CBSE Class 9 Chemistry Revision Worksheet

Question 1. State i) The law of Conservation of mass.
ii) The law of constant proportions
Answer: i) The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction.
ii) The Law of Constant Proportions (also known as the Law of Definite Proportions) states that in a chemical substance, the elements are always present in definite proportions by mass.
In simple words: Total mass remains constant during a chemical reaction, and a pure chemical compound always contains the same elements combined in the same fixed mass ratio.

Exam Tip: Remember that the law of conservation of mass implies that the total mass of reactants equals the total mass of products in any closed system.

Question 2. Define i) atomic mass unit ii) molecule iii) an ion
Answer: i) Atomic mass unit (u): It is defined as exactly one-twelfth (\(\frac{1}{12}\)) the mass of an unbound carbon-12 atom at rest.
ii) Molecule: A molecule is generally a group of two or more atoms that are chemically bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction and retain its chemical properties.
iii) An ion: An electrically charged atom or group of atoms formed by the gain or loss of one or more electrons.
In simple words: An atomic mass unit is a standard unit for measuring atomic mass, a molecule is a cluster of bonded atoms, and an ion is a charged particle.

Exam Tip: State units clearly when defining atomic mass unit (1 u = \(1.66 \times 10^{-24}\) g).

Question 3. Write the symbols for the following
i) Cobalt ii) Zinc iii) Potassium iv) Bromine v) Iron
vi) Boron vii) Tin viii) Lithium ix) Lead x) Silver
Answer: i) Cobalt: Co
ii) Zinc: Zn
iii) Potassium: K
iv) Bromine: Br
v) Iron: Fe
vi) Boron: B
vii) Tin: Sn
viii) Lithium: Li
ix) Lead: Pb
x) Silver: Ag
In simple words: Chemical symbols are standard shorthand abbreviations derived from the English or Latin names of the elements.

Exam Tip: Pay attention to capitalization in chemical symbols - the first letter is always uppercase and the second letter (if present) is always lowercase (e.g., Co for Cobalt, not CO which represents carbon monoxide).

Question 4. Construct the formulae for the following:
i) Calcium phosphate ii) Ammonium Nitrate
iii) Aluminium chloride iv) potassium sulphite
v) Baking Powder vi) Iron (II) Chloride
Answer: i) Calcium phosphate: \(\text{Ca}_3(\text{PO}_4)_2\)
ii) Ammonium Nitrate: \(\text{NH}_4\text{NO}_3\)
iii) Aluminium chloride: \(\text{AlCl}_3\)
iv) potassium sulphite: \(\text{K}_2\text{SO}_3\)
v) Baking Powder (Sodium hydrogen carbonate / Sodium bicarbonate): \(\text{NaHCO}_3\)
vi) Iron (II) Chloride: \(\text{FeCl}_2\)
In simple words: Chemical formulas are constructed by balancing the positive and negative valencies of the constituent ions.

Exam Tip: Cross the valencies of the ions to write correct chemical formulas, and enclose polyatomic ions in brackets if the subscript is greater than 1.

Question 5. Calculate the molecular mass of the following.
i) Calcium carbonate ii) Zinc Chloride
iii) Ethyl alcohol (\(\text{C}_2\text{H}_5\text{OH}\)) iv) Lime water (\(\text{Ca(OH)}_2\))
\(\text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{H} = 1, \text{Zn} = 65, \text{Cl} = 35.5\)
Answer: i) Calcium carbonate (\(\text{CaCO}_3\)):
Molecular mass = \((1 \times \text{Atomic mass of Ca}) + (1 \times \text{Atomic mass of C}) + (3 \times \text{Atomic mass of O})\)
\(= (1 \times 40) + (1 \times 12) + (3 \times 16)\)
\(= 40 + 12 + 48 = 100 \text{ u}\)

ii) Zinc Chloride (\(\text{ZnCl}_2\)):
Molecular mass = \((1 \times \text{Atomic mass of Zn}) + (2 \times \text{Atomic mass of Cl})\)
\(= (1 \times 65) + (2 \times 35.5)\)
\(= 65 + 71 = 136 \text{ u}\)

iii) Ethyl alcohol (\(\text{C}_2\text{H}_5\text{OH}\)):
Molecular mass = \((2 \times \text{Atomic mass of C}) + (6 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of O})\)
\(= (2 \times 12) + (6 \times 1) + (1 \times 16)\)
\(= 24 + 6 + 16 = 46 \text{ u}\)

iv) Lime water (\(\text{Ca(OH)}_2\)):
Molecular mass = \((1 \times \text{Atomic mass of Ca}) + (2 \times \text{Atomic mass of O}) + (2 \times \text{Atomic mass of H})\)
\(= (1 \times 40) + (2 \times 16) + (2 \times 1)\)
\(= 40 + 32 + 2 = 74 \text{ u}\)
In simple words: The molecular mass is found by adding up the atomic masses of all individual atoms present in the molecule.

Exam Tip: Always include the unit 'u' (atomic mass unit) with your final calculated molecular mass values.

Question 6. Calculate the number of moles in 20 g of Calcium oxide.
Answer: Chemical formula of Calcium oxide = \(\text{CaO}\)
Atomic mass of Calcium (\(\text{Ca}\)) = 40 u
Atomic mass of Oxygen (\(\text{O}\)) = 16 u
Molar mass of \(\text{CaO} = 40 + 16 = 56 \text{ g/mol}\)
Given mass = 20 g
Number of moles (\(n\)) = \(\frac{\text{Given mass}}{\text{Molar mass}}\)
\(n = \frac{20}{56} = \frac{5}{14} = 0.357 \text{ moles}\)
In simple words: Divide the given mass of the substance by its molar mass to get the number of moles.

Exam Tip: Write the formula \(n = \frac{m}{M}\) explicitly before substituting values to secure full steps marks.

Question 7. What is the mass of 0.5 moles of oxygen gas.
Answer: Oxygen gas exists as diatomic molecules, \(\text{O}_2\).
Atomic mass of Oxygen (\(\text{O}\)) = 16 u
Molar mass of \(\text{O}_2 = 2 \times 16 = 32 \text{ g/mol}\)
Number of moles = 0.5 moles
Mass = Number of moles \(\times\) Molar mass
Mass = \(0.5 \times 32 = 16 \text{ g}\)
In simple words: Multiply the number of moles by the molar mass of oxygen gas (\(\text{O}_2\), not atomic oxygen O) to find the total mass.

Exam Tip: A common mistake is using 16 g/mol instead of 32 g/mol for oxygen gas; always remember gases like oxygen, nitrogen, and hydrogen are diatomic (\(\text{O}_2, \text{N}_2, \text{H}_2\)).

Question 8. What are Canal rays?
Answer: Canal rays are positively charged radiations discovered by E. Goldstein in 1886 using a modified discharge tube. These rays consist of positively charged particles called protons, which are formed by the ionization of residual gas atoms inside the discharge tube.
In simple words: Canal rays are streams of positively charged particles moving in the opposite direction of cathode rays inside a discharge tube.

Exam Tip: Mention Goldstein and that these rays led to the discovery of protons.

Question 9. Draw a diagram for Thomson's model of an atom. And write two proposals of Thomson.
Answer: Proposals of Thomson's model of an atom:
1. An atom consists of a positively charged sphere in which the electrons are embedded.
2. The negative and positive charges are equal in magnitude, so the atom as a whole is electrically neutral.


In simple words: Thomson pictured an atom like a watermelon or Christmas pudding, where positive charge is spread all over and negative electrons are stuck inside it.

Exam Tip: Clearly label the positive sphere and embedded electrons in your diagram.

Question 10. Write two postulates of Bohr's model of an atom.
Answer: 1. Only certain special orbits known as discrete orbits of electrons are allowed inside the atom.
2. While revolving in discrete orbits, the electrons do not radiate energy.
In simple words: Bohr stated that electrons revolve around the nucleus only in fixed orbits called energy levels without losing energy.

Exam Tip: Mention the term 'discrete orbits' or 'energy shells' for full marks.

Question 11. Write the features of Rutherford from alpha-scattering experiment.
Answer: Observations and features derived from Rutherford's alpha-scattering experiment:
1. Most of the space inside the atom is empty because most of the alpha particles passed straight through the gold foil without deflection.
2. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies a very little space.
3. A very small fraction of alpha particles were deflected by 180 degrees, indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom (the nucleus).
In simple words: The experiment proved that atoms have a tiny, dense, positively charged center called the nucleus, while the rest of the space is empty.

Exam Tip: Structure your answer clearly with the three main observations and their corresponding conclusions.

Question 12. How will you find the valency of i) Chlorine ii) Calcium iii) Sulphur iv) Sodium
Answer: Valency is determined by the number of valence electrons present in the outermost shell, or by how many electrons need to be gained, lost, or shared to achieve a stable octet (8 electrons in the outermost shell, or 2 for the innermost shell).
i) Chlorine (Atomic number = 17): Electronic configuration = 2, 8, 7. It needs 1 electron to complete its octet. Valency = 8 - 7 = 1.
ii) Calcium (Atomic number = 20): Electronic configuration = 2, 8, 8, 2. It has 2 valence electrons which it can lose. Valency = 2.
iii) Sulphur (Atomic number = 16): Electronic configuration = 2, 8, 6. It needs 2 electrons to complete its octet. Valency = 8 - 6 = 2.
iv) Sodium (Atomic number = 11): Electronic configuration = 2, 8, 1. It has 1 valence electron which it can lose. Valency = 1.
In simple words: If valence electrons are 1, 2, 3, or 4, the valency equals the number of valence electrons. If valence electrons are 5, 6, 7, or 8, the valency is calculated as 8 minus the number of valence electrons.

Exam Tip: Always write the electronic configuration first before determining the valency.

