NCERT Solutions for class 9 science chapter 9 - Atomic Foundations of Matter

Download Atomic Foundations of Matter Class 9 Science Chapter 9 Questions and Answers PDF

Students can refer to the button given below to download the NCERT Solutions for Class 9 Science chapter 9 Atomic Foundations of Matter Questions and Answers PDF. Students should download this PDF and prepare effectively for their exams using these NCERT Solutions for Class 9 Science.

Download PDF

NCERT Solution for Class 9 Science Atomic Foundations of Matter (In-text Exercise)

Refer to the detailed Class 9 Science NCERT Solutions Chapter 9 Atomic Foundations of Matter. These NCERT Solutions help students understand the concepts easily.

Class 9 Science Chapter 9 Atomic Foundations of Matter (Think it Over Questions)

Refer to the detailed Think it over questions of NCERT Class 9 Science Chapter 9 Atomic Foundation of Matter.

Page 162

Question: Water can be obtained from various sources. Are all these samples of water chemically identical?

Answer:

Yes, all samples of water are identical because of the same chemical formula $\mathrm{H}_2 \mathrm{O}$.

Question: Oxygen is sometimes represented as O and sometimes as $\mathrm{O}_2$. What is the difference between these symbols?

Answer:

O represents only 1 oxygen atom while $\mathrm{O}_2$ represents 1 oxygen molecule which contains 2 oxygen atoms. Oxygen exists as $\mathrm{O}_2$ in nature.

Question: Why does dissolved salt in water conduct electricity, but sugar does not?

Answer:

The chemical formula of salt is NaCl when it dissolves in water it produces ions there ions move freely in solutions and conducts electricity while sugar does not produce ions while dissolving. Hence, sugar does not conduct electricity.

Class 9 Science Chapter 9 Atomic Foundations of Matter (Pause and Pounder)

Below are detailed pause and ponder questions from the atomic foundations of matter. They are prepared by experts to help students understand the concepts easily.

Page- 166

Question 1: A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is left in the beaker. Does this mean the Law of Conservation of Mass is violated? Explain.

Answer:

According to the law of conservation of mass, mass can never be created and nor be destroyed; it can only transfer from one form to another. In this case when ethanol burns it reacts with atmospheric oxygen to form carbon dioxide and water. These products are gases and escape in the environment due to which no residue is left in the beaker.

$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$

Question 2: When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the Law of Conservation of Mass?

Answer:

According to the law of conservation of mass, the total mass of the reactant is equal to the total mass of the product.

Mass of reactant = Mass of Hydrogen + Mass of Oxygen

Mass of reactant = 20+160

Mass of reactant = 180

Mass of reactant = Mass of Product

Mass of water = 180g

Page - 167

Question 3: A compound consists of 40% sulfur and 60% oxygen by mass. In a sample of the same compound containing 20 g of sulfur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?

Answer:

Sulphur by mass = 40%

Oxygen by mass =60%

According to the law of constant proportions compounds always contain its elements in the same fixed ratio

Sulphur : Oxygen = 40:60

Ratio = 2:3

In a sample mass of sulphur is 20g

Mass of oxygen = $\frac{3}{2} \times 20$

Mass of oxygen = 30g

Question 4: Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3:4. How much oxygen will combine with 9 g of carbon to form carbon monoxide?

Answer:

Mass ratio = 3:4

3 g of carbon combines with 4g of oxygen to form CO

Mass of carbon = 9 g

Oxygen required to form CO = $\frac{4}{3} \times 9$

= 12g

Question 5: The Law of Definite Proportions holds true for compounds but not for mixtures. Give a reason.

Answer:

Elements in compounds always combine in a fixed ratio by mass while in mixtures compounds are present in any proportions and not chemically combined. The Law of Definite Proportions holds true for compounds but not for mixtures.

Question 6: Students X and Y, both prepared an oxide of copper by combining copper and oxygen in the ratios of 4:1 and 8:2, respectively. Do their results justify the Law of Constant Proportions? Explain.

