Maharashtra Board Class 9 Science Chapter 4 Measurement of Matter Solutions

Std 9 Science Chapter 4 Measurement Of Matter Question Answer Maharashtra Board

Class 9 Science Chapter 4 Measurement Of Matter Question Answer Maharashtra Board

Question 1. Give examples.
(a) Positive radicals
Answer: Na+ - Sodium ion, K+ - Potassium ion
In simple words: Positive radicals are ions that carry a positive charge, formed when an atom loses electrons. Sodium and potassium ions are common examples.

🎯 Exam Tip: Remember common positive radicals like alkali metal ions and alkaline earth metal ions as they frequently appear in questions on chemical formulas.

 

Question 1. Give examples.
(b) Basic radicals
Answer: Na+ - Sodium ion, K+ - Potassium ion, Ag+ - Silver ion
In simple words: Basic radicals are positively charged ions, often metals, that combine with acidic radicals to form salts. They are called "basic" because they behave like the base part of a compound.

🎯 Exam Tip: Basic radicals are typically derived from metals. Identifying them correctly is crucial for writing accurate chemical formulas.

 

Question 1. Give examples.
(c) Composite radicals
Answer: SO₄^2-, NH₄⁺
In simple words: Composite radicals are groups of two or more atoms that act as a single unit and carry an electrical charge. They are also known as polyatomic ions.

🎯 Exam Tip: Familiarize yourself with common composite radicals, their names, and their charges, as they are fundamental for understanding chemical bonding.

 

Question 1. Give examples.
(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]

(c) Mercury (Hydragyrum)
(i) Hg⁺ - Mercurous [Mercury -1]
(ii) Hg²⁺ - Mercuric [Mercury - II]
In simple words: Some metals can exhibit more than one valency, meaning they can form different types of ions depending on the chemical reaction. This property is called variable valency.

🎯 Exam Tip: When dealing with metals exhibiting variable valency, always specify the valency in Roman numerals (e.g., Iron (II), Iron (III)) to avoid confusion in naming and writing formulas.

 

Question 1. Give examples.
(e) Bivalent acidic radicals
Answer: O^2- - Oxide, S^2- - Sulphide, CO₃^2- - Carbonate
In simple words: Bivalent acidic radicals are negatively charged ions formed from non-metals that carry a charge of -2. They accept two electrons to achieve a stable electronic configuration.

🎯 Exam Tip: Acidic radicals typically originate from non-metals and gain electrons. Knowing their valency (e.g., bivalent means valency 2) is key for chemical formula derivation.

 

Question 1. Give examples.
(f) Trivalent basic radicals
Answer: Al³⁺ - Aluminium, Cr³⁺ - Chromium, Fe³⁺ - Ferric.
In simple words: Trivalent basic radicals are positively charged ions, typically metals, that carry a charge of +3. They donate three electrons to form stable ions.

🎯 Exam Tip: Basic radicals usually form cations. Understanding their valency (e.g., trivalent means valency 3) is important for predicting chemical compounds.

 

Question 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:

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

In simple words: Elements can form ions (radicals) by gaining or losing electrons, acquiring a positive or negative charge. The charge on the radical indicates the number of electrons gained or lost.

🎯 Exam Tip: Pay close attention to elements that exhibit variable valencies (like Mercury and Copper) and remember the corresponding names for each charge state (e.g., Mercurous/Mercuric).

 

Question 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)
Na SO₄
Step - 2 : To write the valency below the respective radical.
Na SO₄
1 2
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सोडियम (Na) और सल्फेट (SO₄) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 1 और 2 हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं जहाँ प्रत्येक रेडिकल दूसरे की संयोजकता प्राप्त करता है।
Step - 3: To cross-multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Na और SO₄ रेडिकल्स की संयोजकता के क्रॉस-गुणा को दर्शाता है। Na रेडिकल को SO₄ की संयोजकता 2 मिलती है, जबकि SO₄ रेडिकल को Na की संयोजकता 1 मिलती है।
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.
K NO₃
1 1
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पोटेशियम (K) और नाइट्रेट (NO₃) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 1 और 1 हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross-multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र K और NO₃ रेडिकल्स की संयोजकता के क्रॉस-गुणा को दर्शाता है। K रेडिकल को NO₃ की संयोजकता 1 मिलती है, और NO₃ रेडिकल को K की संयोजकता 1 मिलती है।
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.
Fe PO₄
3 3
Dividing by
1 1 common factor
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फेरिक (Fe) और फॉस्फेट (PO₄) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 3 और 3 हैं। दोनों की संयोजकता को सामान्य गुणनखंड से विभाजित करने के बाद, वे 1 और 1 हो जाते हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross-multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Fe और PO₄ रेडिकल्स की संशोधित संयोजकता के क्रॉस-गुणा को दर्शाता है। Fe रेडिकल को PO₄ की संशोधित संयोजकता 1 मिलती है, और PO₄ रेडिकल को Fe की संशोधित संयोजकता 1 मिलती है।
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.
Ca O
2 2
Dividing by common factor
1 1
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कैल्शियम (Ca) और ऑक्सीजन (O) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 2 और 2 हैं। दोनों की संयोजकता को सामान्य गुणनखंड से विभाजित करने के बाद, वे 1 और 1 हो जाते हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross-multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Ca और O रेडिकल्स की संशोधित संयोजकता के क्रॉस-गुणा को दर्शाता है। Ca रेडिकल को O की संशोधित संयोजकता 1 मिलती है, और O रेडिकल को Ca की संशोधित संयोजकता 1 मिलती है।
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.
Al OH
3 1
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एल्यूमीनियम (Al) और हाइड्रॉक्साइड (OH) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 3 और 1 हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross-multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Al और OH रेडिकल्स की संयोजकता के क्रॉस-गुणा को दर्शाता है। Al रेडिकल को OH की संयोजकता 1 मिलती है, और OH रेडिकल को Al की संयोजकता 3 मिलती है।
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.
Ca CO₃
2 2 ...... Dividing by
1 1 common factor
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कैल्शियम (Ca) और कार्बोनेट (CO₃) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 2 और 2 हैं। दोनों की संयोजकता को सामान्य गुणनखंड से विभाजित करने के बाद, वे 1 और 1 हो जाते हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross-multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Ca और CO₃ रेडिकल्स की संशोधित संयोजकता के क्रॉस-गुणा को दर्शाता है। Ca रेडिकल को CO₃ की संशोधित संयोजकता 1 मिलती है, और CO₃ रेडिकल को Ca की संशोधित संयोजकता 1 मिलती है।
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.
Na Cr₂O₇
1 2
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सोडियम (Na) और डाइक्रोमेट (Cr₂O₇) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 1 और 2 हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross-multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Na और Cr₂O₇ रेडिकल्स की संयोजकता के क्रॉस-गुणा को दर्शाता है। Na रेडिकल को Cr₂O₇ की संयोजकता 2 मिलती है, और Cr₂O₇ रेडिकल को Na की संयोजकता 1 मिलती है।
Step - 4 : To write down the chemical formula of the compound.
Na₂Cr₂O₇
(Sodium dichromate)
In simple words: Deducing chemical formulae involves writing the symbols of radicals, indicating their valencies, and then cross-multiplying the valencies to determine the subscripts in the formula, simplifying if necessary.

🎯 Exam Tip: Mastering the cross-multiplication method and knowing the valencies of common radicals are fundamental skills for writing correct chemical formulae and will be frequently tested.

 

Question 4. Write answers to the following questions and explain your answers.
(a) Explain how the element sodium is monovalent.
Answer:
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 1e- from its outermost orbit to form Na⁺ ion with stable electronic configuration (2, 8).
3. As sodium atom gives away 1e- and a cation of sodium is formed, hence the valency of sodium is 1 and therefore, the element sodium is monovalent.
In simple words: Sodium is monovalent because its atomic configuration (2,8,1) shows it has one electron in its outermost shell, which it readily donates to achieve a stable octet, resulting in a +1 charge.

🎯 Exam Tip: Valency is directly related to an element's electron configuration, specifically the number of valence electrons. Elements tend to achieve a stable octet (8 valence electrons).

 

Question 4. Write answers to the following questions and explain your answers.
(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.
M SO₄
2 2
[Dividing valency
1 1 by common factor]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र धातु M (द्विसंयोजी) और सल्फेट (SO₄) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 2 और 2 हैं। दोनों की संयोजकता को सामान्य गुणनखंड से विभाजित करने के बाद, वे 1 और 1 हो जाते हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र धातु M और SO₄ रेडिकल्स की संशोधित संयोजकता के क्रॉस-गुणा को दर्शाता है। M रेडिकल को SO₄ की संशोधित संयोजकता 1 मिलती है, और SO₄ रेडिकल को M की संशोधित संयोजकता 1 मिलती है।
Step - 4: To write down the chemical formula of the compound.
MSO₄

(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.
M PO₄
2 3
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र धातु M (द्विसंयोजी) और फॉस्फेट (PO₄) रेडिकल को दर्शाता है जिनकी संयोजकताएँ क्रमशः 2 और 3 हैं। तीर क्रॉस-गुणा विधि को दर्शाते हैं।
Step - 3: To cross multiply as shown by arrows the number of radicals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र धातु M और PO₄ रेडिकल्स की संयोजकता के क्रॉस-गुणा को दर्शाता है। M रेडिकल को PO₄ की संयोजकता 3 मिलती है, और PO₄ रेडिकल को M की संयोजकता 2 मिलती है।
Step - 4: To write down the chemical formula of the compound.
M₃(PO₄)₂
In simple words: To find the chemical formula of a compound with a bivalent metal 'M', you cross-multiply its valency (2) with the valencies of sulphate (2) and phosphate (3), simplifying the ratio for sulphate, and using parentheses for polyatomic ions like phosphate.

🎯 Exam Tip: Always remember to simplify the valency ratio if possible (as with MSO₄) and use parentheses around polyatomic ions when their subscript is greater than 1 (as with M₃(PO₄)₂).

