CBSE Class 9

In this page, we are providing Atoms and Molecules Class 9 Extra Questions and Answers Science Chapter 3 pdf download. NCERT Extra Questions for Class 9 Science Chapter 3 Atoms and Molecules with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 3 Extra Questions and Answers Atoms and Molecules

Extra Questions for Class 9 Science Chapter 3 Atoms and Molecules with Answers Solutions

Atoms and Molecules Class 9 Extra Questions Very Short Answer Type

Question 1.
Define atomic mass unit.
Answer:
Atomic mass unit of an element is one twelfth (1/12th) of the mass of one atom of carbon-12.

Question 2.
Write the valency of sulphur in H₂S, SO₂ and SO₃.
Answer:
The valency of sulphur in H₂S, SO₂ and SO₃ are 2, 4, and 6 respectively.

Question 3.
What is the difference between 2H and H₂?
Answer:
2H means two atoms of hydrogen, H₂ means a molecule of hydrogen which contains two atoms.

Question 4.
Where do we use the words mole and mol?
Answer:
We use the word mole to define the number of atoms, molecules, ions or particles having a mass equal to its atomic or molecular mass in grams, while as a unit, we call it mol.

Question 5.
The valency of an element A is 4. Write the formula of its oxide.
Answer:
The formula of its oxide is A₂O₂ or AO₂.

Question 6.
Why are Dalton’s symbols not used in chemistry?
Answer:
Dalton was the first scientist to use the symbol for the name of the elements in a specific sense but it was difficult to memorise and in use, so Dalton’s symbols are not used in chemistry.

Question 7.
Which postulate of Dalton’s Atomic theory is the basis of law of conservation of mass?
Answer:
“Atoms can neither be created nor destroyed a physical or a chemical change”.

Question 8.
Calculate the formula unit mass of CaCl₂. [NCERT Exemplar]
Answer:
Atomic mass of Ca + (2 × atomic mass of Cl)
= 40 + 2 × 35.5 = 40 + 71 = 111 u.

Question 9.
What does the symbol ‘u’ represent?
Answer:
The symbol ‘u’ represents unified mass.

Question 10.
Avogadro’s number represents how many particles?
Answer:
Avogadro’s numbers (N₀) represents 6.022 × 10²³ particles.

Question 11.
Formula of the carbonate of a metal M is M₂CO₃. Write the formula of its chloride.
Answer:
The valency of the metal (M) in M₂CO₃ is (1+) i.e., metal exists as M⁺ ion. Therefore, the formula of metal chloride is MCl.

Question 12.
Sample A contains one gram molecule of oxygen molecules and Sample B contains one mole of oxygen molecules. What is the ratio of the number of molecules in both the samples?
Answer:
One gram molecules is the same as one gram mole of a substance. Therefore, both the samples A and B contain the same number of molecules (6.022 × 10)²³ and the ratio is 1: 1.

Question 13.
Name the compound Al₂(SO₄)₃ and mention the ions present in it.
Answer:
The compound is called aluminium sulphate; cation: Al³⁺; anion: (SO₄)^2-

Question 14.
Chemical symbol of sodium is Na. What is its Latin name?
Answer:
Natrium.

Question 15.
Give one example of a polyatomic cation.
Answer:
NH₄⁺ (Ammonium ion).

Question 16.
MNO₃ is the formula of nitrate of metal M. Write the formula of its oxide.
Answer:
M₂O

Question 17.
A vessel contains W molecules of oxygen at a certain temperature and pressure. How many molecules of sulphur dioxide can the vessel accommodate at the same temperature and pressure?
Answer:
N molecules.

Question 18.
What do you understand by a polyatomic ion? Give two examples.
Answer:
A group of atoms carrying positive or negative charge is called a polyatomic ion, e.g., NO₃^–, SO₄^2-, NH₄⁺

Question 19.
Give two examples each of bivalent cations and bivalent anions.
Answer:
Bivalent cations = Zn²⁺, Ca²⁺. Bivalent anions = SO₄^2-, C₃^2-.

Question 20.
What is ‘molar volume’? What is its value?
Answer:
The volume occupied by one mole of a gas under standard conditions of temperature and pressure, i.e., STP conditions (0°C and 1 atmosphere i.e., pressure) is called molar volume of the gas. Its value is 22.4 L at STP.

Question 21.
What is the valency of calcium in CaCO₃?
Answer:
The valency of Ca in CaCO₃ is 2⁺ [i.e. Ca²⁺].

Question 22.
Can we regard sodium as a monoatomic element?
Answer:
No, sodium is a metal and does not exist as a single atom.

Question 23.
What happens to an element ‘A’ if its atom gains two electrons?
Answer:
It changes to a divalent anion [A^2-].

Question 24.
Do atoms in some of the elements have actually fractional mass?
Answer:
No, the atoms do not have fractional mass. Only their average comes out to be in fraction in elements which exists as isotopes.

Atoms and Molecules Class 9 Extra Questions Short Answer Type 1

Question 1.
Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol^-1 and 32 g mol^-1 respectively. [NCERT Exemplar]
Answer:
Molar mass of HgS = 200.6 + 32 = 232.6 g mol^-1
Mass of Hg in 232.6 g of HgS = 200.6 g
Mass of Hg in 225 g of HgS = \(\frac { 200.6 }{ 232.6 }\) × 225 = 194.05 g.

Question 2.
A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold? [NCERT Exemplar]
Answer:
One gram of gold sample will contain = \(\frac { 90 }{ 100 }\) = 0.9 g of gold
Atomic mass of gold =
= \(\frac { 0.9 }{ 197 }\)
One mole of gold contains N₀ atoms = 6.022 × 10²³
0.0046 mole of gold will contain = 0.0046 × 6.022 × 10²³
= 2.77 × 10²¹

Question 3.
A sample of ethane (C₂H₆) gas has the same mass as 1.5 × 10²⁰ molecules of methane (CH₄). How many C₂H₆ molecules does the sample of the gas contain? [NCERT Exemplar]
Answer:

Question 4.
Write the cations and anions present (if any) in the following compounds
(a) CH₃COONa
(b) NaCl
(c) H₂
(d) NH₄NO₃
Answer:

Question 5.
How many molecules of water are present in a drop of water which has a mass of 50 mg?
Answer:
We know that:
1 mole of a compound = 6.023 × 10²³ atoms
= Gram molecular mass
Gram molecular mass of H₂O = 18 g
18 g = 6.023 × 10²³ atoms

Question 6.
Find the number of protons and neutrons in the nucleus of an atom of an element X which is represented as ²⁰⁷ X₈₂.
Answer:
The element is ²⁰⁷ X₈₂.
Now, 82 = Atomic Number
207 = Mass Number.

(a) Atomic number = Number of protons
82 = Number of protons.

(b) Mass number = Number of protons + Number of neutrons
207 = 82 + Number of neutrons
207 – 82 = Number of neutrons
125 = Number of neutrons.

Question 7.
Calculate the molar mass of Na₂SO₄ and CaCO₃.
Answer:
Molar Mass of Na₂SO₄ =
2 × Mass of sodium + 1 × Mass of sulphur + 4 × Mass of oxygen
= 2 × 23 + 1 × 32 + 4 × 16
= 46 + 32 + 64
= 142 a.m.u.

Molar Mass of CaCO₃ =
1 × Mass of calcium + 1 × Mass of carbon + 3 × Mass of oxygen
= 40 + 12 + 3 × 16
= 40 + 12 + 48
= 100 a.m.u.

Question 8.
What is wrong with the statement ‘1 mol of hydrogen’?
Answer:
The statement is not correct. We must always write whether hydrogen is in atomic form or molecular form. The correct statement is 1 mole of hydrogen atoms or one mole of hydrogen molecules.

Question 9.
(a) Calculate the relative molecular mass of water (H₂O). [NCERT Exemplar]
(b) Calculate the molecular mass of HNO₃.
Answer:
(a) Atomic mass of hydrogen = 1 u,
oxygen = 16 u
So the molecular mass of water, which contains two atoms of hydrogen and one atom of oxygen is
= 2 × 1 + 1 × 16 = 18u

(b) The molecular mass of HNO₃ = the atomic mass of H + the atomic mass of N + 3 × the atomic mass of O
= 1 + 14 + 48 = 63 u

Question 10.
Identify the cations and anions in the following compounds:
(a) CH₃COONa
(b) NH₃
(c) NH₄
(d) SrCl₂
Answer:
(a) Na⁺, C₃H₃COO^–
(b) It is a molecular compound
(c) NH₄⁺, Cl^–
(d) Sr²⁺, 2Cl^–

Question 11.
Calculate the total number of electrons present in 1.6 g of methane.
Answer:
Molecular mass of CH₄ = 12+ 1 × 4 = 16 g
16 g = 1 mole
1.6 g = 0.1 mole
1 mole of CH₄ has 6.02 × 10²³ molecules.
0.1 mole has 6.02 × 10²² molecules.
1 molecule of methane contains (6 + 1 × 4) = 10 electrons
∴ 6.02 × 10²² molecule will contain = 10 × 6.02 × 10²² = 6.02 × 10²³ electrons.

Question 12.
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.
Answer:
% oi any element in a compound =
% of boron = \(\frac { 0.096 }{ 0.24 }\) × 100 = 40%
% of oxygen = \(\frac { 0.144 }{ 0.24 }\) × 100 = 60%.

Question 13.
The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound.(Work upto two decimal places).
(Atomic masses: Na = 23, P = 31, O = 16)
Answer:

Empirical formula = Na₃PO₄.

Question 14.
A sample of vitamin C is known to contain 2.58 × 10²⁴ oxygen atoms. How many moles of oxygen atoms are present in the sample?
Answer:
1 mole of oxygen atom = 6.023 × 10²³ atoms
Number of moles of oxygen atoms = \(\frac{2.58 \times 10^{24}}{6.022 \times 10^{23}}\) = 4.28 mol
4.28 moles of oxygen atoms.

