Question 13: Fill the following tables.

Step-by-step solution provided.

Question 14: Complete the following chart.

Step-by-step solution provided.

(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

Maharashtra State Board Class 9 (SSC) Science & Technology Solutions Chapter 4 Measurement of Matter - Digest Answers | Maharashtra Board Solutions

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

Class 9 Science Chapter 4 Measurement of Matter Textbook Questions and Answers

1. Give examples.

a. Positive radicals
Answer:
Na⁺– Sodium ion, K⁺– Potassium ion

b. Basic radicals
Answer:
Na⁺– Sodium ion, K⁺– Potassium ion, Ag⁺– Silver ion

c. Composite radicals
Answer:

d. Metals with variable valency
Answer:
(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]

e. Bivalent acidic radicals
Answer:
O^2-– Oxide, S^2-– Sulphide, – Carbonate

f. Trivalent basic radicals
Answer:
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
Answer:

3. Write the steps in deducing the chemical formulae of the following compounds.
Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide
Answer:
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)

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.
Na₂SO₄
(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.
{array}{cc}
K & NO_{3} \\
1 & 1
{array}
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)
Ca O
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
Step – 2 : To write the valency below the respective radical.
{array}{cc}
Na & Cr_{2} O_{7} \\
1 & 2
{array}
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.

Answer:

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.
Answer:
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.
Answer:

(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.

d. What is meant by Unified Atomic Mass.
Answer:

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

5. Write the names of the following compounds and deduce their molecular masses.
Na₂SO₄, K₂CO₃, CO₂, MgCl₂, NaOH, AlPO₄, NaHCO₃
Answer:

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.
Answer:
(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 g
mass of given sample = 7 g
mass of constituent Ca in sample’m’ = 5 g
mass 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 g
Mass of constituent Ca in sample ‘n’ = 1.0 g
Mass of constituent O in sample ‘n’ = 0.4 g
This 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.

Class 9 Science Chapter 4 Measurement Of Matter Notes Question 1.
32g oxygen
Answer:
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 O₂contains 6.022 x 10²³molecules that is 32 g of 02 contains 6.022 * 10²³molecules of O₂.
32g of oxygen contains 6.022 x 10²³molecules of oxygen.

Class 9 Science Chapter 4 Measurement Of Matter Answers Question 2.
90g water
Answer:
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) x 2 + (Atomic mass of O) x 1
∴ Molecular mass of (H₂O) M = 1 x 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 x 10²³molecules.
5 mol of H₂O contains 5 x 6.022 x 10²³molecules. = 30.11 x 10²³molecules, that is 90g of H₂O contains 30.11 x 10²³molecules of H20.
90g of water contains 30.11 x 10²³molecules of water.

4 Measurement Of Matter Exercise Question 3.
8.8g carbon dioxide
Answer:
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 x 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 CO₂contains 6.022 x 10²³molecules.
∴ 0.2 mol of CO₂contains 0.2 x 6.022 x 10²³molecules.
= 1.2044 x 10²³molecules,
that is 8.8g of CO₂contains 1.2044 x 10²³molecules of CO₂.
8.8g of CO₂contains 1.2044 x 10²³molecules of CO₂.

Class 9 Science Solutions Maharashtra Board Question 4.
7.1g chlorine
Answer:
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 x 2 = 71
According to the formula, Number of moles in the given Cl₂(n)

∴ 1 mol of Cl₂contains 6.022 x 10²³molecules.
∴ 0.1 mol of Cl₂contains 0.1 x 6.022 x 10²³molecules.
= 0.6022 x 10²³molecules,
that is 7.1g of Cl₂contains 0.6022 x 10²³molecules of Cl_2.
7.1g of Cl₂contains 0.6022 x 10²³molecules 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
Answer:
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

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Maharashtra State Board Class 9 Science Solutions Question 1.
What is the type of chemical bond in NaCl and MgCl₂?
Answer:

9th Class Science Chapter 4 Measurement Of Matter Question 2.
Determine the valencies of H, Cl, O and Na from the molecular formulae H₂, HC1, H₂O and NaCl.
Answer:
(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.

Measurement Of Matter Class 9 Exercise Answers Question 3.
How is an element indicated in Chemistry?
Answer:
In chemistry an element is indicated by its symbol.

