Ch 3 Atoms and Molecules Ncert Solutions

1. 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 carbonate.

Ans.

Sodium carbonate	+	Ethanoic acid	→	Sodium ethanoate	+	Carbon dioxide	+	Water
5.3 g	+	6 g	→	8.2 g	+	2.2 g	+	0.9 g

LHS RHS

11.3 g = 11.3 g

(Mass of reactant) (Mass of product)

This shows, that during a chemical reaction mass of reactact = mass of product.

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?

Ans. Ratio of H : O by mass in water is:

Hydrogen : Oxygen → H₂O

1 : 8 = 3 : x

x = 8 × 3

∴ 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

3. Which postulate of Dalton's atomic theory the result of the law of conservation of mass?

Ans. 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.

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

Ans. The relative number and kinds of atoms are constant in a given compound.

1. Define the atomic mass unit.

Ans. One atomic mass unit is equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12. The relative atomic masses of all elements have been found with respect to an atom of carbon-12.

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

Ans. Atom is too small to be seen with naked eyes. It is measured in nanometres.

1 m = 10⁹ nm

1. Write down the formulae of

(i) Sodium oxide

(ii) Aluminium chloride

(iii) Sodium sulphide

(iv) Magnesium hydroxide

Ans. The formulae. are

(i) Formula bf Sodium Oxide

Symbol → Na O

Charge → +1 –2

Formula → Na₂O

(iii) Formula of Sodium Oxide

Symbol → Na S

Charge → + 1 –2

Formula → Na₂S

(ii) Formula of aluminium chloride

Symbol → Al Cl

Charge → +3 –1

Formula → A1C1₃

(iv) Formula of magnesium hydroxide

Symbol → Mg OH

Charge 1

Formula → Mg(OH)₂

2. Write down the names of compounds represented btthe following formulae:

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

(v) CaCO₃

Ans. (i) A1₂(SO₄)₃ → Aluminium sulphate

(ii) CaC1₂ → Calcium chloride

(iii) K∴SO₄ → Potassium sulphate

(iv) KNO₃ → Potassium nitrate

(v) CaCO₃ → Calcium carbonate

3. What is meant by the term chemical formula?

Ans. The chemical formula of the compound is a symbolic representation of its composition, e.g., chemical formula of sodium chloride is NaCl.

4. How many atoms are present in a

(i) H₂S molecule and

(ii) PO₄ ^3– ion?

Ans. (i) H₂S → 3 atoms are present

(ii) PO₄ ^3– → 5 atoms are present

1. Calculate the molecular masses of H₂, O₂, C1₂, CO₂, CH4, C₂H₆, C₂H₄, NH₃, CH₃OH.

Ans. The molecular masses are:

H₂ ⇒ 1 × 2 → 2u

O₂⇒ 16 × 2 → 32u

Cl₂ ⇒ 35.5 × 2 → 71 u

CO₂ ⇒ 1 × 12 + 2 × 16 = 12 +32 = 44 u

CH₄ ⇒ 1 × 12 + 4 × 1 = 16 u

C₂H6 ⇒�n₂ × 12 + 6 × 1 = 30 u

C₂H₄ ⇒ (2 × 12) + (4 × 1) = 28 u

NH₃ ⇒ (1 × 14) + (3 × 1) = 17 u

CH₃OH ⇒ 12 + (3 × 1) + 16 + 1 = 32 u

2. Calculate the formula unit masses of ZnO, Na₂O, K₂Co₃ given atomic masses of Zn=65 u, Na = 23 u, K = 39 u, C = 12u, and O = 16 u.

Ans. The formula unit mass of

(i) ZnO = 65 u + 16 u =.81 u

(ii) Na₂O = (23 u × 2) + .16 u : = 46u+ 161.u.

= 62 u

(iii) K₂CO3 = (39 u × 2) + 12 u + 16 u × 3

= 75 u + 12u = 48u = 138u

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

Ans. 1 mole of carbon atoms 6.022 × 1023 atoms = 12 g

Mass of 1 atom = ?

∴ Mass of 1 atom of carbon

=1.99 × 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)?

Ans. 23 g of Na = 6.022 × 1023 atoms (1 mole).

100 g of Na = ?

= 26.182 ×10²³ = 2.6182×10²⁴ atoms

56 g of Fe = 6.022 × 10²³ atoms

100 g o Fe = ?

100 g of Na contain → 2.618 × 10²⁴ atoms

100 g of Fe contain → 1.075 × 10²⁴ atoms

∴ 100 g of Na contains more atoms.

QUESTIONS FROM NCERT TEXTBOOK

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.

Ans. Boron and oxygen compound → Boron + Oxygen

0.24 g → 0:096g + 0.144g

Percentage composition of the compound

For boron: 0.24g → 0.096 g

100 g → ?