Question 15.
The empirical formula of a compound is C₂H₄O. Its relative molecular mass is 88. Find the molecular formula.
Answer:
Let the molecular formula be (C₂H₄O)n.
Molecular mass = n × empirical formula mass 88 = n × (2 × 12 + 4 × 1 + 16)
n = \(\frac { 88 }{ 44 }\) = 2
The molecular formula is (C₂H₄O)₂ = C₄H₈O₂.

Question 16.
What is the mass in grams of one molecule of caffeine (C₈H₁₀N₄O₂)?
Answer:
1 mole of caffeine molecules = Gram molecular mass of C₈H₁₀N₄O₂
= 8 × 12 + 10 × 1 + 4 × 14 + 2 × 16g
= 96 + 10 + 56 + 32 g = 194 g
Also, 1 mole of caffeine molecules = 6.022 × 10²³ molecules
Thus, 6.022 × 1023 molecules of caffeine have mass = 194 g
1 molecule of caffeine will have mass = \(\frac{194}{6.022 \times 10^{23}}\) g = 3.22 × 10^-22

Question 17.
What is the difference between cation and anion?
Answer:
Cation:

• It is positively charged, exam: Na⁺
• It is formed from metal atoms.
• On passing electric current, it moves towards cathode.
• It is smaller in size than its parent atom.

Anion:

• It is negatively charged, example : Cl^–
• It is formed from non-metal atoms.
• On passing electric current, it moves towards anode.
• It is larger in size than its parent atom.

Question 18.
Differentiate between:
(i) Atom and molecule.
(ii) Molecular mass and formula of unit mass.
Answer:
(i) Atoms cannot exist independently but molecules can.

(ii) Molecular mass in the sum of masses of the atoms in the molecule whereas, formula unit mass is sum of atomic masses of the atoms in its empirical formula.

Atoms and Molecules Class 9 Extra Questions Short Answer Type 2

Question 1.
Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
CaCl₂(aq) → Ca²⁺ (aq) + 2Cl^–(aq)
Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. [NCERT Exemplar]
Answer:
1 mole of calcium chloride = 111 g
222 g of CaCl₂ is equivalent to 2 moles of CaCl₂ Since 1 formula unit of CaCl₂ gives 3 ions, therefore, 1 mol of CaCl₂ will give 3 moles of ions.
2 moles of CaCl₂ would give 3 × 2 = 6 moles of ions.
No. of ions = No. of moles of ions × Avogadro number
= 6 × 6.022 × 10²³
= 36.132 × 10²³
= 3.6 132 × 10²⁴ ions.

Question 2.
Give the formulae of the compounds formed from the following sets of elements. [NCERT Exemplar]
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen
Answer:
(a) CaF₂
(b) H₂S
(c) NH₃
(d) CCl₄
(e) Na₂O
(f) CO, CO₂

Question 3.
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/mol. Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallisation.
Answer:
Zn : S : O : H = \(\frac { 22.65 }{ 65 }\) : \(\frac { 11.15 }{ 32 }\) : \(\frac { 61.32 }{ 16 }\) : \(\frac { 4.88 }{ 1 }\)
= 0.3485 : 0.3484 : 3.833 : 4.88
To obtained an integral ratio, we divide by smallest number
= \(\frac { 0.3485 }{ 0.3484 }\) : \(\frac { 0.3484 }{ 0.3484 }\) : \(\frac { 3.833 }{ 0.3484 }\) : \(\frac { 4.88 }{ 0.3484 }\)
= 1 : 1 : 11 : 14
∴ Empirical formula is Zn SO₁₁H₁₄.
Let molecules formula be (ZnSO₁₁H₁₄)n.
RMM for the molecules = n [65 + 32 + (11 × 16) + 14]
Formula = 287
287 n = 287
n = 1
∴ Molecular formula is ZnSO₁₁H₁₄.

Question 4.
What are the failures of Dalton Atomic theory?
Answer:
Failures of Dalton Atomic Theory are:

1. Atom is not the smallest particle as it is made up of protons, neutrons and electrons.
2. Atom’s mass can be transformed to energy (E = mc²) and hence can be created and destroyed.
3. Atoms of one element have been changed into atoms of another element through artificial transmutation of elements.
4. Atoms of same element need not resemble each other in all respects as isotopes (different forms of same element) exist.
5. Atoms of different elements need not differ in all respects as isobars (same forms of different elements) exist.

Question 5.
The following questions are about one mole of sulphuric acid [H₂SO₄]?
(a) Find the number of gram atoms of hydrogen in it.
(b) How many atoms of hydrogen does it have?
(c) How many atoms (in grams) of hydrogen are present for every gram atom of oxygen in it?
(d) Calculate the number of atoms in H₂SO₄?
1 mole of H₂SO₄ = Gram Molecular mass = 6.023 × 10²³ molecules
Answer:
(a) In H₂SO₄ → 2 gram atoms of hydrogen are present

(b) One mole of H₂SO₄ contains = 6.023 × 10²³ molecules
= 2 × 6.023 × 10²³ atoms of H + 6.023 × 10²³ atoms of S + 4 × 6.023 ×10²³ atoms of O
∴ 2H = 2 × 6.023 × 10²³
= 12.046 × 10²³

(c) In H₂SO₄;
For every 2 hydrogen there are oxygen atoms
So for 1 hydrogen = \(\frac { 4 }{ 2 }\) = oxygen are present
= 2 oxygen atoms are present o
For 1 oxygen = \(\frac { 2 }{ 4 }\) hydrogen atoms are present
= 0.5 hydrogen atoms are present.

(d) 1 mole of H₂SO₄ = 6.023 × 10²³ atoms

Question 6.
Write the formulae for the following and calculate the molecular mass for each one of them. [NCERT Exemplar]
(а) Caustic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt.
Answer:
(a) KOH
[39 + 16 + 1] = 56 g mol^-1

(b) NaHCO₃
23 + 1 + 12 + (3 × 16) = 84 mol^-1

(c) CaCO₃
40 + 12 + (3 × 16) = 100 g mol^-1

(d) NaOH
23 + 16 + 1 = 40 g mol^-1

(e) C₂H₆OH = C₂H₆O
2 × 12 + (6 × 1) + 16 = 46 g mol^-1

(f) Nacl
23 + 35.5 = 58.5 g mol^-1

Question 7.
Calculate the mass of the following: [NCERT Exemplar]
(i) 0.5 mole of N₂ gas (mass from mole of molecule)
(ii) 0.5 mole of N atoms (mass from mole of atom).
(iii) 3.011 × 10²³ number of N atoms (mass from number)
(iv) 6.022 × 10²³ number of N₂ molecules mass from number.
Answer:
(i) mass = molar mass x number of moles
⇒ m = M × n = 28 x 0.5 = 14 g

(ii) mass = molar mass x number of moles
⇒ m = Mxn = 14x 0.5 = 7 g

(iii) The number of moles, n

(iv) n = \(\frac{N}{N_{0}}\)
⇒ m = M x \(\frac{N}{N_{0}}\) = 28 x \(\frac{6.022 \times 10^{2}}{6.022 \times 10^{2}}\)
= 28 x 1 = 28 g.

Question 8.
Calculate the number of particles in each of the following:
(i) 46 g of Na atoms (number from mass)
(ii) 8 g O₂ molecules (number of molecules from mass)

(iii) 0.1 mole of carbon atoms (number from given moles)
Answer:
(i) The numbers of atoms

⇒ N = \(\frac { m }{ M }\) x N₀
⇒ N = \(\frac { 46 }{ 23 }\) x 6.022 x 10²³
⇒ N = 12.044 x 10²³

(ii) The number of molecules

Atomic mass of oxygen = 16 u
molar mass of O₂ molecules
= 16 x 2 = 32 g
⇒ N = \(\frac { 8 }{ 32 }\) x 6.022 x 10²³
⇒ N = 1.5055 x 10²³
= 1.51 x 10²³

(iii) The number of particles (atom) = numbers of moles of particles x Avogadro number
N = n x N₀ = 0.1 x 6.022 x 10²³.
= 6.022 x 10²².

Question 9.
A flask contains 4.4 g of CO₂ gas. Calculate
(a) How many moles of CO₂ gas does it contain?
(b) How many molecules of CO₂ gas are present in the sample?
(c) How many atoms of oxygen are present in the given sample? (Atomic mass of C = 12 u, O = 16 u)
Answer:
(a)

(b) 1 mole of CO₂ has molecules
= 6.022 x 10²³
0. 1 mole of CO₂ has molecules
= 6.022 x 10²³ x (0.1) = 6.022 x 10²²

(c) 1 mole of CO₂ has oxygen atoms = 2 x N₀
0. 1 mole of CO₂ has oxygen atoms = 2 x N₀ x 0.1
= 2 x 6.022 x 10² x 0.1 = 1.204 x 10²⁴ atoms.