Question 4 Maharashtra Board Solution

Write down the symbols of the elements you know.

Solution & Step-by-Step Answer:

Symbols of some elements are

Question 5 Maharashtra Board Solution

Write down the symbols for the following elements. Antimony, Iron, Gold, Silver, Mercury, Lead, Sodium

Solution & Step-by-Step Answer:

The symbols of given elements are as follows:

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)

Question 1 Maharashtra Board Solution

Molecular formula – NaCl

Solution & Step-by-Step Answer:

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

Question 2 Maharashtra Board Solution

Molecular formula – MgCl2

Solution & Step-by-Step Answer:

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?

Question 3 Maharashtra Board Solution

Molecular formula – KNO3

Solution & Step-by-Step Answer:

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

Question 4 Maharashtra Board Solution

Molecular formula – H2O2

Solution & Step-by-Step Answer:

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.

Question 5 Maharashtra Board Solution

Molecular formula – A1C13

Solution & Step-by-Step Answer:

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

Question 6 Maharashtra Board Solution

Molecular formula – Ca(OH)2

Solution & Step-by-Step Answer:

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

Question 7 Maharashtra Board Solution

Molecular formula – MgO

Solution & Step-by-Step Answer:

Molecular mass of MgO (M) = (Atomic mass of Mg) x 1 + (Atomic mass of 0)xl = (24 x 1) + (16 x 1) = 24 + 16 = 40 Molecular mass of MgO (M) = 40

Question 8 Maharashtra Board Solution

Molecular formula – H2S04

Solution & Step-by-Step Answer:

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

Question 9 Maharashtra Board Solution

Molecular formula – HN03

Solution & Step-by-Step Answer:

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 HNOs (M) = 63

Question 10 Maharashtra Board Solution

Molecular formula – NaOH

Solution & Step-by-Step Answer:

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

Question 11 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

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 1 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 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.

Question 12 Maharashtra Board Solution

How many molecules of H2S04 are there in a 49 g sample?

Solution & Step-by-Step Answer:

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 H2SO4 contains 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.

Question 13 Maharashtra Board Solution

Fill the following tables.

Solution & Step-by-Step Answer:

Question 14 Maharashtra Board Solution

Complete the following chart.

Solution & Step-by-Step Answer:

Question 15 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Question 16 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Simple radicals	Composite radicals
Ag⁺
Mg²⁺
Cl
Fe²⁺
Br^–
Na⁺
Cu⁺

Question 17 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Basic Radical	Acidic Radical
(i)Ag⁺	(i) Cl^–
(ii) Cu²⁺	(ii) I^–
(iii) Fe³⁺
(iv) Ca²⁺
Wnh;
(vi) K⁺
(vii) Na⁺

Give examples:

Question 1 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Problem-based questions

Answer the following questions:

Question 1 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

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

Question 2 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

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

Question 3 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

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

Answer the following:

Question 1 Maharashtra Board Solution

What is the Dalton’s atomic theory?

Solution & Step-by-Step Answer:

Dalton’s Atomic theory-

Question 2 Maharashtra Board Solution

How are compounds formed?

Solution & Step-by-Step Answer:

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

Question 3 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

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

Question 4 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Question 5 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

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

Q.3. Maharashtra Board Solution

2. Answer the following questions:

Question 1 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

block bg-white dark:bg-slate-900/80 border-l-4 border-emerald-500 rounded-r-xl p-5 my-3 shadow-xs border border-slate-200/60 dark:border-slate-800">

Solution & Step-by-Step Answer:

(1) Magnesium Chloride (MgCl2) Magnesium atom (Mg): Electronic configuration Magnesium ion Mg2+ (2,8). Chlorine atom (Cl). Electronic configuration Chloride ion Ch (2,8,8). ∴ Mg2+ + 2CT → MgCl2 (Magnesium Chloride)

(2) Calcium Oxide (CaO)
Calcium atom (Ca): Electronic configuration
Calcium ion Ca²⁺(2,8,8).
Oxygen atom (O). Electronic configuration (2,6)
Oxygen ion O^2-(2,8).
∴ Ca²⁺+ O^2-→ CaO

Question 2 Maharashtra Board Solution

Solution & Step-by-Step Answer:

(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.

