Ch.4: Measurement of Matter

Measurement of Matter | Class 9 | Science | Chapter 4 | Maharashtra State Board

Explore key concepts Dalton’s atomic theory, laws of chemical combination, molecular mass, the mole concept, Avogadro's number, and chemical formulae. Learn through real-life experiments, clear explanations, and practical examples.

Questions & Answers

1. Give examples.

a. Positive radicals

Ans.

Na⁺ (Sodium ion), K⁺ (Potassium ion)

b. Basic radicals

Ans.

Na⁺ (Sodium ion), K⁺ (Potassium ion), Ag⁺ (Silver ion)

c. Composite radicals

Ans.

SO₄ ²-, NH₄⁺

d. Metals with variable valency

Ans.

(a) Iron (Ferrum)

(i) Fe²⁺ (Ferrous [Iron – II])

(ii) Fe³⁺ (Ferric [Iron – III])

(b) Copper (Cuprum)

(i) Cu⁺  (Cuprous [Copper -1])

(ii) Cu²⁺ (Cupric [Copper – II])

(c) Mercury (Hydragyrum)

(i) Hg⁺ (Mercurous [Mercury -1])

(ii) Hg²⁺ (Mercuric [Mercury – II])

e. Bivalent acidic radicals

Ans. O²- (Oxide), S²- (Sulphide), CO₃ ²- (Carbonate)

f. Trivalent basic radicals

Ans.

Al³+ (Aluminium), Cr³⁺ (Chromium), Fe³⁺ (Ferric).

2. Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals.

Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen

Ans.

Elements	Symbols	Radicals	Charge of Radicals
Mercury	Hg	Hg⁺ (Mercurous) Hg²⁺ (Mercuric)	+1 +2
Potassium	K	K⁺ (Potassium)	+1
Nitrogen	N	N³⁻ (Nitride)	-3
Copper	Cu	Cu⁺ (Cuprous) Cu²⁺(Cupric)	+1 +2
Sulphur	S	S²⁻ (Sulphide)	-2
Carbon	C	-	-
Chlorine	Cl	Cl⁻ (Chloride)	-1
Oxygen	O	O²⁻ (Oxide)	-2

3. Write the steps in deducing the chemical formulae of the following compounds.

Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide

Ans.

In order to write the chemical formulae of compounds, it is necessary to know the symbols and valency of various radicals.

1. Sodium Sulphate:

Step – 1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)

Na SO₄

Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.

Na₂ SO₄

(Sodium sulphate)

2. Potassium Nitrate:

Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)K NO₃

Step – 2 : To write the valency below the respective radical. KNO₃

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.

KNO₃

(Potassium nitrate)

3. Ferric phosphate:

Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)Fe PO₄

Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound. FePO₄ (Ferric phosphate)

4. Calcium oxide:

Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radicals on the right) CaO

Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.CaO (Calcium oxide)

5. Aluminium hydroxide:

Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radical on the right)Al OH

Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound. Al(OH)₃ (Aluminium hydroxide)

6. Calcium carbonate:

Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radicals on the right) Ca CO₃

Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.

CaCO₃

(Calcium Carbonate)

7. Sodium dichromate:

Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radical on the right)

Na Cr₂O₇

1 2

Step – 2 : To write the valency below the respective radical. Na1Cr₂O₇

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound. Na2Cr₂O₇ (Sodium dichromate)

4. Write answers to the following questions and explain your answers.

a. Explain how the element sodium is monovalent.

Ans.

1. The number of protons or electrons (atomic number) in Sodium (Na) atom is 11. Therefore the electronic configuration of sodium atom is (2, 8,1).

2. In chemical reaction, sodium atom has the capacity to give away le_ from its outermost orbit to form Na+ ion with stable electronic configuration (2, 8).

3. As sodium atom gives away le- and a cation of sodium is formed, hence the valency of sodium is 1 and therefore, the element sodium is monovalent.

b. M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate.

Ans.

M is a bivalent metal. Following are the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate:

(i) Compound of metal ‘M’ with radical sulphate

Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)

M SO₄

Step – 2: To write the valency below the respective radical.

Step – 3: To cross multiply as shown by arrows the number of radicals.

Step – 4: To write down the chemical formula of the compound.

M SO₄

(ii) Compound of metal ‘M’ with radical phosphate.

Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)

M PO₄

Step – 2: To write the valency below the respective radical.

Step – 3: To cross multiply as shown by arrows the number of radicals.

Step – 4: To write down the chemical formula of the compound.

M₃ (PO₄)₂

c. Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.

Ans.

• The mass of an atom is concentrated in its nucleus and it is due to the protons (p) and neutrons (n) in it.

• Since an atom is very very tiny, it was not possible to measure atomic mass accurately. Therefore, the concept of relative mass of an atom was formed.

• To express relative mass of an atom, reference of atom is considered. The two reference atoms were as follows:

(a) Hydrogen (H) atom: The hydrogen atom is the lightest. The relative mass of a hydrogen atom is 1 which has only 1 proton in its nucleus. On this scale, the relative atomic mass of many elements comes out to be fractional. Therefore, carbon was selected as a reference atom.

(b) Carbon (C) atom: The carbon atom is selected as reference atom. In this scale, the relative mass of a carbon atom is accepted as 12.

• The relative atomic mass of 1 hydrogen (H) atom compared to the carbon (C) atom becomes

d. What is meant by Unified Atomic Mass.

Ans.

• During earlier time, relative mass of an atom was considered for measuring the mass of an atom directly. But since the founding of unified mass, relative mass is not accepted henceforth.

• Unified atomic mass is the unit of atomic mass called as Dalton.