Question 10.
Calculate the number of moles for the following: [NCERT Exemplar]
(ii) 12.044 x 10²³ number of He atoms (finding mole from number of particles).
Answer:
No of moles = n
Given mass = m
Molar mass = M
Given number of particles = N
Avogadro number of particles = N₀

(i) Atomic mass of He = 4 u
Molar mass of He = 4 g
Thus, the number of moles =
= n = \(\frac { m }{ M }\) = \(\frac { 52 }{ 4 }\) = 13

(ii) We know,
1 mole = 6.022 x 10²³
The number of moles

Question 11.
Write the molecular formulae of all the compounds that can be formed by the combination of following ions:
Cu²⁺, Na⁺, Fe³⁺, Cb, SO₄^2-, PO₄^3-
Answer:
(i) CuCl₂, CuSO₄, Cu₃(PO₄)₂
(ii) NaCl, Na₂SO₄, Na₃PO₄
(iii) FeCl₃, Fe₂(SO₄)₃, FePO₄

Question 12.
Give the names of any three elements whose names have been derived from Latin. Give their Latin names and symbols.
Answer:

Question 13.
What is the simplest formula of the compound which has the following percentage composition:
Carbon 80%, Hydrogen 20%
If the molecular mass is 30, calculate its molecular formula.
Answer:
Calculation of empirical formula:

∴ Empirical formula is CH₃.
Calculation of molecular formula:
Empirical formula mass = 12 x 1 + 1 x 3 = 15

\(\frac { 30 }{ 15 }\) = 2
Molecular formula = Empirical formula x 2 = CH₃ x 2 = C₂H₆.

Question 14.
On heating, potassium chlorate decomposes to potassium chloride and oxygen. In one experiment 30.0 g of potassium chlorate generates 14.9 g of potassium chloride and 9.6 g of oxygen. What mass of potassium chlorate remains undecomposed?
Answer:
Potassium chlorate → Potassium chloride + Oxygen
Total mass of potassium chlorate before reaction = 30.0 g
After reaction, mass of potassium chloride = 14.9 g Mass of oxygen = 9.6 g
Let mass of undecomposed potassium chlorate = x g
Total mass after reaction = (x + 14.9 + 9.6) g
According to law of conservation of mass the total mass before and after the reaction remains constant.
∴ 30.0 = x + 14.9 + 9.6
x = 30 – 14.9 – 9.6 = 5.5
∴ Mass of undecomposed potassium chlorate
= 5.5 g

Question 15.
Naturally occurring Boron consists of two isotopes whose atomic mass are 10.01 and 11.01. The atomic mass of natural Boron is 10.81. Calculate the percentage of each isotope in natural Boron.
Answer:
Suppose percentage of isotope with atomic mass 10.01 = x
Then percentage of isotope with atomic mass 11.01 = 100 – x

∴ % of isotope with atomic mass 10.01 = 20%
∴ % of isotope with atomic mass 11.01 = 80%

Question 16.
Which amongst the following has more number of atoms, 11.5 g of sodium or 15 g of calcium?
How? (Given atomic mass of Na = 23, Ca = 40)
Answer:
23 g of Na = 1 mole
11.5 g of Na = 11.5/23 mole
Now, 1 mole of Na contains = 6.022 x 10²³ atoms
11.5/23 mole of Na contains = 6.022 x 10²³ x 11.5²³ mole
= 3.011 x 10²³ atoms
40 g of Ca = 1 mole
15 g of Ca = 15/40 moles
Now, 1 mole of Ca contains = 6.022 x 10²³ atoms
15/40 mole of Ca contains = 6.022 x 10²³ x 15/40 mole
= 2.258 x 10²³ atoms
Na has more number of atoms.

Question 17.
(i) Name the body which approves the nomenclature of elements and compounds.
(ii) The symbol of sodium is written as Na and not as S. Give reason.
(iii) Name one element which forms diatomic and one which forms tetraatomic molecules.
Answer:
(i) International Union of Pure and Applied Chemistry.

(ii) The symbol of Na (sodium) is derived from its Latin name Natrium.

(iii) (a) Element forming diatomic molecule: H₂, O₂, N₂ (only one).
(b) Element forming tetratomic molecule: P₄.

Atoms and Molecules Class 9 Extra Questions Long Answer Type Questions

Question 1.
(a) Define empirical formula and molecular formula of a compound. How are molecular formula and empirical formula related to each other?
(b) A hydrocarbon contains 10.5 g of carbon per gram of hydrogen. Calculate the empirical formula.
Answer:
(a) Empirical formula is the chemical formula which gives us the simplest whole number ratio between the atoms of various elements present in one molecule of a compound. Molecular formula gives us the actual number of atoms of various elements present in one molecule of a compound.
Molecular formula = n x Empirical formula
Where n is a simple whole number

(b) In the hydrocarbon,
1 g of hydrogen combines with 10.5 g carbon
∴ % of H in hydrocarbon = \(\frac { 1 }{ 11.5 }\) x 100 = 8.696%
% of C = 100 – 8.696 = 91.304%

Therefore, Empirical formula = CH

Question 2.
(a) What is meant by mole concept?
(b) Calculate the mass of:
(i) 10²² atoms of sulphur
(ii) 0.1 mole of carbon dioxide.
[Atomic mass S = 32, C = 12, O = 16 and Avogadro’s number = 6 x 10²³]
Answer:
(a) 1 mole of a compound has a mass equal to its relative molecular mass expressed in grams.
1 mole = 6.022 x 10²³ number
= Relative mass in grams.

(b) (i) v 6.022 x 10²³ atoms of sulphur weighs = 32 g
∴ 10²² atoms of sulphur weighs = \(\frac{32}{6.022 \times 10^{23}}\) x 10²² = 0.531 g

(ii) Gram molar mass of CO₂ = 12 + 2 x 16 = 44 g
∵ 1 mole of carbon dioxide weighs = 44 g
∴ 0.1 mole of carbon dioxide weighs = \(\frac { 44 g }{ 1 }\) x 0.1 = 4.4 g.

Question 3.
(a) The ratio of hydrogen and oxygen in water is 1: 8 by mass, find out their ratio by number of atoms, in one molecule of water. (Atomic mass: H = 1 u, O – 16 u)
(b) Write the formulae of the following compounds:
(i) Ammonium sulphate
(ii) Magnesium chloride
(Given , ammonium = NH₄²⁺ , sulphate = SO₄^2-, Magnesium = Mg²⁺, chloride = Cl^–)
Answer:
(a)

Ratio of number of atoms for water is H : O = 2 : 1

(b)
(i) Ammonium sulphate: (NH₄)₂SO₄
(ii) Magnesium chloride : MgCl₂

Question 4.
A compound of carbon and sulphur has a composition of 15.8% carbon and 84.2% sulphur.
(a) Find the empirical formula.
(b) The relative molecular mass of the compound is 76. Find the molecular formula.
Answer:
(a)

The empirical formula is CS₂.

(b) Let the molecular formula be (CS₂)_n.
Molecular mass = n x empirical formula mass
76 = n x (12 + 2 x 32)
n = \(\frac { 76 }{ 76 }\) = 1
The molecule formula is (CS₂)₁ = CS₂.

Question 5.
What do the following formulae stand for:
(a) (i) 20
(ii) O₂
(iii) O₃
(iv) H₂O

(b) Give the chemical formulae of the following compounds?
(i) Potassium carbonate
(ii) Calcium chloride

(c) Calculate the formula unit mass of Al₂(SO₄)₃.
(Given atomic mass of Al = 27u, S = 32 u, 0 = 16 u)
Answer:
(a)
(i) Two atoms of oxygen 20
(ii) Diatomic oxygen → O₂ molecule
(iii) Triatomic oxygen → O₃ molecule
(iv) Two atoms of hydrogen and 1 atom of oxygen forming one molecule of water.

(b) K₂CO₃ : Potassium carbonate
CaC₂ : Calcium chloride

(c) Al₂(SO₄)₃
Al: 27 x 2 = 54 u
S : 32 x 3 = 96 u
O : 16 x 12 = 192 u
Formula unit mass = 342 u.

Question 6.
(a) Write the chemical symbols of two elements:
(i) Which are formed from the first letter of the elements name.
(ii) Whose name has been taken from the names of the elements in Latin.

(b) Define Avogadro’s constant and molar mass. How are they related to one mole of an atom, molecule, or ion?
Answer:
(a)
(i) C, N, O, etc.
(ii) Na, Fe, K, etc.

(b) Avogadro’s constant:
6.22 x 10²³ is defined as the Avogadro’s constant. It is number of atoms in exactly 12 g of carbon-12.

Molar mass : Mass of 1 mole of substance.

Question 7.
(a) Calculate the mass of 0.5 mole of sulphuric acid. Atomic mass (H = 1 u, S = 32 u, O = 16 u).
(b) Find the number of atoms in 12 g of carbon.
(c) How many atoms are present in (i) H₂S molecule (ii) PO_3-⁴ ions.
(d) Write the names of elements present in (i) quicklime (ii) hydrogen bromide.
Answer:
(a) 1 mole of H₂SO₄ = 98
∴ 1/2 mole of H₂SO₄ = 49 g

(b) 6.023 x 10²³ atoms

(c) In H₂S molecules, number of atoms = 3
In PO_3-⁴ ion, number of atoms = 5.

(d) (i) Elements in quicklime (CaO)-calcium, oxygen.
(ii) Elements in hydrogen bromide (HBr)-hydrogen, bromine.

Question 8.
(a) Define polyatomic ions. Write an example.
(b) Calculate the formula unit mass of CaCO₃.
(Atomic mass of C = 12 u, Ca = 40 u, O = 16 u)
(c) Calculate the molecular mass of the following:
(i) HNO₃ (ii) CH₃COOH
(Atomic mass of H = 1 u, N = 14 u, O = 16 u, C = 12 u)
Answer:
(a) Cluster of atoms that acts as an ion are called polyatomic ions. example : NH₄⁺, PO_3-⁴, SO_2-⁴, etc.

(b) CaCO₃ = Mass of Ca + Mass of C + Mass of O
= 40 u + 12 u + 16 u x 3 = 100 u.