Question 3 Maharashtra Board Solution

Solution & Step-by-Step Answer:

Question 4 Maharashtra Board Solution

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.

Solution & Step-by-Step Answer:

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.

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

(A) Select the correct option:

Question 1 Maharashtra Board Solution

The proportion by weight of hydrogen and oxygen in water is ………………………... (a) 8 : 1 (b) 2 : 1 (c) 1 : 2 (d) 1 : 8

Solution & Step-by-Step Answer:

(d) 1: 8

Question 2 Maharashtra Board Solution

The proportion by weight of carbon and oxygen in carbon dioxide is ………………………... (a) 8 : 3 (b) 3 : 8 (c) 3 : 2 (d) 2 : 3

Solution & Step-by-Step Answer:

(b) 3 : 8

Question 3 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

(a) protons; neutrons

Question 4 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

radius

Question 5 Maharashtra Board Solution

Atomic radius is expressed in ………………………... (a) milimetres (b) centimetres (c) nanometres (d) picometres

Solution & Step-by-Step Answer:

Question 6 Maharashtra Board Solution

The atomic size depends on the number of ………………………… in the atom. (a) protons (b) nucleus (c) neutrons (d) electron orbits

Solution & Step-by-Step Answer:

(d) electron orbits

Question 7 Maharashtra Board Solution

The mass of an atom is concentrated in its ………………………... (a) protons (b) nucleus (c) neutrons (d) electrons

Solution & Step-by-Step Answer:

(b) nucleus

Question 8 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Question 9 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

(a) Mole

Question 10 Maharashtra Board Solution

Avogadro’s number is denoted by the symbol ………………………... (a) NG (b) Nv (c) NA (d) ND

Solution & Step-by-Step Answer:

Question 11 Maharashtra Board Solution

A mole of any substance stands for ………………………… molecules. (a) 60.22 x 1023 (b) 6.022 x 1022 (c) 6.022 x 1023 (d) 60.22 x 1022

Solution & Step-by-Step Answer:

(a) 60.22 x 1023

Question 12 Maharashtra Board Solution

The capacity of an element to combine is called its ………………………... (a) valency (b) electronic configuration (c) atomic number (d) volence electrons

Solution & Step-by-Step Answer:

(a) valency

Question 13 Maharashtra Board Solution

Electronic configuration of sodium atom is ………………………... (a) (2, 8, 3) (b) (2, 8, 7) (c) (2, 8, 2) (d) (2, 8,1)

Solution & Step-by-Step Answer:

(d) (2,8,1)

Question 14 Maharashtra Board Solution

Electronic configuration of chlorine atom is ………………………... (a) (2, 8, 3) (b) (2, 8, 7) (c) (2, 8, 2) (d) (2, 8, 1)

Solution & Step-by-Step Answer:

(b) (2, 8, 7)

Question 15 Maharashtra Board Solution

Positively charged ions are called as ………………………... (a) cations (b) anions (c) nucleous (d) protons

Solution & Step-by-Step Answer:

(a) cations

Question 16 Maharashtra Board Solution

Negatively charged ions are called as ………………………... (a) cations (b) anions (c) nucleus (d) electrons

Solution & Step-by-Step Answer:

(b) anions

Question 17 Maharashtra Board Solution

Iron (Fe) exhibits the variable valencies ………………………... (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 (d) 2 and 4

Solution & Step-by-Step Answer:

(b) 2 and 3

Question 18 Maharashtra Board Solution

Cationic radicals are called as ………………………… radicals. (a) basic (b) acidic (c) neutral (d) mixed

Solution & Step-by-Step Answer:

(a) basic

Question 19 Maharashtra Board Solution

Anionic radicals are called as ………………………… radicals. (a) basic (b) acidic (c) neutral (d) mixed

Solution & Step-by-Step Answer:

(b) acidic

Question 20 Maharashtra Board Solution

The unit Dalton is used to express ………………………… (a) atomic mass (b) atomic radius (c) atomic number (d) mass number

Solution & Step-by-Step Answer:

(a) atomic mass

Question 21 Maharashtra Board Solution

The valency of element with electronic configuration ………………………… is 2. (a) (2,5) (b) (2, 4) (c) (2, 6) (d) (2, 7)

Solution & Step-by-Step Answer:

Question 22 Maharashtra Board Solution

The symbol of Avogadro’s number is ………………………... (a) ND (b) N0 (c) NB (d) NA

Solution & Step-by-Step Answer:

(d) NA

Question 23 Maharashtra Board Solution

………………………. is bicarbonate radical.