 

Question 4. Write answers to the following questions and explain your answers.
(c) Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.
Answer:
• 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
In simple words: A reference atom is needed to measure the extremely small masses of atoms accurately by comparison, leading to the concept of relative atomic mass. Hydrogen and later Carbon-12 served as such reference points.

🎯 Exam Tip: Understanding the evolution from Hydrogen to Carbon-12 as a reference for atomic mass highlights the importance of choosing a stable and universally agreed-upon standard in chemistry.

 

Question 4. Write answers to the following questions and explain your answers.
(d) What is meant by Unified Atomic Mass.
Answer:
• 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'. 1u = 1.66053904 x 10^-27 kg.
In simple words: Unified Atomic Mass (u) is a standard unit for expressing atomic and molecular masses, replacing earlier relative mass concepts. It's defined as one-twelfth the mass of a carbon-12 atom.

🎯 Exam Tip: The unified atomic mass unit (u) is a crucial concept for quantitative chemistry, allowing for consistent measurement of masses at the atomic and molecular level.

 

Question 4. Write answers to the following questions and explain your answers.
(e) Explain with examples what is meant by a 'mole' of a substance.
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.
• 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.
In simple words: A mole is a unit that represents a specific quantity of a substance, where its mass in grams is numerically equivalent to its molecular mass in Daltons. It links the microscopic world of atoms and molecules to the macroscopic world we can measure.

🎯 Exam Tip: The concept of the mole is central to stoichiometry. Remember that 1 mole of any substance contains Avogadro's number (6.022 x 10²³) of particles.

 

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

Name of compound	Molecule	Constituent elements	Atomic mass (u)	Number of atoms in molecule	Atomic mass x number of atoms	Mass of the constituents u
Sodium sulphate	Na2SO4	Sodium	23	2	23 x 2	46
		Sulphur	32	1	32 x 1	32
		Oxygen	16	4	16 x 4	64
Molecular mass = Sum of constituent atomic masses						Molecular Mass 142
Molecular mass of (Na2SO4) = (Atomic mass of Na) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4

In simple words: To deduce molecular mass, sum the atomic masses of all atoms present in a molecule. For Na₂SO₄, this means adding the mass of two Sodium atoms, one Sulfur atom, and four Oxygen atoms.

🎯 Exam Tip: Always clearly list the atomic masses of each element and their count in the molecule to avoid calculation errors when determining molecular mass.

 

Question 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 = 5g
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 contains 5 g of calcium (Ca) and 2 g of oxygen (O); and 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'.
In simple words: This experiment demonstrates the Law of Constant Proportion, which states that a chemical compound always contains the same elements in the same proportion by mass, regardless of its source or method of preparation. Both samples of calcium oxide (CaO) show a Ca:O ratio of 5:2.

🎯 Exam Tip: The Law of Constant Proportion (also known as the Law of Definite Proportions) is a foundational principle in chemistry, ensuring that compounds have consistent elemental compositions. Clearly state the law and show consistent ratios in calculations.

 

Question 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.

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)
\[ n = \frac{\text{Mass of O}_{\text{2}}\text{ in grams (m)}}{\text{Molecular mass of O}_{\text{2}}\text{ (M)}} \]
\( n = \frac{32}{32} = 1 \) mol
1 mol of O₂ contains 6.022 x 10²³ molecules that is 32 g of O₂ contains 6.022 x 10²³ molecules of O₂.
32g of oxygen contains 6.022 x 10²³ molecules of oxygen.
In simple words: To find the number of molecules, first calculate the number of moles by dividing the given mass by the molecular mass. Then, multiply the number of moles by Avogadro's number (6.022 x 10²³) to get the total number of molecules.

🎯 Exam Tip: Always remember Avogadro's number (6.022 x 10²³) as it's key to converting moles to the number of particles (atoms or molecules).

 

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)
\[ n = \frac{\text{Mass of H}_{\text{2}}\text{O in grams (m)}}{\text{Molecular mass of H}_{\text{2}}\text{O (M)}} \]
\( n = \frac{90}{18} = 5 \) mol
1 mol of H₂O contains 6.022 x 10²³ molecules.
5 mol of H₂O contains 5 x 6.022 x 10²³ molecules. = 30.11 x 10²³ molecules, that is 90g of H₂O contains 30.11 x 10²³ molecules of H₂O.
90g of water contains 30.11 x 10²³ molecules of water.
In simple words: Calculate the moles of water by dividing its given mass (90g) by its molecular mass (18u). Then, multiply this molar quantity by Avogadro's number to determine the total number of water molecules.

🎯 Exam Tip: Always calculate the molecular mass correctly for the given compound before applying the mole concept formula to find the number of molecules.

 

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)
\[ n = \frac{\text{Mass of CO}_{\text{2}}\text{ in grams (m)}}{\text{Molecular mass of CO}_{\text{2}}\text{ (M)}} \]
\( n = \frac{8.8}{44} = 0.2 \) mol
1 mol of CO₂ contains 6.022 x 10²³ molecules.
0.2 mol of CO₂ contains 0.2 x 6.022 x 10²³ molecules.
= 1.2044 x 10²³ molecules,
that is 8.8g of CO₂ contains 1.2044 x 10²³ molecules of CO₂.
8.8g of CO₂ contains 1.2044 x 10²³ molecules of chlorine.
In simple words: To find the number of carbon dioxide molecules, divide the given mass (8.8g) by its molecular mass (44u) to find the moles, then multiply by Avogadro's number.

🎯 Exam Tip: Be careful with calculations involving decimal numbers and scientific notation. Double-check your molecular mass and final number of molecules.

 

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)
\[ n = \frac{\text{Mass of Cl}_{\text{2}}\text{ in grams (m)}}{\text{Molecular mass of Cl}_{\text{2}}\text{ (M)}} \]
\( n = \frac{7.1}{71} = 0.1 \) mol
1 mol of Cl₂ contains 6.022 x 10²³ molecules.
0.1 mol of Cl₂ contains 0.1 x 6.022 x 10²³ molecules.
= 0.6022 x 10²³ molecules,
that is 7.1g of Cl₂ contains 0.6022 x 10²³ molecules of Cl₂.
7.1g of Cl₂ contains 0.6022 x 10²³ molecules of chlorine.
In simple words: To find the number of chlorine molecules, divide the given mass (7.1g) by its molecular mass (71u) to calculate the moles, and then multiply by Avogadro's number.

🎯 Exam Tip: Remember that chlorine exists as a diatomic molecule (Cl₂) for molecular mass calculations, not as a single atom (Cl).

 

Question 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,
\[ 0.2 = \frac{\text{Mass of NaCl in grams (m)}}{\text{Molecular mass of NaCl (M)}} \]
\[ 0.2 = \frac{\text{Mass of NaCl in grams (m)}}{58.5} \] 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
In simple words: To find the mass of a substance given its moles, rearrange the mole formula: mass equals moles multiplied by the molecular mass.

🎯 Exam Tip: Ensure you accurately calculate the molecular mass for each compound. This is a common reversal of mole calculations, so practice both directions.

 

Class 9 Science Chapter 3 Current Electricity Intext Questions And Answers

Question 1. What is the type of chemical bond in NaCl and MgCl₂?
Answer:
• The type of chemical bond in NaCl and MgCl₂ is ionic bond.
In simple words: NaCl and MgCl₂ form ionic bonds because they involve the transfer of electrons from a metal (Sodium or Magnesium) to a non-metal (Chlorine), creating charged ions that are attracted to each other.

🎯 Exam Tip: Ionic bonds typically form between a metal and a non-metal due to a large electronegativity difference, leading to complete electron transfer.

 

Question 2. Determine the valencies of H, Cl, O and Na from the molecular formulae H₂, HCl, H₂O and NaCl.
Answer:
(i) In the molecular formula HCl
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हाइड्रोजन (H) और क्लोरीन (Cl) तत्वों के बीच क्रॉस-गुणा को दर्शाता है, जहाँ प्रत्येक की संयोजकता 1 है, जिससे हाइड्रोजन और क्लोरीन की संयोजकता 1 निर्धारित होती है।
Valency 1 1
The valency of H is 1 and Cl is 1.

(ii) In the molecular formula H₂O
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हाइड्रोजन (H) और ऑक्सीजन (O) तत्वों के बीच क्रॉस-गुणा को दर्शाता है, जहाँ H की संयोजकता 1 और O की संयोजकता 2 है, जिससे ऑक्सीजन की संयोजकता 2 निर्धारित होती है।
Valency 1 2
The valency of H is 1 and O is 2.

(iii) In the molecular formula NaCl
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सोडियम (Na) और क्लोरीन (Cl) तत्वों के बीच क्रॉस-गुणा को दर्शाता है, जहाँ प्रत्येक की संयोजकता 1 है, जिससे सोडियम और क्लोरीन की संयोजकता 1 निर्धारित होती है।
Valency 1 1
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 = 1.
In simple words: Valency can be determined from molecular formulae by observing the combining capacity of each element. For H₂, HCl, H₂O, and NaCl, the valencies of H, Cl, O, and Na are deduced as 1, 1, 2, and 1 respectively.

🎯 Exam Tip: Valency represents an element's combining capacity. In simple binary compounds, it can often be found by swapping the subscripts of the elements after they combine.

 

Question 3. How is an element indicated in Chemistry?
Answer:
In chemistry an element is indicated by its symbol.
In simple words: Elements are represented in chemistry by unique one- or two-letter symbols, derived from their English or Latin names, which serve as a universal shorthand.

🎯 Exam Tip: Knowing the symbols for common elements is fundamental to understanding chemical formulas and reactions.

 

Question 4. Write down the symbols of the elements you know.
Answer:
Symbols of some elements are
• Hydrogen - H
• Helium - He
• Boron - B
• Carbon - C
• Aluminium - Al
In simple words: Elements are represented by specific symbols, usually the first letter or first two letters of their name, or their Latin name.