(c) (i) HNO₃ = Mass of H + Mass of N + Mass of O
= 1 u + 14 u + 16 u x 3 = 63 u

(ii) CH₃COOH = Mass of C + Mass of H + Mass of O
= 12 u x 2 + 1 u x 4 + 16 u x 2
= 24 u + 4 u + 32 u = 60 u

Atoms and Molecules Class 9 Extra Questions HOTS

Question 1.
In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C₆H₁₂O₆. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be 1 g cm^-3. [NCERT Exemplar]
Answer:

1 mole of glucose needs 6 moles of water
180 g of glucose needs (6 x 18) g of water
1 g of glucose will need \(\frac { 108 }{ 180 }\) g of water.
18 g of glucose would need \(\frac { 108 }{ 180 }\) x 18 g of water = 10.8 g
Volume of water used = \(\frac { Mass }{ Density }\) = \(\frac{10.8 \mathrm{g}}{1 \mathrm{g} \mathrm{cm}^{-3}}\) = 10.8 cm³

Question 2.
Fill in the missing data in the Table 3.2. [NCERT Exemplar]
Table 3.2

Answer:

Question 3.
Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight. [NCERT Exemplar]
(a) Whose container is heavier?
(b) Whose container has more number of atoms?
Answer:
(a) Mass of sodium atoms carried by Krish = (5x 23) g = 115 g while mass of carbon atom carried by Raunak = (5 x 12) g = 60 g. Thus, Krish’s container is heavy.

(b) Both the bags have same number of atoms as they have same number of moles of atoms.

Question 4.
How many molecules are present in 1 ml. of water?
Answer:
Molecular mass of H₂O = 18 g
Mass of 1 mole of water = 18 g
18 g of water contain = 6.022 x 10²³ molecules
Density of water = 1 g/ml, therefore 1 ml of water weighs 1 g.
1.0 g of water contain = (6.022 x 10²³)/18 molecules
= 3.34 x 10²² molecules.

Question 5.
A flask P contains 0.5 mole of oxygen gas. Another flask Q contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms?
Answer:
1 molecule of oxygen (O₂) = 2 atoms of oxygen
1 molecule of ozone (O₃) = 3 atoms of oxygen
In flask P:1 mole of oxygen gas = 6.022 x 10²³ molecules
0.4 mole of oxygen gas = 6.022 x 10²³ x 0.5 molecules
= 6.022 x 10²³ x 0.5 x 2 atoms = 6.022 x 10²³ atoms
In flask Q: 1 mole of ozone gas = 6.022 x 10²³ molecules
0.4 mole of ozone gas = 6.022 x 10²³ x 0.4 molecules
= 6.022 x 10²³ x 0.4 x 3 atoms = 7.23 x 10²³ atoms
∴ Flask Q has a greater number of oxygen atoms as compared to flask P.

Question 6.
Complete the following table:

Answer:

Question 7.
What is the valency of underlined element in the following compounds?
(i) MnO₂ (ii) Ca₃N₂ (iii) ZnSO₄ (iv) CCl₄
Answer:
(i) + 4
(ii) + 2
(iii) + 2
(iv) + 4

Question 8.
A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10^-12 g. Calculate the number of carbon atoms in the signature.
Answer:
Mass of carbon = 10^-12 g
No. of moles of carbon = \(\frac{10^{-12}}{12}\) moles
1 mole of C has = 6.022 x 10²³
∵ 1 mole of C has atoms = 6.022 x 10²³
∴ \(\frac{10^{-12}}{12}\) mole of C will have = \(\frac{6.022 \times 10^{23}}{12}\) x 10¹⁰

Question 9.
Rudra took 5 moles of carbon atoms in a container and Ayush also took 5 moles of sodium atoms in another container of some weight.
(a) Whose container is heavier?
(b) Whose container has more number of atoms?
Answer:
(a) 1 mole of C atoms = 12 g
∴ 5 moles of C atoms = 5 x 12 g = 60 g
1 mole of Na atoms = 23 g
∴ 5 moles of Na atoms = 5 x 23 = 115 g
Thus, Ayush’s container is heavier.

(b) 1 mole of C atoms = 6.022 x 10²³ atoms
1 mole of Na atoms = 6.022 x 10²³ atoms
∴ 5 moles of each will contain the same number of atoms.

Question 10.
10²² atoms of an element ‘X’ are found to have a mass of 930 mg. Calculate the molar mass of the element ‘X’.
Answer:
Molar mass of an element is the mass of Avogadro’s number of atoms.
∴6.022 x 10²³ atoms will have mass = \(\frac{0.930}{10^{22}}\) x 6.022 x 10²³ g = 56.0 g
∴Molar mass of the element = 56 g mol^-1.

Question 11.
Which will contain large number of atoms, 1 g of gold or 1 g of silver? Explain with reason. (Atomic masses of Gold = 197 u, silver = 108 u).
Answer:
No. of moles (n) in 1 g of gold = \(\frac { m }{ M }\) = \(\frac { 1 }{ 197 }\)
No. of moles (n) in 1 g of silver = \(\frac { m }{ M }\) = \(\frac { 1 }{ 108 }\)
Greater the number of moles, greater is the number of atoms present. As \(\frac { 1 }{ 197 }\) > \(\frac { 1 }{ 108 }\), silver will contain greater number of atoms.

In this page, we are providing Is Matter Around Us Pure Class 9 Extra Questions and Answers Science Chapter 2 pdf download. NCERT Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 2 Extra Questions and Answers Is Matter Around Us Pure

Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers Solutions

Is Matter Around Us Pure Class 9 Extra Questions Very Short Answer Type

Question 1.
What is the reason for the difference in properties of solutions, colloids and suspensions?
Answer:
Due to different particle size.

Question 2.
Is rain water or distilled water a pure substance?
Answer:
Yes, because it contains particles (molecules) of only water.

Question 3.
Can we separate a mixture of alcohol and water by a separating funnel?
Answer:
No, the two liquids are miscible.

Question 4.
Give two examples of metalloids.
Answer:
Silicon and germanium.

Question 5.
Give an example of a solution in which solid is a solute as well as the solvent.
Answer:
Alloys are solid in solid solutions. For example, brass contains about 30% zinc and 70% copper. Here, zinc is the solute while copper is the solvent.

Question 6.
What type of liquid mixture will kerosesne oil and water form? How will you separate it?
Answer:
Immiscible. We can separate this mixture by using a separating funnel.

Question 7.
Define the term heterogeneous.
Answer:
Heterogeneous means that the substance does not have the same properties or characteristics throughout its bulk.

Question 8.
Write the constituent element of potassium hydroxide and sodium chloride.
Answer:
The constituent element of potassium hydroxide is K, H and O and sodium chloride is Na and Cl.

Question 9.
On the basis of composition, how is matter classified?
Answer:

1. Pure substance
2. Mixture.

Question 10.
What are different categories of pure substances?
Answer:

1. Elements
2. Compounds.

Question 11.
What are the different kinds of mixture?
Answer:

1. Homogeneous mixture
2. Heterogeneous mixture.

Question 12.
What are the constituents of brass?
Answer:
Brass is an alloy and is a mixture of zinc (30%) and copper (70%).

Question 13.
How are elements further classified?
Answer:
Metals, non-metals, metalloids.

Question 14.
A solution of water and alcohol contains 30 g of water and 60 g of alcohol. What is the concentration of solution?
Answer:
\(\frac { 30 }{ 30 + 60 }\) x 100 = \(\frac { 30 }{ 90 }\) x 100 = 33.3%.

Question 15.
What are aqueous solutions?
Answer:
Solutions in which water is the solvent are called aqueous solutions, e.g., sugar solution, in which sugar is dissolved in water.

Question 16.
What is an unsaturated solution?
Answer:
A solution in which some more solute can be dissolved at any fixed temperature is called an unsaturated solution.

Question 17.
How many gram of water is needed to make 8% mass by mass percentage of sodium carbonate solution if 4 g of sodium carbonate is a variable to make a solution?
Answer:
8% means 8 g in 100 g of solution. So if 4 g Na₂CO₃ is present, it means solution must be 50 g.

Question 18.
What are the conditions required to convert air into liquid air?
Answer:
200 atmospheric pressure and – 200°C.

Question 19.
Define the term inter-conversion of matter.
Answer:
The phenomenon of change of one state of matter into another and back to the original state is called inter-conversion of matter.

Question 20.
Why do fish go in deep water during day light?
Answer:
During day time, the shallow water is warmed and hence it has less dissolved oxygen. Therefore fish tend to go in deep water during day time.

Question 21.
Out of colloid, solution and a suspension which one can be separated by filtration.
Answer:
Suspension.

Question 22.
Out of colloid, solution and a suspension which has the smallest particle?
Answer:
Solution.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 1

Question 1.
Suggest separation technique(s) one would need to employ to separate the following mixtures. [NCERT Exemplar]
(a) Mercury and water
(b) Potassium chloride and ammonium chloride
(c) Common salt, water apd sand
(d) Kerosene oil, water and salt.
Answer:
(a) Separation by using separating funnel
(b) Sublimation
(c) Filtration followed by evaporation
Or
Centrifugation followed by evaporation/distillation
(d) Separation by using separating funnel to separate kerosene oil followed by distillation.

Question 2.
A salt can be recovered from its solution by evaporation. Suggest some other technique for the same? [NCERT Exemplar]
Answer:
Crystallisation.

Question 3.
The ‘sea water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Answer:

• Homogeneous – mixture of salts and water only.
• Heterogeneous – contains salts, water, mud, decayed plant, etc.

Question 4.
Fill in the following blanks:
(i) Milk is a ……………… solution but vinegar is a solution.
(ii) Milk is a colloidal solution in which ……………… is the dispersed phase and ……………… is the dispersion medium.
Answer:
(i) colloidal, true
(ii) fat, water.

Question 5.
(a) Classify Brass and Diamond as element and mixture.
(b) How is a chemical change different from a physical change?
Answer:
(a) Brass is homogeneous mixture also called alloy. The constituents are Cu and Zn. Diamond is an element. It is an allotropic form of carbon.

