NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

 Chapter Name NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules Topics Covered NCERT Intext QuestionsNCERT Exercise Solutions Related Study NCERT Solutions for Class 9 ScienceNCERT Revision Notes for Class 9 ScienceImportant Questions for Class 9 ScienceMCQ for Class 9 ScienceNCERT Exemplar Questions For Class 9 Science

NCERT Intext Questions

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 water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid
→ sodium ethanoate + carbon dioxide + water

Solution

In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water. Sodium carbonate + Ethanoic acid → Sodium ethanoate + Carbon dioxide + Water
Mass of sodium carbonate = 5.3 g (Given)
Mass of ethanoic acid = 6 g (Given)
Mass of sodium ethanoate = 8.2 g (Given)
Mass of carbon dioxide = 2.2 g (Given)
Mass of water = 0.9 g (Given)
Now, total mass before the reaction = (5.3 + 6) g = 11.3 g
And, total mass after the reaction = (8.2 + 2.2 + 0.9) g = 11.3 g
Total mass before the reaction = Total mass after the reaction
Hence, the given observations are in agreement with the law of conservation of mass.

2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Solution

It is given that the ratio of hydrogen and oxygen by mass to form water is 1 : 8. Then, the mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g. Therefore, the mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3 g = 24 g.

3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Solution

The postulate of Dalton’s atomic theory, “Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction” is the result of the law of conservation of mass.

4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Solution

The postulate of Dalton, “The relative number and kinds of atoms are fixed in a given compound” can explain the law of definite proportions.

5. Define atomic mass unit.

Solution

Mass unit equal to exactly one-twelfth the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’.

6. Why is it not possible to see an atom with naked eyes?

Solution

The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.

7. Write down the formulae of :
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide

Solution

(i) Na2O
(ii) AlCl3
(iii) Na2S
(iv) Mg(OH)2

8. Write down the names of compounds represented by the following formulae :
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3

Solution

(i) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate

9. What is meant by the term chemical formula?

Solution

The chemical formula of a compound is a symbolic representation of its composition.

10. How many atoms are present in a (i) H2S molecule and (ii) PO43- ion?

Solution

(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO43- ion, five atoms are present; one of phosphorus and four of oxygen.

11. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Solution

Molecular mass of H2 = 2 × Atomic Mass of H = 2 × 1 = 2u
Molecular mass of O2 = 2 × Atomic mass of O
= 2 × 16 = 32 u
Molecular mass of Cl2 = 2×Atomic mass of Cl
= 2 × 35.5 = 71 u
Molecular mass of CO2 = Atomic mass of C + 2×Atomic mass of O
= 12 + 2 ×16 = 44 u
Molecular mass of CH4 = Atomic mass of C + 4×Atomic mass of H
= 12 + 4 × 1 = 16 u
Molecular mass of C2H6 = 2×Atomic mass of C + 6×Atomic mass of H
= 2 × 12 + 6×1 = 30 u
Molecular mass of C2H4 = 2×Atomic mass of C + 4×Atomic mass of H
= 2 × 12 + 4 ×1 = 28 u
Molecular mass of NH3 = Atomic mass of N + 3×Atomic mass of H
= 14 + 3× 1 = 17 u
Molecular mass of CH2OH = Atomic mass of C + 3×Atomic mass of H + Atomic mass of O + Atomic mass of H
= 12 + 3 × 1 + 8 + 1 = 24 u

12. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

Solution

Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16 = 81 u
Formula unit mass of Na2O
= 2 × Atomic mass of Na + Atomic mass of O =2 × 23+16 = 62 u
Formula unit mass of K2CO3
= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of 0
= 2 × 39 + 12 + 3 × 16 = 78 + 12 + 48 = 138 u

13. If one mole of carbon atoms weighs 12 grams, what is the mass (in gram) of 1 atom of carbon?

Solution

One mole of carbon atoms weighs 12 g (Given)
i.e., mass of 1 mole of carbon atoms = 12 g
Then, mass of 6.022 × 1023 number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon = 12/(6.022 × 1023)
= 1.9926 × 10-23 g

14. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Solution

Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = 6.022 × 1023 number of atoms
Thus, 100 g of Na contains
= (6.022 × 1023)/23 × 100 number of atoms
= 2.6182 × 1024 number of atoms
Again, atomic mass of Fe = 56 u (Given)
Then, gram atomic mass of Fe = 56 g
Now, 56 g of Fe contains = 6.022× 1023 number of atoms
Thus, 100 g of Fe contains = (6.022× 1023)/56 ×100 number of atoms
= 1.0753 × 1024 number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

NCERT Exercises Solutions

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

Solution

Total mass of compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)
Thus, percentage of boron by weight in the compound
= 0.096/0.24 × 100% = 40%
And, percentage of oxygen by weight in the compound
= 0.144/0.24 × 100% = 60%

2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?

Solution

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 g of carbon dioxide. If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11 g of carbon dioxide will be formed. The above answer is governed by the law of constant proportions.

3. What are polyatomic ions? Give examples.

Solution

A polyatomic ion is a group of atoms carrying a charge(positive or negative). For example : Nitrate NO3-), hydroxide ion (OH).

4. Write the chemical formulae of the following :
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate

Solution

(a) MgCl2
(b) CaO
(c) Cu(NO3)3
(d) AlCl3
(e) CaCO3

5. Give the names of the elements present in the following compounds :
(a) Quicklime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate

Solution

(a) Calcium and oxygen (CaO)
(b) Hydrogen and bromine (HBr)
(c) Sodium, hydrogen, carbon, and oxygen (NaHCO3)
(d) Potassium, sulphur, and oxygen (K2SO4)

6. Calculate the molar mass of the following substances :
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3

Solution

(a) Molar mass of ethyne,
C2H2 = 2 × 12 + 2 × 1 = 26 g
(b) Molar mass of sulphur molecule,
S8 = 8 × 32 = 256 g
(c) Molar mass of phosphorus molecule,
P4 = 4 × 31 = 124 g
(d) Molar mass of hydrochloric acid,
HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid,
HNO3 = 1 + 4 + 3 × 16 = 63 g

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)?

Solution

(a) The mass of 1 mole of nitrogen atoms is 14 g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108 g
(c) The mass of 10 moles of sodium sulphite(Na2SO3) is 10 × [ 2× 23 + 32 + 3 × 16)g = 10 × 126 g = 1260 g

8. Convert into mole :
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide

Solution

(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas = 12/32 mole = 0.375 mole
(b) 18 g of water = 1 mole, then, 20 g of water = 20/18 mole = 1.111 mole
(c) 44 g of carbon dioxide = 1 mole, then, 22 g of carbon dioxide = 22/44 mole = 0.5 mole

9. What is the mass of :
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Solution

(a) Mass of one mole of oxygen atoms = 16 g, then, mass of 0.2 mole of oxygen atoms = 0.2 × 16 g = 3.2 g
(b) Mass of one mole of water molecule = 18 g Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Solution

1 mole of solid sulphur (S8) = 8 × 32 g = 256 g
i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16 g of solid sulphur contains = 6.022 × 1023/256 × 16 molecules = 3.76375 × 1022 molecules

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)

Sol. Mole of aluminium oxide (Al2O3)
= 2 × 27 + 3 × 16 = 102 g
i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of
Al2O3
Then, 0.051 g of Al2O3 contains
= 6.022 × 1023/102 × 0.051 molecules = 3.011 × 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020 = 6.022 × 1020