Answer:

Student X combines copper and oxygen in the ratio of 4:1

Student Y combines copper and oxygen in the ratio of 8:2

By simplifying ration of student Y

8:2 = 4:1

The ratio for both X and Y is the same means compounds always contain elements in fixed ration. Hence, the Law of Constant Proportions is justified.

Page -168

Question 7: Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water.

Reason (R): According to Dalton’s Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds.

Choose the correct option:

(i) Both A and R are true, and R is the correct explanation of A.

(ii) Both A and R are true, but R is not the correct explanation of A.

(iii) A is true, but R is false.

(iv) A is false, but R is true.

Answer:

The assertion is true because according to the law of conservation of mass when 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water.

The reason is true because according to Dalton’s Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds.

But reason does not correctly explain assertion because assertion is based on conservation of mass not on whole number ratios.

Both A and R are true, but R is not the correct explanation of A.

Hence, the correct answer is option (ii)

Page - 170

Question 8: Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule ($\mathrm{N}_2$ ).

Answer:

Nitrogen has 5 valence electrons it requires 3 more electrons to complete its octet. To do so two nitrogen atoms share 3 pairs of electrons.

Structure of Nitrogen molecule

$: N \equiv N:$

Question 9: The atomic number of fluorine is 9. Explain the formation of the fluorine molecule ($\mathrm{F}_2$ )

Answer:

Atomic number of fluorine = 9

Electronic configuration = 2,7

Fluorine requires one more electron to complete its octet.

Two fluorine atoms share 1 electron to form a single covalent bond to complete their octet.

$: F-F:$

Page -171

Question 10: Show the formation of the following molecules:

(i) Carbon dioxide $\left(\mathrm{CO}_2\right)$

(ii) Hydrogen sulfide $\left(\mathrm{H}_2 \mathrm{~S}\right)$

(iii) Ammonia $\left(\mathrm{NH}_3\right)$

Answer:

(i) No. of Valence electrons of Carbon = 4

No. of valenced electrons of Oxygen = 6

Carbon requires 4 electrons to complete its octet while oxygen requires 2 more electrons to complete its octet.

Carbon shares two pairs of electrons with oxygen atoms to form a double bond.

O=C=O

(ii) No. of Valence electrons of sulfur = 6

No. of Valence electrons of Hydrogen = 1

Sulfur requires 2 more electrons to complete its octet while hydrogen requires 1 more electron to complete its octet.

Sulfur shares 1 each with 2 hydrogen.

H-S-H

(iii) No. of Valence electrons of Nitrogen = 5

No. of Valence electrons of Hydrogen = 1

Nitrogen requires 3 more electrons to complete its octet while hydrogen requires 1 more electron to complete its octet.

Nitrogen shares 1 each with 3 hydrogen.

Question 11: Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.

Answer:

Atomic number of Neon = 10

Electronic configuration = 2,8

The outermost shell of Neon contains 8 electrons, meaning the octet for neon is completed so it does not require to gain or lose any atom to complete its octet. That's why Neon neither transfers nor shares its valence electrons

Page -174

Question 12: What kind of ion will oxygen (O) form?

Answer:

Atomic number of oxygen = 8

Electronic configuration = 2,6

Oxygen requires 2 electrons to complete its octet.

Oxygen tends to gain 2 electrons to complete its octet in this process it forms a negatively charged oxide ion.

$O+2 e^{-} \rightarrow O^{2-}$

Question 13: Fill in the blanks.

Among magnesium and chlorine, magnesium atoms can give two electrons to become $\mathrm{Mg}^2+$. However, chlorine can take only one electron to become ____________. Now, __________ ion of magnesium and __________ ions of chlorine combine to give magnesium chloride.

Answer:

$\mathrm{Cl}^{-}$, 1, 2

Question 14: Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.

