Chapter 3 Atoms and Molcules

By | June 14, 2018
Class 9 | Science | Chapter 3 |Atoms and Molecules| NCERT Solutions

Page 32, 33

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

Answer:
In a 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 = 5.3g (Given)

Mass of ethanoic acid = 6g (Given)
Mass of sodium ethanoate = 8.2g (Given)
Mass of carbon dioxide = 2.2 (Given)
Mass of water = 0.9g (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
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.

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?

Answer:
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 1g of hydrogen gas is 8g. Therefore, the mass of oxygen gas required to react completely with 3g of hydrogen gas is 8 × 3g = 24g.

3. 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 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”.

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

Answer:
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”.

Page 35

1. Define the atomic mass unit.

Answer:
An atomic mass unit is a unit of mass used to express weights of atoms and molecules.One atomic mass is equal to 1/12th the mass of one carbon-12 atom.

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

Answer:
The size of an atom is so small that it can not be seen with naked eyes.Also, atom of an element does not exist independently.

Page 39

1. Write down the formulae of
(i) sodium oxide

Answer: Na2O

(ii) aluminium chloride

Answer: AlCl3

(iii) sodium sulphide

Answer: Na2S

(iv) magnesium hydroxide

Answer: Mg(OH)2

2. Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3

Answer: Aluminium sulphate
(ii) CaCl2

Answer: Calcium chloride

(iii) K2SO4

Answer: Potassium sulphate

(iv) KNO3

Answer: Potassium nitrate

(v) CaCO3

Answer: Calcium carbonate

3. What is meant by the term chemical formula?

Answer:
The chemical formulae of a compound means the symbolic representation of the composition of the compound.

4. How many atoms are present in a
(i) H2S molecule

Answer:
In an H2S molecule three atoms are present.

2 atoms of hydrogen and 1 atom of sulphur are present.

(ii)PO43– ion

Answer:
In PO43- ion 5 atoms are present.

1 atom of phosphorus and 4 atoms of oxygen are present.

Page 40

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

Answer:
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 = 1 × atomic mass of C + 2 ×atomic mass of O = 1 × 12 + 2 ×16 = 44 u
Molecular mass of CH4 = 1 × atomic mass of C + 4 ×atomic mass of H = 1×12 + 4×1 = 12+4= 16 u

Molecular mass of C2H6= 1 × atomic mass of C + 2 ×atomic mass of H = 2×12 + 6× 1= 24+6= 30 u

Molecular mass of C2H4= 2×atomic mass of C + 4 ×atomic mass of H = 2×12 + 4×1 = 24+ 4 =28u

Molecular mass of NH3 = 1 × atomic mass of N + 3 × atomic mass of H = 1×14+3×1 = 14+3 =17 u

Molecular mass of CH3OH= 1× atomic mass of C+ 4 × atomic mass of H + 1 × atomic mass of O= 12+4+ 8=24u

2. 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.

Formula unit mass of ZnO = Atomic mass of Zn + atomic mass of O = 65u + 16u = 81u

Formula unit mass of Na2O= Atomic mass of Na + atomic mass of O=(23u x 2) + 16u = 46u + 16u = 62u

Formula unit of K2CO3 = Atomic mass of K + atomic mass of C + atomic mass of O = (39u x 2) + 12u +( 16u x 3)=138u

Page 42

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

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

2. 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)?

Answer:
Atomic mass of Na = 23u

Then, gram atomic mass of Na = 23 g
Now, 23g of Na contains = 6.022×1023 number of atoms
Thus, 100g of Na contains = (6.022×1023×100)/23  number of atoms
= 2.6182 ×1024 number of atoms

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

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.

Answer:
Mass of boron = 0.096g (Given)

Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound =(0.096×100)/0.24 % = 40%
Thus, percentage of oxygen by weight in the compound = (0.144×100)/ 0.24%= 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 44 SCIENCE formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your ?

Answer:
3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt

in 50g of oxygen, then 3g 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 11g of carbon dioxide will be formed. The
above answer is governed by the law of constant proportions.

3. What are polyatomic ions? Give examples.

Answer:
Those ions which are formed from group of atoms joined together are polyatomic ions.They carry positive or negative charge.

For ex: Ammonium ion ( NH4), Carbonate ion (CO32-), Sulphate ion (SO42-)

4. Write the chemical formulae of the following.

(a) Magnesium chloride

Answer: MgCl2

(b) Calcium oxide

Answer: CaO

(c) Copper nitrate

Answer: Cu(NO3)2

(d) Aluminium chloride

Answer: AlCl3

(e) Calcium carbonate.

Answer: CaCO3

5. Give the names of the elements present in the following compounds.
(a) Quick lime

Answer: Carbon and oxygen

(b) Hydrogen bromide

Answer: Hydrogen and bromide

(c) Baking powder

Answer: Sodium, hydrogen, carbon and oxygen

(d) Potassium sulphate.

Answer: Potassium, sulphur, and oxygen,

6. Calculate the molar mass of the following substances.

(a) Ethyne, C2H2

Answer: Molar mass of C2H2= 2 ×atomic mass of C + 2 × atomic mass of H = 2 × 12 + 2 × 1 = 24 + 2 = 26 g

(b) Sulphur molecule, S8

Answer: Molar mass of S8

= 8 ×32= 256g
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)

Answer: Molar mass of P4 = 4 ×31 = 124 g

(d) Hydrochloric acid, HCl

Answer:
Molar mass of HCl = Atomic mass of H +atomic mass of Cl = 1+ 35.5 = 36.5 g

(e) Nitric acid, HNO3

Answer:
Molar mass of HNO3
= Atomic mass of H + atomic mass of N + 3 × Atomic mass of O = 1 + 14+ 3 ×16 = 63 g

7. What is the mass of—
(a) 1 mole of nitrogen atoms?

Answer:
The mass of 1 mole of nitrogen atom is 14 g.

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

Answer:
The mass of 4 moles of aluminium atom is 4 × 27 = 108 g

(c) 10 moles of sodium sulphite (Na2SO3)?

Answer:
Mass of 10 moles of Na2SO3 = 10 × [2 × 23+ 1 × 32 + 16 × 3 ] = 1260 g

8. Convert into mole.
(a) 12 g of oxygen gas

Answer:
32g of oxygen gas= 1 mole
12 g of oxygen gas = 12/32 mol= 0.375 mol

(b) 20 g of water

Answer:
18 g of water = 1 mole
20 g of water=20/18 mol= 1.111 mol

(c) 22 g of carbon dioxide.

Answer:
44 g of carbon dioxide = 1 mol
22g of carbon dioxide = 22/44 mol= 0.5 mol

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

Answer:
Mass of 1 mol of oxygen atom = 16 g
Mass of 0.2 mole of oxygen atom= 3.2 g

(b) 0.5 mole of water molecules?

Answer:
Mass of 1 mol of water molecule = 18 g
Mass of 0.5 mol of water molecule = 0.5 × 18 =9 g

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

Answer:
1 mole of solid sulphur = 8 32= 256 g
256 g of sulphur contains 6.022 x 1023 molecule
Then 16 g of sulphur contain =6.022 x 1023/256 = 16 molecules = 3.76 × 1022

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:
1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g

i.e., 102g 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 molecules of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020 = 6.022 × 1020

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