KSEEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Students can Download Science Chapter 3 Atoms and Molecules Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Science helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 9 Science Chapter 3 Atoms and Molecules

KSEEB Class 9 Science Atoms and Molecules Intext Questions and Answers

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6g of ethanoic acid the products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 gins of sodium than a show that these observations are in agreement with the law of conservation of mass.
Answer :
Sodium carbonate + ethanoic acid → Sodium ethanoate + carbon dioxide + water
Reaction [Mass of Reactant]
5.3 + 6.0 = 11.3 gm
Mass of products
2.2 + 0.9 + 8.2 = 11.3 gm
∴ Mass of Reactants = Mass of products

Question 2.
Hydrogen and oxygen combine in the ratio of 1.8 by mass to form water, What mass of oxygen gas would be required to react completely with 3g of hydrogen gas.
Answer :
Hydrogen : Oxygen
Ratio of mass 1 : 8
mass of hydrogen = 3 gm
mass of oxygen = ?
x → mass of oxygen
x = 24 gm of oxygen is required,

Question 3.
Which postulate of Dalton’s atomic (the theory is the result of the law of conservation of mass).
Answer :
Atomic theory provided an explanation for the law of conservation of mass.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer :
Dalton’s atomic theory gave an explanation for the law of definite proportions.

Question 5.
Write down the formulae of
(i) Sodium oxide
(ii) aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide.
Answer :
(i) Sodium oxide – Na₂O
(ii) aluminium chloride – AlCl₃
(iii) Sodium sulphide – Na₂S
(iv) Magnesium hydroxide – Mg(OH)₃

Question 6.
Write down the names of compounds represented by the following formulae.
Answer :

Chemical compound	Formula
1. A12(SO4)2	Aluminium Sulphate
2. CaCl2	Calcium chloride
3. K2SO4	Pottassium sulphate
4. kNO3	Pottasium Nitrate
5. CaCO3	Calcium carbonate

Question 4.
What is meant by the term chemical formula?
Answer :
It is a symbolic representation of the composition of a compound in a definite integral ratio.

flow many atoms are present in a
(i) H₂S molecule?
H → 2 atoms
S – 1 atoms
3 atoms

(ii) \(\mathbf{P} \mathbf{O}_{4}^{3-}\) ion?
P → 1 atom
O → 4 atom.

Question 5.
Calculate the molecular masses H₂, O₂, Cl₂, CO₂, CH₄, C₂H₈, C₂H₄, NH₃, CH₃OH
Answer :
1. H₂ = 2 × atomic mass of Hydrogen = 2 × 1 = 2
2. O₂ = 2 × atomic mass of o×ygen =2 × 16 = 32
3. Cl₂ = 2 × atomic mass of chlorine =2 × 35.5 = 71
4. CO₂ = (1 × atomic mass of carbon) + 2(atomic mass of o×ygen) = (1 × 12)+ (2 × 16) = 12 + 32 = 44
5. CH₄ = (1 × atomic mass of carbon) + 4 × atomic mass of hydrogen = (1 × 12) + (4 × 1) = 12 + 4 = 16
6. C₂H₈ = (2× atomic mass of carbon) +(8× atomic mass of hydrogen) (2 × 12) + (8 × 1) = 24 + 8 = 32
7. C₂H₄ =(2×atoinic mass of carbon) +(4× atomic mass of hydrogen) = (2 × 12)+ (4 × 1) = 24 + 4 = 28
8. NH₃ = (1 × atomic mass of Nitrogen) +(3× atomic mass of hydrogen) = (1 × 14) + (3 × 1) = 14 + 3 = 17
9. CH₃OH = (1 × 12) + (H × 4) + (0 × 10) = 12 + 4 + 16 = 32.

Question 6.
Calculate the Molecular formula unit mass of ZnO, Na₂O, K²CO₃, given atomic masses of Zn = 654, Na = 234, K = 394, and O = 164
Answer :
1. ZnO = (1 × 2n) + (1 × O)
= (1 × 64) + (1 × 16)
= 65 + 16 = 81

2. Na₂O = (2 × Na) +(1 × 0)
= (2 × 23) + (1 × 16)
= 46 + 16 = 62

3. K₂CO₃ = (2 × 1C) + (1 × C) + (3 × O)
= (2 × 39) + (1 × 12) + (3 × 16)
= 78 + 12 + 48
= 90 + 48 = 138

Question 7.
If one mole of carbon atoms weight 12 gms what is the mass (in gms) of 1 atom of carbon.
Answer :
= 12 × Avagadro’s Number
= 12 × 6.022 × 10²³ = 72.264 = 10²³

Question 8.
Which has more number of atoms 100 gms- of Sodium or 100 grams of iron given atomic mass of Na = 234, Fe = 564
Answer :
Sodium – Iron
100 × 23 – 56 × 100
2300 × 6.023 × 10²³ – 5600 × 6.023 × 10²³

Question 9.
Define the atomic mass unit?
Answer :
It is a ratio of mass of an element to the mass of \(\frac{1}{12}^{\text {th }}\) the mass of carbon – 12 atoms.

Question 10.
Why it is not possible to see an atom with naked eye.
Answer :
Atom is the smallest particle, it can be seen through the naked eye, hence it is to less resolving power.