(b) In a chemical change, a new substance is formed as a result of chemical reaction. No new substance is formed in a physical change.

Question 6.
Identify the dispersed phase and dispersion medium in the following examples of colloids:
(a) Fog
(b) Cheese
(c) Coloured gemstone.
Answer:
(a) Fog: Liquid (water drops) acts as dispersed phase and gas (air) as the dispersion medium.
(b) Cheese: Solid (fat) acts as the dispersed phase and water (liquid) as the dispersion medium.
(c) Coloured gemstone: Solids act the dispersed phase as well as the dispersion medium.

Question 7.
Explain, why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do. [NCERT Exemplar]
Answer:
Particle size in a suspension is larger than those in a colloidal solution. Also molecular interaction in a suspension is not strong enough to keep the particles suspended and hence they settle down.

Question 8.
Smoke and fog both are aerosols. In what way are they different? [NCERT Exemplar]
Answer:
Both fog and smoke have gas as the dispersion medium. The only difference is that the dispersed phase in fog is liquid and in smoke it is a solid.

Question 9.
How will you bring about the following separation?
(i) Fine mud particles floating in water.
(ii) Carbon particles present in smoke.
Answer:
(i) By coagulation using alum and then filtering.
(ii) By passing smoke through electric plates maintained at a high potential difference. The colloidal particles of carbon get neutralised and fall down while air escapes out.

Question 10.
Classify the following as Physical or chemical properties. [NCERT Exemplar]
(a) The composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(c) Metallic sodium is soft enough to be cut with a knife.
(d) Most metal oxides form alkalies on interacting with water.
Answer:
Physical properties: (a) and (c)
Chemical properties: (b) and (d)

Question 11.
Suggest a suitable separation technique for the following: [NCERT Exemplar]
(a) Mercury and water
(b) Coloured components from blue ink
(c) Ammonium chloride and potassium chloride
(d) Mixture of alcohol and water.
Answer:
(a) The separation can be done by the use of a separating funnel. Mercury forms the lower layer (heavier) and water the upper layer (lighter).
(b) The separation can be done with the help of chromatography.
(c) Process of sublimation can be used. Ammonium chloride collects as the sublimate while potassium chloride remains in the dish.
(d) Process of fractional distillation can be used. Alcohol (ethyl alcohol) with lower boiling point (78°C or 351 K) gets distilled leaving behind water with higher boiling point (100°C or 373 K) in the distillation flask.

Question 12.
Name the type of colloids in each of the following giving an example of each.

Answer:

Question 13.
You are given two samples of water labelled as ‘A’ and ‘B’ Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment. [NCERT Exemplar]
Answer:
Sample ‘B’ will not freeze at 0°C because it is not pure water. At 1 atm, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.

Question 14.
Identify colloids from the following: Copper sulphate solution, milk, smoke, muddy water, butter, sugar solution, face cream, lemonade.
Answer:
Colloids: milk, smoke, muddy water, butter, face cream, lemonade.

Question 15.
What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments? [NCERT Exemplar]
Answer:
Pure gold is very soft as compared to gold alloyed with silver or copper. Thus for providing strength to gold, it is alloyed.

Question 16.
12 mL of dettol is added to a beaker containing 500 mL of water and stirred. State four observations that you make.
Answer:
When dettol is added to a beaker containing of water, the following observations are made.

• An emulsion is formed which is of colloidal nature.
• The colour of emulsion is milky.
• It gives characteristic smell of dettol.
• The solution can pass through a filter paper.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 2

Question 1.
During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution. [NCERT Exemplar]
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.
Answer:
(a) No.
\(Mass\%=\frac{\text { Mass of solute }}{\text { Mass of solute }+\text { Mass of solvent }} \times 100\)

(b) Solution made by Ramesh
Mass % = \(\frac { 10 }{ 10 + 100 }\) = \(\frac { 10 }{ 110 }\) x 100 = 9.09%
solution made by sarika
Mass % = \(\frac { 10 }{ 100 }\) x 100 = 10%
The solution prepared by Sarixa has a higher mass % than that prepared by Ramesh.

Question 2.
State which of the following solutions exhibit Tyndall effect:
Starch solution, Sodium chloride solution, Tincture of iodine, Air.
Answer:
(i) Tyndall effect is shown both by starch solution and air which are heterogeneous mixtures and have the capacity to scatter a beam of light as it passes through them.

(ii) Sodium chloride solution and tincture of iodine (iodine crystals dissolved in ethyl alcohol) are both homogeneous in nature and do not exhibit any Tyndall effect.

Question 3.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice. [NCERT Exemplar]
Answer:
Distillation, since acetone is more volatile it will separate out first.

Question 4.
(a) Why is crystallisation technique better than evaporation?
(b) Write any two physical properties of each of metals and non-metals.
(c) Name the technique used to separate butter from curd.
Answer:
(a) Both these techniques are used to separate solid substances from their solutions. But crystallisation is considered better because during evaporation certain solids may decompose or some of them like sugar get charred when the solution is evaporated completely to dryness. As a result of crystallisation, even the shapes of the crystals do not change.

(b) (i) Metals have a shiny surface known as lustre.
(ii) Metals are malleable and ductile.
(iii) Non-metals are mostly poor conductors of electricity.
(iv) Non-metals are generally soft.

(c) Butter can be separated from curd by the process of centrifugation. This is usually done by churning which is very common as well as convenient.

Question 5.
Name the process associated with the following:
(a) Dry ice is kept at room temperature and at one atmospheric pressure.
(b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(d) A acetone bottle is left open and the bottle becomes empty.
(e) Milk is churned to separate cream from it.
(f) Settling of sand when a mixture of a sand and water is left undisturbed for some time.
Answer:
(a) Sublimation
(b) Diffusion
(c) Dissolution/diffusion
(d) Evaporation, diffusion
(e) Centrifugation
(f) Sedimentation

Question 6.
On dissolving chalk powder in water, a suspension is obtained. Give any four reasons to support the fact that the mixture so obtained is a suspension only.
Answer:
It is supported by the following reasons:

• White particles of chalk powder can be seen with the naked eyes.
• The particles can be separated by ordinary filter paper.
• Upon shaking, a white turbidity reappears in solution.
• Light cannot pass through the suspension which shows that it is of opaque nature.

Question 7.
Give an example for each of the following:
(a) Solid-liquid homogeneous mixture
(b) Gas-gas homogeneous mixture
(c) Liquid-liquid heterogeneous mixture.
Answer:
(a) Mixture of sodium chloride in water.
(b) Air. It is a homogeneous mixture of a number of gases.
(c) Emulsion of oil and water.

Question 8.
What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature?
(b) an aqueous sugar solution is heated to dryness?
(c) a mixture of iron filings and sulphur powder is heated strongly?
Answer:
(a) Solid potassium chloride will separate out.
(b) Initially the water will evaporate and then sugar will get charred.
(c) Iron sulphide will be formed.

Question 9.
(a) Arrange solids, liquids and gases in increasing order of the following properties of matter
(i) rigidity
(ii) diffusion
(iii) compressibility.
(b) Write one example from your daily life which is based on diffusion of gases.
Answer:
(a) (i) Rigidity: Gases < Liquids < Solids
(ii) Diffusion: Solids < Liquids < Gases
(iii) Compressibility: Solids < Liquids < Gases.

(b) Smell of aroma or perfume released in one corner of the room soon spreads in the whole room.

Question 10.
Is air a mixture or a compound? Give three reasons.
Answer:
Air is a mixture and not a compound as discussed below:
(i) The properties of a mixture are in between those of its constituents. The two major components of air are oxygen (20.9% by volume) and nitrogen (78.1% by volume). In oxygen, any fuel bums brightly but in nitrogen it gets extinguished. In contrast, in air the fuel bums slowly.

(ii) The components of a mixture can be separated by simple physical methods. For example, the components of air can be separated by fractional distillation of liquid air.

(iii) The composition of a mixture is variable. The composition of air is also variable. It has more oxygen in the country side than in big cities.

(iv) When air is obtained by mixing its constituent gases, no heat is either evolved or absorbed.

(v) Liquid air does not have a fixed boiling point.

Is Matter Around Us Pure Class 9 Extra Questions Long Answer Type

Question 1.
Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the Sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.
Answer:
Physical changes: (a), (b), (e)
Chemical changes: (c), (d)

Question 2.
Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?
Answer:
Part A

Part B

When dilute HCl is added to it, only the iron filings in the mixture react and sulphur remains unreacted
Fe (s) + 2 HCl (aq) → FeCl₂ + H₂ gas
H₂S gas formed has a foul smell and on passing through lead acetate solution, it turns the solution black. Hydrogen gas burns with a pop sound.

Question 3.
Fill in the blanks:
(a) A colloid is a ……………….. mixture and its components can be separated by the technique known as ………………..
(b) Ice, water and water vapour look different and display different ……………….. properties but they are the same.
(c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for sometime. The upper layer in the separating funnel will be of ……………….. and the lower layer will be that of
(d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called ………………..
(e) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the ……………….. of light by milk and the phenomenon is called ……………….. This indicates that milk is a ……………….. solution. [NCERT Exemplar]
Answer:
(a) heterogeneous, centrifugation
(b) physical, chemically
(c) water, chloroform (hint-density of water is less than that of chloroform)
(d) fractional distillation scattering, Tyndall effect, colloidal.

