Atomic Foundations of Matter Class 9 Extra Questions and Answers Science Chapter 9

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Class 9 Science Chapter 9 Atomic Foundations of Matter Extra Questions

Class 9 Science Chapter 9 Extra Questions on Atomic Foundations of Matter

Atomic Foundations of Matter Class 9 Very Short Question Answer

Question 1.
When nitrogen combines with hydrogen to form ammonia, it is found that 14 g of nitrogen react completely with 3 g of hydrogen to form 17 g of ammonia.
Show that these observations are in agreement with
i. the law of conservation of mass, and
ii. the law of constant proportion.
Solution:
i. According to the law of conservation of mass, total mass of reactants equals the total mass of products. Here, 14 g of nitrogen + 3 g of hydrogen = 17 g, which is equal to the mass of ammonia formed. Hence, the law is verified.

ii. According to the law of constant proportion, elements in a compound are always present in a fixed ratio by mass. In ammonia, nitrogen and hydrogen always combine in the ratio of 14 : 3 by mass, proving the law.

Question 2.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + Acetic acid → Sodium acetate + Carbon dioxide + Water
Solution:
Total mass of reactants = Mass of sodium carbonate + Mass of acetic acid
= 5.3 + 6.0 = 11.3 g
Total mass of products = Mass of sodium acetate + Mass of carbon dioxide + Mass of water
= 8.2 + 2.2 + 0.9 = 11.3 g

Since, the sum of masses of reactants is equal to the sum of masses of products, therefore the observation made is in agreement with the law of conservation of mass.

Question 3.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Solution:
Since, H and O combine in the ratio of 1 : 8 by mass.
Therefore, \(\frac{\text { Mass of H }}{\text { Mass of O }}=\frac{1}{8}\)
Let the mass of oxygen required to react completely with 3 g of hydrogen gas be x.
\(\frac{\text { 3 g of H }}{\text { x g of O }}=\frac{3}{x}\)
= \(\frac{1}{8}\)
or x = 24 g
Therefore, 24 g of oxygen is required to react with 3 g of hydrogen to form water.

Question 4.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Solution:
The postulate which is the result of law of conservation of mass is “atoms are indivisible particles, which can neither be created nor be destroyed in a chemical reaction”.

Question 5.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Solution:
The postulate states that atoms of different elements combine in simple whole number ratios to form compounds. This explains the law of definite proportions, as a particular compound always contains the same elements combined in the same fixed ratio by mass.

Question 6.
Which of the following correctly represents the Mg²⁺ ion (Atomic number of magnesium = 12)?

Solution:
The correct representation is (ii).

The atomic number of Magnesium is 12, so its neutral atom has an electronic configuration of (2, 8, 2). To form an Mg²⁺ ion, it loses two electrons to achieve a stable octet, making the configuration (2, 8). Diagram (ii) correctly shows 2 electrons in the first shell and 8 in the second shell.

Question 7.
Some compounds are formed by transfer of electrons, while others are formed by sharing of electrons. Study the following compounds carefully.
(a) From the given list, select the compound which differs in type of bonding from the others:
NaCl, MgO, CO₂, KBr
(b) Give a suitable reason for your answer.
Solution:
(a) CO₂

(b) CO₂ is a covalent compound formed by sharing of electrons, whereas NaCl, MgO and KBr are ionic compounds formed by transfer of electrons. Hence, it differs in bonding and properties.

Question 8.
A sample of oxygen gas exists as O₂, while sulphur exists as S₈. What does this indicate about the nature of molecules of elements?
Solution:
Molecules of elements can contain different number of atoms, known as atomicity. Oxygen is diatomic (O₂), while sulphur is polyatomic (S₈). This shows that atoms of the same element combine in different numbers to form stable molecules.

Question 9.
i. How do atoms exist?
ii. Give the chemical formula for ammonium sulphate.
Solution:
i. Atoms exist in the form of molecules or ions.
ii. Chemical formula of ammonium sulphate is (NH₄)₂SO₄.

Question 10.
What is meant by the term chemical formula?
Solution:
Chemical formula is the shortest way to represent a compound with the help of symbols and valencies (charge) of elements.
e.g.

Question 11.
How many atoms are present in a
i. H₂S molecule and
ii. \(P \mathrm{O}_4^{3-}\) ion ?
Solution:
i. In H₂S molecule, three atoms
[i.e. 2 atoms of H and 1 atom of S] are present.

ii. In \(P \mathrm{O}_4^{3-}\) ion, five atoms
[i.e. 1 atom of P and 4 atoms of O] are present.

Question 12.
Write down the formulae of
i. sodium oxide
ii. aluminium chloride
iii. sodium sulphide
iv. magnesium hydroxide
Solution:
i. Sodium oxide

ii. Aluminium chloride

iii. Sodium sulphide

iv. Magnesium hydroxide

Question 13.
List the elements present in
i. Quicklime
ii. Sodium hydrogen carbonate
Solution:
i. The chemical name of quicklime is calcium oxide (CaO), the elements present in it are calcium and oxygen.
ii. The chemical formula of sodium hydrogen carbonate is NaHCO₃.
The elements present in it are sodium, hydrogen, carbon and oxygen.

Question 14.
Ram dissolved common salt in water and also tried dissolving it in kerosene. He observed that it dissolved only in water. Explain this observation.
Solution:
Common salt dissolves in water because they have a similar chemical nature, following the rule “like dissolves like”. Since kerosene has a different nature than salt, it cannot break the strong attraction between the salt particles to dissolve them.