• Its symbol is ‘u’. lu = 1.66053904 × 10⁻²⁷ kg.

e. Explain with examples what is meant by a ‘mole’ of a substance.

Ans.

• A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.

• For example: Atomic mass of oxygen atom (O) is 16u. Thus, the molecular mass of oxygen molecule (O2) is 16 x 2 = 32u. Therefore, 32 g of oxygen is 1 mole of oxygen.

5. Write the names of the following compounds and deduce their molecular masses.

Na₂SO₄, K₂CO₃, CO₂, MgCl₂, NaOH, AlPO₄, NaHCO₃

Ans.

6. Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows:

‘sample m’ mass : 7g

Mass of constituent oxygen : 2g

Mass of constituent calcium : 5g ‘sample n’ mass : 1.4g

Mass of constituent oxygen : 0.4g Mass of constituent calcium : 1.0g

Which law of chemical combination does this prove? Explain.

Ans.

(i) The expected proportion by weight of the constituent elements of quick lime that is calcium oxide would be from its known molecular formula CaO. The atomic mass of Ca and O are 40 and 16 respectively. This means, the proportion by weight of the constituent elements Ca and O in the compound CaO is 40 :16 which is 5 : 2.

(ii) Now, for the given sample’m’ of CaO = 5 gmass of given sample = 7 gmass of constituent Ca in sample’m’ = 5 gmass of constituent O in sample’m’ = 2 g

(iii) This means that 7 g of calcium oxide contairis 5 g of calcium (Ca) and 2 g of oxygen (O); apd the proportion by weight of calcium and oxygen in it is 5 : 2.

(iv) Now, for the given sample ‘n’ of CaO mass of given sample CaO = 1.4 gMass of constituent Ca in sample ‘n’ = 1.0 gMass of constituent O in sample ‘n’ = 0.4 gThis means that 1.4g of calcium oxide contains 1.0 g of calcium (Ca) and 0.4 g of oxygen (O); and the proportion by weight of calcium and oxygen in it is 5 : 2.

(v) Above samples’m’ and ‘n’ of calcium oxide (CaO) shows that the proportion by weight of the constituent elements in different samples of a compound is always constant that is the proportion by weight of calcium (Ca) and oxygen (O) in different samples of calcium oxide (CaO) is constant.

(vi) The experimental value of proportion by weight of the constituent elements matched with the expected proportion calculated by molecular mass. This proves and verifies the law of constant proportion.

The law states that ‘The proportion by weight of the constituent elements in the various samples of a compound is fixed’.

7. Deduce the number of molecules of the following compounds in the given quantities.

32g oxygen, 90g water, 8.8g carbon dioxide, 7.1g chlorine.

32g oxygen

Ans.

Given : Mass of oxygen (O2) m = 32g

To find : Number of molecules in 32g of oxygen.

Solution : Atomic mass of oxygen (O) = 16

∴ Molecular mass of oxygen (O2) M = 16 x 2 = 32

According to the formula, Number of moles in the given O2 (n)

1 mol of O₂ contains 6.022 × 10²³ molecules that is 32 g of 0₂ contains 6.022 ×

10²³ molecules of O₂.

32g of oxygen contains 6.022 × 10²³ molecules of oxygen.

90g water

Ans.

Given : Mass of water (H₂O) m = 90g.

To find : Number of molecules in 90g of water.

Solution : Molecular mass of (H₂O) M = (Atomic mass of H) × 2 + (Atomic mass of O) x 1

∴ Molecular mass of (H₂O) M = 1 × 2 +16

∴ Molecular mass of (H₂O) M = 18

According to the formula,

Number of moles in the given H₂O (n)

1 mol of H₂O contains 6.022 × 10²³ molecules.

5 mol of H₂O contains 5 × 6.022 × 10²³ molecules. = 30.11 × 10²³ molecules, that is 90g of H₂O contains 30.11 × 10²³ molecules of H₂0.

90g of water contains 30.11 × 10²³ molecules of water.

8.8g carbon dioxide

Ans. 

Given : Mass of Carbon dioxide (CO₂)m = 8.8g.

To find : Number of molecules in 8.8g of carbon dioxide.

Solution :

Molecular mass of (CO₂)M = (Atomic mass of C) x 1 + (Atomic mass of O) x 2

∴ Molecular mass of (CO₂)M = 12 × 1 + 16 x 2 = 12 + 32

Molecular mass of (CO₂)M = 44 According to the formula, Number of moles in the given CO₂ (n)

∴ 1 mol of CO₂ contains 6.022 × 1023 molecules.

∴ 0.2 mol of CO₂ contains 0.2 × 6.022 × 10²³ molecules.

= 1.2044 × 1023 molecules,

that is 8.8g of CO₂ contains 1.2044 × 10²³ molecules of CO₂.

8.8g of CO₂ contains 1.2044 × 10²³ molecules of CO₂.

7.1g chlorine

Ans.

Given : Mass of Chlorine (Cl₂)m = 7.1g.

To find : Number of molecules in 7.1g of chlorine.

Solution : Atomic mass of (Cl) = 35.5

∴ Molecular mass of chlorine (Cl₂)M = 35.5 × 2 = 71

According to the formula, Number of moles in the given Cl₂ (n)

∴ 1 mol of Cl₂ contains 6.022 × 10²³ molecules.

∴ 0.1 mol of Cl₂ contains 0.1 × 6.022 × 10²³ molecules.

= 0.6022 × 10 molecules,

that is 7.1g of Cl₂ contains 0.6022 × 10²³ molecules of Cl₂.