Question 4.
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures. [NCERT Exemplar]
(a) A volatile and a non-volatile component.
(b) Two volatile components with appreciable difference in boiling points.
(c) Two immiscible liquids.
(d) One of the components changes directly from solid to gaseous state.
(e) Two or more coloured constituents soluble in some solvent.
Answer:
(a) Evaporation or distillation
(b) Distillation
(c) By using a separating funnel
(d) Sublimation
(e) Chromatography

Question 5.
Which separation techniques you will apply for the separation of the following mixtures?
(a) Oil from water
(b) Camphor from sand
(c) Sodium chloride from its solution in water
(d) Cream from milk
(e) Metal pieces from engine oil of a car.
Answer:
(a) By the use of a separating funnel.
(b) With the help of sublimation technique
(c) By evaporation crystallisation technique.
(d) By the use of a centrifuge
(e) By the use of filtration technique

Is Matter Around Us Pure Class 9 Extra Questions HOTS

Question 1.
Which of the tubes shown below will be more effective as a condenser in the distillation apparatus? [NCERT Exemplar]

Answer:
The presence of beads in tube (a) would provide a larger surface area for cooling. So, this tube will be more effective as a condenser.

Question 2.
Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non- sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name a non-metal other than carbon which shows allotropy.
(f) Name a non-metal which is required for combustion.
Answer:
(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, phosphorus
(f) Oxygen.

Question 3.
The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
Answer:
‘C’ has made the desired solution.

= \(\frac { 50 }{ 100 }\) x 100
= 50% mass by volume.

Question 4.
On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
(a) Is this a physical or a chemical change?
(b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
Answer:
(a) Chemical change

(b) Acidic and basic solutions can be prepared by dissolving the products of the above process in water
CaO + H₂O → Ca(OH)₂ (basic solution)
CO₂ + H₂O → H₂CO₃ (acidic solution)

Question 5.
Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl + 100 g water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d) 0.10 g of NaCl + 99.90 g of water
Answer:
(c) 0.01 g of NaCl + 99.99 g of water

= \(\frac { 0.01 }{ 0.01 + 99.99 }\) x 100
= \(\frac { 0.01 }{ 100 }\) x 100
= 0.01 g

Question 6.
A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the figure. The filter paper was removed when the water moved near the top of the filter paper. [NCERT Exemplar]

1. What would you expect to see, if the ink contains three different coloured components?
2. Name the technique used by the child.
3. Suggest one more application of this technique.

Answer:

1. Three different bands will be observed.
2. Chromatography
3. To separate the pigments present in chlorophyll.

Question 7.
A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the figure. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

1. Explain why the milk sample was illuminated. Name the phenomenon involved.
2. Same results were not observed with a salt solution. Explain.
3. Can you suggest two more solutions which would show the solution?

Answer:

1. Milk is a colloid and would show Tyndall effect.
2. Salt solution is a true solution and would not scatter light.
3. Detergent solution, sulphur solution.

Question 8.
Sudha tested the solubility of four salts X, Y, Z and T at different temperatures and collected the following data. (Solubility refers to the amount in grams dissolved in 100 g of water to give a saturated solution.)

Answer the following questions from the table:

1. Which salt has the highest and lowest solubility at 323 K?
2. A student prepared a saturated solution of X at 323 K and then added 25 g water to it. What mass of X must be added to again make the solution saturated?
3. The solubility of which salt is least affected by increase in temperature?
4. What mass of T would be required to make saturated solution in 200 g of water at 290 K?

Answer:
1. At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility.

2. By definition of saturated solution,
100 g of water at 323 K contain salt = 40 g
125 g of water at 323 K contain salt = \(\frac { 40 g }{ 100 g }\) x (125 g) = 50 g
∴ Mass of salt to be added to make the solution again saturated = (50 – 40) = 10 g

3. The data show that the solubility of the salt Y is least affected with increase in temperature.

4. At 290 K, mass of T required to make a saturated solution in 200 g of water = \(\frac { 25 g }{ 100 g }\) x (200 g) = 50 g

Is Matter Around Us Pure Class 9 Extra Questions Value Based (VBQs)

Question 1.
Mallika’s mother was suffering from cold and cough. Mallika prepared tea for her mother. She boiled water in a pan, then she added tea leaves, sugar and milk to it. She filtered the tea in a cup and served it to her mother.
(a) Explain the values shown by Mallika.
(b) Identify solute, solvent, residue and filtrate in this activity.
Answer:
(a) Mallika used the knowledge of chemistry to provide relief to her mother. Actually she prepared an extract of tea leaves which is helpful in curing cold and cough and gives warmth to the body.

(b) Solute: Tea leaves and sugar.
Solvent: water and milk.
Filtrate: homogeneous mixture of water, milk, sugar and extract of tea leaves.

Question 2.
Amit was asked by his teacher to separate a liquid mixture of acetone and ethyl alcohol. He set up a distillation apparatus and tried to distil the mixture. To his surprise, both the liquids got distilled. Teacher told Amit to repeat the experiment by using a fractionating column in the distillation flask. Amit followed the advice of the teacher and he was able to separate the two liquids.

1. Why was Amit not successful in separating the liquid mixture earlier?
2. Why did teacher ask him to use the fractionating column?
3. Which liquid was distilled first?
4. As a student of chemistry, what value based information you have gathered?

Answer:
1. The difference in boiling point temperatures of acetone (56°C) and ethyl alcohol (78°C) is only 22°C.
Therefore, process of simple distillation fails in this case.

2. Fractionating column is quite effective in this case because it obstructs the distillation of ethyl alcohol which is high boiling and at the same time helps in the distillation of acetone which is low boiling.

3. Acetone was distilled first since it has comparatively low boiling point.

4. The process of simple distillation can be used only in case, the liquids present in the mixture differ in their boiling point by 25°C or more.

In this page, we are providing Matter in Our Surroundings Class 9 Extra Questions and Answers Science Chapter 1 pdf download. NCERT Extra Questions for Class 9 Science Chapter 1 Matter in Our Surroundings with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 1 Extra Questions and Answers Matter in Our Surroundings

Extra Questions for Class 9 Science Chapter 1 Matter in Our Surroundings with Answers Solutions

Matter in Our Surroundings Class 9 Extra Questions Very Short Answer Type

Question 1.
Define the term ‘matter’.
Answer:
Matter is defined as anything that occupies some space and has definite mass.

Question 2.
What is Law of Conservation of Mass?
Answer:
Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction.

Question 3.
Define melting point.
Answer:
It is the temperature at which a solid becomes liquid at atmospheric pressure by absorbing heat.

Question 4.
Out of water and alcohol, which is more volatile?
Answer:
The boiling point of alcohol (78°C or 351K) is lower than that of water (100°C or 373K), therefore, alcohol is more volatile than water.

Question 5.
What is sublimation?
Answer:
Direct conversion of a solid into vapour and vice-versa (i.e., vapour into solid) is called sublimation.

Question 6.
Is dry ice the same thing as ordinary ice?
Answer:
No, dry ice is solid carbon dioxide while ordinary ice is solid water.

Question 7.
Define latent heat of fusion.
Answer:
It is the heat energy required to convert 1 kg of solid into liquid at its melting point at atmospheric pressure.

Question 8.
Define vapourisation.
Answer:
The process of change from liquid state to gaseous (vapour) state is called vapourisation.

Question 9.
Give the important properties on the basis of which the three states of matter can be distinguished.
Answer:
The three states of matter can distinguished on the basis of shape, volume, compressibility, packing of molecules, number of free surfaces, etc.

Question 10.
Name the term used for the solid which is directly formed from the gas.
Answer:
Sublimate.

Question 11.
Define the term volatile liquid.
Answer:
Those liquids which can change into vapour easily are termed as volatile liquids.

Question 12.
What is the effect of pressure on boiling point?
Answer:
Boiling point increases with increase in pressure.

Question 13.
Name any two substances which sublime.
Answer:
Camphor, napthalene, iodine, ammonium chloride.

Question 14.
Define condensation.
Answer:
The change of a gaseous state to a liquid state on cooling is known as condensation.

Question 15.
State the effect of surface area on rate of evaporation.
Answer:
If the surface area is increased, the rate of evaporation increases.

Question 16.
Define evaporation.
Answer:
Evaporation is a physical process in which a liquid changes to its gaseous state, at a temperature lower than its boiling point.

Question 17.
What are the ways in which a gas can be liquefied?
Answer:
Applying pressure and reducing temperature can liquefy gases.

Question 18.
What is plasma?
Answer:
It is a state of matter which consists of super energetic and super excited particles. These particles are in the form of ionised gases.

Question 19.
How do solids, liquids and gases differ in shape and volume?
Answer:
Solids have a definite shape and a fixed volume, liquids have a definite volume but no fixed shape while gases neither have a definite volume nor a definite shape.

Question 20.
Kelvin scale of temperature is regarded as better scale than Celsius. Why?
Answer:
As it has a wide range of measurement and temperature in kelvin scale always has a positive sign, hence regarded as better scale than Celsius.

Matter in Our Surroundings Class 9 Extra Questions Short Answer Type 1

Question 1.
What are characteristics of particles of matter?
Answer:
The particles of matter have following characteristics:

• Particles of matter are very very small.
• Particles of matter have space between them.
• Particles of matter attract each other.
• Particles of matter are constantly moving.

Question 2.
Write four main characteristics of solid state of matter.
Answer:

• Solids have definite mass, volume and shape.
• The particles in solid state are closely packed and empty spaces in them are negligible.
• Solids are rigid.
• Solids can have a number of free surfaces.

Question 3.
Write four main characteristics of liquid state of matter.
Answer:

• Liquids have a definite mass and volume.
• A liquid can take the shape of a container.
• Liquids have only one free surface.
• Liquids show the property of diffusion.

Question 4.
Write four characteristics of gaseous state of matter.
Answer:

• A gas has definite mass but it has neither definite shape nor definite volume.
• Gases can occupy the whole of the space available to them.
• There are larger vacant spaces between the molecules of a gas.
• Gases are highly compressible.