Answer:

Atomic number of Potassium = 19

Electronic configuration = 2,8,8,1

$K \rightarrow K^{+}+e^{-}$

Atomic number of chlorine = 17

Electronic configuration = 2,8,7

$\mathrm{Cl}+e^{-} \rightarrow \mathrm{Cl}^{-}$

$K^{+}$ and $\mathrm{Cl}^{-}$ combines to form KCl

$\mathrm{K}^{+}+\mathrm{Cl}^{-} \rightarrow \mathrm{KCl}$

Formation of Calcium Chloride

Atomic number of Calcium = 20

Electronic configuration = 2,8,8,2

$C l+e^{-} \rightarrow C l^{-}$

Atomic number of chlorine = 17

Electronic configuration = 2,8,7

Chlorine requires 1 electron to complete its octet.

$\mathrm{Cl}+e^{-} \rightarrow \mathrm{Cl}^{-}$

$\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_2$

Question 15. Illustrate how sodium sulfide $\left(\mathrm{Na}_2 \mathrm{~S}\right)$ is formed.

Answer:

Atomic number of sodium = 11

Electronic configuration = 2,8,1

Sodium can lose 1 electron to complete its octet.

$\mathrm{Na} \rightarrow \mathrm{Na}^{+}+e^{-}$

Atomic number of Sulfur = 16

Electronic configuration = 2,8,6

Sulfur needs to electrons to complete its octet

$S+2 e^{-} \rightarrow S^{2-}$

One sodium atom gives only one electron so two sodium atoms are required to provide two electrons to sulfur.

$2 \mathrm{Na}^{+}+\mathrm{S}^{2-} \rightarrow \mathrm{Na}_2 \mathrm{~S}$

Page - 177

Question 16: Name the following:

(i) $\mathrm{CO}_2$ _______________________________

(ii) $\mathrm{NO}_2$ _______________________________

(iii) $\mathrm{SF}_6$_______________________________

(iv) $\mathrm{PCl}_3$ _______________________________

Answer:

(i) Carbon dioxide

(ii) Nitrogen dioxide

(iii) Sulfur hexafluoride

(iv) Phosphorus trichloride

Question 17: Write the formula for the following:

(i) Sodium hydrogencarbonate _____________________

(ii) Sulfur dioxide ____________________________________

(iii) Ferric chloride ___________________________________

(iv) Cuprous oxide ___________________________________

Answer:

(i) $\mathrm{NaHCO}_3$

(ii) $\mathrm{SO}_2$

(iii) $\mathrm{FeCl}_3$

(iv) $\mathrm{Cu}_2 \mathrm{O}$

Question 18: Write the formulae for the compounds formed from the following pairs of ions:

(i) $\mathrm{Fe}^{3+}$ and $\mathrm{OH}^{-}$

(ii) $\mathrm{K}^{+}$and $\mathrm{CO}_3^{2-}$

Answer:

(i) $\mathrm{Fe}^{3+}$ ions combine with three $\mathrm{OH}^{-}$ ions to form $\mathrm{Fe}(\mathrm{OH})_3$

(ii) Two $\mathrm{K}^{+}$ ions required to balance one $\mathrm{CO}_3^{2-}$ ion

$\mathrm{K}_2 \mathrm{CO}_3$

Page - 179

Question 19: What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?

Answer:

When an ionic bond is present in a solid compound then it does not conduct electricity in the solid state but but conducts electricity when dissolved in water. Because when solid compounds dissolved in water, the ions become free to move and can conduct electricity.

Question 20: Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water. Predict its:

(i) formula

(ii) type of bond

(iii) electrical conductivity of its aqueous solution.

Answer:

No. of valence electrons of M = 2

M can lose 2 electrons to form $M^{2+}$ ion

Oxygen requires 2 electrons to complete its octet. Oxygen forms $O^{2-}$ ion by gaining 2 electrons.

M has 2 valenced electrons and O requires 2 electrons to complete its octet.

(i) Formula of compounds

MO

(ii) Ionic bonds formed because electrons are transferred from metal to oxygen.

(iii) Aqueous solution of compound will conduct electricity because compound dissociated into free ions in water.

Question 21. Find the molecular mass of nitric acid $\left(\mathrm{HNO}_3\right)$.