🎯 Exam Tip: Practice memorizing the symbols of the first 20 elements, as they are frequently used in basic chemistry.

 

Question 5. Write down the symbols for the following elements. Antimony, Iron, Gold, Silver, Mercury, Lead, Sodium
Answer:
The symbols of given elements are as follows:
• Antimony - Sb
• Iron - Fe
• Gold - Au
• Silver - Ag
• Mercury - Hg
• Lead - Pb
• Sodium - Na
In simple words: Many elements have symbols derived from their Latin names, such as Antimony (Stibium), Iron (Ferrum), Gold (Aurum), Silver (Argentum), Mercury (Hydrargyrum), Lead (Plumbum), and Sodium (Natrium).

🎯 Exam Tip: Elements with symbols derived from Latin names are common. Make a dedicated effort to learn these, as they are often encountered.

 

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(1), O(16), N(14), C(12), K(39), S(32) Ca(40), Na(23), Cl(35.5), Mg(24), Al(27)

 

Question 1. Molecular formula - NaCl
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
In simple words: The molecular mass of NaCl is calculated by summing the atomic mass of one Sodium atom and one Chlorine atom.

🎯 Exam Tip: Remember to use the correct atomic masses for each element when calculating molecular mass, and sum them according to the formula.

 

Question 2. Molecular formula - MgCl2
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?
In simple words: The molecular mass of MgCl2 is found by adding the atomic mass of one Magnesium atom to twice the atomic mass of a Chlorine atom.

🎯 Exam Tip: Pay attention to subscripts in chemical formulas, as they indicate the number of atoms of each element to be included in the molecular mass calculation.

 

Question 3. Molecular formula - KNO3
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
In simple words: To get the molecular mass of KNO3, sum the atomic masses of one Potassium, one Nitrogen, and three Oxygen atoms.

🎯 Exam Tip: Accuracy in atomic mass values and careful multiplication by subscripts are crucial for correct molecular mass determination.

 

Question 4. Molecular formula - H2O2
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.
In simple words: The molecular mass of H2O2 is calculated by adding the atomic masses of two Hydrogen atoms and two Oxygen atoms.

🎯 Exam Tip: Ensure all elements in the formula are accounted for and their respective atomic masses are multiplied by their subscripts before summing.

 

Question 5. Molecular formula - AlCl3
Answer:
Molecular mass of AlCl3 (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 AlCl3 (M) = 133.5
In simple words: The molecular mass of AlCl3 is the sum of one Aluminum atom's atomic mass and three Chlorine atoms' atomic masses.

🎯 Exam Tip: Be meticulous with calculations, especially when dealing with decimal atomic masses like chlorine's 35.5.

 

Question 6. Molecular formula - Ca(OH)2
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
In simple words: To find the molecular mass of Ca(OH)2, sum the atomic mass of one Calcium atom with twice the combined atomic mass of an Oxygen and a Hydrogen atom.

🎯 Exam Tip: For polyatomic ions in parentheses with a subscript, multiply the entire group's atomic mass sum by that subscript.

 

Question 7. Molecular formula - MgO
Answer:
Molecular mass of MgO (M)
= (Atomic mass of Mg) x 1 + (Atomic mass of O)x1
= (24 x 1) + (16 x 1)
= 24 + 16
= 40
Molecular mass of MgO (M) = 40
In simple words: The molecular mass of MgO is the sum of the atomic masses of one Magnesium atom and one Oxygen atom.

🎯 Exam Tip: Simple binary compounds require summing the atomic masses of each constituent atom present in the formula.

 

Question 8. Molecular formula - H2SO4
Answer:
Molecular mass of H2SO4 (M)
= (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4
= (1 x 2) + (32x1) + (16x4)
= 2 + 32 + 64
= 98
Molecular mass of H2SO4 (M) = 98
In simple words: To calculate the molecular mass of H2SO4, add the atomic masses of two Hydrogen, one Sulfur, and four Oxygen atoms.

🎯 Exam Tip: Always double-check your arithmetic, especially when multiple atoms of different types are involved.

 

Question 9. Molecular formula - HNO3
Answer:
Molecular mass of HNO3 (M)
= (Atomic mass of H) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3
= (1x1)+ (14x1)+ (16x3)
= 1 + 14 + 48
= 63
Molecular mass of HNO3 (M) = 63
In simple words: The molecular mass of HNO3 is obtained by summing the atomic masses of one Hydrogen, one Nitrogen, and three Oxygen atoms.

🎯 Exam Tip: Remember to clearly identify the number of each atom from the chemical formula before multiplying by their atomic masses.

 

Question 10. Molecular formula - NaOH
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) + (1 x 1)
= 23 + 16 + 1
= 40
Molecular mass of NaOH (M) = 40
In simple words: The molecular mass of NaOH is found by adding the atomic masses of one Sodium, one Oxygen, and one Hydrogen atom.

🎯 Exam Tip: Practice with various chemical formulas to become proficient in quickly determining molecular masses.

 

Question 11. How many molecules of water are there in 36 g water?
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)
\[ = \frac{\text{Mass of H₂O in grams (m)}}{\text{Molecular mass of H₂O (M)}} \]
\[ = \frac{36}{18} = 2 \]
∴ n = 2 mol
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.
In simple words: First, calculate water's molecular mass, then find the number of moles in 36g, and finally multiply by Avogadro's number to get the total molecules.

🎯 Exam Tip: This question tests your understanding of molecular mass, mole concept, and Avogadro's number, all crucial for stoichiometry problems.

 

Question 12. How many molecules of H2SO4 are there in a 49 g sample?
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)
\[ = \frac{\text{Mass of H₂SO₄ in grams (m)}}{\text{Molecular mass of H₂SO₄ (M)}} \]
\[ = \frac{49}{98} = \frac{1}{2} = 0.5 \]
∴ n = 0.5 mol
∴ 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.
In simple words: Determine the molecular mass of H2SO4, then calculate the moles in 49g, and multiply by Avogadro's number to find the total molecules.

🎯 Exam Tip: Precision in calculating molecular mass is critical as any error will propagate through the mole and molecule count calculations.

 

Question 13. Fill the following tables.
Answer:

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

In simple words: This table lists common elements along with their standard atomic masses.

🎯 Exam Tip: Memorizing common atomic masses can save time during calculations, but a periodic table is always a reliable reference.

 

Question 14. Complete the following chart.
Answer:

Element	Atomic number	Electronic Configuration	Valence Electrons	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

In simple words: This chart provides a summary of atomic number, electronic configuration, valence electrons, and valency for various elements.

🎯 Exam Tip: Understanding electronic configuration is fundamental to predicting an element's chemical behavior and valency.

 

Question 15. The relative atomic masses of some elements in the chart below are given. You have to find the relative atomic masses of the others.
Answer:

Element	Atomic mass	Element	Atomic mass	Element	Atomic 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	39.9
Boron	11	Magnesium	24	Potassium	39
Carbon	12	Aluminium	27	Calcium	40
Nitrogen	14	Silicon	28

In simple words: This table provides the relative atomic masses for a selection of common elements.

🎯 Exam Tip: Knowing the atomic masses of common elements is useful for solving numerical problems in chemistry.

 

Question 16. Classify the following radicals into simple radicals and composite radicals: (Use your brain power; Ag+, Mg2+, Cl-, SO₄²-, Fe2+, ClO₃-, NH₄+, Br-, NO₃-, Na+, Cu+
Answer:

Simple radicals	Composite radicals
Ag+	SO₄²-
Mg2+	ClO₃-
Cl-	NH₄+
Fe2+	NO₃-
Br-
Na+
Cu+

In simple words: Simple radicals consist of a single atom with a charge, while composite radicals are made of two or more atoms covalently bonded together and carrying a charge.

🎯 Exam Tip: Identifying simple vs. composite radicals is crucial for correctly writing chemical formulas and understanding ionic bonding.

 

Question 17. Which are the basic radicals and which are the acidic radicals among the following? Ag+, Cu2+, Cl-, I-, SO₄²-, Fe3+, Ca2+, NO3-, S²-, NH₄+, K+, MnO₄-, Na+
Answer:

Basic Radical	Acidic Radical
(i)Ag+	(i) Cl-
(ii) Cu2+	(ii) I-
(iii) Fe3+	(iii) SO₄²-
(iv) Ca2+	(iv) NO₃-
(v) NH₄+	(v) S²-
(vi) K+	(vi) MnO₄-
(vii) Na+

In simple words: Basic radicals are positively charged ions, typically formed from metals or ammonium, while acidic radicals are negatively charged ions, often derived from non-metals or polyatomic groups.

🎯 Exam Tip: Basic radicals usually donate electrons (cations) and acidic radicals accept electrons (anions) in forming ionic compounds.

 

Give Examples:

 

Question 1. Make a list of elements in the monoatomic and in the diatomic molecular state. (Make a list and discuss;
Answer:
• 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).
In simple words: Monoatomic elements exist as single atoms, like noble gases, while diatomic elements exist as molecules of two atoms bonded together, such as H2 or O2.

🎯 Exam Tip: Knowing which elements naturally occur as monoatomic or diatomic is important for writing correct chemical equations.

 

Problem-based Questions

 

Answer the following questions:

 

Question 1. Is it possible to weigh one molecule using a weighing balance?
Answer:
No, it is not possible to weigh one molecule using a weighing balance.
In simple words: No, a single molecule is far too small to be weighed by any standard weighing balance due to its extremely tiny mass.

🎯 Exam Tip: This question highlights the macroscopic limitation of weighing balances for submicroscopic particles like molecules.

 

Question 2. Will the number of molecules be the same in equal weights of different substances?
Answer:
No, the number of molecules will not be the same in equal weights of different substances.
In simple words: No, because different substances have different molecular masses, equal weights will contain different numbers of molecules.

🎯 Exam Tip: This concept relates to the mole and Avogadro's number, where equal moles (not equal masses) of different substances contain the same number of molecules.