7.1g of Cl₂ contains 0.6022 × 10²³ molecules of chlorine.

8. If 0.2 mol of the following substances are required how many grams of those substances should be taken?

Sodium chloride, magnesium oxide, calcium carbonate

Ans.

Given : Number of moles of sodium chloride (NaCl) n = 0.2 mol

To find : Mass in grams of 0.2 mol of NaCl

Solution:Molecular mass of (NaCl)M

= (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1

= 23 x 1 + 35.5 x 1= 23 + 35.5

Molecular mass of (NaCl) M = 58.5

According to the formula,

Number of moles in the given NaCl (n)

Mass of NaCl in grams (m) = 0.2 x 58.5

Mass of NaCl in grams (m) = 11.7 g

Mass of 0.2 mole of NaCl is 11.7g

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E X T R A

Intext Questions and Answers

1. What is the type of chemical bond in NaCl and MgCl₂?

Ans.

The type of chemical bond in NaCl and MgCl₂ is ionic bond.

2. Determine the valencies of H, Cl, O and Na from the molecular formulae H₂, HC1, H₂O and NaCl.

Ans.

(i) In the molecular formula HCl

∴ The valency of H is 1 and Cl is 1.

(ii) In the molecular formula H₂O

∴ The valency of H is 1 and O is 2.

(iii) In the molecular formula NaCl

∴ The valency of Na is 1 and Cl is 1.

∴ From all the above, the valencies of the given elements are as follows : H = 1, Cl = 1, O = 2 and Na = l.

3. How is an element indicated in Chemistry?

Ans.

In chemistry an element is indicated by its symbol.

4. Write down the symbols of the elements you know.

Ans.

Symbols of some elements are

• Hydrogen – H

• Helium – He

• Boron – B

• Carbon – C

• Aluminium – A1

5. Write down the symbols for the following elements.

Antimony, Iron, Gold, Silver, Mercury, Lead, Sodium

Ans. 

The symbols of given elements are as follows:

• Antimony – Sb

• Iron – Fe

• Gold – Au

• Silver – Ag

• Mercury – Hg

• Lead – Pb

• Sodium – Na

2. Following are atomic masses of a few elements in Daltons and the molecular formulae of some compounds. Deduce the molecular masses of those compounds:

Atomic masses – H(l), 0(16), N(14), C(12), K(39), S(32) Ca(40), Na(23), Cl(35.5), Mg(24), Al(27)

1. Molecular formula – NaCl

Ans.

Molecular mass of NaCl (M)

= (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1

= (23 x 1) + (35.5 x 1)

= 23 + 35.5= 58.5

∴ Molecular mass of NaCl (M)

= 58.5

2. Molecular formula – MgCl₂

Ans.

Molecular mass of MgCl₂ (M)

= (Atomic mass of Mg) x 1 + (Atomic mass of Cl) x 2

= (24 x 1) + (35.5 x 2)

= 24 + 71= 95

∴ Molecular mass of MgCl₂ (M) = 95

3. Molecular formula – KNO₃

Ans.

Molecular mass of KNO₃ (M)

= (Atomic mass of K) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3

= (39 x 1) + (14 x 1) + (16 x 3)

= 39 + 14 + 48

= 101

Molecular mass of KNO₃ (M) = 101

4. Molecular formula – H₂O₂

Ans.

Molecular mass of H₂O₂ (M)

= (Atomic mass of H) x 2 + (Atomic mass of O) x 2

= (1 x 2) + (16 x 2)

= 2 + 32

= 34

∴ Molecular mass of H₂O₂ (M) = 34.

5. Molecular formula – A1C1₃

Ans.

Molecular mass of A1C1₃ (M)

= (Atomic mass of Al) x 1 + (Atomic mass of Cl) x 3

= (27 x 1) + (35.5 x 3)

= 27 + 106.5= 133.5

∴ Molecular mass of A1C1₃ (M) = 133.5

6. Molecular formula – Ca(OH)₂

Ans.

Molecular mass of Ca(OH)₂ (M)

= (Atomic mass of Ca) x 1 + (Atomic mass of O + Atomic Mass of H) x 2

= (40 x 1) + (16 + 1) x 2

= 40 + (17 x 2)= 40 + 34

= 74

∴ Molecular mass of Ca(OH)₂ (M)= 74

7. Molecular formula – MgO

Ans.

Molecular mass of MgO (M)

= (Atomic mass of Mg) x 1 + (Atomic mass of 0) x 1

= (24 x 1) + (16 x 1)

= 24 + 16

= 40

Molecular mass of MgO (M) = 40

8. Molecular formula – H₂S0₄

Ans.

Molecular mass of H₂S0₄ (M)

= (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4

= (1 x 2) + (32xl) + (16×4)

= 2 + 32 + 64

= 98

Molecular mass of H₂S0₄ (M) = 98

9. Molecular formula – HN0₃

Ans.

Molecular mass of HN0₃ (M)

= (Atomic mass of H) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3

= (lxl)+ (14xl)+ (16×3)

= 1 + 14 + 48= 63

Molecular mass of HNO₃ (M) = 63

10. Molecular formula – NaOH

Ans.

Molecular mass of NaOH (M)

= (Atomic mass of Na) x 1 + (Atomic mass of O) x 1 + (Atomic mass of H) x 1

= (23 x 1) + (16 x 1) + (l x l)

= 23 + 16 + 1= 40

Molecular mass of NaOH (M) = 40

11. How many molecules of water are there in 36 g water?

Ans.