Atomic mass H = 1 u; N = 14 u; O = 16 u.

Answer:

Mass of H = 1u

Mass of N = 14u

Mass of O = 16u

Molecular mass of $\mathrm{HNO}_3$ = (1x1) + ( 1x14) + (3 x 16)

Molecular mass of $\mathrm{HNO}_3$ = 1+14+48

Molecular mass of $\mathrm{HNO}_3$ = 63

Question 22. Find the molecular mass of methane $\left(\mathrm{CH}_4\right)$.

Atomic mass - C = 12 u; H = 1 u.

Answer:

Mass of C = 12u

Mass of H = 1u

Molecular Mass of $\left(\mathrm{CH}_4\right)$ = (1x12) + (4x1)

Molecular Mass of $\left(\mathrm{CH}_4\right)$ = 12+4

Molecular Mass of $\left(\mathrm{CH}_4\right)$ = 16

Page - 180

Question 23: Find the formula unit mass of potassium chloride (KCl).

Atomic mass  K = 39 u; Cl = 35.5 u.

Answer:

Atomic mass of K = 39

Atomic mass of Cl = 35.5

Formula unit mass of KCl = 39 + 35.5

Formula unit mass of KCl = 74.5

Question 24: Find the formula unit mass of magnesium hydroxide, $\mathrm{Mg}(\mathrm{OH})_2$.

Atomic mass Mg = 24 u; O = 16 u; H = 1 u

Answer:

Atomic mass of Mg = 24u

Atomic mass of O = 16u

Atomic mass of H = 1u

Formula unit mass of $\mathrm{Mg}(\mathrm{OH})_2$ = 24+ (16+1)2

Formula unit mass of $\mathrm{Mg}(\mathrm{OH})_2$ =58u

Class 9 Science Chapter 9 Atomic Foundations of Matter Question Answer (Revise, Reflect, Refine)

Students can refer to the detailed solutions given below for end chapter exercise questions of NCERT Class 9 Science Chapter 9 Atomic foundations of matter.

Question 1: A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell.

(i) How many electrons does A tend to give or take to become stable?

(ii) What kind of ion would it form?

(iii) How many electrons does B tend to give or take to become stable?

(iv) What kind of ion would it form?

(v) If A and B were to combine, what kind of bond would be formed?

(vi) What would be the formula for the compound thus formed?

Answer:

Element A has 1 electron in 3rd shell

Electronic configuration of A = 2,8,1

Element B has 6 electron in 2nd shell

Electronic configuration of B = 2,6

(i) A tends to give 1 electron to become stable

(ii) It will form positive ion

(iii) B has 6 electrons; it requires 2 more to complete its octet. So, B tends to gain 2 electrons to become stable

(iv) It form negative ion

(vi) Combination of A and B will form ionic bonds because electrons are transferred.

(vii) Two atoms of A are required to provide two electrons to one atom of B.

Formula of compound B is $A_2 B$

Question 2: An element X has six electrons in its outer shell and forms a diatomic molecule.

(i) Why would that be so?

(ii) What kind of bond would it form?

(iii) Draw the structure of the molecule it would form.

(iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.

Answer:

There are 6 electrons in the outer shell of the element X and it forms a diatomic molecule.

(i) Two X atoms share 2 pairs of electrons to complete their octet and form a diatomic molecule $X_2$.

(ii) It will form a covalent bond.

(iii) $: X=X:$

(iv) Compound Y has 2 electrons in the second shell. So it can lose 2 electrons and form $Y^{2+}$.

Element X gains 2 electrons to form $X^{2-}$.

An ionic compound YX is formed.

Question 3: You want to design a new ionic compound, where the total positive charge is +6 and the total negative charge is -6. Which of the following combinations gives the correct number of ions?