 

Question 3. If we want equal number of molecules of different substances, will it work to take equal weights of those substances.
Answer:
No, if we want equal number of molecules of different substances, it will not work to take equal weights of those substances.
In simple words: To have an equal number of molecules for different substances, you must take quantities proportional to their molecular masses, not equal weights.

🎯 Exam Tip: This reinforces the mole concept; to have equal numbers of particles, you must have equal numbers of moles, which corresponds to different masses for different substances.

 

Answer the following:

 

Question 1. What is the Dalton's atomic theory?
Answer:
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.
In simple words: Dalton's theory states that matter is made of indestructible atoms, all atoms of an element are identical, compounds form from combining different atoms, and chemical reactions involve atom rearrangement.

🎯 Exam Tip: Understand these postulates as they form the historical foundation of modern atomic theory, even if some have been refined.

 

Question 2. How are compounds formed?
Answer:
Compounds are formed by a chemical combination of two or more different kinds of atoms.
In simple words: Compounds are created when two or more distinct types of atoms chemically bond together.

🎯 Exam Tip: The key idea here is "chemical combination," which implies new bonds and properties, unlike a simple mixture.

 

Question 3. What are the molecular formulae of salt, slaked lime, water, lime, limestone?
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
In simple words: This lists the common chemical formulas for everyday substances: NaCl for salt, Ca(OH)2 for slaked lime, H2O for water, CaO for lime, and CaCO3 for limestone.

🎯 Exam Tip: Familiarize yourself with the chemical names and formulas of common compounds, as they frequently appear in chemistry problems.

 

Question 4. From which experiments was it discovered that atoms have an internal structure? When?
Answer:
• 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.
In simple words: Ernest Rutherford's gold foil experiment in 1911 revealed that atoms have an internal structure, consisting of a dense nucleus.

🎯 Exam Tip: The Gold Foil experiment is a landmark in atomic theory; remember its name, year, and key finding about the atomic nucleus.

 

Question 5. What are the two parts of an atom? What are they made up of?
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.
In simple words: An atom has two main parts: the nucleus, containing protons and neutrons, and the extranuclear part, where electrons orbit.

🎯 Exam Tip: Clearly distinguish between the nucleus (protons, neutrons) and the electron cloud, along with their respective charges.

 

Open-Ended Questions

 

Q.3. 2. Answer the following questions:

 

Question 1. How will the compounds, MgCl2 and CaO be formed from their elements?
Answer:
(1) Magnesium Chloride (MgCl2)
Magnesium atom (Mg): Electronic configuration
\( (2,8,2) \xrightarrow{-2e^-} \) Magnesium ion Mg2+ (2,8).
Chlorine atom (Cl). Electronic configuration (2, 8, 7) \( \xrightarrow{+1e^-} \) Chloride ion Cl- (2,8,8).
∴ Mg2+ + 2Cl- → 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 1e- 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 2Cl- each and the compound MgCl2 is formed.

(2) Calcium Oxide (CaO)
Calcium atom (Ca): Electronic configuration
\( (2,8,8,2) \xrightarrow{-2e^-} \) Calcium ion Ca2+ (2,8,8).
Oxygen atom (O). Electronic configuration (2,6)
\( \xrightarrow{+2e^-} \) 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.
In simple words: MgCl2 and CaO are formed through ionic bonding where Mg and Ca atoms lose electrons to become positive ions, and Cl and O atoms gain electrons to become negative ions, resulting in stable compounds due to electrostatic attraction.

🎯 Exam Tip: Focus on explaining electron transfer for ionic bond formation, ensuring the octet rule is met for both cation and anion, and the overall charge neutrality of the compound.

 

Question 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.
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.
In simple words: When calcium oxide reacts with water, calcium hydroxide forms, and the total mass of the reactants (56g + 18g = 74g) equals the total mass of the product (74g), demonstrating the Law of Conservation of Matter.

🎯 Exam Tip: This experiment perfectly illustrates the Law of Conservation of Mass; ensure you can clearly state the law and explain how the experiment supports it.

 

Question 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.
Answer:
• In this activity, a white precipitate of CaSO4 in NaCl 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.
In simple words: This experiment shows that even when a precipitate forms, the total mass of the sealed system remains unchanged, thereby confirming the Law of Conservation of Matter.

🎯 Exam Tip: Emphasize the sealed container to highlight that no matter escapes, allowing the accurate verification of mass conservation.

 

Question 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.
Answer:

	Ions/Radicals
Basic Radicals		Acidic Radicals
H+ Hydrogen	Al3+ Aluminium	H- Hydride	MnO₄- Permanganate
Na+ Sodium	Cr3+ Chromium	F- Fluoride	ClO₃- Chlorate
K+ Potassium	Fe3+ Ferric	Cl- Chloride	BrO₃- Bromate
Ag+ Silver	Au3+ Gold	Br- Bromide	IO₃- Iodate
Cu+ Cuprous	Sn4+ Stannic	I- Iodide	CO₃2- Carbonate
Hg+ Mercurous	NH₄+ Ammonium	O2- Oxide	SO₄2- Sulphate
Cu2+ Cupric/Copper		S2- Sulphide	SO₃2- Sulphite
Mg2+ Magnesium		N3- Nitride	CrO₄2- Chromate
Ca2+ Calcium			Cr₂O₇2- Dichromate
Ni2+ Nickel		OH- Hydroxide	PO₄3- Phosphate
Co2+ Cobalt		NO₃- Nitrate
Hg2+ Mercuric		NO₂- Nitrite
Mn2+ Manganese		HCO₃- Bicarbonate
Fe2+ Ferrous (Iron II)		HSO₄- Bisulphate
Sn2+ Stannous		HSO₃- Bisulphite
Pt2+ Platinum

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.
In simple words: By using the provided table of ions and applying the cross-multiplication method, the chemical formulas for these compounds are derived, ensuring charge neutrality.

🎯 Exam Tip: Mastering the cross-multiplication method with valencies is essential for correctly writing chemical formulas for ionic compounds.

 

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

 

(A) Select The Correct Option:

 

Question 1. The proportion by weight of hydrogen and oxygen in water is ........................
(a) 8:1
(b) 2:1
(c) 1:2
(d) 1:8
Answer: (d) 1:8
In simple words: Water (H2O) is composed of hydrogen and oxygen in a fixed mass ratio of 1:8, reflecting their atomic masses.

🎯 Exam Tip: This question tests knowledge of the Law of Constant Proportions and the atomic masses of Hydrogen (1) and Oxygen (16).

 

Question 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
Answer: (b) 3:8
In simple words: In carbon dioxide (CO2), the mass ratio of carbon to oxygen is 3:8, based on their atomic masses.

🎯 Exam Tip: To calculate this ratio, use the atomic masses: Carbon (12) and Oxygen (16), so for CO2, it's 12:(2x16) = 12:32, which simplifies to 3:8.

 

Question 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
Answer: (a) protons; neutrons
In simple words: The atomic nucleus contains positively charged protons and electrically neutral neutrons.

🎯 Exam Tip: Remember the charge and location of all three subatomic particles: protons (+ve, nucleus), neutrons (neutral, nucleus), and electrons (-ve, orbits).

 

Question 4. The size of an atom is determined by its ........................
Answer:
radius
In simple words: An atom's size is primarily defined by its atomic radius, which measures the distance from the nucleus to its outermost electron shell.

🎯 Exam Tip: Atomic radius is a fundamental property used to compare the sizes of different atoms and understand periodic trends.

 

Question 5. Atomic radius is expressed in ........................
(a) milimetres
(b) centimetres
(c) nanometres
(d) picometres
Answer: (c) nanometres
In simple words: Atomic radii are typically expressed in nanometres (nm) or picometres (pm) due to their extremely small scale.

🎯 Exam Tip: Understand that atoms are measured in very small units like nanometers (10^-9 m) or picometers (10^-12 m).

 

Question 6. The atomic size depends on the number of ........................ in the atom.
(a) protons
(b) nucleus
(c) neutrons
(d) electron orbits
Answer: (d) electron orbits
In simple words: The size of an atom is determined by the number of electron orbits it possesses; more orbits generally mean a larger atomic radius.

🎯 Exam Tip: The atomic size increases down a group in the periodic table due to the addition of new electron shells or orbits.

 

Question 7. The mass of an atom is concentrated in its ........................
(a) protons
(b) nucleus
(c) neutrons
(d) electrons
Answer: (b) nucleus
In simple words: Almost all of an atom's mass is found in its nucleus, which contains the heavier protons and neutrons.

🎯 Exam Tip: While electrons determine atomic volume, their mass is negligible compared to protons and neutrons in the nucleus.

 

Question 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
Answer: (c) atomic mass number
In simple words: The sum of protons and neutrons in an atom's nucleus is known as the atomic mass number.

🎯 Exam Tip: Distinguish between atomic number (number of protons) and atomic mass number (protons + neutrons).

 

Question 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
Answer: (a) Mole
In simple words: A mole is a unit representing a specific quantity of a substance, where its mass in grams numerically matches its molecular mass in Daltons.

🎯 Exam Tip: The definition of a mole is fundamental to quantitative chemistry; understand its relationship to molecular mass and Avogadro's number.

 

Question 10. Avogadro's number is denoted by the symbol ............................
(a) NG
(b) Nv
(C) NA
(d) ND
Answer: (C) NA
In simple words: Avogadro's number, a fundamental constant in chemistry representing the number of particles in one mole, is universally symbolized as \( \text{N}_\text{A} \).

🎯 Exam Tip: Knowing standard scientific symbols is crucial for clear communication and avoiding confusion in chemistry problems.

 

Question 11. A mole of any substance stands for ............................ molecules.
(a) \( 6.022 \times 10^{23} \)
(b) \( 6.022 \times 10^{22} \)
(c) \( 6.022 \times 10^{23} \)
(d) \( 60.22 \times 10^{22} \)
Answer: (a) \( 6.022 \times 10^{23} \)
In simple words: One mole of any substance contains Avogadro's number of particles, which is approximately \( 6.022 \times 10^{23} \) molecules.