Given : Mass of water (H₂O) m = 36g

To find : Number of molecules in 36g of water

Solution :

Molecular mass of (H₂O) M

= (Atomic mass of H) x 2 + (Atomic mass of O) x 1 Molecular mass of (H₂O) M

= (1 x 2) + 16 x 1Molecular mass of (H₂O) M = 18

According to the formula,

Number of moles in the given H₂O (n)

1 mol of H₂O contains 6.022 x 10²³ molecules.

∴ 2 mol of H₂O contains 2 x 6.022 x 10²³ molecules.

= 12.044 x 10²³ molecules,

that is 36g of H₂O contains 12.044 x 10²³ molecules of H₂O.

36 g of water contains 12.044 x 10²³ molecules of water.

12. How many molecules of H₂SO₄ are there in a 49 g sample?

Ans.

Given : Mass of Sulphuric acid (H₂SO₄) m = 49g

To find : Number of molecules in 49g of H₂SO₄

Solution:

Molecular mass of (H₂SO₄) M

= (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4

Molecular mass of (H₂SO₄)M = (1 x 2) + (32 x 1) + (16 x 4)

= 2 + 32 + 64= 98.

According to the formula,

Number of moles in the given H₂SO₄ (n)

∴ 1 mol of H₂SO₄contains 6.022 x 10²³ molecules.

∴ 0.5 mol of H₂SO₄ contains 0.5 x 6.022 x 10²³ molecules.

= 3.011 x 10²³ molecules,

that is 49g of H₂SO₄ contains 3.011 x 10²³ molecules of H₂SO₄.

49 g of Sulphuric acid contains 3.011 x 10²³ molecules of H₂SO₄.

3. Fill the following tables.

Ans.

Element	Atomic Mass
Oxygen	16
Sodium	23
Aluminium	27
Phosphorus	31
Argon	39.9
Potassium	39

4. Complete the following chart.

Ans.

Element	Automic number	Electronic Configuration	Valence Electron	Valency
Lithium	3	2, 1	1	1
Beryllium	4	2, 2	2	2
Boron	5	2, 3	3	3
Carbon	6	2, 4	4	4
Nitrogen	7	2, 5	5	3
Oxygen	8	2, 6	6	2
Fluorine	9	2, 7	7	1
Neon	10	2, 8	8	0
Sodium	11	2, 8, 1	1	1
Magnesium	12	2, 8, 2	2	2
Aluminium	13	2, 8, 3	3	3
Silicon	14	2, 8, 4	4	4

5. The relative atomic masses of some elements in the chart below are given. You have to find the relative atomic masses of the others.

Ans.

Element	Atomic mass	Element	Atomic mass	Element	Automic mass
Hydrogen	1	Oxygen	16	Phosphorus	31
Helium	4	Fluorine	19	Sulphur	32
Lithium	7	Neon	20	Chlorine	35.5
Beryllium	9	Sodium	23	Argon	39.3
Boron	11	Magnesium	24	Potassium	39
Carbon	12	Aluminium	27	Calcium	40
Nitrogen	14	Silicon	28

6. Classify the following radicals into simple radicals and composite radicals: (Use your brain power;

Ag⁺, Mg²⁺, Cl⁻, SO²⁻₄, Fe²⁺, ClO⁻₃, NH⁺₄ ,Br⁻, NO⁻₃, Na⁺, Cu⁺

Ans.

Simple radicals	Composite radicals
Ag⁺	SO²⁻₄
Mg²⁺	ClO⁻₃
Cl⁻	NH⁺₄
Fe²⁺	NO⁻₃
Br⁻
Na⁺
Cu⁺

7. Which are the basic radicals and which are the acidic radicals among the following?

Ag⁺, Cu²⁺, Cl⁻, I⁻, SO²⁻₄, Fe³⁺, Ca²⁺, NO₃⁻, S²⁻, NH⁺₄, K⁺, MnO₄⁻, Na⁺

Ans.

Basic Radical	Acidic Radical
(i) Ag⁺	(i) Cl⁻
(ii) Cu²⁺	(ii) I⁻
(iii) Fe³⁺	(iii) SO²⁻₄
(iv)  Ca²⁺	(iv)  NO₃⁻
(v) NH⁺₄	(v) S²⁻
(vi) K⁺	(vi) MnO₄⁻
(vii) Na⁺

8. Give examples:

1. Make a list of elements in the monoatomic and in the diatomic molecular state. (Make a list and discuss)

Ans.

• Elements in the monoatomic molecular state are: Helium (He), Neon (Ne), Argon (Ar), Sodium (Na), Copper (Cu),

• Elements in the diatomic molecular state are: Oxygen (O₂), Nitrogen (N₂), Hydrogen (H₂), Chlorine (Cl₂), Fluorine (F₂).

Problem-based questions

1. Answer the following questions:

1. Is it possible to weigh one molecule using a weighing balance?

Ans.

No, it is not possible to weigh one molecule using a weighing balance.

2. Will the number of molecules be the same in equal weights of different substances?

Ans.

No, the number of molecules will not be the same in equal weights of different substances.

3. If we want equal number of molecules of different substances, will it work to take equal weights of those substances.

Ans.

No, if we want equal number of molecules of different substances, it will not work to take equal weights of those substances.

2. Answer the following:

1. What is the Dalton’s atomic theory?

Ans. Dalton’s Atomic theory-

• All matter is made of atoms. Atoms are indivisible and indestructible.

• All atoms of a given element are identical in mass and properties.

• Compounds are formed by a combination of two or more different kinds of atoms.

• A chemical reaction is a rearrangement of atoms.

2. How are compounds formed?

Ans.

Compounds are formed by a chemical combination of two or more different kinds of atoms.