(i) $2 \mathrm{Al}^{3+}$ and $3 \mathrm{Cl}^{-}$

(ii) $3 \mathrm{Mg}^{2+}$ and $1 \mathrm{PO}_4^{3-}$

(iii) $2 \mathrm{Fe}^{3 +}$ and $3 \mathrm{O}^{2-}$

(iv) $3 \mathrm{Ca}^{2+}$ and $2 \mathrm{SO}_4^{2-}$

Answer:

(i) Positive charge on $2 \mathrm{Al}^{3+}$ = 2x3=6

Negative charge on $3 \mathrm{Cl}^{-}$= 3x(-1) = -3

Charges are not equal

(ii) Positive charge on $3 \mathrm{Mg}^{2+}$ = 3x2 =6

Negative charge on $1 \mathrm{PO}_4^{3-}$ = 1x(-3) = -3

Charges are not equal

(iii) Positive charge on $2 \mathrm{Fe}^{3 +}$ = 2x3 = 6

Negative charge on $3 \mathrm{O}^{2-}$ = 3x(-2) = -6

Charges are equal

(iv) Positive charge on $3 \mathrm{Ca}^{2+}$ = 3x2 =6

Negative charge on $2 \mathrm{SO}_4^{2-}$ = 2x(-2) = -4

Charges are not equal

Hence, the correct answer is option (iii)

Question 4: Choose the correct statement(s) and correct the false statement(s).

(i) Elements are made up of molecules and compounds are made up of atoms.

(ii) The molecule of a compound is always made up of two or more atoms of the same kind.

(iii) One molecule of nitrogen gas contains three nitrogen atoms.

(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.

Answer:

(i) False, because Elements are made up of atoms and compounds are made up of molecules.

(ii) False, because the molecule of a compound is made up of two or more atoms of the different elements.

(iii) False, because one molecule of nitrogen gas contains two nitrogen atoms.

(iv) True

Question 5. Write the chemical formulae for the following compounds.

(i) Aluminium nitrate

(ii) Calcium oxide

(iii) Ferric oxide

Answer:

(i) $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$

(ii) CaO

(iii) $\mathrm{Fe}_2 \mathrm{O}_3$

Question 6. Write the formulae of the compounds formed from the following pairs of ions.

(i) $\mathrm{Ca}^{2+}$ and $\mathrm{Br}^{-}$

(ii) $\mathrm{Al}^{3+}$ and $\mathrm{CO}_3^{2-}$

(iii) $\mathrm{K}^{+}$and $\mathrm{SO}_4^{2-}$

(iv) $\mathrm{NH}_4$ and $\mathrm{Cl}^{-}$

Answer:

(i) $\mathrm{CaBr}_2$

(ii) $\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3$’

(iii) $\mathrm{K}_2 \mathrm{SO}_4$

(iv) $\mathrm{NH}_4 \mathrm{Cl}$

Question 7: Which of the following, in Fig. 9.18, correctly represents $\mathrm{Cl}^{-}$ ion (Atomic number of chlorine = 17).

Answer:

Atomic number of Chlorine = 17

Electronic configuration of Cl = 2,8,7

Electronic configuration of $\mathrm{Cl}^{-}$ = 2,8,8

Correct representation of $\mathrm{Cl}^{-}$ is (ii)

Hence, the correct answer is option (ii)

Question 8: Determine the formula unit mass of the following substances.

(i) Ammonium nitrate ($\left.\mathrm{NH}_4 \mathrm{NO}_3\right)$, used as a nitrogen fertilizer, which is essential for plant growth.

(ii) Phosphoric acid $\left(\mathrm{H}_3 \mathrm{PO}_4\right)$, used to make phosphate fertiliser and detergents.

(iii) Sodium hydrogencarbonate $\left(\mathrm{NaHCO}_3\right)$, used to relieve acidity and helps in digestion.

