Atoms and Molecules Class 9 Important Questions: CBSE Science Chapter 3

Very Short Answer Questions (1 Mark)

1. Atomic radius is measured in nanometers and

1. $1nm={{10}^{-10}}m$

2. $1m={{10}^{-10}}nm$

3. $1nm={{10}^{-9}}m$

4. $1m={{10}^{-9}}nm$

Ans: (c) $1nm={{10}^{-9}}m$

2. Symbol of Iron is –

1. $Ir$

2. $I$

3. $Fe$

4. None of these

Ans:  (c) $Fe$

3. Atomicity of Chlorine and Argon is

1. Diatomic and Monoatomic

2. Monoatomic and Diatomic

3. Monoatomic and Monoatomic

4. Diatomic and Diatomic

Ans: (a) Diatomic and Monoatomic

4. Molecular mass of water $({{H}_{2}}O)$ is

1. $18g$

2. $8g$

3. $33g$

4. $34g$

Ans:  (a) $18g$

5. It is said that 1 mole of a compound contains –

1. $6.023\times {{10}^{23}}atoms$

2. $6.023\times {{10}^{24}}atoms$

3. $60.23\times {{10}^{23}}atoms$

4. $6.023\times {{10}^{25}}atoms$

Ans: (a) $6.023\times {{10}^{23}}atoms$

6. Oxygen is –

1. Monovalent

2. Bivalent

3. Trivalent

4. Tetravalent

Ans: (b) Bivalent

7. What is the molecular formula for Calcium Hydroxide?

1. $CaO{{H}_{2}}$

2. $Ca{{(OH)}_{2}}$

3. $C{{a}_{2}}OH$

4. $Ca{{H}_{2}}$

Ans: (b) $Ca{{(OH)}_{2}}$

8. Neutron is

1. Chargeless and Massless

2. Chargeless and has Mass

3. Has charge and Mass

4. Has charge and Massless.

Ans: (b) Chargeless and has Mass

9. Which of the following statements is correct?

1. Cathode rays travel in a straight line and have momentum.

2. Cathode rays travel in a straight line and have no momentum

3. Cathode rays do not travel in a straight line but have Momentum.

4. Cathode rays do not travel in a straight line and have no momentum.

Ans: (a) Cathode rays travel in a straight line and have momentum.

10.  How are \[\beta \]–particles represented?

1. $e_{-1}^{0}$

2. ${{e}_{+1}}$

3. $e_{-1}^{1}$

4. $e_{0}^{1}$

Ans: (a) $e_{-1}^{0}$

11.  Elements $Ar_{18}^{40}$ and $Ca_{20}^{40}$ are

1. Isotopes

2. Isobars

3. Isotones

4. Both b and c

Ans:  (b) Isobars

12. The maximum number of electrons in L shell is

1. $8$

2. $18$

3. $28$

4. $38$

Ans:  (a) $8$

Short Answer Questions (3 Marks)

1. Define the atomic mass unit.

Ans: One atomic mass unit is a mass unit equal to exactly one-twelfth (${1}/{12}\;th$) the mass of one atom of $carbon-12$. The relative atomic masses of all elements have been found with respect to an atom of $carbon-12$.

According to the latest IUPAC (International Union of Pure and Applied Chemistry) recommendations, the atomic mass unit (written as ‘u’ – unified mass) is equal to the mass of one-twelfth (${1}/{12}\;th$) of$carbon-12$ atom.

$1\text{ }amu={1}/{12}\;th\text{ }Mass\text{ }Of\text{ }C_{6}^{12}$

2. Write down the formulae of

1. Sodium oxide

Ans:  Sodium oxide – $N{{a}_{2}}O$

2. Aluminium chloride

Ans:  Aluminium chloride – $AlC{{l}_{3}}$

3. Sodium sulphide

Ans:  Sodium sulphide – $N{{a}_{2}}S$

4. Magnesium hydroxide

Ans:  Magnesium hydroxide – $Mg{{(OH)}_{2}}$

3. Write down the names of compounds represented the following formulae:

1. $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$

Ans: $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$  - Aluminium sulphate

2. $CaC{{l}_{2}}$

Ans: $CaC{{l}_{2}}$ - Calcium chloride

3. \[{{K}_{2}}S{{O}_{4}}\]

Ans:  \[{{K}_{2}}S{{O}_{4}}\] - Potassium sulphate

4. \[KN{{O}_{3}}\]

Ans:  \[KN{{O}_{3}}\] - Potassium nitrate

5. $CaC{{O}_{3}}$

Ans:  $CaC{{O}_{3}}$ - Calcium carbonate

4. What is meant by the term chemical formula?

Ans: The term chemical formula of a compound is said to be the symbolic representation of its composition or it is a notation that shows the type and number of atoms in a molecule of a compound with the help of atomic symbols and numbers.