3. What are the molecular formulae of salt, slaked lime, water, lime, limestone?

Ans.

The molecular formulae for

Salt – Sodium chloride – NaCl

Slaked lime – Calcium hydroxide Ca(OH)₂

Water – H₂O

Lime – Calcium oxide – CaO

Lime stone – Calcium carbonate – CaCO₃

4. From which experiments was it discovered that atoms have an internal structure? When?

Ans.

• In 1911, Earnest Rutherford conducted a well known experiment called as ‘Gold foil experiment’.

• From this experiment it was discovered that atoms have internal structure.

5. What are the two parts of an atom? What are they made up of?

Ans.

The two parts of atoms are nucleus and extra nuclear part. Nucleus is made up of positively charged protons and electrically neutral neutrons and the extra nuclear part is made up of negatively charged electrons revolving around the nucleus in different orbits.

Open-ended questions

Answer the following questions:

1. How will the compounds, MgCl₂ and CaO be formed from their elements?

Ans. 

(1) Magnesium Chloride (MgCl₂)

Magnesium atom (Mg): Electronic configuration

(2,8,2)⟶−2e− Magnesium ion Mg²⁺ (2,8).

Chlorine atom (Cl). Electronic configuration (2,8,7)⟶+1e− Chloride ion Ch (2,8,8).

∴ Mg²⁺ + 2CT → MgCl₂ (Magnesium Chloride)

• A Magnesium atom gives away 2e– and a cation of Magnesium (Mg²⁺) is formed, hence, the valency of magnesium is two.

• Two chlorine atoms takes le– each and forms two anions of chlorine (2Cl–) (chloride), and thus, the valency of chlorine is one.

• After the give and take of electrons is over, the electronic configuration of all the resulting ions has a complete octet.

• Due to the attraction between the unit but opposite charges on all the ions, one chemical bond known as ionic bond is formed between Mg²⁺ and 2C1– each and the compound MgCl₂ is formed.

(2) Calcium Oxide (CaO)

Calcium atom (Ca): Electronic configuration

(2,8,8,2)⟶−2e− Calcium ion Ca²⁺ (2,8,8).

Oxygen atom (O). Electronic configuration (2,6)

⟶+2e− Oxygen ion O²⁻ (2,8).

∴ Ca²⁺ + O²⁻ → CaO

• A calcium atom gives away 2e- and a cation of calcium (Ca²⁺) is formed, hence, the valency of calcium is two.

• An oxygen atom takes 2e– and forms anions of oxygen (O²⁻) (oxide), and thus, the valency of oxygen is two.

• After the give and take of electrons is over, the electronic configuration of both the resulting ions has a complete octet.

• Due to the attraction between the unit but opposite charges on the two ions, one chemical bond known as ionic bond is formed between Ca²⁺ and O²⁻ and the compound CaO is formed.

2. Take 56 g calcium oxide in a large conical flask and put 18 g water in it.

• Observe what happens.

• Measure the mass of the substance formed.

• What similarity do you find? Write your inference.

Ans.

(i) When 18 g of water is added to 56 g of calcium oxide, calcium oxide combines with water to form calcium hydroxide Ca(OH)₂

(ii) The mass of calcium hydroxide formed is 74 g.?

(iii) In this activity the total mass of reactants,

Calcium oxide + Water = 56 g +18 g = 74 g.

It is equal to the mass of the product formed. Ca(OH)₂ = 74g.

This activity verifies the Law of Conservation of Matter, i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

3. Take a solution of calcium chloride in a conical flask and a solution of sodium sulphate in a test tube.

• Tie a thread to the test tube and insert it in the conical flask.

• Seal the conical flask with an airtight rubber cork.

• Weigh the conical flask using a balance.

• Now tilt the conical flask so that the solution in the test tube gets poured in the conical flask.

• Now weigh the conical flask again.

Ans.

• In this activity, a white precipitate of CaS04 in NaOl is seen in the conical flask after the reaction.

• There is no change in the weight of the flask before and after the reaction.

• This activity verifies the Law of Conservation of Matter i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

4. Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds : Calcium carbonate, Sodium bicarbonate, Silver chloride, Calcium hydroxide, Magnesium oxide, Ammonium phosphate, Cuprous bromide, Copper sulphate, Potassium nitrate, Sodium dichromate.

Ans.

Calcium carbonate – CaCO₃ Sodium bicarbonate – NaHCO₃ Silver chloride – AgCl, Calcium hydroxide – Ca(OH)₂, Magnesium oxide – MgO, Ammonium phosphate – (NH4)₃PO₄, Cuprous bromide – CuBr, Copper sulphate – CuSO₄, Potassium nitrate – KNO₃, Sodium dichromate – Na₂Cr₂O₇.

Additional Important Questions and Answers

(A) Select the correct option:

1. The proportion by weight of hydrogen and oxygen in water is ……………………….. .

(a) 8 : 1

(b) 2 : 1

(c) 1 : 2

(d) 1 : 8

Ans.

(d) 1: 8

2. The proportion by weight of carbon and oxygen in carbon dioxide is ……………………….. .

(a) 8 : 3

(b) 3 : 8

(c) 3 : 2

(d) 2 : 3

Ans.

(b) 3 : 8

3. A nucleus of an atom is made up of positively charged ………………………… and electrically neutral ……………………….. .

(a) protons; neutrons

(b) electrons; neutrons

(c) neutrons; protons

(d) neutrons; electrons

Ans.