Question 15.
i. Calculate the molar mass of sugar (C₁₂H₂₂O₁₁).
[Atomic mass of C = 12u, O = 16u and H = 1 u]
ii. The valency of an element X is 4. What is the formula of its oxide?
Solution:
i. C₁₂H₂₂O₁₁ = (12 × 12) u + (22 × 1) u + (11 × 16) u = 342 u.
ii. The formula of the oxide will be XO or XO₂.

Question 16.
i. If the valency of carbon is 4 and that of sulphur is 2. What is the formula of the compound formed between carbon and sulphur atoms.
ii. Determine the molecular mass of NH₄OH.
Solution:
i. The formula of compound can be written by exchanging the valencies. Therefore, the formula is C₂S₄ or CS₂.
ii. Molecular mass of NH₄OH 14 + (5 × 1) + 16 = 35 u.

Question 17.
i. If an element X has its valence equal to 3, what will be its formula with carbonate ion?
ii. Calculate the formula unit mass of NaHCO₃. [Atomic mass of Na = 23 u, H = 1 u, C = 12 u, O = 16 u]
Solution:
i.  = X₂(CO₃)₃
ii. NaHCO₃ = (Atomic mass of Na) + (Atomic mass of H) + (Atomic mass of C) + (3 × Atomic mass of O)
= (23 + 1 + 12 + 3 × 16) = 84 u.

Question 18.
Write the name of the compound PCl₃ and explain the basis of its naming.
Solution:
The compound is phosphorus trichloride. In covalent compounds, prefixed aree used to indicate the number of atoms; “tri” shows three chlorine atoms.
The first element is named normally, and the second end with “ide”.

Question 19.
Why do ionic compounds form crystalline solids with a definite shape?
Solution:
Ionic compounds consist of oppositely charged ions arranged in a regular repeating pattern called crystal lattice. This orderly arrangement gives them a definite shape and structure.

Question 20.
Why do noble gases not usually form compounds with other elements?
Solution:
Nobel gases have a completely filled outermost shell (stable electronic configuration). Hence, they neither gain, lose nor share electrons and generally do not form compounds.

Question 21.
i. An element has a valency of 3. Write the simplest formula for sulphide of the element.
ii. Name the compound represented by formula K₂SO₄.
Solution:
i. The simplest formula for sulphide of the element is X₂S₃.
ii. The compound represented by the formula is potassium sulphate.

Atomic Foundations of Matter Class 9 Short Question Answer

Question 1.
i. What mass of silver nitrate will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate?
ii. On which law is the above reaction based and state the law?
Solution:
i. Silver nitrate + Sodium chloride → Silver chloride + Sodium nitrate
5.85 g xg 14.35 g 8.5 g
Total mass of reactants = Total mass of products
x + 5.85 = 14.35 + 8.5
⇒ x + 5.85 = 22.85
⇒ x = 22.85 – 5.85
⇒ x = 17 g
Therefore, silver nitrate is 17 g.

ii. It is based on law of conservation of mass which states that matter can neither be created nor be destroyed in a chemical reaction.

Question 2.
Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide. When 10 g of calcium carbonate is decomposed completely then 5.6 g of calcium oxide is formed? Calculate the mass of carbon dioxide formed. Which law of chemical combination will you use in solving this problem? State the law.
Solution:
The reaction occurs as follows

According to the law of conservation of mass,
Total mass of reactant(s) = Total mass of products(s)
⇒ 10 g = 5.6 g + Mass of CO₂
⇒ Mass of CO₂ = 10 – 5.6 = 4.4g
This problem is solved using law of conservation of mass according to which mass can neither be created nor be destroyed during a chemical reaction.

Question 3.
i. State the law of constant proportion with example.
ii. In a compound carbon and oxygen react in a ratio of 3:8 by mass to form carbon dioxide. What mass of oxygen is required to react completely with 9 g carbon?
Solution:
i. Law of constant proportion states that, “a pure chemical compound always consists of the same elements that are combined together in a fixed (or definite) proportion by mass”, e.g. in a compound such as water, the ratio of the mass of hydrogen to the mass of oxygen is always 1 : 8, whatever the source of water.

ii. Carbon : oxygen (by mass) = 3 : 8,
i.e. 3 g of carbon requires 8 g of oxygen to form carbon dioxide.
\(\frac{3}{8}=\frac{9}{x}\) ;
x = \(\frac{8 \times 9}{3}\) = 24 g
∴ 9 g of carbon requires 24 g of oxygen to form carbon dioxide.

Question 4.
State three points of differences between an atom and a molecule.
Solution:

Atom	Molecule
i. An atom is the smallest indivisible particle of an element that takes part in a chemical reaction.	A molecule is the smallest particle of an element or compound which has the properties of that element or compound.
ii. An atom may or may not exist independently.	A molecule is capable of independent existence.
iii. Examples: hydrogen (H), oxygen (O).	Examples: hydrogen molecule (H2), oxygen molecule (O2), water molecule (H2O).

Question 5.
Carefully observe the given diagram illustrating the formation of a molecule and answer the questions that follow.