They provide information on the elements that constitute the molecules of a compound and the ratio in which the atoms of those elements combine to form the molecules.

Example: A molecule of water, which is a compound, contains two molecules of hydrogen and one molecule of oxygen. Its chemical formula is \[{{H}_{2}}O\].

5. What are polyatomic ions? Give examples.

Ans:  Polyatomic ions are a group of atoms carrying a charge. They are typically clusters of atoms that act as an ion, which carry a fixed charge on them.

Examples:

• Ammonium – \[N{{H}_{4}}^{+}\]

• Hydroxide – \[O{{H}^{-}}\]

• Nitrate – \[N{{O}_{3}}^{-}\]

• Hydrogen carbonate – \[HC{{O}_{3}}^{-}\]

6. Write the chemical formulae of the following.

1. Magnesium chloride

Ans:  Magnesium chloride – $MgC{{l}_{2}}$

2. Calcium oxide

Ans: Calcium oxide –$CaO$

3. Copper nitrate

Ans:  Copper nitrate –\[CuN{{O}_{3}}\]

4. Aluminium chloride

Ans:  Aluminium chloride –$AlC{{l}_{3}}$

5. Calcium carbonate

Ans:  Calcium carbonate – $CaC{{O}_{3}}$

7. Give the names of the elements present in the following compounds.

1. Quick lime

Ans: Quick lime –$CaO$

Elements present – Calcium, Oxygen

2. Hydrogen bromide

Ans: Hydrogen bromide –$HBr$

Elements present – Hydrogen, Bromine

3. Baking powder

Ans: Baking powder –$NaHC{{O}_{3}}$

Elements present – Sodium, Hydrogen, Carbon, Oxygen

4. Potassium sulphate

Ans:  Potassium sulphate –\[{{K}_{2}}S{{O}_{4}}\]

Elements present – Potassium, Sulphur, Oxygen

8. Calculate the molar mass of the following substances.

Atomic mass of –

$C=12u,\text{ }H=1u,\text{ }S=32u,\text{ }P=31u,\text{ }Cl=35.5u,\text{ }N=14u,\text{ }O=16u$

1. Ethyne – ${{C}_{2}}{{H}_{2}}$

Ans: ${{C}_{2}}{{H}_{2}}=(12\times 2)+(1\times 2)=24+2=26u=26{g}/{mole}\;$

2. Sulphur molecule –${{S}_{8}}$

Ans: ${{S}_{8}}=32\times 88=256u=256{g}/{mole}\;$

3. Phosphorus molecule – ${{P}_{4}}$ (Atomic mass of phosphorus is $31$)

Ans: ${{P}_{4}}=31\times 4=124u=124{g}/{mole}\;$ 

4. Hydrochloric acid – $HCl$

Ans: $HCl=(1\times 1)+(35.5\times 1)=1+35.5=36.5u=36.5{g}/{mole}\;$

5. Nitric acid – \[HN{{O}_{3}}\]

Ans: \[HN{{O}_{3}}=(1\times 1)+(14\times 1)+(16\times 3)=1+14+48=63u=63{g}/{mole}\;\]

9. What is the mass of –

Atomic mass of –

$S=32u,\text{ Al}=27u,\text{ Na}=23u,\text{ }N=14u,\text{ }O=16u$

1. 1 mole of nitrogen atoms?

Ans:  Given its atomic mass, the mass of 1 mole of nitrogen atoms is $14g$

2. 4 moles of aluminium atoms (Atomic mass of aluminium is $27$)?

Ans:  Given its atomic mass, the mass of 1 mole of aluminium atoms is $27g$

Thus, the mass of 4 moles of aluminium atoms is $27\times 4=108g$

3. 10 moles of sodium sulphite (\[N{{a}_{2}}S{{O}_{3}}\] )?