Question 5.
Explain evaporation and its cooling effect in terms of kinetic energy of particles.
Answer:
During evaporation, the molecules which possess higher kinetic energy leave the liquid and go into the space above the liquid as vapour. The remaining molecules possessing lower kinetic energy are left in the liquid state. Consequently, the average kinetic energy decreases which results in the fall of temperature of the liquid.

Question 6.
How is heat transferred when a solid sublimes?
Answer:
Certain solids like iodine, naphthalene, solid CO₂ sublimes on heating. Heat is absorbed by the molecules of these solids rapidly which provides enough kinetic energy to show phase change into gaseous state.

Question 7.
Why do gases diffuse rapidly?
Answer:
Gases diffuse rapidly due to high speed of the particles and large space between them.

Question 8.
For any substance, why does the temperature remain constant during the change of state?
Answer:
On increasing the temperature of a substance, for example a solid, the kinetic energy of the particles increases which is used to overcome the forces of attraction between the particles therefore the temperature remains constant during the change of state.

Question 9.
Explain compressibility in gases with an example.
Answer:
Liquefied petroleum gas (LPG) cylinders are used in our homes for cooking, contains gases in the compressed state. Similarly, compressed natural gas (CNG) is used as a fuel in vehicles. Large volume of gases can be compressed in small cylinders and are transported to distant places.

Question 10.
Why solids cannot be compressed like gases?
Answer:
The particles in solids are so tightly packed that there are no or little interparticle spaces left among them. Therefore solids are not compressible like gases. Gases which have large interparticle spaces are therefore compressible.

Question 11.
Define boiling. Why is boiling considered as bulk phenomenon?
Answer:
Rapid and breaking of bubbles in the bulk of a liquid being heated is called boiling. During boiling particles from the bulk of liquid gain enough energy to get converted to vapour. Therefore it is a bulk phenomenon.

Question 12.
Why do we see water droplets on the outer surface of a glass containing ice-cold water?
Answer:
The water vapour present in air, on coming in contact with the cold glass of water, loses energy and gets converted to liquid state, which we see as water droplets.

Question 13.
Why do we sprinkle water on the roof or open ground in summer?
Answer:
During hot summer evenings, we often sprinkle water on the roof of the house or open ground in front of our house. The water evaporates by absorbing heat from the ground and the surrounding air. By losing heat, the ground becomes cool and we feel comfortable.

Question 14.
Why is ice rubbed on a burnt part of the skin?
Answer:
When a finger or some part of our body gets burnt, we rub the burnt portion with an ice cube. The reason being that due to burning, the temperature of the injured skin increases. When ice is rubbed, the excess heat from the skin is taken away by large latent heat of fusion of water. As a result, the temperature of the injured skin decreases and we feel less pain.

Question 15.
How will you demonstrate that particles of matter are continuously moving?
Answer:
When an incense stick is lit in one corner of a room, we get the smell while sitting at a distance from the stick. This is because the particles of matter are continuously moving. Because of their random motion, the particles of incense mix with the particles of air rapidly and the smell of the incense reaches us even when we are sitting at a distance from the incense stick.

Question 16.
Why do solids expand a bit on heating and contract a bit on cooling?
Answer:
The solid molecules do not have sufficient intermolecular (or interparticle) space thus its expands a bit on heating. The interparticle forces of attraction are very strong which do not let solid particles leave their mean positions. Therefore solid contracts a bit on cooling.

Question 17.
Draw diagram to shown interconversion among states of matter.
Answer:

Question 18.
Why is light not considered matter?
Answer:
Matter occupies space and has mass. Light has neither of the two and that is why it is not considered as matter. It is considers as a form of energy and electromagnetic radiation.

Question 19.
Convert the following temperatures:
(a) – 78.0°C to kelvin
(b) 775 K to °C
(c) 489 K to °C
(d) 24°C to kelvin
Answer:
(a) – 78 + 273 = 195 K
(b) 775 – 273 = 502°C
(c) 489-273 = 216°C
(d) 24 + 273 = 297 K

Question 20.
Mention the difference between gas and vapour.
Answer:
Gas – The gas is a substance which exists in the gaseous state at a temperature equal to or more than the boiling point of its liquid state. For example oxygen, hydrogen, nitrogen, etc.

Vapour – A vapour is a substance which exists in the gaseous state such that its temperature is lower than that of boiling point of its liquid state. For example, water vapour, iodine vapour, etc.

Question 21.
A sample of water under study was found to boil at 102″C at normal temperature and pressure. Is the water pure? Will this water freeze at 0°C? Comment.
Answer:
It’s freezing point will be below 0°C due to the presence of a non-volatile impurity in it.

Question 22.
Water as ice has a cooling effect, whereas water as steam may cause severe burns. Explain these observations. [NCERT Exemplar]
Answer:
In case of ice, the water molecules have low energy while in the case of steam the water molecules have high energy. The high energy of water molecules in steam is transformed as heat and may cause burns. On the other hand, in case of ice, the water molecules take energy from the body and thus gave a cooling effect.

Question 23.
It is a hot summer day, Priyanshi and Ali are wearing cotton and nylon clothes respectively. Who do you think would be more comfortable and why? [NCERT Exemplar]
Answer:
Cotton being a better absorber of water than nylon helps in absorption of sweat followed by evaporation which leads to cooling. So Priyanshi would be more comfortable than Ali.

Question 24.
You want to wear your favourite shirt to a party, but the problem is that it is still wet after a wash. What steps would you take to dry it faster? [NCERT Exemplar]
Answer:
Conditions that can increase the rate of evaporation of water are:

• An increase of surface area by spreading the shirt
• An increase in temperature by putting the shirt under the Sun
• An increase the wind speed by spreading it under the fan.

Matter in Our Surroundings Class 9 Extra Questions Short Answer Type 2

Question 1.
What is evaporation? Why does evaporation cause cooling?
Answer:
The process in which a liquid changes into its vapour state at a temperature below the boiling point is called evaporation. Evaporation is an endothermic process i.e., the liquid absorbs heat during evaporation. This heat may be provided either by the surrounding or by the liquid itself. When the evaporating liquid takes the required heat from other parts of the liquid, the rest of the liquid cools down.

On the other hand, if the liquid takes heat from the surroundings, it causes cooling of the surroundings. For example, on a hot day (sunny day) we perspire. When this sweat evaporates, it absorbs the required heat from our body, and we feel cool.

Question 2.
What factors affect the rate of evaporation?
Answer:
Factors that affect the rate of evaporation are:

• Temperature: Evaporation increases with increase in temperature.
• Humidity: Evaporation decreases with an increase in humidity.
• Wind speed: Evaporation increases with an increase in wind speed.

Question 3.
What is a dry ice and what are its properties?
Answer:
Solid carbon dioxide is known as dry ice. It is stored under high pressure. Solid CO₂ gets converted directly to gaseous state on decrease of pressure to 1 atmosphere without passing through the liquid state (i.e., sublimes). This is the reason that solid carbon dioxide is also known as dry ice.

It is mainly used as a cooling agent because its temperature is very low than ice formed from water. Dry ice is commonly used in theaters and in movies to produce the effect of fog.

Question 4.
Why should we wear cotton clothes in summer?
Answer:
During summer, we perspire more because of the mechanism of our body which keeps us cool. We know that during evaporation, the particles at the surface of the liquid gain energy from the surroundings or body surface and change into vapour. The heat energy equal to latent heat of vapourisation is absorbed from the body leaving the body cool. Cotton, being a good absorber of water helps in absorbing the sweat and exposing it to the atmosphere for easy evaporation.

Question 5.
Give the main postulates of kinetic theory of matter.
Answer:
The main postulates of kinetic theory are:

• All matter is made up of a large number of extremely small particle called molecules.
• The molecules are always in a state of rapid random motion.
• The molecules possess kinetic energy.
• There are attractive forces between the molecules.
• The kinetic energy of molecules increases with increase in temperature.

Question 6.
Identify each of the following changes of state as evaporation, boiling or condensation. Give reason for your answer.
(a) Wet clothes dry when spread on wire.
(b) After a hot shower, your bathroom mirror is covered with water.
(c) Lava flows into the ocean and forms steam.
Answer:
(a) Evaporation, because conversion of liquid water to vapour occur at room temperature.
(b) Condensation, because hot water vapour condense to form liquid water.
(c) Boiling, because heat of lava makes liquid water boil and hence steam is formed.

Question 7.
Why do surgeons often spray some ether on the skin before performing minor surgery?
Answer:
Quite often doctors spray ether on a portion of skin the before performing minor surgery. The reason being that a ether has very low boiling point (308 K). Therefore, it evaporates quite rapidly. The heat energy needed for evaporation is taken from the skin. As a result, the temperature of the skin becomes so low that it almost becomes numb. Due to this numbness, the patient does not feel much pain when a minor cut is made in the skin in order to perform surgery.

Similarly, when a player gets injured during a game, ethyl chloride on the injured portion of the body. Since the boiling point of ethyl chloiide (285.5K or 12.5°C) is very low, it quickly evaporates. The heat energy needed for evaporation is taken from the skin. By losing heat, temperature of the skin becomes so low that it almost becomes numb. Due to this numbness, the player does not feel much pain.

Question 8.
Comment upon the following:
(i) Rigidity
(ii) Compressibility
(iii) Fluidity
Answer:
(i) Rigidity means tendency to maintain shape when some outside force is applied due to strong interparticle force.

(ii) Compressibility means tendency to decrease volume when some outside force is applied. Due to large interparticle distances in gases their volume decreases when some pressure is applied on them therefore, gases possess high compressibility.

(iii) Fluidity means tendency to flow. Due to large interparticle distances and weak forces of attraction gases have highest fluidity.