(a) protons; neutrons

4. The size of an atom is determined by its ……………………….. .

Ans.

radius

5. Atomic radius is expressed in ……………………….. .

(a) milimetres

(b) centimetres

(c) nanometres

(d) picometres

Ans.

(c) nanomet res

6. The atomic size depends on the number of ………………………… in the atom.

(a) protons

(b) nucleus

(c) neutrons

(d) electron orbits

Ans.

(d) electron orbits

7. The mass of an atom is concentrated in its ……………………….. .

(a) protons

(b) nucleus

(c) neutrons

(d) electrons

Ans.

(b) nucleus

8. The total number of protons and neutrons in the atomic nucleus is called the ……………………….. .

(a) atomic number

(b) electronic configuration

(c) atomic mass number

(d) valency

Ans.

(c) atomic mass number

9. A ………………………… is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.

(a) mole

(b) dalton

(c) dozen

(d) gross

Ans.

(a) Mole

10. Avogadro’s number is denoted by the symbol ……………………….. .

(a) NG

(b) Nv

(c) NA

(d) ND

Ans.

(c) NA

11. A mole of any substance stands for ………………………… molecules.

(a) 60.22 x 10²²

(b) 6.022 x 1022

(c) 6.022 x 10²³

(d) 60.22 x 10²²

Ans.

(a) 60.22 x 10²³

12. The capacity of an element to combine is called its ……………………….. .

(a) valency

(b) electronic configuration

(c) atomic number

(d) volence electrons

Ans.

(a) valency

13. Electronic configuration of sodium atom is ……………………….. .

(a) (2, 8, 3)

(b) (2, 8, 7)

(c) (2, 8, 2)

(d) (2, 8,1)

Ans.

(d) (2,8,1)

14. Electronic configuration of chlorine atom is ……………………….. .

(a) (2, 8, 3)

(b) (2, 8, 7)

(c) (2, 8, 2)

(d) (2, 8, 1)

Ans.

(b) (2, 8, 7)

15. Positively charged ions are called as ……………………….. .

(a) cations

(b) anions

(c) nucleous

(d) protons

Ans.

(a) cations

16. Negatively charged ions are called as ……………………….. .

(a) cations

(b) anions

(c) nucleus

(d) electrons

Ans.

(b) anions

17. Iron (Fe) exhibits the variable valencies ……………………….. .

(a) 1 and 2

(b) 2 and 3

(c) 3 and 4

(d) 2 and 4

Ans.

(b) 2 and 3

18. Cationic radicals are called as ………………………… radicals.

(a) basic

(b) acidic

(c) neutral

(d) mixed

Ans.

(a) basic

19. Anionic radicals are called as ………………………… radicals.

(a) basic

(b) acidic

(c) neutral

(d) mixed

Ans.

(b) acidic

20. The unit Dalton is used to express …………………………

(a) atomic mass

(b) atomic radius

(c) atomic number

(d) mass number

Ans.

(a) atomic mass

21. The valency of element with electronic configuration ………………………… is 2.

(a) (2,5)

(b) (2, 4)

(c) (2, 6)

(d) (2, 7)

Ans.

(c) (2, 6)

22. The symbol of Avogadro’s number is ……………………….. .

(a) ND

(b) N0

(c) NB

(d) NA

Ans.

(d) NA

23. ………………………. is bicarbonate radical.

(a)HCO²⁻₃

(b)CO₋₃

(c)HCO₋₃

(d)CO²⁻₃

Ans.  

(c) HCO₋₃

24. Molecular formula of sodium sulphate is ……………………….. .

(a) Na(SO4)₂

(b) Na₂SO₄

(c)Na₂(SO₄)2

(d)NaSO₄

Ans.

(b) Na₂SO₄

25. ………………………… is a composite radical.

(a) Fe³⁺

(b) Ca²⁺

(c) NH⁴⁺

(d) S²⁻

Ans.

(c) NH⁴⁺

26. A mole of any substance stands for ………………………… molecules.

(a) 6.022 x 10²³

(b) 6.022 x 10²²

(c) 60.22 x 10²³

(d) 60.22 x 10²²

Ans.

(a) 6.022 x 10²³

27. The mass of an atom is concentrated in its ………………………… .

(a) nucleus

(b) electrons

(c) extranuclear part

(d) protons

Ans.

(a) nucleus

28. ………………………… g of water make 1 mole of water.

(a) 32

(b) 33

(c) 16

(d) 18

Ans.

(d) 18

2. Complete the analogy:

(1) Electron : extra nuclear part:: Neutron ………………………… .

Ans.

nucleus

(2) Sodium: (2, 8, 1):: Chlorine:: ………………………… .

Ans.

(2, 8, 7)

(3) K : basic radical :: Br₋ : ………………………… .

Ans.

acidic radical

(4) Cut: simple radical:: NH₄₊ : ………………………… .

Ans.

composite radical

(5) Sodium sulphate: Na₂SO₄:: Potassium Sulphate: ………………………… .

Ans.

K₂SO₄

(6) Mercurous: Hg⁺:: Mercuric : ………………………… .

Ans.

Hg₂⁺

(7) Positively charged ion : cation:: Negatively charged ion : ………………………… .

Ans.

anion

(8) 12: 1 dozen :: 144 : ………………………… .

Ans.

1 gross

(9) Hydrogen : ⊙ :: Copper : ………………………… .

Ans.

©

(10) Law of constant proportions : J. L. Proust::Law of conservation of matter : ………………………… .

Ans.

Antoine Lavoisier.

3. Match the columns:

Column 'A’ Example	Column ‘B’ Atomic radius (in metres)
(1) Water molecule	(a) 10⁻¹⁰
(2) Haemoglobin molecule	(b) 10⁻⁹
(3) Hydrogen atom	(c) 10⁻⁸

Ans.