🎯 Exam Tip: Avogadro's number is a key concept for converting between moles and the number of atoms or molecules, essential for stoichiometry calculations.

 

Question 12. The capacity of an element to combine is called its ............................
(a) valency
(b) electronic configuration
(c) atomic number
(d) volence electrons
Answer: (a) valency
In simple words: Valency describes an atom's combining power, indicating how many bonds it can form with other atoms.

🎯 Exam Tip: Understanding valency is fundamental to writing correct chemical formulas and predicting reaction outcomes.

 

Question 13. Electronic configuration of sodium atom is ............................
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8,1)
Answer: (d) (2,8,1)
In simple words: Sodium (Na) has an atomic number of 11, meaning its electrons are arranged in shells as 2 in the first, 8 in the second, and 1 in the outermost shell.

🎯 Exam Tip: Electronic configuration helps determine an element's chemical properties and its tendency to form bonds.

 

Question 14. Electronic configuration of chlorine atom is ............................
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8, 1)
Answer: (b) (2, 8, 7)
In simple words: Chlorine (Cl) has an atomic number of 17, so its electrons are distributed as 2 in the first shell, 8 in the second, and 7 in the outermost shell.

🎯 Exam Tip: The number of electrons in the outermost shell (valence electrons) dictates an element's reactivity and valency.

 

Question 15. Positively charged ions are called as ............................
(a) cations
(b) anions
(c) nucleous
(d) protons
Answer: (a) cations
In simple words: Atoms that lose electrons become positively charged ions, known as cations.

🎯 Exam Tip: Distinguishing between cations and anions is fundamental to understanding ionic bonding and electrolyte behavior.

 

Question 16. Negatively charged ions are called as ............................
(a) cations
(b) anions
(c) nucleus
(d) electrons
Answer: (b) anions
In simple words: Atoms that gain electrons become negatively charged ions, known as anions.

🎯 Exam Tip: Cations are typically formed by metals, while anions are usually formed by non-metals.

 

Question 17. Iron (Fe) exhibits the variable valencies
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 2 and 4
Answer: (b) 2 and 3
In simple words: Iron can exist in two common ionic forms: \( \text{Fe}^{2+} \) (ferrous) and \( \text{Fe}^{3+} \) (ferric), demonstrating variable valency.

🎯 Exam Tip: Many transition metals exhibit variable valency, which affects the formulas of their compounds; remember the common valencies for important elements like iron.

 

Question 18. Cationic radicals are called as ............................ radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer: (a) basic
In simple words: Cationic radicals are positively charged ions, which often act as the basic component in salts, combining with acidic radicals.

🎯 Exam Tip: Basic radicals are typically formed from metals or ammonium ions and determine the basic nature of a compound.

 

Question 19. Anionic radicals are called as ............................ radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer: (b) acidic
In simple words: Anionic radicals are negatively charged ions that typically combine with basic radicals to form salts, contributing to the acidic part of the compound.

🎯 Exam Tip: Acidic radicals are typically formed from non-metals or polyatomic ions and determine the acidic nature of a compound.

 

Question 20. The unit Dalton is used to express ............................
(a) atomic mass
(b) atomic radius
(c) atomic number
(d) mass number
Answer: (a) atomic mass
In simple words: The Dalton (u) is the standard unit used to quantify the mass of atoms and molecules, also known as the unified atomic mass unit.

🎯 Exam Tip: The Dalton unit provides a convenient way to express the incredibly small masses of atomic and subatomic particles.

 

Question 21. The valency of element with electronic configuration ............................ is 2.
(a) (2,5)
(b) (2, 4)
(c) (2, 6)
(d) (2, 7)
Answer: (c) (2, 6)
In simple words: An element with an electronic configuration of (2, 6) has 6 valence electrons and needs to gain 2 more electrons to achieve a stable octet, thus its valency is 2.

🎯 Exam Tip: Valency is often determined by the number of electrons an atom needs to gain, lose, or share to achieve a stable outer electron shell, typically 8 electrons (octet rule).

 

Question 22. The symbol of Avogadro's number is ............................
(a) ND
(b) NO
(c) NB
(d) NA
Answer: (d) NA
In simple words: The internationally recognized symbol for Avogadro's number, representing the count of particles in one mole, is \( \text{N}_\text{A} \).

🎯 Exam Tip: Consistent use of correct symbols is essential for scientific accuracy and universal understanding in chemistry.

 

Question 23. ............................ is bicarbonate radical.
(a) \( \text{HCO}_3^- \)
(b) \( \text{CO}_3^{2-} \)
(c) \( \text{HCO}_3^- \)
(d) \( \text{CO}_3^{2-} \)
Answer: (c) \( \text{HCO}_3^- \)
In simple words: The bicarbonate radical, also known as hydrogen carbonate, is a polyatomic ion with the chemical formula \( \text{HCO}_3^- \).

🎯 Exam Tip: Memorizing common polyatomic ions and their charges is crucial for correctly writing chemical formulas and understanding acid-base chemistry.

 

Question 24. Molecular formula of sodium sulphate is ............................
(a) Na(\( \text{SO}_4 \))2
(b) \( \text{Na}_2\text{SO}_4 \)
(c) \( \text{Na}_2(\text{SO}_4)_2 \)
(d) \( \text{NaSO}_4 \)
Answer: (b) \( \text{Na}_2\text{SO}_4 \)
In simple words: Sodium has a valency of +1, and the sulphate radical has a valency of -2; therefore, two sodium ions combine with one sulphate ion to form sodium sulphate.

🎯 Exam Tip: Use the criss-cross method of valencies to correctly derive molecular formulas for ionic compounds, balancing the total positive and negative charges.

 

Question 25. ............................ is a composite radical.
(a) \( \text{Fe}^{3+} \)
(b) \( \text{Ca}^{2+} \)
(c) \( \text{NH}_4^+ \)
(d) \( \text{S}^{2-} \)
Answer: (c) \( \text{NH}_4^+ \)
In simple words: A composite radical, also known as a polyatomic ion, is an ion composed of two or more atoms covalently bonded together and carrying a net charge, such as ammonium.

🎯 Exam Tip: Simple radicals consist of a single atom with a charge (like \( \text{Fe}^{3+} \)), while composite radicals involve multiple atoms (like \( \text{NH}_4^+ \)).

 

Question 26. A mole of any substance stands for ............................ molecules.
(a) \( 6.022 \times 10^{23} \)
(b) \( 6.022 \times 10^{22} \)
(c) \( 60.22 \times 10^{23} \)
(d) \( 60.22 \times 10^{22} \)
Answer: (a) \( 6.022 \times 10^{23} \)
In simple words: One mole of any substance always contains \( 6.022 \times 10^{23} \) particles, a value known as Avogadro's constant.

🎯 Exam Tip: Be precise with powers of ten when writing Avogadro's number, as a single digit error can lead to a vastly different quantity.

 

Question 27. The mass of an atom is concentrated in its ............................
(a) nucleus
(b) electrons
(c) extranuclear part
(d) protons
Answer: (a) nucleus
In simple words: Almost all the mass of an atom is found in its dense central nucleus, which contains protons and neutrons.

🎯 Exam Tip: The nucleus accounts for the vast majority of an atom's mass, while electrons contribute negligibly due to their much smaller mass.

 

Question 28. ............................ g of water make 1 mole of water.
(a) 32
(b) 33
(c) 16
(d) 18
Answer: (d) 18
In simple words: The molecular mass of water (H2O) is 18 u, so 18 grams of water constitute one mole.

🎯 Exam Tip: To find the mass of one mole of a substance, calculate its molecular mass in atomic mass units (u) and express it in grams per mole.

 

Complete The Analogy:

Question. Complete the analogy:
(1) Electron: extra nuclear part:: Neutron ............................
(2) Sodium: (2, 8, 1):: Chlorine:: ............................
(3) K: basic radical :: Br- : ............................
(4) Cu+: simple radical:: \( \text{NH}_4^+ \): ............................
(5) Sodium sulphate: \( \text{Na}_2\text{SO}_4 \):: Potassium Sulphate: ............................
(6) Mercurous: \( \text{Hg}^+ \):: Mercuric : ............................
(7) Positively charged ion : cation:: Negatively charged ion : ............................
(8) 12: 1 dozen :: 144 : ............................
(9) Hydrogen : \( \odot \) :: Copper : \( \odot \)
(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) \( \text{K}_2\text{SO}_4 \)
(6) \( \text{Hg}^{2+} \)
(7) anion
(8) 1 gross
(9) \( \odot \)
(10) Antoine Lavoisier.
In simple words: This section tests understanding of fundamental chemical and physical relationships, including atomic structure, electronic configurations, types of radicals, chemical formulas, and historical scientific laws.

🎯 Exam Tip: Analogies require identifying the relationship between the first pair of terms and applying that same relationship to complete the second pair; precise knowledge of definitions and classifications is key.

 

Match The Columns:

Question 1. Match the columns:

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

Answer:
(1-b),
(2-с),
(3 - a)
In simple words: This question matches examples of molecules or atoms with their typical atomic radii in meters, demonstrating the varying scales of different particles.

🎯 Exam Tip: Remember common atomic/molecular size scales to quickly identify the correct order of magnitude for different substances.

 

Question 2. Match the columns:

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)
In simple words: This matching exercise connects various elements with their respective atomic masses, which are essential values in chemical calculations.

🎯 Exam Tip: Familiarity with the atomic masses of common elements speeds up molecular mass calculations and other quantitative problems.

 

Question 3. Match the columns:

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)
In simple words: This task requires matching chemical molecules to their molecular masses expressed in grams, which is equivalent to their molar mass.

🎯 Exam Tip: To find the molecular mass in grams (molar mass), calculate the sum of the atomic masses of all atoms in the molecule and attach the unit 'g/mol'.