For oxygen:

0.24 g → 0.144 g of oxygen

100 g → ?

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 formed when. 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemicai combination will govern your answer?

Ans. The reaction of burning of carbon in oxygen may be written as:

It shows that 12 g of carbon burns in 32 g oxygen to form 44 g of carbon dioxide. Therefore 3 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide. It is given that 3.0 g of carbon is burnt with 8 g of oxygen to produce 11.0 g of CO2. Consequently 11.0 g of carbon dioxide will be formed when 3.0 g of C is burnt in 50 g of oxygen consuming 8 g of oxygen, leaving behind 50 �V 8 = 42 g of 02. The answer governs the law of constant proportion.

3. What are polyatomic ions? Give eaamples:

Ans. The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit are called polyatomic ions e.g., OH^–, SO^2– ₄, CO₃ ^2–.

4. Write the chemical formulae of the following:

(a) Magnesium chloride (b) Caldum oxide

(e) Calcium carbonate.

Ans. (a) Magnesium chloride

Symbol → Mg Cl

Change → +2 –1

Formula → MgCl₂

(b) Calcium oxide

Symbol → Ca O

Charge → +2 –2

Formula → CaO

Symbol → Cu NO₃

Change → +2 –1

Formula → Cu(NO₃)₂

(d) Aluminium chloride

Symbol → Al Cl

Change → +3 –1

Formula → A1Cl₃

(d) Calcium carbonate

Symbol → Ca CO3

Change → +2 –2

Formula → CaCO₃

5. Give the names of the elements present in the following compounds:

(a) Quick time (b) Hydrogen bromide

Ans. (a) Quick lime → Calcium oxide

Elements → Calcium and oxygen

(b) Hydrogen bromide

Elements → Hydrogen and bromine

Elements → Calcium, hydrogen, carbon and oxygen

(d) Potassium sulphate

Elements → Potassium, sulphur and oxygen

6. Calculate the molar mass of the following substances.

(a) Ethyne, C₂H₂

(b) Sulphur molecule, S₈

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO₃

Ans. The molar mass of the following: [Unit is 'g']

(a) Ethyne, C₂H₂ = 2 × 12 + 2 × 1 = 24 + 2 = 26 g

(b) Sulphur molecule, S₈ = 8 × 32 = 256 g

(d) Hydrochloric acid, HCl = 1 × 1 + 1 × 35.5

= 1 + 35.5 = 36.5 g

(e) Nitric acid, HNO₃ = 1 × 1 + 1 × 14 + 3 × 16

= 1 + 14 + 48 = 63 g

7. What is the mass of

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms

Ans. (a) Mass of 1 mole of nitrogen atoms = 14g

(b) 4 moles of aluminium atoms

Mass of 1 mole of aluminium atoms = 27 g

Mass of 4 moles of aluminium atoms = 27 × 4 = 108 g

Mass of 1 mole of Na₂SO₃ = 2 × 23 + 32 + 3 ×16

=46 + 32 + 48 = 126 g

Mass of 10 moles of Na₂SO₃ = 126 × 10

= 1260 g

8. Convert into mole.

(a) 12 g of oxygen gas

(b) 20 g of water

Ans. (a) Given mass of oxygen gas = 12 g

Molar mass of oxygen gas (O2) = 32 g

Mole of oxygen gas 12 mole

(b) Given mass of water = 20 g

Molar mass of water (H₂O) = (2 × 1) + 16 = 18 g

Mole of water mole

Molar mass of carbon dioxide (CO₂) = (1 × 12) + (2 × 16)

= 12 + 32 = .44 g

Mole of carbon dioxide mole

9. What is the mass of

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Ans. (a) Mole of Oxygen atoms = 0.2 mole

Molar mass of oxygen atoms = 16 g

Mass of oxygen atoms = 16 × 0.2 = 3.2 g

(b) Mole of water molecule = 0.5 mole

Molar mass of water molecules = 2 × 1 + 16 = 18 g

Mass of H2O = 18 × 0.5 = 9 g

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

Ans. Molar mass of S8 sulphur = 256 g = 6.022 × 1023 molecule

Given mass of sulphur = 16 g

= 0.376 × 10²³

= 3.76 × 10²² molecules

11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

Ans. Molar mass of aluminium oxide Al₂O₃

= (2 × 27) + (3 × 16)

= 54 + 48 = 102g.

102 g of Al2O3 contains = 2 × 6.022 × 1023 aluminium ions

= 0.006022 × 10²³

= 6.022 × 10²⁰ Al³⁺ ions

Ch 3 Atoms and Molecules Ncert Solutions

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Ch 3 Atoms and Molecules Ncert Solutions

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