Solution & Step-by-Step Answer:

Question 24 Maharashtra Board Solution

Molecular formula of sodium sulphate is ………………………... (a) Na(SO4)2 (b) Na2SO4 (c)Na2(SO4)2 (d)NaSO4

Solution & Step-by-Step Answer:

(b) Na2SO4

Question 25 Maharashtra Board Solution

………………………… is a composite radical. (a) Fe3+ (b) Ca2+ (c) NH4+ (d) S2-

Solution & Step-by-Step Answer:

Question 26 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

(a) 6.022 x 1023

Question 27 Maharashtra Board Solution

The mass of an atom is concentrated in its …………………………. (a) nucleus (b) electrons (c) extranuclear part (d) protons

Solution & Step-by-Step Answer:

(a) nucleus

Question 28 Maharashtra Board Solution

………………………… g of water make 1 mole of water. (a) 32 (b) 33 (c) 16 (d) 18

Solution & Step-by-Step Answer:

(d) 18

Complete the analogy:

(1) Electron : extra nuclear part:: Neutron ………………………….
(2) Sodium: (2, 8, 1):: Chlorine:: ………………………….
(3) K : basic radical :: Br^–: ………………………….
(4) Cut: simple radical:: NH₄⁺: ………………………….
(5) Sodium sulphate: Na₂SO₄:: Potassium Sulphate: ………………………….
(6) Mercurous: Hg⁺:: Mercuric : ………………………….
(7) Positively charged ion : cation:: Negatively charged ion : ………………………….
(8) 12: 1 dozen :: 144 : ………………………….
(9) Hydrogen : :: Copper : ………………………….
(10) Law of constant proportions : J. L. Proust::
Law of conservation of matter : ………………………….
Answer:
(1) nucleus
(2) (2, 8, 7)
(3) acidic radical
(4) composite radical
(5) K₂SO₄
(6) Hg²⁺
(7) anion
(8) 1 gross
(9) ©
(10) Antoine Lavoisier.

Match the columns:

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

Answer:
(1-b),
(2- c),
(3 – a)

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

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b)

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

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Column ‘A’	Column B’
Radicals	Names
(1)	(a) Carbonate
(2)	(b) Chromate
(3)	(c) Dichromate
(4)	(d) Chlorate

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Answer the following in one sentence:

Question 1 Maharashtra Board Solution

What are valence electrons?

Solution & Step-by-Step Answer:

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

Question 2 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

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

Question 3 Maharashtra Board Solution

What are basic radicals? Give examples.

Solution & Step-by-Step Answer:

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

Question 4 Maharashtra Board Solution

What are acidic radicals? Give examples.

Solution & Step-by-Step Answer:

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

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

(1) Molecular state of oxygen is monoatomic.
(2) The capacity of an element to combine is called its valency.
(3) Anionic radicals are basic radicals.
(4) The magnitude of charge on any radical is its atomic number.
(5) In a chemical reaction, mass of original matter and mass of matter newly formed as a result of chemical change are equal.
(6) The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 5.
(7) Relative mass of hydrogen is 1.
(8) The number of molecules in a given quantity of a substance is determined by its atomic mass.
(9) Avogadro’s number is 6.022 x 10²³
(10) Valency of sodium is 2.
Answer:
(1) False. Molecular state of oxygen is diatomic:
(2) True
(3) False. Anionic radicals are acidic radicals.
(4) False. Magnitude of charge on any radical is its valency.
(5) True
(6) False. The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 8.
(7) True
(8) False. The number of molecules in a given quantity of a substance is determined by its molecular mass.
(9) True
(10) False. Valency of sodium is 1.

Name the following:

Question 1 Maharashtra Board Solution

Scientist who gave Law of Conservation of Matter.

Solution & Step-by-Step Answer:

Antoine Lavoisier

Question 2 Maharashtra Board Solution

Scientist who gave Law of Constant Proportion.