Answer:

(i) $\mathrm{NH}_4 \mathrm{NO}_3$

Mass of nitrogen = 14u

Mass of Hydrogen = 1u

Mass of Oxygen = 16u

Formula unit mass of $\mathrm{NH}_4 \mathrm{NO}_3$ = (2x14)+(4x1)+(3x16)

Formula unit mass of $\mathrm{NH}_4 \mathrm{NO}_3$ = 80u

(ii) $\left(\mathrm{H}_3 \mathrm{PO}_4\right)$

Mass of Hydrogen = 1u

Mass of Oxygen = 16u

Mass of Phosphorous = 31u

Formula unit mass of $\left(\mathrm{H}_3 \mathrm{PO}_4\right)$ = (3x1)+(1x31)+(4x16)

Formula unit mass of $\left(\mathrm{H}_3 \mathrm{PO}_4\right)$ = 98u

(iii) $\left(\mathrm{NaHCO}_3\right)$

Mass of Hydrogen = 1u

Mass of Oxygen = 16u

Mass of Carbon = 12u

Mass of Sodium = 32u

Formula unit mass of $\left(\mathrm{NaHCO}_3\right)$ = 23+12+1+(3x16)

Formula unit mass of $\left(\mathrm{NaHCO}_3\right)$ = 84u

Question 9: Write the formulae for the compounds formed by the reaction of:

(i) Magnesium and nitrogen

(ii) Lithium and nitrogen

(iii) Sodium and sulfur

(iv) Aluminium and oxygen

Answer:

(i) When magnesium and nitrogen react it will form $M g_3 N_2$

$3 M g+N_2 \rightarrow M g_3 N_2$

(ii) When Lithium and nitrogen reacts it will form $\mathrm{Li}_3 \mathrm{~N}$

$6 L i+N_2 \rightarrow 2 L i_3 N$

(iii) When Sodium and sulfur it will form $\mathrm{Na}_2 \mathrm{~S}$

$2 \mathrm{Na}+\mathrm{S} \rightarrow \mathrm{Na}_2 \mathrm{~S}$

(iv) When Aluminium and oxygen reacts it will form $\mathrm{Al}_2 \mathrm{O}_3$

$4 \mathrm{Al}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3$

Question 10: Complete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. $\mathrm{LiNO}_3$ is given as an example.

Answer:

Row 1:

$\mathrm{NH}_4^{+}+\mathrm{NO}_3^{-} \rightarrow \mathrm{NH}_4 \mathrm{NO}_3$

Two $\mathrm{NH}_4^{+}$ ions balance one $\mathrm{SO}_4^{2-}$ ion

Three $\mathrm{NH}_4^{+}$ions balance one $\mathrm{PO}_4^{3-}$ ion

Row 2:

$\mathrm{Li}^{+}+\mathrm{NO}_3^{-} \rightarrow \mathrm{LiNO}_3$

Two $\mathrm{Li}^{+}$ balances one $\mathrm{SO}_4^{2-}$ one

Three $\mathrm{Li}^{+}$ ions balance one phosphate ion

Row 3:

One $\mathrm{Al}^{3+}$ ions combines with 3 $\mathrm{NO}_3^{-}$

Two aluminium ions and three sulfate ions balance charge

One $\mathrm{Al}^{3+}$ and one $\mathrm{PO}_4^{3-}$ balances

Row 4:

One copper ion combines with two nitrate ions

One copper ion balances one sulfate ion

Three copper ions and two phosphate ions balance charges

Question 11: 5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of conservation of mass is valid.

Answer:

According to the law of conservation of mass, total mass of reactants is equal to the total mass of products.

Mass of sodium carbonate = 5.3g

Mass of acetic acid = 6.0 g

Mass of carbon dioxide = 2.2 g

Mass of water = 0.9 g

Mass of sodium acetate = 8.2 g

Mass of reactants = Mass of sodium carbonate + Mass of acetic acid

Mass of reactants = 5.3 + 6.0

Mass of reactants = 11.3g

Mass of Product = Mass of carbon dioxide + Mass of water + Mass of sodium acetate

Mass of Product = 2.2 + 0.9 + 8.2

Mass of Product = 11.3g

Hence, mass of reactants and products are equal

Thus law of conservation of mass is valid
Question 12: If a species has 11 protons, 12 neutrons and 10 electrons then

(i) what is its atomic number and mass number?

(ii) is it neutral, a cation or an anion? Explain.

(iii) write its electronic configuration.

