Chapter 3 Class 9 - Atoms And Molecules

Question 1:

In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water, and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

Sodium carbonate + Acetic acid → Sodium acetate + Carbon dioxide + Water

Solution:

In the given reaction, sodium carbonate reacts with acetic acid to produce sodium acetate, carbon dioxide, and water.

Mass of sodium carbonate = 5.3 g (Given)
Mass of acetic acid = 6 g (Given)
Mass of sodium acetate = 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 g + 6 g = 11.3 g
And, total mass after the reaction = 8.2 g + 2.2 g + 0.9 g = 11.3 g

Therefore, total mass before the reaction = Total mass after the reaction

Hence, the given observations are in agreement with the law of conservation of mass.

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 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.

Question 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 which is a result of the law of conservation of mass is:

Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.

Question 4:

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

Solution:

The postulate of Dalton’s atomic theory which can explain the law of definite proportion is:

The relative number and kind of atoms in a given compound remains constant.

Question 5:

Define the atomic mass unit.

Solution:

The atomic mass unit (amu) is defined as a mass unit equal to exactly one-twelfth (1/12) of the mass of an atom of carbon-12. It is written as 'u'.

Question 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, atoms do not exist independently; they form molecules or combine with other atoms to form larger structures, making them invisible to the naked eye.

Question 7:

Write down the formulae of:

1. Sodium oxide
2. Aluminium chloride
3. Sodium sulphide
4. Magnesium hydroxide

Solution:

1. Sodium oxide → Na₂O
2. Aluminium chloride → AlCl₃
3. Sodium sulphide → Na₂S
4. Magnesium hydroxide → Mg(OH)₂

Question 8:

Write down the names of compounds represented by the following formulae:

1. Al₂(SO₄)₃
2. CaCl₂
3. K₂SO₄
4. KNO₃
5. CaCO₃

Solution:

1. Al₂(SO₄)₃ → Aluminium sulphate
2. CaCl₂ → Calcium chloride
3. K₂SO₄ → Potassium sulphate
4. KNO₃ → Potassium nitrate
5. CaCO₃ → Calcium carbonate

Question 9:

What is meant by the term chemical formula?

Solution:

The chemical formula of a compound means the symbolic representation of the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For example, from the chemical formula CO₂ of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

Question 10:

How many atoms are present in a:

1. H₂S molecule and
2. PO₄^3- ion?

Solution:

1. In an H₂S molecule, three atoms are present; two of hydrogen and one of sulphur.
2. In a PO₄^3- ion, five atoms are present; one of phosphorus and four of oxygen.

Question 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 = 2 u
• 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 CH₃OH = Atomic mass of C + 4 × Atomic mass of H + Atomic mass of O = 12 + 4 × 1 + 16 = 32 u

Question 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 O = 2 × 39 + 12 + 3 × 16 = 138 u

Question 13:

If one mole of carbon atoms weighs 12 grams, what is the mass in grams of 1 atom of carbon?

Solution:

One mole of carbon atoms weighs 12 g. Given that,

• Mass of 1 mole of carbon atoms = 12 g
• Mass of a number of carbon atoms = 12 g

Therefore, the mass of 1 atom of carbon is:

Mass of 1 atom of carbon = \( \frac{12 \{ g}}{6.022 \times 10^{23}} \)

Question 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:

Given:

• Atomic mass of Na = 23 u
• Atomic mass of Fe = 56 u

Number of atoms in 100 grams of sodium:

\( \text{Number of atoms} = \frac{100 \{ g}}{23 \{ g/mol}} \times 6.022 \times 10^{23} \{ atoms/mol} \)

Number of atoms in 100 grams of iron:

\( \text{Number of atoms} = \frac{100 \{ g}}{56 \{ g/mol}} \times 6.022 \times 10^{23} \{ atoms/mol} \)

Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Question 15

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.

Solution:

Given that:

Percentage of boron = \( \frac{0.096 \, \text{g}}{0.24 \, \text{g}} \times 100\% = 40\% \)

Percentage of oxygen = \( \frac{0.144 \, \text{g}}{0.24 \, \text{g}} \times 100\% = 60\% \)

Question 16

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:

Given that:

From the law of constant proportions, we know that regardless of the amount of carbon burnt, the ratio of carbon to oxygen to form carbon dioxide remains constant.

So, if 3.0 g of carbon reacts with 8.00 g of oxygen to form 11.00 g of carbon dioxide, then the same mass of carbon dioxide will be formed when 3.0 g of carbon reacts with 8.00 g of oxygen, regardless of the excess oxygen present.

Question 17

What are polyatomic ions? Give examples?

Solution:

Polyatomic ions are groups of atoms carrying a net positive or negative charge. These ions are covalently bonded together but act as a single charged unit in chemical reactions.

Examples of polyatomic ions include:

• Ammonium ion (( NH_4^+ ))
• Hydroxide ion (( OH^- ))
• Carbonate ion (( CO_3^{2-} ))
• Sulfate ion (( SO_4^{2-} ))

Question 18

Question 19

Question 20

Calculate the molar mass of the following substances:

1. Molar mass of ethyne, C₂H₂ = (2 times 12 + 2 times 1 = 26 ,{g/mol}\)
2. Molar mass of sulphur molecule, S₈ = (8 times 32 = 256 ,{g/mol}\)
3. Molar mass of phosphorus molecule, P₄ = (4 times 31 = 124 ,{g/mol}\)
4. Molar mass of hydrochloric acid, HCl = (1 + 35.5 = 36.5 ,{g/mol}\)
5. Molar mass of nitric acid, HNO₃ = (1 + 14 + 3 times 16 = 63\,{g/mol}\)