 

Question 4. Match the columns:

Column 'A'	Column B'
Radicals	Names
(1) \( \text{Cr}_2\text{O}_7^{2-} \)	(a) Carbonate
(2) \( \text{ClO}_3^- \)	(b) Chromate
(3) \( \text{CO}_3^{2-} \)	(c) Dichromate
(4) \( \text{CrO}_4^{2-} \)	(d) Chlorate

Answer:
(1 - c),
(2 - d),
(3 - a),
(4 - b)
In simple words: This question tests knowledge of common polyatomic radicals by matching their chemical formulas with their correct names.

🎯 Exam Tip: Accurate identification of radicals and their names is vital for writing correct chemical formulas and naming compounds systematically.

 

Answer The Following In One Sentence:

Question 1. What are valence electrons?
Answer: The electrons present in the outermost orbit of an atom are called valence electrons.
In simple words: Valence electrons are the electrons located in the outermost electron shell of an atom, responsible for its chemical bonding behavior.

🎯 Exam Tip: The number of valence electrons often determines an element's group in the periodic table and its valency.

 

Question 2. Give the formula to determine the number of moles of a substance.
Answer: The formula to determine the number of moles of a substance is as given below.
\[ \text{Number of moles (n)} = \frac{\text{Mass of substance in grams (m)}}{\text{Molecular mass of substance (M)}} \]
In simple words: To find the number of moles, divide the given mass of the substance by its molecular (or molar) mass.

🎯 Exam Tip: This formula is fundamental for all stoichiometric calculations, allowing conversion between mass and moles.

 

Question 3. What are basic radicals? Give examples.
Answer: The radicals which are formed by removal of electrons from the atoms of metals are called as basic radicals, e.g., Na+, Cu2+
In simple words: Basic radicals are positively charged ions, typically formed when metal atoms lose electrons, and they are the basic part of a compound.

🎯 Exam Tip: Basic radicals are also known as cations and usually originate from metallic elements or the ammonium ion.

 

Question 4. What are acidic radicals? Give examples.
Answer: The radicals which are formed by adding electrons to the atoms of non-metals are called as acidic radicals, e.g., Cl-, S2-
In simple words: Acidic radicals are negatively charged ions, usually formed when non-metal atoms gain electrons, and they constitute the acidic part of a compound.

🎯 Exam Tip: Acidic radicals are also known as anions and can be monatomic or polyatomic, such as chloride (\( \text{Cl}^- \)) or sulfate (\( \text{SO}_4^{2-} \)).

 

State Whether The Following Statement Is 'True' Or 'False'. Correct The False Statement.

Question. 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 \( 6.022 \times 10^{23} \)
(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.
In simple words: This section assesses knowledge across various chemical principles, including molecular states, valency, types of radicals, chemical laws, and fundamental constants, requiring corrections for false statements.

🎯 Exam Tip: When correcting false statements, ensure the correction directly addresses the incorrect part of the statement, providing accurate scientific information.

 

Name The Following:

Question 1. Scientist who gave Law of Conservation of Matter.
Answer: Antoine Lavoisier
In simple words: Antoine Lavoisier, a French chemist, is credited with formulating the Law of Conservation of Mass, stating that mass is neither created nor destroyed in a chemical reaction.

🎯 Exam Tip: Associating key scientific laws with their discoverers helps in understanding the historical development of chemistry.

 

Question 2. Scientist who gave Law of Constant Proportion.
Answer: J. L. Proust
In simple words: Joseph Proust established the Law of Definite Proportions, which states that a given chemical compound always contains its component elements in fixed ratio by mass.

🎯 Exam Tip: Understanding the contributions of scientists like Proust highlights how empirical observations led to fundamental chemical laws.

 

Question 3. What are protons and neutrons present in nucleus together called as?
Answer: Nucleons
In simple words: Protons and neutrons, the particles found within an atom's nucleus, are collectively referred to as nucleons.

🎯 Exam Tip: Nucleons determine the atomic mass of an element, as electrons contribute negligibly to the overall mass.

 

Question 4. Unit used to express atomic radius.
Answer: Nanometre
In simple words: The nanometre (nm) is a common unit for expressing atomic radii due to the extremely small size of atoms.

🎯 Exam Tip: Remember that 1 nanometre is \( 10^{-9} \) meters, providing a suitable scale for atomic dimensions.

 

Question 5. The number (p + n) in the atomic nucleus is called as?
Answer: Atomic mass number
In simple words: The total count of protons (p) and neutrons (n) in an atom's nucleus is defined as its atomic mass number.

🎯 Exam Tip: Atomic mass number is crucial for identifying isotopes of an element, as it represents the total number of nucleons.

 

Question 6. Name the unit of atomic mass.
Answer: Dalton (u)
In simple words: The standard unit for measuring atomic mass is the Dalton, also known as the unified atomic mass unit (u).

🎯 Exam Tip: The Dalton is defined as approximately the mass of one proton or one neutron, making it convenient for expressing atomic and molecular weights.

 

Question 7. Write molecular formula of two ionic compounds containing chlorine.
Answer: NaCl, MgCl2
In simple words: Sodium chloride (NaCl) and Magnesium chloride (MgCl2) are examples of ionic compounds formed between metals and chlorine.

🎯 Exam Tip: When writing formulas for ionic compounds, ensure the total positive charge equals the total negative charge to maintain electrical neutrality.

 

Question 8. Give two monoatomic radicals.
Answer: Na+, Cl-
In simple words: Monoatomic radicals are ions composed of a single atom that has gained or lost electrons, such as the sodium ion or chloride ion.

🎯 Exam Tip: Remember that the charge on a monoatomic radical reflects the number of electrons gained or lost by the single atom.

 

Question 9. Give two examples of simple radicals.
Answer: Ag+, O2-
In simple words: Simple radicals are individual atoms that have gained or lost electrons, resulting in a positive or negative charge, like silver ion or oxide ion.

🎯 Exam Tip: Simple radicals can be cations (e.g., metal ions) or anions (e.g., non-metal ions) and participate in ionic bonding.

 

Give Scientific Reasons:

Question 1. An atom is electrically neutral though it contains charged particles.
Answer:
* 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.
In simple words: An atom is electrically neutral because the number of positively charged protons in its nucleus is exactly equal to the number of negatively charged electrons surrounding the nucleus, cancelling out the overall charge.

🎯 Exam Tip: This balance between protons and electrons is a defining characteristic of a neutral atom; a change in this balance results in an ion.

 

Question 2. Neon is chemically inert element.
Answer:
* 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.
In simple words: Neon is chemically inert because its outermost electron shell is completely filled with eight electrons (a stable octet), making it highly unreactive.

🎯 Exam Tip: Elements with stable electron configurations, particularly a full outermost shell, are known as noble gases and are highly unreactive.

 

Question 3. The valency of sodium (Na) is one.
Answer:
* 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.
In simple words: Sodium has one electron in its outermost shell and readily loses it to achieve a stable electron configuration, resulting in a valency of one.

🎯 Exam Tip: Metals typically lose electrons to achieve a stable octet, and their valency is equal to the number of electrons lost.

 

Question 4. The valency of chlorine (Cl) is one.
Answer:
* 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.
In simple words: Chlorine has seven electrons in its outermost shell and readily gains one electron to achieve a stable octet, giving it a valency of one.

🎯 Exam Tip: Non-metals typically gain electrons to achieve a stable octet, and their valency is equal to the number of electrons gained.

 

Question 5. The valency of Magnesium (Mg) is two.
Answer:
* 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.
In simple words: Magnesium has two valence electrons and achieves stability by losing both, forming a \( \text{Mg}^{2+} \) ion and thus exhibiting a valency of two.

🎯 Exam Tip: For elements with 1, 2, or 3 valence electrons, their valency is usually equal to the number of electrons they lose to become stable.

 

Question 6. Valency is always a whole number.
Answer:
* 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.
In simple words: Valency is defined by the discrete number of electrons an atom exchanges or shares, which cannot be fractional, hence it's always a whole number.

🎯 Exam Tip: Valency reflects fundamental atomic interactions involving whole electrons, ensuring that chemical bonds are formed in fixed, integral ratios.

 

Question 7. Atomic size of potassium is bigger than atomic size of sodium.
Answer:
* 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.
In simple words: Potassium atoms are larger than sodium atoms because potassium has four electron shells compared to sodium's three, meaning its outermost electrons are further from the nucleus.

🎯 Exam Tip: Atomic size generally increases down a group in the periodic table due to the addition of new electron shells.

 

Question 8. The atomic size of sodium is bigger than atomic size of Magnesium.
Answer:
* 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.
In simple words: Sodium atoms are larger than magnesium atoms because, despite being in the same period, magnesium has a higher nuclear charge, pulling its electron shells closer and reducing its atomic radius.

🎯 Exam Tip: Across a period, atomic size generally decreases due to increasing nuclear charge pulling the electron shells more tightly, even if the number of shells remains the same.

 

Write The Names Of The Following Compounds And Deduce Their Molecular Masses:

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

Question 1. Molecular mass of K2CO3
Answer:

Name of compound	Molecule	Constituent elements	Atomic mass (u)	Number of atoms in molecule	Atomic mass \( \times \) number of atoms	Mass of the constituents u
Potassium carbonate	\( \text{K}_2\text{CO}_3 \)	Potassium	39	2	39 \( \times \) 2	78
		Carbon	12	1	12 \( \times \) 1	12
		Oxygen	16	3	16 \( \times \) 3	48

Molecular mass = Sum of constituent atomic masses
Molecular mass of \( (\text{K}_2\text{CO}_3) \) = (Atomic mass of K) \( \times \) 2 + (Atomic mass of C) \( \times \) 1 + (Atomic mass of O) \( \times \) 3
Molecular Mass = 138
In simple words: The molecular mass of potassium carbonate (\( \text{K}_2\text{CO}_3 \)) is calculated by summing the atomic masses of two potassium atoms, one carbon atom, and three oxygen atoms, resulting in 138 u.

🎯 Exam Tip: Always multiply the atomic mass of each element by its subscript in the chemical formula before summing them to find the molecular mass.

