# Tamil Nadu 10th Standard Science Book Solution | Lesson 7 – ATOMS AND MOLECULES

## Lesson 7 > ATOMS AND MOLECULES

 Formulae Relative Atomic Mass (RAM) (Ar) = Average mass of the isotopes /1/12th mass of one Carbon-12 atom Relative Molecular Mass (RMM) Mass of one molecules of the substance/1/12th mass of one C-12 atom Number of moles Mass / Atomic Mass Mass / Molecular mass Number of Atoms / 6.023 x 1023 Number of Molecules / 6.023 x 1023 Volume/Molar Volume Mass % of an element mass of that element in the compound/ molar mass of compound x 100 Atomicity Molecular mass/Atomic mass Vapour Density (V.D) Mass of given volume of gas or vapour at S.T.P/Mass of same volume of hydrogen 2 x vapour density Relative molecular mass of a gas Atomic Number Z Number of protons = number of electrons Atomic Mass A Number of protons = number of neutrons

### I. Choose the best answer.

1. Which of the following has the smallest mass?

1. 6.023 × 1023 atoms of He
2. 1 atom of He
3. 2 g of He
4. 1 mole atoms of He

Ans ; 1 atom of He

2. Which of the following is a triatomic molecule?

1. Glucose
2. Helium
3. Carbon dioxide
4. Hydrogen

Ans ; Carbon dioxide

3. The volume occupied by 4.4 g of CO2 at S.T.P

1. 22.4 litre
2. 2.24 litre
3. 0.24 litre
4. 0.1 litre

Ans ; 2.24 litre

4. Mass of 1 mole of Nitrogen atom is

1. 28 amu
2. 14 amu
3. 28 g
4. 14 g

Ans ; 28 g

5. Which of the following represents 1 amu?

1. Mass of a C – 12 atom
2. Mass of a hydrogen atom
3. 1/12th of the mass of a C – 12 atom
4. Mass of O – 16 atom

Ans ; 1/12th of the mass of C – 12 atom

6. Which of the following statement is incorrect?

1. One gram of C – 12 contains Avogadro’s number of atoms.
2. One mole of oxygen gas contains Avogadro’s number of molecules.
3. One mole of hydrogen gas contains Avogadro’s number of atoms.
4. One mole of electrons stands for 6.023 × 1023 electrons.

Ans ; One mole of electrons stands for 6.023 × 1023 electrons.

7. The volume occupied by 1 mole of a diatomic gas at S.T.P is

1. 11.2 litre
2. 5.6 litre
3. 22.4 litre
4. 44.8 litre

Ans ; 22.4 litre

8. In the nucleus of 20Ca40, there are

1. 20 protons and 40 neutrons
2. 20 protons and 20 neutrons
3. 20 protons and 40 electrons
4. 40 protons and 20 electrons

Ans ; 20 protons and 20 neutrons

9. The gram molecular mass of oxygen molecule is

1. 16 g
2. 18 g
3. 32 g
4. 17 g

Ans ; 32 g

10. 1 mole of any substance contains ____ molecules.

1. 6.023 × 1023
2. 6.023 × 10-23
3. 3.0115 × 1023
4. 12.046 × 1023

Ans ; 6.023 × 1023

### II. Fill in the blanks

1. Atoms of different elements having __________ mass number, but __________ atomic numbers are called isobars.

Ans ; same, different

2. Atoms of different elements having same number of __________ are called isotones.

Ans ; neutrons

3. Atoms of one element can be transmuted into atoms of other element by __________.

Ans ; artificial transmutation

4. The sum of the numbers of protons and neutrons of an atom is called its __________.

Ans ; Mass number

5. Relative atomic mass is otherwise known as __________.

Ans ; Standard atomic weight

6. The average atomic mass of hydrogen is __________ amu.

Ans ; 1.008

7. If a molecule is made of similar kind of atoms, then it is called __________ atomic molecule.

Ans ; Homo

8. The number of atoms present in a molecule is called its __________.

Ans ; Atomicity

9. One mole of any gas occupies __________ ml at S.T.P.

Ans ; 22400

10. Atomicity of phosphorous is __________.

Ans ; 4

### III. Match the following

 1. 8 g of O2 4 moles 2. 4 g of H2 0.25 moles 3. 52 g of He 2 moles 4. 112 g of N2 0.5 moles 5. 35.5 g of Cl2 13 moles Ans ; 1 – b, 2 – c, 3 – e, 4 – a, 5 – d

### IV. True or False: (If false give the correct statement)

1. Two elements sometimes can form more than one compound. ( True )

2. Noble gases are Diatomic. ( False )

• Noble gases are Monoatomic.