Ans:  Given its atomic mass, the mass of 1 mole of sodium sulphite (\[N{{a}_{2}}S{{O}_{3}}\]) is $(23\times 2)+(32\times 1)+(16\times 3)=46+32+48=126u=126{g}/{mole}\;$

Thus, the mass of 10 moles of sodium sulphite (\[N{{a}_{2}}S{{O}_{3}}\]) is $126\times 10=1260g$

10. Convert into mole.

Atomic mass of – $C=12u,\text{ }H=1u,\text{ }O=16u$

1. 12 g of oxygen gas

Ans: Molar mass of ${{O}_{2}}=(16\times 2)=32{g}/{mole}\;$

$\Rightarrow 1\text{ }mole\text{ }of\text{ }{{O}_{2}}=32g$

$\Rightarrow 1g\text{ }of\text{ }{{O}_{2}}=\dfrac{1}{32}\text{ }moles\text{ }$

$\Rightarrow 12g\text{ }of\text{ }{{O}_{2}}=12\times \dfrac{1}{32}\text{ =0}\text{.375}moles\text{ }$ 

2. 20 g of water

Ans: Molar mass of ${{H}_{2}}O=(1\times 2)+(16\times 1)=18{g}/{mole}\;$

$\Rightarrow 1\text{ }mole\text{ }of\text{ }{{H}_{2}}O=18g$

$\Rightarrow 1g\text{ }of\text{ }{{H}_{2}}O=\dfrac{1}{18}\text{ }moles\text{ }$

$\Rightarrow 20g\text{ }of\text{ }{{H}_{2}}O=20\times \dfrac{1}{18}\text{ =1}\text{.11}moles\text{ }$

3. 22 g of carbon dioxide

Ans: Molar mass of $C{{O}_{2}}=(12\times 1)+(16\times 2)=44{g}/{mole}\;$

$\Rightarrow 1\text{ }mole\text{ }of\text{ C}{{O}_{2}}=44g$

$\Rightarrow 1g\text{ }of\text{ C}{{O}_{2}}=\dfrac{1}{44}\text{ }moles\text{ }$

$\Rightarrow 12g\text{ }of\text{ }{{O}_{2}}=22\times \dfrac{1}{44}\text{ =0}\text{.5}moles\text{ }$

11. State the Postulates of Dalton Theory?

Ans: Dalton’s atomic theory states that all matter, be it an element, a compound, or a mixture is composed of small particles called atoms. 

The postulates of the theory are:

1. All matter is made of very tiny particles called atoms, which participate in chemical reactions. 

2. Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. 

3. Atoms of a given element are identical in mass and chemical properties. 

4. Atoms of different elements have different masses and chemical properties. 

5. Atoms combine in the ratio of small whole numbers to form compounds.

6. The relative number and kinds of atoms are constant in a given compound.

12. Find the percentage of water of crystallization in \[FeS{{O}_{4}}.7{{H}_{2}}O\].

Ans: Atomic mass of –

$Fe=55.9u,\text{ S}=32u,\text{ H}=1u,\text{ }O=16u$

Molar mass of$FeS{{O}_{4}}.7{{H}_{2}}O=(55.9\times 1)+(32\times 1)+(16\times 4)+7\times \left[ (1\times 2)+(16\times 1) \right]$

$=55.9+32+64+7\times \left[ 18 \right]{=151.9+126=227.9g}/{mole}\;$

Thus, we can say that \[227.9{g}/{mole}\;\]of $FeS{{O}_{4}}$contains $126g$water of crystallization.

So, \[1g\] of \[FeS{{O}_{4}}\]contains $\dfrac{126}{277.6}g$ water of crystallization.

Converting this fraction into percentage –

$\dfrac{126}{277.6}=0.4534g$ water of crystallization

Thus, we get $\dfrac{126}{277.6}\times 100=0.4534\times 100=45.34\%$

The percentage of water of crystallization in \[FeS{{O}_{4}}.7{{H}_{2}}O\] is $45.34\%$.

13.  \[2.42g\] of copper gave \[3.025g\] of a black oxide of copper, \[6.49g\] of a black oxide, on reduction with hydrogen, gave \[5.192g\] of copper. Show that these figures are in accordance with the law of constant proportion?

Ans: Given:

Case A –

Mass of copper: \[2.42g\]

Mass of copper oxide: \[3.025g\]

Case B –

Mass of black copper oxide: \[6.49g\]

Mass of copper obtained after reduction: \[5.192g\]

Verification: To prove the law of constant proportions, we need to find out the percentage of copper in copper oxide in both cases A and B.

Percentage of copper in Case A $=\dfrac{Mass\text{ }of\text{ }Copper}{Mass\text{ }of\text{ }Copper\text{ }Oxide}\times 100\%$

$=\dfrac{2.42}{3.025}\times 100\%=0.8\times 100\%=80\%$

Percentage of copper in Case B $=\dfrac{Mass\text{ }of\text{ }Copper}{Mass\text{ }of\text{ }Copper\text{ }Oxide}\times 100\%$

$=\dfrac{5.192}{6.49}\times 100\%=0.8\times 100\%=80\%$

It is clear from the above two calculations that the percentage of copper in copper oxide in both cases A and B is the same. This proves the law of constant proportions – copper always combines with oxygen in the same proportion.

14. A compound was found to have the following percentage composition by mass \[Zn=22.65\%\], \[S=11.15\%\], \[H=4.88\%\], \[O=61.32\%\]. The relative molecular mass is \[287{g}/{mole}\;\]. Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallization.

Ans: Given:

\[Zn=22.65\%\]

\[S=11.15\%\]

\[H=4.88\%\]

\[O=61.32\%\]

Relative molecular mass: \[287{g}/{mole}\;\]

To find: Molecular formula of the compound.

Atomic mass of –

$Zn=65.4u,\text{ S}=32u,\text{ H}=1u,\text{ }O=16u$

To find the formula, we need to find the proportion in which these atoms have combined.

It is known that – $Percentage\text{ }of\text{ }element\text{ present }in\text{ }a\text{ }compound=\dfrac{number\text{ }of\text{ }atoms\times atomic\text{ }mass}{mass\text{ }of\text{ }compound}\times 100\%$$\Rightarrow Number\text{ }of\text{ }atoms=\dfrac{percentage\text{ }of\text{ }element\text{ present }in\text{ }a\text{ }compound\times mass\text{ }of\text{ }compound}{atomic\text{ }mass\times 100}$Using the formula above,

\[Number\text{ }of\text{ }Zn\text{ }atoms=\dfrac{22.65\times 287}{65.4\times 100}=0.99=1\]

\[Number\text{ }of\text{ S }atoms=\dfrac{11.15\times 287}{32\times 100}=1.00=1\]

\[Number\text{ }of\text{ H }atoms=\dfrac{4.88\times 287}{1\times 100}=14\]

\[Number\text{ }of\text{ O }atoms=\dfrac{61.32\times 287}{16\times 100}=10.99=11\]

Here, all the Hydrogen atoms belong to the water of crystallization.

Water has the molecular formula – \[{{H}_{2}}O\]with two molecules of Hydrogen and one molecule of oxygen. 

Since we have $14$ atoms of Hydrogen, we can say that there are 7 molecules of water in this compound.

That leaves one atom of zinc, one atom of sulphur, and 4 atoms of oxygen (out of $11$, $7$ atoms of oxygen are in the water of crystallization). It is clear that the compound is Zinc Sulphate with the formula – \[ZnS{{O}_{4}}\]

Thus, the formula of the compound is \[ZnS{{O}_{4}}.7{{H}_{2}}O\]

15. Which element will be more reactive and why – the element whose atomic number is 10 or the one whose atomic number is 11?

Ans:   The element with atomic number $11$ is more reactive than the element with atomic number $10$. This is because of the electronic configuration of the atoms. 

The element with the atomic number $11$, has the configuration of $(2,8,1)$, which means it can easily lose an electron to attain stability. Thus, before losing the electron, it is not stable and is said to be more reactive.

While the element with the atomic number $10$, has the configuration of $(2,8)$, which means it is already stable with a completely filled L shell and does not have to gain or lose electrons to attain stability. Thus, it is said to be less reactive.

16. What are the failures of Dalton's Atomic theory?

Ans:   

1. It does not account for subatomic particles: It stated that atoms were the smallest unit of matter. But, the discovery of subatomic particles namely, protons, electrons, and neutrons disproved this postulate.

2. It does not account for isotopes: For example hydrogen $H_{1}^{1}$, deuterium $H_{1}^{2}$, and tritium$H_{1}^{3}$, have the same atomic number, but different mass numbers.

3. It does not account for isobars. Example: $Ar_{18}^{40}$ and$Ca_{20}^{40}$, they have different atomic numbers, but the same mass number.

4. Elements need not combine in simple, whole-number ratios to form compounds: There are complex organic compounds that do not combine in simple ratios of constituent atoms. Example: sugar/sucrose (${{C}_{11}}{{H}_{22}}{{O}_{11}}$).

5. It does not account for allotropes: The differences in the properties of diamond and graphite, even though they contain only carbon, cannot be explained by Dalton’s atomic theory.

17. Calculate the Molecular Mass of

Atomic mass of – $S=32u,\text{ H}=1u,\text{ C}=12u,\text{ }N=14u,\text{ }O=16u$

1. Ammonium sulphate ${{(N{{H}_{4}})}_{2}}S{{O}_{4}}$

Ans: Molar mass of \[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}=2\times \left[ (14\times 1)+(1\times 4) \right]+(32\times 1)+(16\times 4)\]

\[=2\times \left[ (14)+(4) \right]+(32)+(64)=(2\times 18)+96=36+96=132{g}/{mole}\;\]

2. Penicillin ${{C}_{16}}{{H}_{18}}{{N}_{2}}S{{O}_{4}}$

Ans: Molar mass of \[{{C}_{16}}{{H}_{18}}{{N}_{2}}S{{O}_{4}}=(12\times 16)+(1\times 18)+(14\times 2)+(32\times 1)+(16\times 4)\]

\[=(192)+(18)+(28)+(32)+(64)=334{g}/{mole}\;\]

3. Paracetamol ${{C}_{8}}{{H}_{9}}NO$

Ans: Molar mass of \[{{C}_{8}}{{H}_{9}}NO=(12\times 8)+(1\times 9)+(14\times 1)+(16\times 1)\]

\[=(96)+(9)+(14)+(16)=135{g}/{mole}\;\]

18.  Answer the following questions are about one mole of sulphuric acid ${{H}_{2}}S{{O}_{4}}$

Atomic mass of – \[S=32u,\text{ H}=1u,\text{ }O=16u\]

1. Find the number of gram atoms of hydrogen in it?

Ans: Mass of $1$ mole of ${{H}_{2}}S{{O}_{4}}=(1\times 2)+(32\times 1)+(16\times 4)=2+32+64=98{g}/{mole}\;$

It is known that $1$ mole of any substance contains $6.023\times {{10}^{23}}\text{ atoms/molecules}$.

Thus, $1$ mole of ${{H}_{2}}S{{O}_{4}}=98{g}/{mole}\;=6.023\times {{10}^{23}}\text{ molecules}$

From the molecular formula, we can say that ${{H}_{2}}S{{O}_{4}}$ has two atoms of hydrogen.

i.e. $(2\times atomic\text{ mass of H)}=2\times 1=2g$

Thus, the number of gram atoms of hydrogen in ${{H}_{2}}S{{O}_{4}}$is, $2g$

2. How many atoms of hydrogen does it have?

Ans: Number of atoms of H $=number\text{ of atoms of H in }{{H}_{2}}S{{O}_{4}}\times AvogadroNumber$

$=2\times 6.023\times {{10}^{23}}\text{ =12}\text{.046}\times {{10}^{23}}atoms$

3. How many atoms (in grams) of hydrogen are present for every gram atom of oxygen in it?

Ans: From the molecular formula, we can say that ${{H}_{2}}S{{O}_{4}}$

has two atoms of hydrogen for every four atoms of oxygen.

i.e. $2H:4O$

$\dfrac{2}{4}H:1O\Rightarrow \dfrac{1}{2}H:1O\Rightarrow 0.5H:1O$

Thus, for one atom of oxygen we get $0.5$hydrogen atoms (in grams).  

4. Calculate the number of atoms in ${{H}_{2}}S{{O}_{4}}$?

Ans: $1$ mole of ${{H}_{2}}S{{O}_{4}}$ contains $6.023\times {{10}^{23}}\text{ molecules}$

19. Write an experiment to show that cathode rays travel in a straight line?

Ans: An experiment to show that cathode rays travel in a straight line can be performed using a fluorescent coated discharge tube and a source of cathode rays, an opaque object, and a high voltage source.

Set-up for the experiment:

• In a discharge tube coated with a fluorescent substance initiate the production of cathode rays using a high voltage source.

• In the path of the cathode rays, place an opaque object and observe the fluorescence phenomena.

• When cathode rays strike against the screen, they produce fluorescence. But due to the placement of the opaque object, we will observe a sharp shadow being formed on the screen in the shape of the object. 

• This shadow of the object can be formed if and only if the cathode rays travel in a straight line and do not bend around the edges of the object.

• This experiment shows that cathode rays travel in a straight line.

20. What is radioactivity? What are the applications of radioisotopes?

Ans: Radioactivity is defined as the spontaneous emission of radiation in the form of particles or high-energy photons that are a result of a nuclear reaction. It is the release of energy from the decay of the nucleus of atoms and/or isotopes.

Applications of radioisotopes:

• The isotope of $Co-60$ emits $\gamma $-radiation that is used to treat cancer.

• $I-131$ is used in the diagnosis and treatment of thyroid gland diseases.

• $P-32$ is used in the treatment of leukemia and the identification of malignant tumors.

• $C-14$ is used to study biochemical processes.

21. There are two elements C and B. C emits an $\alpha $ – particle and B emits a $\beta $ – particle. How will the resultant elements charge?

Ans: When an element emits $\alpha $ particle, its atomic number decreases by $2$ , and its mass number decreases by $4$. This is because alpha particles are positively charged nuclei of Helium with two protons and two neutrons.

Thus, in the case of element C that emits $\alpha $particle, its atomic number decreases by $2$ and its mass number decreases by $4$.

When an element emits $\beta $ particle, its atomic number increases by $1$ and its mass number remains the same. This is because a beta particle is essentially an electron.

Thus, in the case of element B that emits $\beta $ particle, its atomic number increases by $1$ and its mass number remains the same.

22. What are isotopes? Name the isotopes of hydrogen and draw the structure of their atoms.

Ans: Isotopes are defined as the atoms of the same element that have different mass numbers; i.e. elements having the same atomic number but different mass numbers.

Example – Isotopes of Hydrogen: 

• Hydrogen $H_{1}^{1}$

• Deuterium $H_{1}^{2}$

• Tritium$H_{1}^{3}$

Structure of Isotopes of Hydrogen:

Long Answer Questions (5 Marks)

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.

\[\mathbf{Sodium}\text{ }\mathbf{carbonate}\text{ + }\mathbf{Ethanoic}\text{ }\mathbf{acid}\to \mathbf{Sodium}\text{ }\mathbf{ethanoate}\text{ }+\text{ }\mathbf{Carbondioxide}\text{ }+\text{ }\mathbf{Water}\]

Ans: The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. This means that the mass of the constituents of a closed chemical reaction will remain the same before and after the reaction.

Mathematically - $\text{Mass of reactants = Mass of products}$

Here, the reactants are Sodium carbonate and Ethanoic acid. 

The products are Sodium ethanoate, carbon dioxide and water.

To prove the law of conservation of mass, we need to prove the mass of reactants is equal to the mass of the products.

Given:

Mass of Sodium carbonate: $5.3g$

Mass of Ethanoic acid: $6g$

Mass of Sodium ethanoate: $8.2g$

Mass of Carbon Dioxide: $2.2g$

Mass of Water: $0.9g$

The reaction – 

\[\mathbf{Sodium}\text{ }\mathbf{carbonate}\text{ + }\mathbf{Ethanoic}\text{ }\mathbf{acid}\to \mathbf{Sodium}\text{ }\mathbf{ethanoate}\text{ }+\text{ }\mathbf{Carbondioxide}\text{ }+\text{ }\mathbf{Water}\]

Now,

\[Mass\text{ of reactants = Mass of }\mathbf{Sodium}\text{ }\mathbf{carbonate}\text{ + Mass of }\mathbf{Ethanoic}\text{ }\mathbf{acid}\]$=5.3+6=11.3g$

\[Mass\text{ of products = Mass of }\mathbf{Sodium}\text{ ethanoate + Mass of carbondioxide + Mass of Water}\]$=8.2+2.2+0.9=11.3g$

It is clear from the above calculations that – 

$\text{Mass of reactants = Mass of products=11}\text{.3g}$

Thus this proves the law of conservation of mass.

2. Calculate the molecular masses of

Atomic mass of – $\text{H}=1u,\text{ C}=12u,\text{ }N=14u,\text{ }O=16u,\text{ }Cl=35.5u$

1. ${{H}_{2}}$

Ans: Molar mass of ${{H}_{2}}=(1\times 2)=2u$

2. ${{O}_{2}}$

Ans: Molar mass of ${{O}_{2}}=(16\times 2)=32u$

3. $C{{l}_{2}}$

Ans: Molar mass of $C{{l}_{2}}=(35.5\times 2)=71u$

4. $C{{O}_{2}}$

Ans: Molar mass of $C{{O}_{2}}=(12\times 1)+(16\times 2)=12+32=44u$

5. $C{{H}_{4}}$

Ans: Molar mass of $C{{H}_{4}}=(12\times 1)+(1\times 4)=12+4=16u$

6. ${{C}_{2}}{{H}_{6}}$

Ans: Molar mass of \[{{C}_{2}}{{H}_{6}}=(12\times 2)+(1\times 6)=24+6=30u\]

7. ${{C}_{2}}{{H}_{4}}$

Ans: Molar mass of ${{C}_{2}}{{H}_{4}}=(12\times 2)+(1\times 4)=24+4=28u$

8. $N{{H}_{3}}$

Ans: Molar mass of $N{{H}_{3}}=(14\times 1)+(1\times 3)=14+3=17u$

9. $C{{H}_{3}}OH$

Ans: Molar mass of $C{{H}_{3}}OH=(12\times 1)+(1\times 3)+(16\times 1)+(1\times 1)=12+3+16+1=32u$

3. If one mole of carbon atoms weighs $12$ grams, what is the mass (in grams) of one atom of carbon?

Ans:

It is known that $1$ mole of any substance contains$6.023\times {{10}^{23}}\text{ atoms/molecules}$.

Thus, $1$ mole of $C=6.023\times {{10}^{23}}\text{ C-atoms}$

It is given that one mole of carbon atoms weighs $12$ grams

Combining these two observations,

$1\text{ }mole\text{ }of\text{ }C=12g=6.023\times {{10}^{23}}\text{ C-atoms}$

We need to find the mass of one carbon atom.

Since – $12g\text{ }of\text{ }C=6.023\times {{10}^{23}}\text{ C-atoms}$ i.e. $12$ grams contain $6.023\times {{10}^{23}}\text{ C-atoms}$

Now for the mass of one carbon atom –

$6.023\times {{10}^{23}}\text{ C-atoms=}12g\text{ }of\text{ }C$

$1\text{ C-atom = }\dfrac{12}{6.023\times {{10}^{23}}}g\text{ =1}\text{.993}\times {{10}^{-23}}g$

Thus, the mass of one carbon atom is $\text{1}\text{.993}\times {{10}^{-23}}g$

4. A $0.24g$ sample of compound of oxygen and boron was found by analysis to contain $0.096g$ of boron and $0.144g$ of oxygen. Calculate the percentage composition of the compound by weight.

Ans: Given:

Mass of sample compound: $0.24g$

Mass of boron in the sample: $0.096g$

Mass of oxygen in the sample: $0.144g$

To find: Percentage composition of boron and oxygen in the compound by weight.

$Percentage\text{ }of\text{ }element\text{ present }in\text{ }a\text{ }compound=\dfrac{mass\text{ }of\text{ }element\text{ }in\text{ }compound}{mass\text{ }of\text{ }compound}\times 100\%$

Thus,

$Percentage\text{ }of\text{ Boron }in\text{  }compound=\dfrac{mass\text{ }of\text{ Boron }in\text{ }compound}{mass\text{ }of\text{ }compound}\times 100\%$

$=\dfrac{0.096}{0.24}\times 100\%=0.4\times 100\%=40\%$

\[Percentage\text{ }of\text{ Oxygen }in\text{ }compound=\dfrac{mass\text{ }of\text{ Oxygen }in\text{ }compound}{mass\text{ }of\text{ }compound}\times 100\%\]

$=\dfrac{0.144}{0.24}\times 100\%=0.6\times 100\%=60\%$

The percentage of Boron by weight in the compound is $40\%$ and the percentage of Oxygen by weight in the compound is $60\%$.

Topics Covered For Class 9 Science Ch 3 Important Questions

The ch 3 science class 9 important questions notes have been prepared by the experts who have been teaching in their respective subjects for quite some time. They know all the ins and outs of the topics and the chapters. Also, with these notes, you will get some insights about the topics which you might not find in your textbook or from your class teacher. Below are some of the topics covered in this chapter, and we have discussed them briefly to give students a revision about the topics they need to learn to answer these questions.

Chemical Reactions 

When a chemical reaction occurs between two or more molecules, then a new compound is formed, and then the two molecules are called reactants whereas the newly formed compound will be called products. 

During the chemical reaction, a chemical change needs to occur which can be seen by a physical change like precipitation, or production of heat, or in some cases change of colour. 

Law of Conservation of Mass

A matter can neither be created nor destroyed in any chemical reaction; it only remains conserved; this is the law of conservation of mass. 

On the other hand, the mass of the given reactants in the chemical reaction will be equal to the mass of the products formed from the chemical reaction. 

Law of Constant Proportions 

A chemical compound that is said to be pure contains the same amount of elements combined in a fixed proportion by mass is said to be the law of definite proportions. A great example of the law of constant proportions is that when we take out water from the lake present in the mountains and take out the water from the ocean both have the same number of oxygen and hydrogen molecules present in them. 

Atoms 

Atoms are said to be the building blocks of the world that we see around us. They are present everywhere, and they are present inside our body. An atom is the most fundamental part of the element, and it cannot be broken by any chemical means. 

Dalton's Atomic Theory 

• The matter present everywhere in the universe is made up of tiny particles that cannot be divisible into other smaller particles, and they are called atoms. 

• When comparing the properties of the atoms of a given element, they tend to be the same, meaning their mass is also the same. As a result, we can state that an element's atoms have the same mass and chemical properties. On the other hand, atoms of a different element have different mass, showing different chemical properties. 

• Compounds are formed when the atoms of different elements combine in fixed ratios. 

• Atoms are the particles that can neither be created, nor they can be destroyed. The formation of the new compounds occurs from the rearrangement of the existing atoms in a chemical reaction. 

• Lastly, in a given compound, the relative number and kinds of atoms are constant. 

Atomic Mass 

Atomic mass is said to be the total mass of all the neutrons, protons, and electrons present in a given atom or a group of atoms. Atomic mass is also said to be the average mass. 

The mass of the atomic particle can be defined as the atomic mass. 

Molecular Mass 

In the important question of atoms and molecules, class 9 molecular mass of a given element can be presented as the sum of the masses of the elements present in the molecule. 

Concept of Molar Mass

In any given substance, the number of atoms, molecules, ions present is defined as a mole. A mole of any substance is said to be 6.022×1023molecules. It is one of the easiest ways to express the number of reactants and products in the reaction. On the other hand, there is an Avogadro's number that approximately has the same value as one mole. It tells us about the number of particles present in one mole. These particles which are represented in mole could be electrons, protons, and neutrons. 

Molar Mass 

Any particle or a substance present in the universe has some mass to it and acquires some space. The molar mass or molecular weight is the sum of the total mass in grams of the atoms present inside the atom that makes up the molecule per mole. The unit of molar mass is grams/mole. 

Concept of Atomicity 

A molecule is the smallest unit of a compound that can represent all the chemical properties of a compound. The atomicity of a given element is measured by the number of atoms present in its one molecule. 

These are some of the important concepts that students need to learn about before they tackle the class 9 science ch 3 important questions that Vedantu has created. 

Important Questions for Class 9 Science Chapter 3 Exam Point of View 

Now you get the idea of the concepts and definitions you need to learn for chapter 3 of the science textbook. Now let's move on to the interesting part, which is the important questions from the exam point of view. Given below, we have ten questions which can help you prepare for your upcoming science exams and clear lots of written concepts in this chapter. 

Q1) Name the scientist who laid the foundation of the chemical sciences and led to chemistry studies. Also, provide the answer to how did he find it?

Q2) Write down the law of conservation of mass by giving the example of a chemical reaction.

Q3) What is the law of constant proportion, and how does it help students perform chemical reactions?

Q4) Which world organization approves all the names of the elements we see in our textbooks and use in our daily lives? Also, write the symbol of the mercury. 

Q5) Write down the symbols of Oxygen and Hydrogen.

Q6) "Atoms of most of the elements tend to not exist on their own" Name the two atoms which can exist as an independent atom. 

Q7) Provide one relevant reason in your answer for why the scientists have chosen 1/16 of the mass of an atom of a naturally found oxygen as the atomic mass unit. 

Q8) Which of the postulates from Dalton's atomic theory results from the law of conservation of mass?

Q9) What are the two drawbacks of Dalton's atomic theory and how were they corrected? 

Q10) How will you be able to differentiate between the molecule of a given element and the molecule of a given compound?

In addition to this, we appreciate students to write down the answer to these questions on their own and don't take help from our solved questions Pdf. But in case there are some questions you might find difficult you can look at its solution without a problem.

Tips to learn Class 9 Science Chapter 3 Atoms and Molecules

• Start with the laws of chemical combination and Dalton's atomic theory to get a clear idea of how atoms behave and combine.

• Learn terms like atomic mass, molecular mass, and mole concept. Use flashcards or notes to revise these regularly.

• Solve problems on calculating molar mass, and atomic mass, and balancing chemical equations to strengthen your understanding.

• Diagrams and flowcharts can help you understand atomic structure and molecular formulas better.

• Think about how molecules like water (H₂O) or carbon dioxide (CO₂) are formed to connect the theory to everyday life.

• Go over important questions, textbook exercises, and NCERT solutions to stay confident before exams.

Benefits of Learning Ch 3 Science Class 9 Important Questions

• Chapter 3 science class 9 important questions Pdf will help students learn the various complicated topics, including the molar mass and valency of the atoms. 

• In addition to this, students get to know step by step solutions of the numerical answers, so they don't lose any extra marks for skipping a step. 

• Likewise, the answer to chapter 3 science class 9 important questions are written by well trained and experienced teachers from all around India. They know the core of the subject and have been teaching it for a while. So when you are looking at the solution of the question, you are reading the answer which is written by an expert in the field of science.

CBSE Class 9 Science Atoms and Molecules Extra Questions

1. Name two scientists who established the law of chemical combination.

2. What is the unit used to measure the size of an atom?

3. Why is Avogadro's number also known as  Avogadro’s constant? 

4. Write 6 postulates of Dalton’s atomic theory?

5. Find the mass percentage of oxygen present in HNO3.

Conclusion

CBSE Class 9 Science Chapter 3 Atoms and Molecules help to understand the structure of an atom and molecules. It introduces you to the basic building blocks of matter and the principles governing their behaviour. Understanding concepts like atomic theory, chemical laws, and molecular structures will not only help you excel in exams but also lay a solid foundation for advanced topics in chemistry. Keep practising, and remember—science is all about exploring how the world around us works!

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