 

Question 2. Molecular mass of CO2
Answer:

Name of compound	Molecule	Constituent elements	Atomic mass (u)	Number of atoms in molecule	Atomic mass \( \times \) number of atoms	Mass of the constituents u
Carbon dioxide	\( \text{CO}_2 \)	Carbon	12	1	12 \( \times \) 1	12
		Oxygen	16	2	16 \( \times \) 2	32

Molecular mass = Sum of constituent atomic masses
Molecular mass of \( (\text{CO}_2) \) = (Atomic mass of C) \( \times \) 1 + (Atomic mass of O) \( \times \) 2
Molecular Mass = 44
In simple words: The molecular mass of carbon dioxide (\( \text{CO}_2 \)) is found by adding the atomic mass of one carbon atom and two oxygen atoms, which totals 44 u.

🎯 Exam Tip: Be careful to correctly identify the number of atoms of each element from the subscript in the molecular formula.

 

Question 3. Molecular mass of MgCl2
Answer:

Name of compound	Molecule	Constituent elements	Atomic mass (u)	Number of atoms in molecule	Atomic mass \( \times \) number of atoms	Mass of the constituents u
Magnesium chloride	\( \text{MgCl}_2 \)	Magnesium	24	1	24 \( \times \) 1	24
		Chlorine	35.5	2	35.5 \( \times \) 2	71

Molecular mass = Sum of constituent atomic masses
Molecular mass of \( (\text{MgCl}_2) \) = (Atomic mass of Mg) \( \times \) 1 + (Atomic mass of Cl) \( \times \) 2
Molecular Mass = 95
In simple words: To calculate the molecular mass of magnesium chloride (\( \text{MgCl}_2 \)), sum the atomic mass of one magnesium atom and two chlorine atoms, resulting in 95 u.

🎯 Exam Tip: For elements with fractional atomic masses (like chlorine), use the given value precisely to ensure an accurate molecular mass calculation.

 

Question 4. Molecular mass of NaOH
Answer:

Name of compound	Molecule	Constituent elements	Atomic mass (u)	Number of atoms in molecule	Atomic mass \( \times \) number of atoms	Mass of the constituents u
Sodium hydroxide	NaOH	Sodium	23	1	23 \( \times \) 1	23
		Oxygen	16	1	16 \( \times \) 1	16
		Hydrogen	1	1	1 \( \times \) 1	1

Molecular mass = Sum of constituent atomic masses
Molecular mass of (NaOH) = (Atomic mass of Na) \( \times \) 1 + (Atomic mass of O) \( \times \) 1 + (Atomic mass of H) \( \times \) 1
Molecular Mass = 40
In simple words: The molecular mass of sodium hydroxide (NaOH) is determined by adding the atomic masses of one sodium, one oxygen, and one hydrogen atom, which totals 40 u.

🎯 Exam Tip: When dealing with compounds containing multiple elements, ensure each element's atomic mass is counted according to its stoichiometry in the formula.

 

Question 5. Molecular mass of AlPO4
Answer:

Name of compound	Molecule	Constituent elements	Atomic mass (u)	Number of atoms in molecule	Atomic mass \( \times \) number of atoms	Mass of the constituents u
Aluminium phosphate	\( \text{AlPO}_4 \)	Aluminium	27	1	27 \( \times \) 1	27
		Phosphorus	31	1	31 \( \times \) 1	31
		Oxygen	16	4	16 \( \times \) 4	64

Molecular mass = Sum of constituent atomic masses
Molecular mass of \( (\text{AlPO}_4) \) = (Atomic mass of Al) \( \times \) 1 + (Atomic mass of P) \( \times \) 1 + (Atomic mass of O) \( \times \) 4
Molecular Mass = 122
In simple words: The molecular mass of aluminium phosphate (\( \text{AlPO}_4 \)) is calculated by adding the atomic masses of one aluminium, one phosphorus, and four oxygen atoms, resulting in a total of 122 u.

🎯 Exam Tip: Pay close attention to polyatomic ions within a compound's formula, correctly accounting for all constituent atoms and their respective numbers.

Numerical

Question 1. Magnesium Oxide:
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)
\[ 0.2 = \frac{\text{Mass of MgO in grams (m)}}{\text{Molecular mass of MgO (M)}} \]
\[ 0.2 = \frac{\text{Mass of MgO in grams (m)}}{40} \] 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 8g
In simple words: To find the mass of 0.2 moles of Magnesium Oxide, we first calculate its molecular mass (Mg: 24, O: 16, so 24+16=40). Then, we multiply the number of moles by the molecular mass to get the total mass in grams.

🎯 Exam Tip: Remember the formula linking moles, mass, and molecular mass for quick calculations in stoichiometry problems.

Question 6. Molecular mass of NaHCO3
Answer:

Name of compound	Molecule	Constituent elements	Atomic mass (u)	Number of atoms in molecule	Atomic mass x number of atoms	Mass of the constituents u
Sodium bicarbonate	NaHCO3	Sodium	23	1	23 x 1	23
		Hydrogen	1	1	1 x 1	1
		Carbon	12	1	12 x 1	12
		Oxygen	16	3	16 x 3	48

Molecular mass = Sum of constituent atomic masses
Molecular mass of (NaHCO3) = (Atomic mass of Na) x 1 + (Atomic mass of H) x 1 + (Atomic mass of C) x 1 + (Atomic mass of O) x 3
Molecular Mass = 84
In simple words: The molecular mass of Sodium Bicarbonate (NaHCO3) is calculated by summing the atomic masses of all atoms in its formula: one Sodium (23u), one Hydrogen (1u), one Carbon (12u), and three Oxygen atoms (3 x 16u = 48u), totaling 84u.

🎯 Exam Tip: Always list atomic masses and the number of atoms for each element clearly to avoid calculation errors when determining molecular mass.

 

Question 2. Calcium Carbonate:
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 1) + (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)
\[ 0.2 = \frac{\text{Mass of CaCO}_3 \text{ in grams (m)}}{\text{Molecular mass of CaCO}_3 \text{ (M)}} \]
\[ 0.2 = \frac{\text{Mass of CaCO}_3 \text{ in grams (m)}}{100} \] 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
In simple words: To find the mass of 0.2 moles of Calcium Carbonate (CaCO3), we first calculate its molecular mass (Ca: 40, C: 12, O: 16x3=48, totaling 100). Then, we multiply the number of moles by the molecular mass to find the total mass in grams.

🎯 Exam Tip: Practice converting between moles and mass using molecular weight. This is a fundamental skill for many chemistry calculations.

State laws/Define the following:

Question 1. Law of Conservation of Matter.
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.
In simple words: The Law of Conservation of Matter states that matter cannot be created or destroyed in a chemical reaction; the total mass of substances before and after the reaction remains constant.

🎯 Exam Tip: This law is crucial for understanding chemical equations, as it explains why equations must be balanced.

Question 2. Law of Constant Proportion.
Answer:
The proportion by weight of the constituent elements in the various samples of a compound is fixed.
In simple words: The Law of Constant Proportion (or Definite Proportions) states that a pure chemical compound always contains the same elements in the same proportion by mass, regardless of its source or method of preparation.

🎯 Exam Tip: This law helps in identifying pure compounds and is foundational to understanding chemical formulas.

 

Question 3. Molecular Mass:
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.
In simple words: Molecular mass is the total mass of all atoms in one molecule of a compound, found by adding up the atomic masses of each constituent atom.

🎯 Exam Tip: Molecular mass is typically expressed in atomic mass units (u) or Daltons. It's essential for converting between mass and moles.

Question 4. Mole
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.
In simple words: A mole is a unit of measurement for the amount of substance, defined as the quantity containing Avogadro's number (6.022 x 10^23) of particles, and its mass in grams is numerically equal to the substance's molecular mass in Daltons.

🎯 Exam Tip: Understanding the mole concept is fundamental in chemistry for calculations involving quantities of reactants and products.

Question 5. Valency
Answer:
The capacity of an element to combine is called its valency.
In simple words: Valency refers to an atom's combining capacity, indicating how many bonds it can form with other atoms.

🎯 Exam Tip: Valency helps predict the chemical formula of compounds and understand how atoms bond together.

Question 6. Electronic definition of Valency
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.
In simple words: Electronically, valency is the number of electrons an atom gains, loses, or shares to achieve a stable electron configuration, typically an octet.

🎯 Exam Tip: Valency is often determined by the number of valence electrons and an atom's tendency to achieve a stable noble gas configuration.

Question 7. Radicals
Answer:
The positively or negatively charged ions that take part independently in chemical reactions are called radicals.
In simple words: Radicals are groups of atoms (or single atoms) that carry a charge (positive or negative) and behave as a single unit in chemical reactions.

🎯 Exam Tip: Recognizing common radicals and their charges is essential for writing correct chemical formulas and understanding ionic compounds.

Question 8. Atomic size determination
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.
In simple words: Atomic size is determined by the atomic radius, which is the distance from the center of the nucleus to the outermost electron shell of an atom.

🎯 Exam Tip: Atomic radius trends across the periodic table (decreasing across a period, increasing down a group) are important for understanding elemental properties.

 

Question 9. Atomic mass number
Answer:
The number of protons and neutrons in the atomic nucleus is called the atomic mass number.
In simple words: The atomic mass number, also known as the mass number, is the total count of protons and neutrons found within an atom's nucleus.

🎯 Exam Tip: The atomic mass number helps identify isotopes of an element, as isotopes have the same number of protons but different numbers of neutrons.

Question 10. Unified mass
Answer:
Unified mass is the standard unit of atomic mass that quantifies mass on an atomi or molecular scale. Its symbol is 'u'.
1 u = 1.66053904 x 10-27 kg.
In simple words: Unified mass (u) is a standard unit for measuring the mass of atoms and molecules, defined as one-twelfth of the mass of a carbon-12 atom.

🎯 Exam Tip: Unified mass (u) is often used interchangeably with Dalton (Da) and provides a convenient way to express very small atomic masses.

Question 11. Molecular mass of a substance
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).
In simple words: Molecular mass is the sum of the atomic masses of all atoms present in one molecule of a substance, expressed in atomic mass units (u) or Daltons.

🎯 Exam Tip: Accurately calculating molecular mass is essential for mole conversions and stoichiometric calculations in chemistry.

Answer the following questions:

Question 1. What is variable valency?
Answer:
- 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.
In simple words: Variable valency is the ability of certain elements, especially transition metals, to exhibit more than one combining capacity by losing or gaining different numbers of electrons under varying chemical conditions.

🎯 Exam Tip: Knowing elements with variable valency helps in naming compounds using Roman numerals (e.g., Iron(II) and Iron(III)) and predicting their chemical behavior.

Complete the following table:

 

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

Base	Cation	Anion	Acid	Cation	Anion
NaOH	Na+	OH-	HCl	H+	Cl-
KOH	K+	OH-	HBr	H+	Br-
Ca(OH)2	Ca2+	OH-	HNO3	H+	NO3-

In simple words: When bases and acids dissolve, they dissociate into positively charged cations and negatively charged anions, which are shown in the table for common compounds like NaOH, HCl, KOH, HBr, Ca(OH)2, and HNO3.

🎯 Exam Tip: Understanding the dissociation of compounds into cations and anions is key to predicting reaction products and writing ionic equations.

Answer the following questions:

Question 1. Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds:
(a) Calcium carbonate
Answer:
Symbol Ca CO3

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कैल्शियम (Ca) और कार्बोनेट (CO3) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। कैल्शियम की संयोजकता 2 है और कार्बोनेट की संयोजकता 2 है। ये दोनों संख्याएँ समान होने के कारण एक सामान्य कारक (2) से विभाजित होती हैं, जिससे दोनों की प्रभावी संयोजकता 1 हो जाती है। अंत में, ये मान क्रॉस-मल्टीप्लाई होते हैं।
Valency 2 2 Dividing by common factor
1 1
CaCO3
\( \implies \) Chemical formula of Calcium carbonate is CaCO3
In simple words: Calcium carbonate (CaCO3) is formed by combining Calcium (Ca) with a valency of 2 and Carbonate (CO3) with a valency of 2; since both valencies are 2, they simplify to 1:1, resulting in CaCO3.

🎯 Exam Tip: When using the cross-multiplication method, always simplify valencies by their greatest common factor before swapping them to get the empirical formula.

(b) Sodium bicarbonate
Answer:
Symbol Na HCO3

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सोडियम (Na) और बाइकार्बोनेट (HCO3) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दिखाता है। सोडियम की संयोजकता 1 है और बाइकार्बोनेट की संयोजकता भी 1 है। इन समान संयोजकताओं को सीधे क्रॉस-मल्टीप्लाई किया जाता है।
Valency 1 1
NaHCO3
\( \implies \) Chemical formula of Sodium bicarbonate is NaHCO3
In simple words: Sodium bicarbonate (NaHCO3) is formed by combining Sodium (Na) with a valency of 1 and Bicarbonate (HCO3) with a valency of 1, leading to the simple 1:1 ratio.

🎯 Exam Tip: Remember common polyatomic ions like bicarbonate (HCO3-) and their charges to quickly deduce chemical formulas.

(c) Silver chloride
Answer:
Symbol Ag Cl

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सिल्वर (Ag) और क्लोराइड (Cl) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। सिल्वर की संयोजकता 1 है और क्लोराइड की संयोजकता भी 1 है। इन समान संयोजकताओं को सीधे क्रॉस-मल्टीप्लाई किया जाता है।
Valency 1 1
AgCl
\( \implies \) Chemical formula of Silver chloride is AgCl
In simple words: Silver chloride (AgCl) is formed when Silver (Ag) with a valency of 1 combines with Chlorine (Cl) with a valency of 1, resulting in a 1:1 ratio.

🎯 Exam Tip: Many common ionic compounds with Group 1 metals and Group 17 non-metals will have a 1:1 valency, simplifying formula writing.

 

(d) Calcium hydroxide
Answer:
Symbol Ca OH

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कैल्शियम (Ca) और हाइड्रॉक्साइड (OH) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। कैल्शियम की संयोजकता 2 है और हाइड्रॉक्साइड की संयोजकता 1 है। ये संख्याएँ क्रॉस-मल्टीप्लाई होती हैं।
Valency 2 1
Ca(OH)2
\( \implies \) Chemical formula of Calcium hydroxide is Ca(OH)2
In simple words: Calcium hydroxide (Ca(OH)2) is formed when Calcium (Ca) with a valency of 2 combines with Hydroxide (OH) with a valency of 1, requiring two hydroxide ions to balance the calcium ion.

🎯 Exam Tip: When a polyatomic ion like hydroxide is involved and its subscript is greater than 1, always enclose the ion in parentheses in the chemical formula.

(e) Magnesium oxide
Answer:
Symbol Mg O

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मैग्नीशियम (Mg) और ऑक्सीजन (O) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। मैग्नीशियम की संयोजकता 2 है और ऑक्सीजन की संयोजकता 2 है। ये दोनों संख्याएँ समान होने के कारण एक सामान्य कारक (2) से विभाजित होती हैं, जिससे दोनों की प्रभावी संयोजकता 1 हो जाती है। अंत में, ये मान क्रॉस-मल्टीप्लाई होते हैं।
Valency 2 2 Dividing by common factor
1 1
MgO
\( \implies \) Chemical formula of Magnesium oxide is MgO
In simple words: Magnesium oxide (MgO) is formed by combining Magnesium (Mg) with a valency of 2 and Oxygen (O) with a valency of 2, simplifying to a 1:1 ratio.

🎯 Exam Tip: Remember to simplify the subscripts in a chemical formula to their lowest whole number ratio after cross-multiplication.

(f) Ammonium phosphate
Answer:
Symbol NH4 PO4

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अमोनियम (NH4) और फॉस्फेट (PO4) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। अमोनियम की संयोजकता 1 है और फॉस्फेट की संयोजकता 3 है। ये संख्याएँ क्रॉस-मल्टीप्लाई होती हैं।
Valency 1 3
(NH4)3PO4
\( \implies \) Chemical formula of Ammonium phosphate is (NH4)3PO4
In simple words: Ammonium phosphate ((NH4)3PO4) is formed by Ammonium (NH4) with a valency of 1 and Phosphate (PO4) with a valency of 3, requiring three ammonium ions to balance one phosphate ion.

🎯 Exam Tip: Be careful with polyatomic ions; if a subscript is needed for a polyatomic ion, it must be enclosed in parentheses.

(g) Cuprous bromide
Answer:
Symbol Cu Br

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्यूप्रस (Cu) और ब्रोमाइड (Br) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। क्यूप्रस की संयोजकता 1 है और ब्रोमाइड की संयोजकता भी 1 है। इन समान संयोजकताओं को सीधे क्रॉस-मल्टीप्लाई किया जाता है।
Valency 1 1
CuBr
\( \implies \) Chemical formula of Cuprous bromide is CuBr.
In simple words: Cuprous bromide (CuBr) is formed by combining Cuprous (Cu+) with a valency of 1 and Bromide (Br-) with a valency of 1, resulting in a 1:1 ratio.

🎯 Exam Tip: Remember that "cuprous" refers to the copper(I) ion, which has a valency of 1, distinguishing it from "cupric" (copper(II), valency 2).

(h) Copper sulphate
Answer:
Symbol Cu SO4

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कॉपर (Cu) और सल्फेट (SO4) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। कॉपर की संयोजकता 2 है और सल्फेट की संयोजकता 2 है। ये दोनों संख्याएँ समान होने के कारण एक सामान्य कारक (2) से विभाजित होती हैं, जिससे दोनों की प्रभावी संयोजकता 1 हो जाती है। अंत में, ये मान क्रॉस-मल्टीप्लाई होते हैं।
Valency 2 2 Dividing by common factor
1 1
CuSO4
\( \implies \) Chemical formula of Copper sulphate is CuSO4.
In simple words: Copper sulphate (CuSO4) is formed by combining Copper (Cu) with a valency of 2 and Sulphate (SO4) with a valency of 2, which simplifies to a 1:1 ratio after dividing by the common factor.

🎯 Exam Tip: When dealing with polyatomic ions like sulphate, remember to treat the entire ion as a single unit with its specific charge.

(i) Potassium nitrate
Answer:
Symbol K NO3

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पोटेशियम (K) और नाइट्रेट (NO3) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। पोटेशियम की संयोजकता 1 है और नाइट्रेट की संयोजकता भी 1 है। इन समान संयोजकताओं को सीधे क्रॉस-मल्टीप्लाई किया जाता है।
Valency 1 1
KNO3
\( \implies \) Chemical formula of Potassium nitrate is KNO3.
In simple words: Potassium nitrate (KNO3) is formed by combining Potassium (K) with a valency of 1 and Nitrate (NO3) with a valency of 1, resulting in a 1:1 ratio.

🎯 Exam Tip: Learn the charges of common polyatomic ions like nitrate (NO3-) to efficiently write chemical formulas for ionic compounds.

(j) Sodium dichromate
Answer:
Symbol Na Cr2O7

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सोडियम (Na) और डाइक्रोमेट (Cr2O7) आयनों के बीच रासायनिक सूत्र बनाने के लिए क्रॉस-मल्टीप्लिकेशन विधि को दर्शाता है। सोडियम की संयोजकता 1 है और डाइक्रोमेट की संयोजकता 2 है। ये संख्याएँ क्रॉस-मल्टीप्लाई होती हैं।
Valency 1 2
Na2Cr2O7
\( \implies \) Chemical formula of Sodium dichromate is Na2Cr2O7.
In simple words: Sodium dichromate (Na2Cr2O7) is formed by Sodium (Na) with a valency of 1 and Dichromate (Cr2O7) with a valency of 2, requiring two sodium ions to balance one dichromate ion.

🎯 Exam Tip: Pay close attention to the subscripts within polyatomic ions, as they are part of the ion's identity and not subject to simplification with other elements' valencies.