(1-b), (2- c), (3 – a)

Column ‘A’ Element	Column ‘B’ Atomic mass
(1) Neon	(a) 35.5
(2) Silicon	(b) 32
(3) Chlorine	(c) 28
(4) Sulphur	(d) 20

Ans.

(1 – d), (2 – c), (3 – a), (4 – b)

Column ‘A’ Molecule	Column ‘B’ Molecular mass in grams
(1) h₂	(a) 32 g
(2) H₂O	(b) 34 g
(3) O₂	(c) 2 g
(4) H₂O₂	(d) 18 g

Ans.

(1 – c), (2 – d), (3 – a), (4 – b)

Column ‘A’ Radicals	Column B’ Names
(1) Cr₂O²⁻₇	(a) Carbonate
(2) ClO⁻₃	(b) Chromate
(3) CO²⁻₃	(c) Dichromate
(4) CrO²⁻₄	(d) Chlorate

Ans.

(1 – c), (2 – d), (3 – a), (4 – b)

4. Answer the following in one sentence:

1. What are valence electrons?

Ans.

The electrons present in the outermost orbit of an atom are called valence electrons.

2. Give the formula to determine the number of moles of a substance.

Ans.

The formula to determine the number of moles of a substance is as given below.

3. What are basic radicals? Give examples.

Ans.

The radicals which are formed by removal of electrons from the atoms of metals are called as basic radicals, e.g., Na+, Cu2+

4. What are acidic radicals? Give examples.

Ans.

The radicals which are formed by adding electrons to the atoms of non-metals are called as acidic radicals, e.g., CT, S2-

5. State whether the following statement is ‘True’ or ‘False’. Correct the false statement.

(1) Molecular state of oxygen is monoatomic.

Ans.

False. Molecular state of oxygen is diatomic:

(2) The capacity of an element to combine is called its valency.

Ans.

True

(3) Anionic radicals are basic radicals.

Ans.

False. Anionic radicals are acidic radicals.

(4) The magnitude of charge on any radical is its atomic number.

Ans.

False. Magnitude of charge on any radical is its valency.

(5) In a chemical reaction, mass of original matter and mass of matter newly formed as a result of chemical change are equal.

Ans.

True

(6) The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 5.

Ans.

False. The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 8.

(7) Relative mass of hydrogen is 1.

Ans.

True

(8) The number of molecules in a given quantity of a substance is determined by its atomic mass.

Ans.

False. The number of molecules in a given quantity of a substance is determined by its molecular mass.

(9) Avogadro’s number is 6.022 x 10²³

Ans.

True

(10) Valency of sodium is 2.

Ans.

(10) False. Valency of sodium is 1.

6. Name the following:

1. Scientist who gave Law of Conservation of Matter.

Ans.

Antoine Lavoisier

2. Scientist who gave Law of Constant Proportion.

Ans.

J. L. Proust

3. What are protons and neutrons present in nucleus together called as?

Ans.

Nucleons

4. Unit used to express atomic radius.

Ans.

Nanometre

5. The number (p + n) in the atomic nucleus is called as?

Ans.

Atomic mass number

6. Name the unit of atomic mass.

Ans.

Dalton (u)

7. Write molecular formula of two ionic compounds containing chlorine.

Ans.

NaCl, MgCl₂

8. Give two monoatomic radicals.

Ans.

Na+, Cl⁻

9. Give two examples of simple radicals.

Ans.

Ag+, O²⁻

7. Give scientific reasons:

1. An atom is electrically neutral though it contains charged particles.

Ans.

• An atom is made up of a nucleus and an extranuclear part. Protons and neutrons are present in the nucleus.

• The nucleus is positively charged. The extranuclear part is made up of negatively charged electrons.

• Protons are positively charged, electrons are negatively charged and neutrons are without any charge.

• The magnitude of their charges is the same when they are equal in number.

• Hence, the negative charge on all the extra, nuclear electrons together balances the positive charge on the

• nucleus.

• Therefore, an atom is electrically neutral though it contains charged particles.

2. Neon is chemically inert element.

Ans.

• Atomic number of neon is 10, so its electronic configuration is (2, 8). There are 8 electrons in its 2nd shell, fulfilling its capacity.

• Thus, neon has a complete octet.

• It has a stable orbit therefore, it does not indulge in chemical reactions. Hence, neon is a chemically inert element.

3. The valency of sodium (Na) is one.

Ans.

• The electronic configuration of sodium (Na) is (2, 8,1). It has 1 electron in its 3rd orbit.

• It tends to give up this electron so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.

• The loss of one electron leads to the formation of sodium ion (Na+) which is positively charged as it has lost one electron.

4. The valency of chlorine (Cl) is one.

Ans.

• The electronic configuration of chlorine (Cl) is (2, 8, 7). It has 7 electrons in its 3rd orbit.

• It tends to take one electron from another atom so that it has 8 electrons in the outermost orbit with electronic configuration (2,8,8) with stable state.

• The gaining of one electron leads to formation of chloride ion (Cl⁻) which is negatively charged as it has gained one electron.

5. The valency of Magnesium (Mg) is two.

Ans.

• The electronic configuration of Magnesium (Mg) is (2,8,2), it has 2 electrons in its 3rd orbit.

• It tends to give these ‘2’ electrons so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.

• The loss of two electrons leads to the formation of Magnesium ion (Mg2+) which is double positively charged as it has lost two electrons.

6. Valency is always a whole number.

Ans.