Question 13. Define valency by taking examples of carbon and oxygen.
Answer: Valency is defined as the combining capacity of an atom of an element, determined by the number of electrons lost, gained, or shared to attain a stable electronic configuration.
- Carbon (Atomic number = 6, Electronic configuration: 2, 4): Carbon has 4 valence electrons and shares 4 electrons with other atoms to complete its octet. Hence, the valency of carbon is 4.
- Oxygen (Atomic number = 8, Electronic configuration: 2, 6): Oxygen has 6 valence electrons and needs 2 electrons to complete its octet, gaining or sharing 2 electrons. Hence, the valency of oxygen is 2.
In simple words: Valency tells us how many bonds an atom can form with other atoms.

Exam Tip: Clearly state the electronic configuration for both Carbon and Oxygen when explaining.

Question 14. Define i) Isotopes ii) Isobars. Give 2 examples for each.
Answer: i) Isotopes: Atoms of the same element having the same atomic number but different mass numbers.
Examples of isotopes: Protium (\(^1_1\text{H}\)) and Deuterium (\(^2_1\text{H}\)) (or Carbon-12 and Carbon-14).
ii) Isobars: Atoms of different elements having different atomic numbers but the same mass number.
Examples of isobars: Argon (\(^{40}_{18}\text{Ar}\)) and Calcium (\(^{40}_{20}\text{Ca}\)).
In simple words: Isotopes are same elements with different masses, while isobars are different elements with the same mass.

Exam Tip: Remember: iso-topes have same atomic number (bottom index same), iso-bars have same mass number (top superscript same).

Question 15. Define i) Atomic number ii) Mass number.
Answer: i) Atomic number: It is defined as the total number of protons present in the nucleus of an atom, denoted by the symbol 'Z'.
ii) Mass number: It is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom, denoted by the symbol 'A'.
In simple words: Atomic number is the count of protons, and mass number is the sum of protons and neutrons.

Exam Tip: Mass number = Number of protons + Number of neutrons (\(A = Z + n\)).

Question 16. Write any two uses of isotopes.
Answer: 1. An isotope of cobalt (\(\text{Co-60}\)) is used in the treatment of cancer.
2. An isotope of iodine (\(\text{I-131}\)) is used in the treatment of goitre.
In simple words: Isotopes have important medical and industrial applications, such as treating diseases like cancer and goitre.

Exam Tip: Mentioning medical applications like Cobalt-60 for cancer therapy is standard and highly scoring.

Question 17. Write the distribution of electrons in Magnesium with the help of energy shells diagram.
Answer: Atomic number of Magnesium (\(\text{Mg}\)) = 12
Electronic configuration = 2, 8, 2
- K shell (n = 1) = 2 electrons
- L shell (n = 2) = 8 electrons
- M shell (n = 3) = 2 electrons


In simple words: Electrons are filled in shells K, L, M with a maximum capacity of 2, 8, and 2 electrons respectively for Magnesium.

Exam Tip: Follow the \(2n^2\) rule for maximum electron capacity per shell (\(2(1)^2 = 2, 2(2)^2 = 8, 2(3)^2 = 18\)).

Question 18. What are the limitations of J. J Thomson's model of an atom.
Answer: 1. Thomson's model could not explain the results of alpha-particle scattering experiments carried out by Rutherford.
2. It failed to explain how the positive charge holds the negatively charged electrons inside the atom and could not account for the atomic stability.
In simple words: Thomson's model could not explain experimental observations like Rutherford's scattering experiment.

Exam Tip: Keep limitations concise by linking them directly to Rutherford's alpha scattering experiment.

Question 19. If bromine atom is available in the form of two isotopes \(^{79}_{35}\text{Br}\) (49.7%) and \(^{81}_{35}\text{Br}\) (50.3%). Calculate the average atomic mass of bromine atom.
Answer: Average atomic mass of bromine = \(\left( 79 \times \frac{49.7}{100} \right) + \left( 81 \times \frac{50.3}{100} \right)\)
\(= (79 \times 0.497) + (81 \times 0.503)\)
\(= 39.263 + 40.743\)
\(= 80.006 \text{ u}\)
In simple words: The average atomic mass is calculated by taking the weighted average of the atomic masses of all its naturally occurring isotopes based on their percentage abundances.

Exam Tip: Always multiply each isotope's mass by its fractional abundance (percentage divided by 100) and sum them up.

Question 20. How does the atmosphere act as a blanket?
Answer: The atmosphere acts as a blanket in the following ways:
1. Air is a bad conductor of heat. The atmosphere keeps the average temperature of the Earth fairly steady during the day and even across the course of the year.
2. It prevents sudden increases in temperature during daylight hours and slows down the escape of heat into outer space during the night.
In simple words: The atmosphere traps warmth and protects the Earth from extreme temperature fluctuations between day and night.

Exam Tip: Mention that air is a poor conductor of heat to explain the insulating property of the atmosphere.