(i) Identify the exact type of chemical bond formed between the oxygen and hydrogen atoms in the newly formed water molecule. Justify your answer.
(ii) After the formation of the water molecule, what noble gas electronic configuration does the central oxygen atom attain? Explain how it achieves this.
(iii) By observing the outermost shell of the central oxygen atom in the final molecule, determine the number of non-bonding electrons (lone pairs) present.
Solution:
(i) Type of bond: Single covalent bond.
Justification:
The bond is formed by the mutual sharing of one pair of electrons (one electron from hydrogen and one from oxygen) in each overlapping region. Since exactly one pair is shared between the two atoms, it is a single covalent bond.

(ii) Noble gas configuration: Neon (Ne) configuration (2, 8).
Initially, oxygen has an electronic configuration of 2, 6 (6 valence electrons). By sharing 2 electrons (one with each hydrogen atom), its outermost shell now contains 8 electrons, completing its octet and achieving the stable configuration of neon.

(iii) There are 4 non-bonding electrons. These 4 electrons make up exactly 2 lone pairs on the outermost shell of the oxygen atom that do not participate in bonding.

Question 6.
i. Give one point of difference between an atom and an ion.
ii. Give one example each of a polyatomic cation and a polyatomic anion.
iii. Write the chemical formula for the chloride of magnesium.
Solution:
i. An atom is always neutral i.e. it does not carry any charge, while an ion carries either positive charge or negative charge.
ii. Ammonium [NH₄]⁺ ion is a polyatomic cation, while sulphate [SO₄]^2- ion is a polyatomic anion.
iii. Chemical formula for the chloride of magnesium is , MgCl₂.

Question 7.
Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
i. Ammonia
ii. Carbon monoxide
iii. Hydrogen chloride
iv. Aluminium fluoride
V. Magnesium sulphide
Solution:

Compound	Chemical Formula	Ratio by Mass of the Combining Elements
i. Ammonia	NH3	N : H = 14 : 3
ii. Carbon monoxide	CO	C : O = 12 : 16 = 3 : 4
iii. Hydrogen chloride	HCl	H : Cl = 1 : 35. 5
iv. Aluminium fluoride	AlF3	Al : F = 27 : 57 = 9 : 19
v. Magnesium sulphide	MgS	Mg : S = 24 : 32 = 3 : 4

Question 8.
An element X has one electron in its outermost shell and reacts with chlorine to form a compound that is soluble in water. Predict
(i) the formula of the compound
(ii) the type of bond formed
(iii) the electrical conductivity of its aqueous solution
Solution:
(i) The formula of the compound is XCl. Element X loses one electron to form X⁺ and chlorine gains one electron to form Cl^–, combining in 1 : 1 ratio.

(ii) The bond formed is ionic. This is because there is transfer of electron from X to chlorine, leading to formation of oppositely charged ions.

(iii) The aqueous solution will conduct electricity. In water, the compound dissociates into free ions (X⁺ and Cl^–), which move and carry electric current.

Question 9.
Write the chemical formulae of following compound, using criss-cross method.
i. Magnesium bicarbonate
ii. Barium nitrate
iii. Potassium nitrate
S0lution:

Formula: Mg(HCO₃)₂

ii.

Formula: Ba(NO₃)₂

iii.

Formula: KNO₃

Question 10.
A student performs an experiment to study the properties of different compounds such as sodium chloride, copper sulphate, sugar and naphthalene. He tests their solubility in water and kerosene, and also checks their electrical conductivity using the given setup (battery, electrodes and bulb). Based on his observations, he notices that some compounds dissolve in water and conduct electricity, while others do not.

Answer the following questions.
(a) Identify which type of compounds (ionic or covalent) will conduct electricity in aqueous solution.
(b) Explain why these compounds do not conduct electricity in solid state but conduct in aqueous solution.
(c) Which type of compounds are likely to dissolve in kerosene and why?
Solution:
(a) Ionic compounds conduct electricity in aqueous solution because they dissociate into free ions. These mobile ions act as charge carriers and allow the flow of electric current.

(b) In solid state, ions are held tightly in a fixed lattice and cannot move. In aqueous solution, the lattice breaks and ions become free to move, enabling conduction of electricity.

(c) Covalent compounds are likely to dissolve in kerosene both have a similar chemical nature, following the rule “like dissolves like”. Since both the compound and the kerosene are alike in their makeup, they mix together easily.

Question 11.
SO₂ is an air pollutant released during burning of fossil fuels and from automobile exhaust.
(a) Write the names of elements present in this gas.
(b) What are the valencies of sulphur in SO₂ and SO₃?
ii. Define the term molecular mass.
iii. Determine the molecular mass of ZnSO₄ [Atomic mass of Zn = 65 u, S = 32 u and O = 16 u].
Solution:
i. (a) Sulphur and oxygen
(b) Valency of sulphur in SO₂ = 4
Valency of sulphur in SO₃ = 6

ii. Molecular mass It is the sum of the atomic masses of all the atoms present in a molecule of the substance.

iii. Molecular mass of ZnSO₄ = 65 + 32 + (4 × 16) = 161 u.

Question 12.
i. State the atomicity of sulphur molecule.
ii. Calculate the formula unit mass of CaCl₂.
Solution:
i. Atomicity of sulphur is 8 i.e. it is a polyatomic molecule as it exists as S₈.
ii. Formula unit mass of CaCl₂ = 1 × (atomic mass of Ca) + (2 × atomic mass of Cl)
= (1 × 40) + (2 × 355) = 111 u.

