Ca}^{2+} \phantom{\nearrow} \text{CO}_3^{1-} Ca 2 + ↗ CO 3 1 −
Ca 2 2 ↘ ↙ CO 3 1 ⇒ Ca 1 1 ↘ ↙ CO 3 2 \overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{CO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{CO}_3} \\[0.5em] Ca 2 2 ↘ ↙ CO 3 1 ⇒ 1 1 Ca ↘ ↙ 2 CO 3
Formula of copper nitrate : C a ( C O 3 ) 2 \bold{Ca(CO_3)_2} Ca ( C O 3 ) 2
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
(a) Quick lime (CaO) — Calcium and oxygen
(b) Hydrogen bromide (HBr) — Hydrogen and bromine
(c) Baking powder (NaHCO 3 ) — Sodium, hydrogen, carbon and oxygen
(d) Potassium sulphate (K 2 SO 4 ) — Potassium, sulphur and oxygen,
Calculate the molar mass of the following substances.
(a) Ethyne, C 2 H 2
(b) Sulphur molecule, S 8
(c) Phosphorus molecule, P 4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO 3
(a) Molar mass of Ethyne C 2 H 2 = (2 x Mass of C) + (2 x Mass of H) = (2 × 12) + (2 × 1) = 24 + 2 = 26 g
(b) Molar mass of Sulphur molecule S 8 = (8 x Mass of S) = 8 x 32 = 256 g
(c) Molar mass of Phosphorus molecule, P 4 = 4 x Mass of P = 4 x 31 = 124 g
(d) Molar mass of Hydrochloric acid, HCl = Mass of H + Mass of Cl = 1 + 35.5 = 36.5 g
(e) Molar mass of Nitric acid, HNO 3 = Mass of H + Mass of N + (3 x Mass of O) = 1 + 14 + (3 × 16) = 15 + 48 = 63 g
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