(iv) name the species.

Answer:

No. of protons = 11

No. of neutrons = 12

No. of electrons = 10

(i) Atomic number = No. of Protons

Atomic number = 11

Mass number = Protons + Neutrons

Mass number = 11+12

Mass number = 23

(ii) Neutral ions have equal number of protons and electrons but in this atom 11 protons and 10 electrons are there.

1 electron is less than protons

It has a net positive charge. Hence, it is cation

(iii) No. of electrons = 10

Electronic configuration = 2,8

(iv) Atomic number is 11

Hence, name of the species is Sodium

Question 13: Two elements, A and B, have the following configurations 

A: 2, 8, 5 B: 2, 8, 7

(i) Which element is more reactive?

(ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing.

(iii) Predict the formula of the compound they would form.

Answer:

Electronic configuration of A = 2,8,5

Electronic configuration of B = 2,8,7

(i) B is more reactive because, it only requires 1 electron to complete its octet

(ii) A and B will form ionic bonds because A tends to gain 1 electron to complete its octet and B tends to lose or share electrons.

(iii) Valency of A is 3

Valency of B is 1

So, 1 A combines with 3 B atoms

Formula of compound would be $\mathrm{AB}_3$

Question 14: Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state.

Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.

Choose the correct option:

(i) Both A and R are true, and R is the correct explanation of A.

(ii) Both A and R are true, but R is not the correct explanation of A.

(iii) A is true, but R is false.

(iv) A is false, but R is true.

Answer:

Assertion is correct because Copper sulfate conducts electricity in the molten state but not in the solid state.

The reason is incorrect because copper and sulfate ions are fixed in the solid state, while in molten state they can move freely.

A is true, but R is false

Hence, the correct answer is option (iii).

Question 15: The species ${ }^{27} \mathrm{Al},{ }^{80} \mathrm{Br}^{-}$and ${ }^{201} \mathrm{Hg}^{2+}$ have 13,35 and 80 protons, respectively. How many electrons and neutrons do they have?

Answer:

For ${ }^{27} \mathrm{Al}$

No. of protons = 13

This is neutral atom hence number of electrons and protons are equal

No. of electrons = 13

No. of Neutrons = Mass no. - Protons

No. of Neutrons = 27-13

No. of Neutrons = 14

For ${ }^{80} \mathrm{Br}^{-}$

No. of protons = 35

Charge = -1

Electrons = No. of protons + 1

Electrons = 36

No. of Neutrons = Mass no. - Protons

No. of Neutrons = 80 - 35

No. of Neutrons = 45

For ${ }^{201} \mathrm{Hg}^{2+}$

No. of protons = 80

Charge = +2

No. of electrons = No. of protons - 2

No. of electrons = 80-2

No. of electrons = 72

No. of Neutrons = Mass no. - Protons

No. of Neutrons = 201-80

No. of Neutrons = 121

Topics Covered in Class 9 Science Chapter 9 Atomic Foundations of Matter

Given below are all the topics and subtopics that are covered in NCERT Class 9 Science Chapter 9 Atomic Foundations of Matter:

9.1 Law of Conservation of Mass

9.2 Law of Constant Proportions

9.3 Dalton’s Atomic Theory

9.4 How Atoms Combine?

• 9.4.1 Bonding by sharing of electrons - Covalent Bond
• 9.4.2 Bonding by electron transfer - Ionic bond

9.5 Writing Chemical Formulae

• 9.5.1 Writing chemical formulae of covalent compounds
• 9.5.2 Writing chemical formulae of ionic compounds

9.6 Properties of the Ionic and the Covalent Compounds

9.7 Molecular Mass of Covalent Compounds

9.8 Formula Unit Mass of Ionic Compounds

Also Check

• NCERT Notes for Class 9 Science
• NCERT Syllabus for Class 9 Science

NCERT Solutions for Class 9 Science

Refer to the links given below for detailed solutions of NCERT Class 9 Science. These solutions are prepared by subject experts to help you understand the chapters well and prepare effectively for exams using NCERT.