Question 13.
i. An element X has a valency of 2. Write the chemical formula for
(a) bromide of the element
(b) oxide of the element
ii. Define formula unit mass of a substance.
Solution:
i. (a) Valency of X = 2
Valency of bromine = 1
∴ The formula of compound

(b) Valency of X = 2
Valeny of oxygen = 2
The formula of compound

ii. Formula unit mass is the sum of atomic masses of all atoms present in a formula unit of compound. It is calculated by adding the atomic masses of all the atoms present in one formula unit.

Question 14.
i. An element \({ }_7^{14} A\) exists as diatomic gas in nature which is relatively inert and forms 78% of earth’s atmosphere.
(a) Identify the gas and write its molecular formula. Write the formulae of nitrite and nitrate ions.
(b) Calculate the molecular mass of
I. NH₄NO₃ and
II.HNO₃
[Given atomic masses N = 14 u, O = 16 u, H = 1 u]
ii. Calculate the formula unit mass of Na₂SO₃.
[Atomic mass of Na = 23 u, S = 32 u, O = 16 u, H = 1 u and NA = 6.022 × 10²³ mol^-1]
Solution:
i. (a) Nitrogen gas (N₂), nitrite ion (\(N \mathrm{O}_2^{-}\)), nitrate ion (\(N \mathrm{O}_3^{-}\))

(b) I. Molecular mass of NH₄NO₃
= 14 + (1 × 4) +14 + (3 × 16) = 80 u

II. Molecular mass of HNO₃
= 1 + 14 + (3 × 16) = 63 u

ii. Formula unit mass of Na₂SO₃
= (2 × 23) + 32 + (3 × 16)
= 46 + 32 + 48 = 126 u.

Question 15.
Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃.
[Given, atomic masses of Zn = 65 u, Na = 23 u, K =39 u, C = 12 u and O = 16 u]
Solution:
i. Formula unit mass of ZnO (zinc oxide)
= 65 + 16 = 81 u

ii. Formula unit mass of Na₂O (sodium oxide)
= (23 × 2) + (16 × 1)
= 46 + 16 = 62 u

iii. Formula unit mass of K₂CO₃ (potassium carbonate)
= (39 × 2) + (12 × 1) + (16 × 3)
= 78 + 12 + 48 = 138 u.

Question 16.
Name any two monovalent cations, divalent cations and trivalent cations. Also name any one compound each one of them make.
Solution:
Monovalent cation: Na⁺, K⁺
NaCl is the compound formed by Na⁺.

Divalent cation: Mg²⁺, Ca²⁺
MgO is the compound formed by Mg²⁺.

Trivalent cation: Al³⁺, Bi³⁺
AlCl3 is the compound formed by Al³⁺.

Question 17.
The symbols of some of the ions are given below Na⁺, Mg²⁺, H⁺, \(C \mathrm{O}_3^{2-}\), Cl^–, S^2-
Using this information, find out the formulae of
i. sodium carbonate
ii. magnesium chloride
iii. hydrogen sulphide
Solution:
i. The formula for sodium carbonate is Na₂CO₃.
ii. The formula for magnesium chloride is MgCl₂.
iii. The formula for hydrogen sulphide is H₂S.

Question 18.
An element X has a valency 1.
i. Write the chemical formula of its phosphide.
ii. Write the chemical formula of its chloride.
iii. Is element X a metal or a non-metal?
Solution:
i. The phosphide of element X (valency = 1) is X₃P, as phosphorus has a valency of 3.
ii. Its chloride is XCl, since chlorine has a valency of 1.
iii. Element X is a metal, because it forms positive ions with non-metals.

Atomic Foundations of Matter Class 9 Long Question Answer

Question 1.
Hydrogen gas reacts with oxygen gas to form water shown in the following reaction,
2H₂ + O₂ → 2H₂O
This experiment was conducted in a closed container. Answer the following questions.
i. Which law of chemical combination is illustrated in this reaction? State the law.
ii. Calculate the total mass of reactants and the total mass of the product. Does this support your answer in part (i)?
iii. What will be the mass ratio of hydrogen to oxygen in water? Which law does this constant ratio support?
iv. Calculate the total number of atoms of hydrogen and oxygen present on the reactants side (left side).
v. Give one more example of a reaction that follows the law of constant proportion and name the compound formed.
Solution:
i. This reaction illustrates the law of conservation of mass.
It is stated as “the mass can neither be created nor be destroyed in a chemical reaction. The total mass of the products is equal to the total mass of the reactants.

ii. Mass of reactants = 4 g (H₂) + 32 g (O₂) = 36 g Mass of product = 36 g (H₂O)
Since the mass of reactants is equal to the mass of the product, this supports the law of conservation of mass.

iii. In water, hydrogen and oxygen combine in the ratio Hydrogen : Oxygen = 4 g : 32 g = 1 : 8
This supports the law of constant proportion, which states that a chemical compound always contains the same elements in a fixed ratio by mass.

iv. Hydrogen atoms = 2 × 2 = 4
Oxygen atoms = 1 × 2 = 2
Thus, there are 4 hydrogen atoms and 2 oxygen atoms in the reactant side or left hand side of the equation.

v. The compound formed is carbon dioxide (CO₂). Carbon and oxygen always combine in a mass ratio of 12 : 32 = 3 : 8, which is constant.
C + O₂ → CO₂

Question 2.
When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.0 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Solution:
First we find the proportion of mass of carbon and oxygen in carbon dioxide.
In CO₂, C : O = 12 : 32 or 3 : 8
In other words, we can say that
∵ 12.00 g carbon reacts with oxygen = 32.00 g
∴ 3.00 g carbon will react with oxygen = 8 g

Therefore, 3.00 g of carbon will always react with 8.0 g of oxygen to form 11 g of carbon dioxide, even if a large amount (50.00 g) of oxygen is present.