3. The gram atomic mass of an element has no unit. ( False )

• Relative atomic mass of an element has no unit.

4. 1 mole of Gold and Silver contain same number of atoms. ( False )

5. Molar mass of CO2 is 42g. ( False )

• Molar mass of CO2 is 44 g.

### V. Assertion and Reason:

1. A and R are correct, R explains the A.
2. A is correct, R is wrong.
3. A is wrong, R is correct.
4. A and R are correct, R doesn’t explains A.

1. Assertion: Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1/12th of the mass of the C – 12 atom.

• Ans : A and R are correct, R doesn’t explains A

2. Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
Reason: The natural abundance of Chlorine isotopes are not equal.

• Ans : A and R are correct, R explains the A

1. Define: Relative atomic mass.
Relative atomic mass of an element in the ratio between the average mass of its isotopes to 1/12th part of the mass of a carbon–12 atoms. It is denoted as A. It is otherwise called Standard Atomic Weight.

An = Average mass of the isotopes of the element / 1/12th of the mass of one carbon –12atom

2. Write the different types of isotopes of oxygen and its percentage abundance.

 Isotope Mass (amu) % abundance 8O16 15.9949 99.757 8O17 16.9991 0.038 8O18 17.9992 0.205

3. Define: Atomicity.

The number of atoms present in the molecule is called its Atomicity.

Atomicity = Molecular Mass / Atomic Mass

4. Give any two examples for heterodiatomic molecules.

HCl, H2O, NH4.

5. What is Molar volume of a gas?

One mole of any gas occupies 22.4 litre or 22400 ml at STP. The volume occupied by one mole of any gas at S.T.P is called molar volume.

6. Find the percentage of nitrogen in ammonia.

 Molar mass of Ammonia% of Nitrogen = 14 + 3 = 17 g.= 14/17 x 100 = 82.35%.

1. Calculate the number of water molecule present in one drop of water which weighs 0.18 g.

 Given MassAvogadro NumberMolecular Mass of water = 0.18 g= 6.023 × 1023= 18 g (H2O = 2(1) + 1(16)= 2 + 16 = 18) No. of water molecules =Avogadro number × given man / Molecular Mass of water =6.023×1023×0.18g/18 g = 6.023×1023×0.18 x100 / 18×102 = 6.023×1023×18/18×102 = 6.023×1023×10–2×18/18 = 6.023×1023×10–2 = 6.023×1021 molecules of water

2. N2 + 3H2 → 2NH3.

(The atomic mass of nitrogen is 14, and that of hydrogen is 1)

1 mole of nitrogen (_______g) +

3 moles of hydrogen ( _________ g) →

2 moles of ammonia ( _________ g)

1 mole of nitrogen (28 g) +

3 moles of hydrogen (3×1 g) →

2 moles of Ammonia (34 g)

28, 3, 34

3. Calculate the number of moles in

i) 27g of Al ii) 1.51 × 1023 molecules of NH4Cl

 i) No. of moles = Mass/Atomic Mass= 27g / 27g= 1 mole. ii) No. of moles = No. of molecules of NH4Cl / Avogadro’s number= 1.51 x 1023 / 6.023 x 1023= 1/4 = 0.25 mole

4. Give the salient features of “Modern atomic theory”.

### Modern Atomic Theory:

• An atom is no longer indivisible (after the discovery of the electron, proton and neutron).
• Atoms of the same element may have different atomic mass (discovery of Isotopes 17Cl3517Cl37
• Atoms of different elements may have same atomic masses (discovery of Isobars 20Ar40, 20Ca40).
• Atoms of one element can be transmuted into atoms of other elements. In otherwords, atom is no longer indestructible (discovery of artificial transmutation).
• Atoms may not always combine in a simple whole number ratio (Eg. Glucose C6H12O6 C:H:O = 6:12:6 or 1:2:1 and Sucrose C12H22O11 C:H:O = 12:22:11).
• Atom is the smallest particle that take part in a chemical reaction.
• The mass of an atom can be converted into energy (E=MC2).

### VIII. HOT question IX. Solve the following problems

5. Derive the relationship between Relative molecular mass and Vapour density.

Relative Molecular Mass : The ratio of Mass of one molecule of gas or vapour to the mass of one atom of hydrogen.