KSEEB Class 9 Science Atoms and Molecules Textbook Exercise Questions and Answers

Question 1.
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 :
0.24 gm sample of a compound of oxygen and boron

Question 2.
When 3.0 g of carbon is burnt in 8.0 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen. Which law of chemical combination will govern your answer?
Answer :

Question 3.
What are polyatomic ions? Give examples.
Answer :
Many numbers of atoms of an element are in the Ionic state.
\(\mathrm{O}_{2}^{-} \mathrm{SO}_{4}^{-}\)

Question 4.
Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer :
(a) MgCl₂
(b) CaO
(c) Cu(NO₃)₂
(d) AlCl₃
(e) CaCO₃

Question 5.
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer :
(a) Calcium and Oxygen
(b) Hydrogen and bromide
(c) Sodium, hydrogen, carbon, oxygen
(d) Potassium, Sulphur and oxygen.

Question 6.
Calculate the molar mass of the following substances.
(a) Ethync, C₂H₂
(b) Sulphur molecule, S_i
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
Answer :
(a) C₂H₂ = (2 × C) + (2 × 4)
= (2 × 12) + (2 × 1)
= 24 + 2 = 26

(b) S₈ =(S × 32) = 8 × 32 = 256
(c) P₄ = (4 × p) = 4 × 31 = 124
(d) HCl = (l × 4) + (l × 4)
= (1 × 1)+ (1 × 35 – 5)
= 1 × 35.5 = 36.5

(e) HNO₃
(1 × 4) + (1 × N) + (3 × O)
(1 × ]) + (1 × 14) + (3 × 16)
1 + 14 + 48 = 63

Question 7.
What is the mass of
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na₂SO₃)?
Answer :
(a) N₂
(2 × N) = 2 × 14 = 28
mole = 28 × 6.023 × 10²³

(b) 4 × 6.023 × 10²³ × 27
= 108 × 6.023 × 10²³ atoms

(c) Na₂SO₃
(Na × 2) + (1 × S) + (3 × O)
(23 × 2) + (1 × 32) + (3 × 16)
46 + 32 + 48 = 126
126 × 10 = 1260 g
116× 10 × 6.023 × 10²³
1160 × 6.023 × 10²³

Question 8.
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Answer :
(a) 12 × 2 × 6.0.23 × 10²³
24 × 6.023 × 10²³
(b) H2O = 18 × 6.022 × 10²³
(c) 22 × 28 × 6.023 × 10²³

9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer :
(a) 0.2 moles
0.2 × 16 = 3 – 2 g
(b) (2 × 1) +16 = 18 g

Question 10.
Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.
Answer :
\(\frac{16 \times 6 \cdot 022 \times 10^{23}}{256}\)
= 3 – 76 × 10²³ molecules

Question 11.
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
Answer :
Al₂O₃ = 102 g
\(\frac{12 \cdot 044 \times 10^{23}}{102} \times 0 \cdot 081\)
6.022 x 10³⁰ ions

Question 12.
Calculate the number of moles for the following:
(i) 52 g of He (finding mole from mass)
(ii) 12.044 x 10²³ number of He atoms (finding mole from the 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 = 4g
Thus, the number of moles
\(=\frac{\text { given mass }}{\text { molar mass }} \Rightarrow n=\frac{m}{M}=\frac{52}{4}=13\)

(ii) We know, 1 mole = 6.022 x 10²³ The number of moles
\(=\frac{\text { given number of particles }}{\text { Avogadro number }}\)
\(\Rightarrow \mathrm{n}=\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}=2\)

Question 13.
Calculate the mass of the following:
(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 numker of N atoms (mass from number)
(iv) 6.022 × 10²³ number of N₂ mole cules (mass front number).
Answer :
(i) mass = molar mass x number of moles
⇒ m = M × n = 28 × 0.5 = 14g

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

(iii) The number of moles, n

Question 14.
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 number of atoms

(ii) The number of molecules

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

KSEEB Class 9 Science Atoms and Molecules Additional Questions and Answers

Question 1.
An element Z forms an oxide with formula Z₂O₃. What is its valency?
Answer :
Valency is 3⁺

Question 2.
Give one example each of polyatomic element and polyatomic ion.
Answer :
Poly atomic element: phosphorous (P₄)
Polyatomic ion : \(\left(\mathrm{SO}_{4}^{2-}\right)\)

Question 3.
Mention the elements present in (1) quick line (2) sodium hydrogen carbonate.,
Answer :
(1) Quick line (calcium oxide) (CaO) element present are calcium and oxygen.
(2) Sodium hydrogen carbonate (NaHCO₃) elements are sodium, hydrogen, hydrogen, carbon and oxygen.

Question 4.
How many moles are present in 11.5 g of sodium?
Answer :
Gram atomic mass of sodium – 23g
23 g of sodium represent – 1 mole
∴ 11-5 g of sodium represent
\(=\frac{(1 \mathrm{mol}) \times 11 \cdot 5}{23 \cdot 0 \mathrm{g}}=0.5 \mathrm{mol}\)
6.022 atoms of element will have mass
= 2 × 10²³ g × 6.022 × 10²³
= 12.044g = 12.0 gram

Question 5.
Calculate the total number of ions in 0.585 g of sodium chloride.
Answer :
Gram formula mass of NaCl = 23 + 35-5 = 58.5 g
58.5 g of NaCl have ions = 2 × N_A
58.5 g of NaCl have ions
\(=2 \times \mathrm{N}_{\mathrm{A}} \times \frac{0.585}{58 \cdot 5}=0 \cdot 02 \times \mathrm{N}_{\mathrm{A}}\)
= 0.02 × 6.022 x 10²³
= 1.20 × 10²² ions