This means when 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be used to produce 11.00 g of carbon dioxide.

The remaining 42.00 g of oxygen will remain as it is. This reaction will be governed by the law of constant proportions.

Question 3.
i. Define Dalton’s atomic theory.
ii. According to Dalton, what are atoms made of? Are they divisible or indivisible?
iii. Give atleast three postulates of Dalton’s atomic theory.
iv. Name two postulates of Dalton’s theory that have been proved incorrect in modern science.
v. How did Dalton’s Atomic Theory contribute to the understanding of chemical reactions?
Solution:
i. Dalton’s atomic theory states that matter is made up of very small indivisible particles called atoms, which take part in chemical reactions and combine in fixed ratios to form compounds.

ii. According to Dalton, atoms are indivisible and indestructible particles of matter. They cannot be created or broken down by chemical means.

iii. Three postulates of Dalton’s atomic theory are
1. All matter is made of indivisible particles called atoms.
2. Atoms of a given element are identical in mass and chemical properties.
3. Atoms combine in simple whole-number ratios to form compounds.

iv. Two postulates now proved incorrect are
1. Atoms are indivisible: Modern science shows atoms can be divided into protons, neutrons, and electrons.
2. Atoms of an element are identical in mass: Isotopes of the same element have different masses.

v. Dalton’s theory helped to understand chemical reactions as rearrangements of atoms, where no atom is created or destroyed, only recombined.

Question 4.
Five ions/elements A, B, C, D and E are given as
A. Chloride
B. Sulphide
C. Nitride
D. Phosphate
E. Sulphite
i. Arrange the given ions/elements in increasing order of their valency.
ii. Which of the following elements are non-metallic elements.
iii. Explain their atomicity with symbol representation.
iv. What are ions?
v. What are the types of ions?
Solution:
i. The given five ions/elements are
(A) Chloride → 1 valency
(B) Sulphide → 2 valency
(C) Nitride → 3 valency
(D) Phosphate → 3 valency
(E) Sulphite → 2 valency
A < B = E < C = D

ii. All the elements involved in these ions (chlorine, Sulphur, nitrogen, phosphorus, and oxygen) are non-metallic elements. In fact, almost all negative ions (anions) are formed by non-metals because they gain electrons.

iii.

Ion/Element	Symbol/Formula	Atomicity	Type
Chloride	Cl–	1	Monoatomic
Sulphide	S2-	1	Monoatomic
Nitride	N3-	1	Monoatomic
Sulphite	\(S \mathrm{O}_3^{2-}\)	4 (1S + 30)	Polyatomic
Phosphate	\(P \mathrm{O}_4^{3-}\)	5 (1P + 40)	Polyatomic

iv. Ions are charged particles formed when atoms or groups of atoms lose or gain electrons. For example, sodium ion (Na⁺), chloride ion (Cl^–), and hydroxide ion (OH^–).

v. An ion can be of two types i.e. negatively or positively charged. A negatively charged ion is called an ‘anion’ e.g. Br^– and the positively charged ion, a ‘cation’, e.g. Na⁺.

Question 5.
i. Classify the following molecules based on their atomicity.
(a) F₂
(b) NO₂
(c) N₂O
(d) C₂H₆
(e) P₄
(f) H₂O₂
ii. Arrange the above compounds in increasing order of their atomicity.
iii. Explain the term atomicity. Give one example each of monoatomic and triatomic molecules not listed above.
iv. Why is P₄ considered polyatomic despite being a single element?
Solution:
i.

Molecules	Atomicity
F2	2 (diatomic)
NO2	3 (triatomic)
N2O	3 (triatomic)
C2H6	8 (polyatomic)
P4	4 (polyatomic)
H2O2	4 (polyatomic)

ii. The increasing order of their atomicity is as
F₂ < NO₂ = N₂O < P₄ = H₂O₂ < C₂H₆

iii. Atomicity is defined as the number of atoms present in one molecule of an element or a compound.
Monoatomic → He
Triatomic → CO₂

iv. P₄ is considered polyatomic because each molecule contains four phosphorus atoms bonded together.

Question 6.
Complete the following table by writing the chemical formulae of the compounds formed by the cations on the left and the anions at the top. CaCl₂ is given as an example.

Solution:
The formulae are written by balancing the charges of cations and anions using the criss-cross method. The total positive and negative charges must be equal to form a neutral compound.
The completed formulae for the given table are as follows:

	Cl–	\(C \mathrm{O}_3^{2-}\)	OH–
Ca2+	CaCl2	CaCO3	Ca(OH)2
Na+	NaCl2	Na2CO3	NaOH
Al3+	AlCl3	Al2(CO3)3	Al(OH)3
Mg2+	MgCl2	MgCO3	Mg(OH)2

Question 7.
Write down the names of compounds represented by the following formulae.
i. Al₂(SO₄)₃
ii. CaCl₂
iii. K₂SO₄
iv. KNO₃
v. CaCO₃
Solution:
i. Aluminium sulphate – [Al₂(SO₄)₃ ]
ii. Calcium chloride – (CaCl₂)
iii. Potassium sulphate – (K₂SO₄)
iv. Potassium nitrate – (KNO₃)
v. Calcium carbonate – (CaCO₃)