• The number of electrons that an atom of an element gives away, takes up or shares forming a bond is called the valency of that element.

• These electrons are always in whole numbers and not in fractions.

• Therefore, valency is always a whole number.

7. Atomic size of potassium is bigger than atomic size of sodium.

Ans.

• The atomic size of an element depends on the number of electron orbits in the atom of that element.

• The greater the number of orbits, the larger the size.

• Atomic number of potassium (K) is 19. Hence, its electronic configuration is (2, 8, 8,1). While atomic number of sodium (Na) is 11. Hence its electronic configuration is (2, 8,1)

• Number of orbits in potassium atom is 4, while that in sodium atom is 3.

• Hence, atomic size of potassium is bigger than atomic size of sodium.

8. The atomic size of sodium is bigger than atomic size of Magnesium.

Ans.

• The atomic size of an element depends on the number of electron orbits in the atom of that element.

• If 2 atoms have the same outermost orbit, then the atom having the larger number of electrons in the outermost orbit is smaller than the one having fewer electrons in the same outermost orbit.

• Atomic number of sodium (Na) is 11. Hence, its electronic configuration is (2, 8, 1) while atomic number of magnesum (Mg) is 12 and hence its electronic configuration is (2, 8, 2).

• As compared to sodium atom Magnesum atom has larger number of electrons n its electronic configuration.

• Therefore, atomic size of sodium is bigger than atomic size of Magnesium.

1. Numerical.

1. Magnesium Oxide:

Ans.

Given : Number of moles of Magnesium oxide (MgO)n = 0.2 mol

To find : Mass in grams of 0.2 mol of MgO

Solution:

Molecular mass of (MgO)M

= (Atomic mass of Mg) x 1 + (Atomic mass of O) x 1

= 24 x 1 + 16 x 1

= 24 + 16

Molecular mass of (MgO)M = 40

According to the formula Number of moles in the given MgO (n)

Mass of MgO in grams (m) = 0.2 x 40

Mass of MgO in grams (m) = 8 g.

Mass of 0.2 mole of MgO is 8 g

2. Calcium Carbonate:

Ans.

Given : Number of moles of Calcium carbonate (CaCO₃) n = 0.2 mol

To find : Mass in grams of 0.2 mol of CaCO₃

Solution

:Molecular mass of (CaCO₃) M

= (Atomic mass of Ca) x 1 + (Atomic mass of C) x 1 + (Atomic mass of O) x 3

.1. = (40 x l) + (12 x 1) +(16 x 3)

= 40+ 12+ 48

Molecular mass of (CaCO₃) M = 100

According to the formula Number of moles in the given CaCO₃ (n)

∴ Mass of CaCO₃ in grams (m) = 0.2 x 100

∴ Mass of CaCO₃ in grams (m) = 20 g

Mass of 0.2 mole of CaCO₃ is 20 g

2. State laws/Define the following:

1. Law of Conservation of Matter.

Ans.

In a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to chemical reaction.

2. Law of Constant Proportion.

Ans. The proportion by weight of the constituent elements in the various samples of a compound is fixed.

3. Molecular Mass:

Ans.

The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance.

4. Mole

Ans.

A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.

5. Valency

Ans.

The capacity of an element to combine is called its valency.

6. Electronic definition of Valency

Ans.

The number of electrons that an atom of an element gives away or takes up while forming an ionic bond is called valency of that element.

7. Radicals

Ans.

The positively or negatively charged ions that take part independently in chemical reactions are called radicals.

8. Atomic size determination

Ans.

The size of an atom is determined by its radius. The atomic radius of an isolated atom is the distance between the nucleus of an atom and its outermost orbit.

9. Atomic mass number

Ans.

The number of protons and neutrons in the atomic nucleus is called the atomic mass number.

10. Unified mass

Ans.

Unified mass is the standard unit of atomic mass that quantifies mass on an atomic or molecular scale. Its symbol is ‘u’.1 u = 1.66053904 x 10⁻²⁷ kg.

11. Molecular mass of a substance

Ans.

The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance. Like atomic mass, molecular mass is also expressed in the unit Dalton (u).

3. Answer the following questions:

1. What is variable valency?

Ans.

• Under different conditions, the atoms of some elements give away or take up a different number of electrons.

• In such cases, those elements exhibit more than one valency.

• This property of elements is called variable valency.

4. Complete the following table:

1. Write down the cations and anions obtained from the compounds in the following chart.

Ans.

5. Answer the following questions:

1. Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds:

(a) Calcium carbonate

Ans.

∴ Chemical formula of Calcium carbonate is CaCO₃

(b) Sodium bicarbonate

Ans.

∴ Chemical formula of Sodium bicarbonate is NaHCO₃

(c) Silver chloride

Ans.

∴ Chemical formula of Silver chloride is AgCl

(d) Calcium hydroxide Ans. Symbol Ca OH

Ans.

∴ Chemical formula of Calcium hydroxide is Ca(OH)₂

(e) Magnesium oxide

Ans.

∴ Chemical formula of Magnesium oxide is MgO

(f) Ammonium phosphate

Ans.

∴ Chemical formula of Ammonium phosphate is (NH₄)₃PO₄

(g) Cuprous bromide

Ans.

∴ Chemical formula of Cuprous bromide is CuBr.

(h) Copper sulphate

Ans.

∴ Chemical formula of Copper sulphate is CuSO₄

(i) Potassium nitrate

Ans.

∴ Chemical formula of Potassium nitrate is KNO₃

(j) Sodium dichromate

Ans.

∴ Chemical formula of Sodium dichromate is Na₂Cr₂O₇.

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