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

# Atoms and Molecules NCERT Solutions for Class 9 Science Chapter 3 with Answers

We have completed the NCERT/CBSE Solutions chapter-wise for Class 9 Science Chapter 3 Atoms and Molecules with Answers by expert subject teacher for latest syllabus and examination. Prepare effectively for the exam taking the help of the Class 9 Science NCERT Solutions PDF free of cost from here. Students also can take a free NCERT Solutions of Atoms and Molecules. Each question has right answer Solved by Expert Teacher. Download the Science NCERT Solutions with Answers for Class 9 Science Pdf and prepare to help students understand the concept very well.

## NCERT Solutions for Class 9 Science Chapter wise

###### Q1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g 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

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.

### Q2. 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?

We know hydrogen and water mix in the ratio 1: 8.

For every 1g of hydrogen, it is 8g of oxygen.
Therefore, for 3g of hydrogen, the quantity of oxygen = 3 x 8 = 24g
Hence, 24g of oxygen would be required for the complete reaction with 3g of hydrogen gas.

## Q3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer: The postulate of Dalton’s atomic theory that is the result of the law of conservation of mass is—the relative number and kinds of atoms are constant in a given compound. Atoms cannot be created nor destroyed in a chemical reaction.

## Q4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer: The law of definite proportions is based on the following postulate of Dalton’s Atomic theory.
“All atoms of a particular element are identical in every respect. This means that they have same mass, same size and also same chemical properties.”

Questions

## Q1. Define atomic mass unit.

Answer: 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’.

## Q2. Why is it not possible to see an atom with naked eyes?

Answer: 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.

Questions

## Q1. Write down the formulae of

(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide

The following are the formulae:

(i) sodium oxide – Na2O
(ii) aluminium chloride – AlCl3
(iii) sodium sulphide – Na2S
(iv) magnesium hydroxide – Mg (OH)2

## Q2. Write down the names of compounds represented by the following formulae:

(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3.

(i) Al2(SO4)3 – Aluminium sulphate
(ii) CaCl2 – Calcium chloride
(iii) K2SO4 – Potassium sulphate
(iv) KNO3 – Potassium nitrate
(v) CaCO3 – Calcium carbonate

## Q3. What is meant by the term chemical formula?

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

### Q4. How many atoms are present in a(i) H2S molecule and(ii) PO43- ion?

The number of atoms present are as follows:

(i) H2S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in totality.
(ii) PO43- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in totality.

## Q2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u, Na = 23 u, K=39u, C = 12u, and O=16u.

(i) Formula unit mass of ZnO (Zinc oxide)

= (1 x Atomic mass of Zn) +(1 x Atomic mass of O)
= (1 x 65 u) + (1 x 16 u) = 81 u.

(ii) Formula unit mass of Na2O (Sodium oxide)

= (2 x Atomic mass of Na) + (1 x Atomic mass of O)
= (2 x 23 u) + (1 x 16 u) = 62 u.

(iii) Formula unit mass of K2CO3 (Potassium carbonate).

= (2 x Atomic mass of K) + (1 x Atomic mass of C) + (3 x Atomic mass of O)
= (2 x 39 u) + (1 x 12 u) + (3 x 16 u) = 138 u

## Exercise

### Q1. A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.

###### Q2. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 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.

## Q3. What are polyatomic ions? Give examples.

Answer: Polyatomic ions are ions that contain more than one atom but they behave as a single unit

Example: CO32-, H2PO4

## Q4. Write the chemical formula of the following.

(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate

(a) Magnesium chloride – MgCl2
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)2
(d) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3

## Q5. Give the names of the elements present in the following compounds.

(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

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

## Q6. 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

Answer: (a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g
(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8 x 32 = 256g
(c) Molar mass of Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g
(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g
(e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+3×16 = 63g

## Q7. 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) Mass of 1 mole of nitrogen atoms = 14 g

(b) 4 moles of aluminium atoms
Mass of 1 mole of aluminium atoms = 27 g
∴ Mass of 4 moles of aluminium atoms = 27 x 4 = 108 g

(c) 10 moles of sodium sulphite (Na2SO3)
Mass of 1 mole of Na2SO3 = 2 x 23 + 32 + 3 x 16 = 46 + 32 + 48 = 126 g
∴ Mass of 10 moles of Na2SO3 = 126 x 10 = 1260 g

## Q8. Convert into mole.

(a) 12g of oxygen gas
(b) 20g of water
(c) 22g of carbon dioxide

## Q9. What is the mass of:

(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

(a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 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

## Q10. Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.

Answer: To calculate molecular mass of sulphur:

Molecular mass of Sulphur (S8) = 8xMass of Sulphur = 8×32 = 256g
Mass given = 16g
Number of moles = mass given/ molar mass of sulphur
= 16/256 = 0.0625 moles

To calculate the number of molecules of sulphur in 16g of solid sulphur:

Number of molecules = Number of moles x Avogadro number
= 0.0625 x 6.022 x 10²³ molecules
= 3.763 x 1022 molecules

## Q11. Calculate the number of aluminium ions present in 0.051g 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 = 27u)

Answer: Molar mass of aluminium oxide Al203
= (2 x 27) + (3 x 16)
= 54 + 48 = 102 g.

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