Question 8.
Write the chemical formulae of the following.
i. Magnesium chloride
ii. Calcium oxide
iii. Copper nitrate
iv. Aluminium chloride
v. Calcium carbonate
Solution:
i. MgCl₂
ii. CaO
iii. Cu(NO₃)₂
iv. AlCl₃
v. CaCO₃

Question 9.
Give the names of the elements present in the following compounds.
i. Quicklime
ii. Hydrogen bromide
iii. Baking powder
iv. Potassium sulphate
Solution:
i. Quicklime (Calcium oxide) – CaO
Elements: Calcium and oxygen

ii. Hydrogen bromide – HBr
Elements: Hydrogen and bromine

iii. Baking powder (Sodium hydrogen carbonate) – NaHCO₃
Elements: Sodium, hydrogen, carbon and oxygen

iv. Potassium sulphate – K₂SO₄
Elements: Potassium, sulphur and oxygen

Question 10.
Calculate the molecular masses of
H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH
Solution:
i. Molecular mass of H₂ (hydrogen) = Atomic mass of hydrogen × 2
= 1 × 2 = 2 u

ii. Molecular mass of O₂ (oxygen) = Atomic mass of oxygen × 2
= 16 × 2 = 32 u

iii. Molecular mass of C₂ (chlorine) = Atomic mass of chlorine × 2
= 35.5 × 2 = 71 u

iv. Molecular mass of CO₂ (carbon dioxide) = (Atomic mass of carbon × 1) + (Atomic mass of oxygen × 2)
= 12 × 1 + (16 × 2)
= 12 + 32 = 44 u

v. Molecular mass of CH₄ (methane) = (Atomic mass of carbon × 1) + (Atomic mass of hydrogen × 4)
= 12 × 1 + (1 × 4)
= 12 + 4 = 16 u

vi. Molecular mass of C₂H₆ (ethane) = (Atomic mass of carbon × 2) + (Atomic mass of hydrogen × 6)
= (12 × 2) + (1 × 6)
= 24 + 6 = 30 u

vii. Molecular mass of C₂H₄ (ethene) = (Atomic mass of carbon × 2) + (Atomic mass of hydrogen × 4)
= (12 × 2) + (1 × 4)
= 24 + 4 = 28 u

viii. Molecular mass of NH₃ (ammonia) = (Atomic mass of nitrogen × 1) + (Atomic mass of hydrogen × 3)
= (14 × 1) + (1 × 3)
= 14 + 3 = 17u

ix. Molecular mass of CH₃OH (methanol or methyl alcohol) = (Atomic mass of carbon × 1) + (Atomic mass of hydrogen × 3) + (Atomic mass of oxygen × 1) + (Atomic mass of hydrogen × 1)
= (12 × 1) + (1 × 3) + (16 × 1) + (1 × 1)
= 12 + 3 + 16 + 1 = 32 u.

Question 11.
Calculate the molar mass of the following substances:
i. Ethyne, C₂H₂
ii. Sulphur molecule, S₈
iii. Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
iv. Hydrochloric acid, HCl
v. Nitric acid, HNO₃
Solution:
i. Molar mass of C₂H₂ = (2 × Atomic mass of C) + (2 × Atomic mass of H)
= (2 × 12) + (2 × 1) = 26 u

ii. Molar mass of S₈ = 8 × Atomic mass of S
= 8 × 32 = 256 u

iii. Molar mass of P₄ = 4 × Atomic mass of P
= 4 × 31 = 124 u

iv. Molar mass of HCl = (Atomic mass of H) + (Atomic mass of Cl)
= 1 + 35.5 = 36.5 u

v. Molar mass of HNO₃ = (Atomic mass of H) + (Atomic mass of N) + (3 × Atomic mass of O)
= 1 + 14 + (3 × 16)
= 15 + 48 = 63 u

Question 12.
i. Write the molecular formulae of all the compounds that can be formed by the combination of following ions :
Cu²⁺, Na⁺, Fe³⁺, Cl^–, \(S \mathrm{O}_4^{2-}\), \(P \mathrm{O}_4^{3-}\) [NCERT Exemplar]
ii. What is the molecular mass of PH₃ and sulphur dioxide?
Solution:
i. Compounds of Cu²⁺

(a) With Cl^– →  = CuCl₂
(b) With \(S \mathrm{O}_4^{2-}\) →  = CuSO₄
(c) With \(P \mathrm{O}_4^{3-}\) →  = Cu₃(PO₄)₂

Compounds of Na⁺

(a) With Cl^– →  = NaCl
(b) With SO₂ →  =Na₂SO₄
(c) With PO₄ →  = Na₃PO₄

Compounds of Fe³⁺

(a) With Cl^– →  = FeCl₃
(b) With SO₄ →  = Fe₂(SO₄)₃
(c) With PO₃ →  = FePO₄

ii. Molecular mass of PH₃ = 31 + 3 = 34
Molecular mass of sulphur dioxide (SO₂) = 2 × atomic mass of O + atomic mass of S × 1
= (2 × 16) + 32 = 64 u.

Question 13.
i. State the two important laws of chemical combination. How Dalton’s atomic theory explains the two Laws?
ii. Write the chemical formulae of nitrates (\(N \mathrm{O}_3^{-}\)) of Na⁺, K⁺, Al³⁺, Mg²⁺, Ca²⁺, Zn²⁺.
Solution:
The two important laws of chemical combination are

1. Law of conservation of mass: Mass is neither be created nor be destroyed in a chemical reaction.
2. Law of constant proportions: A pure chemical compound always contains the same elements in a fixed proportion by mass.

