Atoms and Molecules

INTEXT QUESTIONS

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

Ans. 

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?

Ans. 

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.

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

Ans. 

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.

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

Ans. 

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.

INTEXT QUESTIONS

Q1. Define atomic mass unit.

Ans. 

Mass unit equal to exactly one-twelfth 112th 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?

Ans. 

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.

INTEXT QUESTIONS

Q1. Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Ans.

(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

Ans.

(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?

Ans.    

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.

Q4. How many atoms are present in a

(i) H2S molecule and

(ii) PO43− ion?

Ans.

(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.

INTEXT QUESTIONS

Q1. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Ans. 

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

Q2. 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.

Ans. 

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

INTEXT QUESTIONS

Q1. If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

Ans. 

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 =126.022×1023 g

=1.9926×10-23

Q2. 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)?

Ans. 

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 

=6.022×102323×100 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 =6.022×102356×100 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 TEXT BOOK EXERCISES

Q1. 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.

Ans. 

Mass of boron = 0.096 g (Given)

Mass of oxygen = 0.144 g (Given)

Mass of sample = 0.24 g (Given)

Thus, percentage of boron by weight in the compound =0.0960.24×100%

= 40%

And, percentage of oxygen by weight in the compound =0.1440.24×100%

= 60%

Q2. 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?

Ans. 

Carbon + Oxygen Carbon dioxide

3 g of carbon reacts with 8 g of oxygen to produce 11 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.

Q3. What are polyatomic ions? Give examples?

Ans. 

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion NH4+, hydroxide ion (OH), carbonate ion CO32, sulphate ion SO42 .

Q4. Write the chemical formulae of the following:

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Ans.

(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

Ans.

Compound

Chemical formula

Elements present

Quick lime

CaO

Calcium, Oxygen

Hydrogen bromide

HBr

Hydrogen, Bromine

Baking powder

NaHCO3

Sodium, Hydrogen, Carbon, Oxygen

Potassium sulphate

K2SO4

Potassium, Sulphur, 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

Ans.

(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 + 14 + 3 × 16 = 63 g

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)?

Ans.

(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

Q8. Convert into mole.

(a) 12 g of oxygen gas

(b) 20 g of water

(c) 22 g of carbon dioxide

Ans.

(a) 32 g of oxygen gas = 1 mole

Then, 12 g of oxygen gas = 1232mole

= 0.375 mole

(b) 18 g of water = 1 mole

Then, 20 g of water = 2018mole

= 1.11 moles (approx)

(c) 44 g of carbon dioxide = 1 mole

Then, 22 g of carbon dioxide = 2244mole = 0.5 mole

Q9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Ans.

(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 =9g

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

Ans. 

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

=6.022×1023256×16 molecules 

= 3.76 × 1022 molecules (approx)

Q11. 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)

Ans. 

1 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 = 6.022×1023102×0.051 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

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