 Relative Molecular Mass = Mass of one molecule of gas or vapour/Mass of one atom of hydrogen …….(1)

Vapour density : The ratio of mass of a certain volume of a gas or vapour to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.

 Vapour Density = Mass of 1 volume of gas or vapour / Mass of 1 volume of hydrogen …….(2) VD = Mass of 1 volume of gas or vapour / Mass of 1 volume of hydrogen …….(3) Applying Avogadro’s law, VD = Mass of 1 molecule of gas or vapour / Mass of 1 molecule of hydrogen …….(4) Hence hydrogen is diatomic VD = Mass of 1 molecule of gas or vapour /  Mass of 2 × atoms of hydrogen ……. (5) VD = Mass of 1 molecule of gas or vapour / 2 × mass of 1 atom of hydrogen …….(6) Multiplying ‘2’ on both sides 2 × VD = 2 × Mass of 1 molecule of gas or vapour / 2 × Mass of 1 atom of hydrogen …….(7) 2 × VD = Mass of 1 molecule of gas or vapour / Mass of 1 atom of hydrogen …….(8) 2 × VD = Relative Molecular Mass …….(9) VD RMM/2 or  Molecular Weight/2 …….(10)

### VIII. HOT Questions

1. Calcium carbonate is decomposed on heating in the following reaction.
CaCO3 → CaO + CO2

i) How many moles of Calcium carbonate are involved in this reaction?

One mole of CaCO3 is involved in the reaction.

ii) Calculate the gram molecular mass of calcium carbonate involved in this reaction.

 Gram Molecular Mass of CaCO3 = (40 × 1) + (12 × 1) + (16 × 3)= 40 + 12 + 48 = 100 g One mole of CaCO3 is involved in the reaction, ∴ Gram Molecular Mass of CaCO3 is = 1 × 100 g = 100 g

iii) How many moles of CO2 are there in this equation?

One mole of CO2 is involved in this reaction.

### IX. Solve the following problems.

i) 2 moles of hydrogen molecule, H2

 Molecular massMass = 1 × 2 = 2= 2 × 2 = 4 g

ii) 3 moles of chlorine molecule, Cl2

 Molecular massMass = 35.5 × 2 = 71= 3 × 71 = 213 g

iii) 5 moles of sulphur molecule, S8

 Molecular massMass = 32 × 8 = 256= 5 × 256 = 1280 g

iv) 4 moles of phosphorous molecule, P4

 Molecular massMass = 31 × 4 = 124= 4 × 124 = 496 g

2. Calculate the % of each element in calcium carbonate. (Atomic mass: C-12, O-16, Ca -40)

CaCO3.

 Molar Mass of CaCO3 = 1 (Ca) + 1 (C) + 3 (O) = 1 (40) + 1 (12) + 3 (16) = 40 + 12 + 48 = 100 g. % of Ca in CaCO3 = Mass of Ca / Molar Mass of CaCO3 x 100 = 40 g / 100 g x 100 = 40%. % of Ca in CaCO3 Mass of Carbon/Molar Mass of CaCO3 x 100 = 12 g / 100 g x 100 = 12%. % of O in CaCO3 = Mass of Oxygen / Molar Mass of Calcium × 100 = 48 g / 100 x 100

3. Calculate the % of oxygen in Al2(SO4)3. (Atomic mass: Al-27, O-16, S -32)

 Molecular mass of Al2(SO4)3 = (2 × 27) + (3×(32 +(4 ×16))) = (2 × 27) + (3×96) = 54 + 288 = 342 g % of O in Al2(SO4)3 = 3×4×16 / 342 × 100= 192/342 × 100 = 56.14%

4. Calculate the % relative abundance of B-10 and B-11, if its average atomic mass is 10.804 amu

Let 𝑎1, 𝑎2 be the % abundance of B-10 and B-11 respectively.

 Mass of B-10, m1 = 10 Mass of B-11, m2 = 11 𝑎1+𝑎2 = 100 𝑎1 = 100 − 𝑎2 Average Atomic Mass of B = 10.804 amu Average Atomic Mass = m1 × a1 /100 + m2 ×a1/100 = 10 × (100−𝑎2)/100 + 11 x 𝑎2/100 = 10 − 10𝑎2/100 + 11𝑎2/100 10.804 = 10 + 𝑎2/100 𝑎2/100 =10.804 – 10 = 0.804 𝑎2 = 0.804 × 100 = 80.4 % 𝑎1 = 100 − 80.4 = 19.6 % ∴ % abundance of B-10 = 19.6 % % abundance of B-11 = 80.4%

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