Dalton’s Atomic theory explains these laws

• According to Dalton, atoms are indivisible and indestructible, which supports the law of conservation of mass.
• It also states that compounds are formed when atoms combine in fixed ratios, explaining the law of constant proportions.

ii.

Ion	Chemical Formula of Nitrate
Na+	NaNO3
K+	KNO3
Al3+	Al(NO3)3
Mg2+	Mg(NO3)2
Ca2+	Ca(NO3)2
Zn2+	Zn(NO3)2

Question 14.
i. What is the difference between 2H and H₂ ?
ii. The valency of an element X is 4. Write the formula of its oxide.
iii. Show how ammonium sulphate is formed and write its chemical formula.
iv. Two compounds are given:
A – Calcium hydroxide, Ca(OH)₂
B – Aluminium chloride, AlCl₃
Calculate the molecular masses of both compounds.
Solution:
i. 2H represents two separate atoms of hydrogen.
H₂ represents a molecule of hydrogen made of two atoms chemically bonded together.

ii. The formula of its oxide is XO₂.

iii. Ammonium sulphate

Formula: (NH₄)₂SO₄

iv. Molecular mass of Ca(OH)₂
= 40+ 2 × (16 + 1)
= 40 + (2 × 17)
= 40 + 34 74 u.
Molecular mass of AlCl₃
= 27 + 3 × 355
= 27 + 106.5 = 133.5 u.

Atomic Foundations of Matter Class 9 Case Based Questions

Direction (Question Nos. 1 – 6) Answer the questions on the basis of your understanding of the following passage and related studied concepts:

Question 1.
In order to verify the law of conservation of mass, a student mixed 6.3 g of sodium carbonate and 15 g of ethanoic acid in a conical flask. After the experiment, he weighed the flask again. The weight of the residue in the flask was only 18 g. He approached the teacher who guided him to carry the experiment in a closed flask with a cork. There was no difference in weight of the flask before and after the experiment.
Answer the following questions based on the above information.
A. What was the mistake committed by the student?
B. Why did the measured mass decrease in the first experiment?
C. State and name the law of chemical combination which govern these experiments.
Or
What important lesson did the student learn from this activity?
Solution:
A. The mistake committed by the student is that he carried out the experiment in an open flask.

B. The measured mass decreased in the first experiment because the reaction between sodium carbonate and ethanoic acid released carbon dioxide gas, which escaped from the open flask. As the gas was not trapped, the total mass of the system appeared to decrease.

C. The given experiment is governed by the law of conservation of mass. In this, the total mass of the reactants is equal to the total mass of the products and there is no change in mass during chemical reactions.
Or
The student learnt that a chemical reaction should always be carried out in a closed flask to prevent the loss of any gaseous product. This ensures that the total mass remains constant, correctly demonstrating the law of conservation of mass. The student should also consult the teacher to understand the importance of performing such experiments in a closed system.

Question 2.
Ravi was performing some experiments related to the laws of chemical combination in his science laboratory under the guidance of his chemistry teacher Mr. John. His teacher gave him different samples of reacting species having different masses. Ravi performed the experiments and collected data as

Compounds	Reactant species	Masses of reactant species (in gram)
H2O	H	1
	O	16
CO2	C	12
	O	32
NH3	N	14
	H	3

Answer the following questions based on the above information.
A. Calculate the mass of carbon dioxide gas formed.
B. Write a chemical reaction for the reaction between nitrogen and hydrogen.
C. In a given sample ammonia contains 3 g of hydrogen and 14 g of nitrogen. If another sample contains 5g of hydrogen then how much amount of nitrogen present in second sample?
Or
Name the two laws of chemical combination and state them.
Solution:
A. Mass of reactants = Mass of products
Mass of carbon dioxide = Mass of carbon + Mass of oxygen
= 12 + 32 g = 44g.

B. The chemical reaction for the reaction between nitrogen and hydrogen is as follows
Nitrogen + Hydrogen → Ammonia.

C. The ratio of mass of hydrogen and nitrogen = 3 : 14
If 5g of hydrogen, then nitrogen = \(\frac{5 \times 14}{3}\)
Mass of nitrogen = 23.3 g.
Or
The two laws of chemical combination are as follows

• Law of conservation of mass: This law states that matter can neither be created nor be destroyed in a chemical reaction.
• Law of definite proportions: It states that in any compound formed by two or more elements, the elements combine in a fixed ratio by mass.

Question 3.
A student performed an experiment to under¬stand how compounds are formed from elements. He observed that when a piece of sodium metal reacts with chlorine gas, a white crystalline substance is formed. On studying the process, he found that sodium loses one electron while chlorine gains one electron, leading to the formation of charged particles which arrange themselves in a regular pattern.
Answer the following questions based on the above information.
A. What type of bond is formed between sodium and chlorine?
B. Why does sodium lose an electron while chlorine gains an electron?
C. i. What kind of ions are formed from sodium and chlorine?
ii. What is the charge present on each of these ions?
Or
i. Why is the compound formed electrically neutral?
ii. What holds the ions together in the compound?
Solution:
A. An ionic bond is formed between sodium and chlorine.

B. Sodium loses one electron to achieve a stable electronic configuration, while chlorine gains one electron to complete its octet.

C. i. Sodium forms a cation (Na⁺) and chlorine forms an anion (Cl^–).

ii. Sodium ion has a +1 charge and chloride ion has -1 charge.
Or

i. The compound is electrically neutral because total positive and negative charges are equal.
ii. The ions are held together by strong electrostatic forces of attraction.

Question 4.
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance. It is therefore, the relative mass of a molecule expressed in atomic mass units (u).
Depending upon the number of atoms of same or different elements present in the molecule, it can be monoatomic, diatomic, triatomic, tetra-atomic or polyatomic molecule.
Answer the following questions based on the above information.
A. Calculate the ratio by mass of the combining elements in the compound: Methanol (CH₃OH).
B. What is the relative molecular mass of H₂O?
C. How many kinds of atoms are present in a molecule of copper carbonate (CuCO₃)?
Or
Explain polyatomic molecules with example.
Solution:
A. Methanol (CH₃OH) → C : H : O :: 12 : 4 : 16 = 3 : 1 : 4.

B. The relative molecular mass of water (H₂O) = (2 × 1) + (1 × 16) = 18 u

C. Copper carbonate (CuCO₃), contains three type of atoms, i.e. one copper atom, one carbon atom and three oxygen atoms.
Or
Polyatomic molecules are compounds or elements that are made up of four or more atoms in a stable structure. Sg has more than 4 atoms. Therefore, it is an example of polyatomic molecule (octa-atomic).

Question 5.
In a chemistry lab, four students were asked to determine the formula of magnesium oxide. They heated known masses of magnesium in air, which combined with oxygen to form magnesium oxide. The results obtained by each student are given below.

Student	Mass of Mg (g)	Mass of O2 combined (g)	Mass of MgO formed (g)
A	2.4	1.6	4.0
B	3.0	2.0	5.0
C	4.8	3.2	8.0
D	6.0	4.0	10.0

Answer the following questions based on the above information.
A. In this experiment, a student strongly heats 10.0 g of magnesium in a crucible to form magnesium oxide. If the mass of the magnesium oxide formed is 16.6 g, what can be concluded about the reaction?
i. The mass of oxygen that combined with magnesium is 26.6 g.
ii. The mass of oxygen that is combined with magnesium is 6.6 g.
iii. The reaction does not follow the Law of conservation of mass.
Justify your answer.

B. If a student heats 12.0 g of magnesium in air, what mass of oxygen will combine with it to form magnesium oxide and what will be the mass of magnesium oxide (MgO) formed?
Or
Write the chemical formula of magnesium oxide and calculate its formula unit mass. [Atomic masses: Mg = 24 u, O =16 u]

C. Which of the following statements is correct based on the above data?
i. The ratio of magnesium to oxygen by mass in magnesium oxide is 3 : 2.
ii. The mass of oxygen that combines with 12 g of magnesium is 6 g.
iii. The law of chemical combination is illustrated by the given data.
iv. The formula of magnesium oxide is MgO₂.
Solution:
A. ii. The mass of oxygen that combined with magnesium is 6.6 g.
According to the law of conservation of mass
Mass of magnesium oxide = Mass of magnesium + Mass of oxygen
16.6 g = 10.0 g + Mass of oxygen
So, Mass of oxygen = 16.6 g -10.0 g = 6.6 g.

B. Using the fixed ratio (Mg : O = 3 : 2)
For 12 g Mg,
Mass of oxygen = \(\frac{2}{3}\) × 12 = 8 g
Mass of MgO formed = 12g + 8g = 20g
Or
Chemical formula of magnesium oxide is MgO.
Calculation of Formula Unit Mass:
Atomic mass of magnesium (Mg) = 24 u
Atomic mass of oxygen (O) = 16 u
Formula unit mass of MgO = Mass of Mg + Mass of O
Formula unit mass of MgO = 24 u +16 u = 40 u.

C. The correct statement is (i).
The ratio of magnesium to oxygen by mass in magnesium oxide is 3 : 2.
From the data,
the mass ratio of Mg : O is consistent by 24 : 16 = 3 : 2.

Question 6.
The combining capacities of different elements are compared with that of hydrogen. The valency of hydrogen is taken as 1 and the valencies of all other elements are measured against this standard. An atom of calcium always combines with two atoms of hydrogen to form calcium hydride, a compound of calcium and hydrogen. Hence, the combining capacity or valency of calcium is twice that of hydrogen.

There are some elements which show different valencies in different compounds. Copper shows two valencies, one and two.

In the red oxide of copper, its valency is one, while in the black oxide, its valency is two. Similarly, iron shows valency two and three.
Answer the following questions based on the above information.
A. The element B shows valencies of 4 and 6. Find formulae of its two oxides.
B. The formula of the sulphate of an element X is X₂(SO₄)₃. What is the formula of nitride of the element X?
C. What is meant by the term chemical formula? How is it related to valency of the element?
Or
The formula of oxide of an element Z is Z₂O₃. What is the valency of element Z?
S0lution:
A. For valencies 4 and 6, the oxides are BO₂ (for valency 4) and BO₃ (for valency 6).

B. From X₂SO₄, valency of X is 1, so the formula for its nitride is X₃N.

C. A chemical formula represents the type and number of atoms in a compound which is determined by the valency of the elements. The chemical formula of a compound is written by balancing the valencies of the combining elements.
Or
In oxide Z₂O₃, the valency of Z is 3.