Question 9.
Comment on the following statements:
(а) Evaporation produces cooling.
(b) Rate of evaporation of an aqueous solution decreases with increase in humidity.
(c) Sponge though compressible is a solid. [NCERT Exemplar]
Answer:
(а) For evaporation to occur, heat energy is needed. This heat energy is taken out from the substance or the surroundings. As a result surrounding becomes cool. Thus, evaporation causes cooling.

(b) By humidity we mean, the amount of water vapours present in the air. With increase in humidity the rate of evaporation decreases. If the humidity of air is already high, it can hold only a little more amount of water vapour to reach that optimum level, therefore the rate of evaporation decreases.

(c) Sponge has large number of minute holes in which air is trapped. When we press it, air expelled and sponge is compressed to a small amount of matter which has a definite shape as well as definite volume.

Matter in Our Surroundings Class 9 Extra Questions Long Answer Type

Question 1.
List any five physical properties of liquids.
Answer:

• Liquids do not have fixed shape or boundaries.
• They have fixed volume.
• They exhibit fluidity i.e., they can flow.
• Less compressible as compared to gases but higher than solids.
• Lower density as compared to solids.
• Compared to solids, liquids have higher kinetic energy but less than gases.
• The intermolecular forces of attraction are weaker than those of solids.
• Show the property of intermixing i.e., can diffuse.

Question 2.
Fill in the blanks:
(а) Evaporation of a liquid at room temperature leads to a ………………… effect.
(b) At room temperature the forces of attraction between the particles of solid substances are ………………… than those which exist in the gaseous state.
(c) The arrangement of particles is less ordered in the state. However, there is no order in ………………… the state.
(d) is the change of gaseous state directly to solid state without going through the ………………… state.
(e) The phenomenon of change of a liquid into the gaseous state at any temperature below its boiling point is called ………………… [NCERT Exemplar]
Answer:
(a) cooling
(b) stronger
(c) liquid, gaseous
(d) Sublimation, liquid
(e) evaporation

Question 3.
Classify the following into osmosis/diffusion [NCERT Exemplar]
(a) Swelling up of a raisin on keeping in water.
(b) Spreading of virus on sneezing.
(c) Earthworm dying on coming in contact with common salt.
(d) Shrinking of grapes kept in thick sugar syrup.
(e) Preserving pickles in salt.
Answer:
(a) Osmosis
(b) Diffusion
(c) Osmosis
(d) Osmosis
(e) Osmosis

Question 4.
You are provided with a mixture of naphthalene and ammonium chloride by your teacher. Suggest an activity to separate them with well labelled diagram. [NCERT Exemplar]
Answer:
Naphthalene is insoluble in water but soluble in ether an organic solvent. It is volatile at room temperature. Ammonium chloride is soluble in water and volatile at higher temperature. It decomposes on heating to dryness.

Question 5.
Why does the temperature of a substance remain constant during its melting point or boiling point?
Answer:
The temperature of a substance remains constant at its melting and boiling points untill all the substance melts or boils because, the heat supplied is continuously used up in changing the state of the substance by overcoming the forces of attraction between the particles. This heat energy absorbed without showing any rise in temperature is given the name latent heat of fusion/latent heat of vapourisation.

Matter in Our Surroundings Class 9 Extra Questions HOTS

Question 1.
Arrange the following substances in increasing order of force of attraction between the particles.
(a) Milk
(b) Salt
(c) Oxygen
Answer:
Oxygen < Milk < Salt.

Question 2.
Explain with an experiment to show gases do not have fixed shape or volume.
Answer:
Experiment: To show gases do not have fixed shape or volume.

Method:

• Take two balloons of different shapes. For example, one round and one heart shape or cylindrical.
• Fill the balloons with air.

Observation: Air takes up the shape of balloon.

Conclusion: This shows air has no definite shape or volume. It takes up the shape of the balloon.

Question 3.
Name the change of state during the following changes:
(a) Drying of wet clothes
(b) Melting of wax
(c) Melting of ice
(d) Formation of clouds
Answer:
(a) Liquid to gaseous state
(b) Solid to liquid state
(c) Solid to liquid state
(d) Liquid to gaseous state

Question 4.
With proper explanation, explain whether the following statements are true or false?
(a) Sublimation occurs only when the solid is heated.
(b) A lighter gas can move downwards and a heavier gas can move upwards.
(c) Interconversion of matter is a constant temperature process.
Answer:
(a) Statement is wrong. Sublimation may occur on its own or by heating, e.g., camphor, naphthalene, iodine, etc., sublime slowly at room temperature.

(b) Statement is true. Diffusion occurs against the law of gravitation. Therefore, lighter gases can also diffuse downwards and the heavier gases can also diffuse upwards. However rate of diffusion of lighter gases is faster than those of heavier gases.

(c) Statement is true. During interconversion of state of matter from solid to liquid or from liquid to gas, it tends to reach its melting point or boiling point. At this point, the temperature remains constant unit it has changed in another state.

Question 5.
What is meant by Bose-Einstein Condense?
Answer:
(a) In 1920, Indian scientist Satyendra Nath Bose did some calculations, based on which Albert Einstein predicted that a new state of matter should exist.

(b) This new state was named as Bose-Einstein Condensate (BEC). In 2001, Cornell, Ketterie and Wieman of USA received Noble Prize for actually making this state in laboratory. BEC is made by cooling gas of very low density to super low temperature.

Question 6.
A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of time. Which of the following figure would correctly represent the result? Justify your choice. [NCERT Exemplar]

Answer:
Since ice and water are in equilibrium, the temperature would be zero. When we heat the mixture, energy supplied is utilised in melting the ice and the temperature does not change till all the ice melts because of latent heat of fusion. On further heating, the temperature of the water would increase. Therefore the correct option is (d).

Question 7.
Look at the figure and suggest in which of the vessels A, B, C or D the rate of evaporation will be the highest? Explain. [NCERT Exemplar]
Answer:

(c) The rate of evaporation increases with an increase in the surface area of absorption because evapo-ration in a surface phenomenon. Also, with the increase in air speed, the particles of water vapour will move away with the air, which will increase the rate of evaporation.

Matter in Our Surroundings Class 9 Extra Questions Value Based (VBQs)

Question 1.
In a hot summer day, Rajeev wants to watch a movie in a nearest cineplex. His mother wears a nylon saree so Rajeev suggest her to wear cotton saree.
(i) Why does Rajeev suggest her to wear cotton saree and not nylon saree?
(ii) Mention the values exhibited by Rajeev.
Answer:
(i) On a hot summer day, we sweat a lot. Cotton clothes absorb sweat from the body. As the sweat evaporates it results in cooling giving comfort to the body.

(ii) Caring, use of knowledge of science.

Question 2.
Mohan was getting late for school. He tried to sip tea from the cup. His father advised him to use a plate and asked him to sip the tea from the plate. Mohan followed the advice and finished his tea very easily.
(i) Why is siping off tea easier from a plate?
(ii) Mention the values exhibited by Mohan’s father.
Answer:
(i) A plate has a larger surface area than a cup. Evaporation becomes faster in this case. Since cooling is always caused during evaporation the temperature got lowered. Therefore it became easier to sip tea from a plate.

(ii) Caring, use of knowledge of science, helpful.

NCERT CBSE Important Questions for Class 9 Science: Students who are struggling to find out what are the important question asked in the annual exams? Here is the list of CBSE Important questions for class 9 science chapterwise which are prepared by subject experts as per the latest CBSE syllabus curriculum. All these questions are designed after analyzing the previous questions papers & model papers. So, make sure to include practicing these NCERT extra important science questions and attain good marks in Exams.

Class 9 Science Important Questions with Answers

Access all CBSE NCERT Chapter Wise Important Questions of Class 9 Science with answers and solutions by clicking on the particular chapter link available over here.

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

More Resources

• NCERT Solutions for Class 9 Science
• NCERT Exemplar Solutions for Class 9 Science
• Value Based Questions in Science for Class 9
• HOTS Questions for Class 9 Science

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Extra Questions for Class 9 Maths: Here we are providing NCERT Extra Questions for Class 9 Maths with Solutions Answers Chapter Wise Pdf free download. Students can get Class 9 Maths NCERT Solutions, CBSE Class 9 Maths Important Extra Questions and Answers designed by subject expert teachers.

CBSE Class 9 Maths Extra Questions and Answers is an ultimate revision tool for students who are preparing for board exams. We have already compiled NCERT solutions for class 9 maths on our site. Apart from this important exam resource, CBSE Extra Questions of Maths Class 9 prepared by subjects experts based on the latest NCERT syllabus is essential for efficient preparation. So, we have listed chapter-wise NCERT Class 9 Maths Important Extra Questions with Answers in pdf formats.

Class 9 Maths Extra Questions with Solutions Answers

The NCERT Maths Extra Questions Class 9 with solutions and answers can be accessed from the available chapter-wise pdf links for free of cost.

1. Number Systems Class 9 Extra Questions
2. Polynomials Class 9 Extra Questions
3. Coordinate Geometry Class 9 Extra Questions
4. Linear Equations for Two Variables Class 9 Extra Questions
5. Introduction to Euclid’s Geometry Class 9 Extra Questions
6. Lines and Angles Class 9 Extra Questions
7. Triangles Class 9 Extra Questions
8. Quadrilaterals Class 9 Extra Questions
9. Areas of Parallelograms and Triangles Class 9 Extra Questions
10. Circles Class 9 Extra Questions
11. Constructions Class 9 Extra Questions
12. Heron’s Formula Class 9 Extra Questions
13. Surface Areas and Volumes Class 9 Extra Questions
14. Statistics Class 9 Extra Questions
15. Probability Class 9 Extra Questions

We hope the given NCERT Extra Questions for Class 9 Maths with Solutions Answers Chapter Wise Pdf free download will help you. If you have any queries regarding CBSE Class 9 Maths Important Extra Questions and Answers, drop a comment below and we will get back to you at the earliest.