Solution & Step-by-Step Answer:

J. L. Proust

Question 3 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Nucleons

Question 4 Maharashtra Board Solution

Unit used to express atomic radius.

Solution & Step-by-Step Answer:

Nanometre

Question 5 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Atomic mass number

Question 6 Maharashtra Board Solution

Name the unit of atomic mass.

Solution & Step-by-Step Answer:

Dalton (u)

Question 7 Maharashtra Board Solution

Write molecular formula of two ionic compounds containing chlorine.

Solution & Step-by-Step Answer:

NaCl, MgCl2

Question 8 Maharashtra Board Solution

Give two monoatomic radicals.

Solution & Step-by-Step Answer:

Na+, Cl–

Question 9 Maharashtra Board Solution

Give two examples of simple radicals.

Solution & Step-by-Step Answer:

Ag+, O2-

Give scientific reasons:

Question 1 Maharashtra Board Solution

An atom is electrically neutral though it contains charged particles.

Solution & Step-by-Step Answer:

Question 2 Maharashtra Board Solution

Neon is chemically inert element.

Solution & Step-by-Step Answer:

Question 3 Maharashtra Board Solution

The valency of sodium (Na) is one.

Solution & Step-by-Step Answer:

Question 4 Maharashtra Board Solution

The valency of chlorine (Cl) is one.

Solution & Step-by-Step Answer:

Question 5 Maharashtra Board Solution

The valency of Magnesium (Mg) is two.

Solution & Step-by-Step Answer:

Question 6 Maharashtra Board Solution

Valency is always a whole number.

Solution & Step-by-Step Answer:

Question 7 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Question 8 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Write the names of the following compounds and deduce their molecular masses:

Atomic masses : H(1), 0(16), N(14), C(12), K(39), S(32), Ca(40), Na(23), C1(35.5), Mg(24), A1(27), P(31)

Question 1 Maharashtra Board Solution

Molecular mass of K2CO3

Solution & Step-by-Step Answer:

Question 2 Maharashtra Board Solution

Molecular mass of CO2

Solution & Step-by-Step Answer:

Question 3 Maharashtra Board Solution

Molecular mass of MgCl2

Solution & Step-by-Step Answer:

Question 4 Maharashtra Board Solution

Molecular mass of NaOH

Solution & Step-by-Step Answer:

Question 5 Maharashtra Board Solution

Molecular mass of AIPO4

Solution & Step-by-Step Answer:

Question 6 Maharashtra Board Solution

Molecular mass of NaHCO3

Solution & Step-by-Step Answer:

Numerical.

Question 1 Maharashtra Board Solution

Magnesium Oxide:

Solution & Step-by-Step Answer:

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

Question 2 Maharashtra Board Solution

Calcium Carbonate:

Solution & Step-by-Step Answer:

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 = (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

State laws/Define the following:

Question 1 Maharashtra Board Solution

Law of Conservation of Matter.

Solution & Step-by-Step Answer:

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

Question 2 Maharashtra Board Solution

Law of Constant Proportion.

Solution & Step-by-Step Answer:

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

Question 3 Maharashtra Board Solution

Molecular Mass:

Solution & Step-by-Step Answer:

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

Question 4 Maharashtra Board Solution

Mole

Solution & Step-by-Step Answer:

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.

Question 5 Maharashtra Board Solution

Valency

Solution & Step-by-Step Answer:

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

Question 6 Maharashtra Board Solution

Electronic definition of Valency

Solution & Step-by-Step Answer:

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.

Question 7 Maharashtra Board Solution

Radicals

Solution & Step-by-Step Answer:

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

Question 8 Maharashtra Board Solution

Atomic size determination

Solution & Step-by-Step Answer:

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.

Question 9 Maharashtra Board Solution

Atomic mass number

Solution & Step-by-Step Answer:

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

Question 10 Maharashtra Board Solution

Unified mass

Solution & Step-by-Step Answer:

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.

Question 11 Maharashtra Board Solution

Molecular mass of a substance

Solution & Step-by-Step Answer:

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).

Answer the following questions:

Question 1 Maharashtra Board Solution

What is variable valency?

Solution & Step-by-Step Answer:

Complete the following table:

Question 1 Maharashtra Board Solution

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

Solution & Step-by-Step Answer: