CBSE Class 9 Science Structure of atom Sure Shot Questions Set 02

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1. Which of the following correctly represent the electronic distribution in the Mg atom? [1 Mark]
(a) 3, 8, 1
(b) 2, 8, 2
(c) 1, 8, 3
(d) 8, 2, 2

Answer: (b) 2, 8, 2

1. Magnesium (Mg) has an atomic number of 12.
2. The electrons are distributed in shells as 2 in K shell, 8 in L shell, and 2 in M shell.
3. Hence, the correct option is (b).

Teacher's Note:
a) The maximum capacity of a shell is given by the formula \(2n^2\).
b) Always fill inner shells before populating the outermost valence shell.

2. Rutherford’s ‘alpha (\(\alpha\)) particles scattering experiment’ resulted in to discovery of [1 Mark]
(a) Electron
(b) Proton
(c) Nucleus in the atom
(d) Atomic mass

Answer: (c) Nucleus in the atom

1. In Rutherford's experiment, fast-moving alpha particles were bombarded on a thin gold foil.
2. Most particles passed straight, but a few deflected at large angles, proving a heavy positive center exists.
3. This dense positive center is called the atomic nucleus.
4. Hence, the correct option is (c).

Teacher's Note:
a) Electrons were discovered by J.J. Thomson using cathode ray tubes.
b) Protons were discovered by E. Goldstein through canal rays.

3. The ion of an element has 3 positive charges. Mass number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion? [1 Mark]
(a) 13
(b) 10
(c) 14
(d) 16

Answer: (b) 10

1. Mass number = Number of protons + Number of neutrons = 27.
2. Number of protons = Mass number - Number of neutrons = 27 - 14 = 13.
3. A neutral atom has 13 electrons, but the ion has 3 positive charges, meaning 3 electrons were lost.
4. Number of electrons in the ion = 13 - 3 = 10.
5. Hence, the correct option is (b).

Teacher's Note:
a) Positive charge on an ion indicates a loss of electrons from the neutral atom.
b) The atomic number remains equal to the number of protons regardless of the ionic charge.

4. Elements with valency 1 are [1 Mark]
(a) Always metals
(b) Always metalloids
(c) Either metals or non-metals
(d) Always non-metals

Answer: (c) Either metals or non-metals

1. Elements with valency 1 can be metals like sodium (Na) which lose 1 electron.
2. They can also be non-metals like fluorine (F) or hydrogen (H) which gain or share 1 electron.
3. Hence, the correct option is (c).

Teacher's Note:
a) Valency depends on the combining capacity, not solely on metallic character.
b) Group 1 elements are metals with valency 1, whereas halogens are non-metals with valency 1.

5. The first model of an atom was given by [1 Mark]
(a) N. Bohr
(b) E. Goldstein
(c) Rutherford
(d) J.J. Thomson

Answer: (d) J.J. Thomson

1. J.J. Thomson proposed the plum pudding or watermelon model of the atom in 1898.
2. This was the earliest atomic model proposed before Rutherford or Bohr.
3. Hence, the correct option is (d).

Teacher's Note:
a) Thomson pictured electrons embedded in a positive sphere.
b) Chronologically, Thomson came first, followed by Rutherford, then Bohr.

6. Is it possible for the atom of an element to have one electron, one proton and no neutron? If so, name the element. [1 Mark]

Answer:
1. Yes, it is possible.
2. The element is Hydrogen (specifically the common isotope Protium, \(^1_1\text{H}\)).

Teacher's Note:
a) Protium is the only stable nuclide known to have no neutrons.
b) Atomic mass of this hydrogen isotope is 1u, consisting only of 1 proton and 1 electron.

7. Why did Rutherford select a gold foil in his \(\alpha\)-ray scattering experiment? [1 Mark]

Answer:
1. Gold is extremely malleable, which allowed the preparation of a very thin foil (about 1000 atoms thick).
2. A thin foil ensures that alpha particles do not suffer multiple scattering events while passing through.

Teacher's Note:
a) Thicker sheets would cause excessive deflection and obscure single-scattering results.
b) Gold is a heavy metal with a high atomic number, providing a large positive nuclear charge.

8. Will Cl-35 and Cl-37 have different valences? [1 Mark]

Answer:
1. No, they will not have different valences.
2. Both are isotopes of chlorine and possess the same atomic number (17) and electronic configuration (2, 8, 7), resulting in a valency of 1.

Teacher's Note:
a) Chemical properties of isotopes are identical because they have the same number of electrons.
b) Valency is determined by the number of valence electrons in the outermost shell.

9. Calculate the number of neutrons present in the nucleus of an element X which is represented as \(^{31}_{15}\text{X}\). [1 Mark]

Answer:
1. Mass number (A) = 31
2. Atomic number (Z) = Number of protons = 15
3. Number of neutrons = Mass number - Atomic number = 31 - 15 = 16.

Teacher's Note:
a) The superscript represents the mass number, and the subscript represents the atomic number.
b) Neutron count is found by subtracting atomic number from mass number.

10. The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements? [1 Mark]

Answer:
1. Such pairs of elements are called Isobars.
2. Isobars have the same mass number but different atomic numbers.

Teacher's Note:
a) Calcium ( \(^{40}_{20}\text{Ca}\) ) and Argon ( \(^{40}_{18}\text{Ar}\) ) share the same mass number 40.
b) Do not confuse isobars with isotopes, which have the same atomic number.

11. Why do Helium, Neon and Argon have a zero valency? [1 Mark]

Answer:
1. Helium has a completely filled duplet in its K shell (2 electrons).
2. Neon and Argon have completely filled octets in their outermost valence shells (8 electrons).
3. Hence, they are chemically inert with a combining capacity of zero.

Teacher's Note:
a) Noble gases have stable electronic configurations.
b) They do not need to gain, lose, or share electrons to achieve stability.

12. In what way the Rutherford proposed atomic model? [2 Marks]

Answer:
1. An atom consists of a tiny, dense, positively charged center called the nucleus where almost all the mass is concentrated.
2. Electrons revolve around the nucleus in circular orbits at high speeds, and the size of the nucleus is very small compared to the size of the atom.

Teacher's Note:
a) Rutherford compared the atomic structure to the solar system.
b) Coulombic attraction provides the centripetal force for revolving electrons.

13. In what way the Thomson proposed atomic model? [2 Marks]

Answer:
1. An atom consists of a positively charged sphere in which negatively charged electrons are embedded.
2. The negative and positive charges are equal in magnitude, making the atom electrically neutral as a whole.

Teacher's Note:
a) Popularly known as the plum pudding or watermelon model.
b) It successfully explained the electrical neutrality of atoms but failed experimental scattering results.

14. What were the drawbacks of Rutherford’s model of an atom? [2 Marks]

Answer:
1. According to electromagnetic theory, an accelerated charged particle should radiate energy continuously.
2. Thus, a revolving electron would lose energy and spiral into the nucleus, making the atom unstable, which contradicts reality.

Teacher's Note:
a) This major flaw could not explain the stability of atoms.
b) Bohr later resolved this by introducing fixed stationary orbits.

15. What are the limitations of J.J. Thomson’s model of an atom? [2 Marks]

Answer:
1. It could not explain the results of alpha-particle scattering experiments conducted by Rutherford.
2. It failed to provide any experimental evidence for the distribution of positive and negative charges inside the atom.

Teacher's Note:
a) Thomson's model was purely conceptual and lacked mathematical or experimental backing.
b) It could not account for the nuclear findings.

16. In television picture tube which type of rays are used? [1 Mark]

Answer:
1. Cathode rays (streams of high-speed electrons) are used in television picture tubes.

Teacher's Note:
a) Cathode rays strike a phosphor screen to create images.
b) Also known as a cathode ray tube (CRT) display.

17. Which is heavier, neutron or proton? [1 Mark]

Answer:
1. Neutron is heavier than a proton.

Teacher's Note:
a) A neutron has a mass of approximately \(1.675 \times 10^{-27}\) kg.
b) A proton has a slightly smaller mass of approximately \(1.672 \times 10^{-27}\) kg.

18. If electrons move from K to L shell, will the energy be absorbed or evolved? [1 Mark]

Answer:
1. Energy will be absorbed.

Teacher's Note:
a) Moving from an inner shell (K) to an outer shell (L) requires excitation to a higher energy level.
b) Energy is released only when electrons fall from an outer shell to an inner shell.

19. Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have? [1 Mark]

Answer:
1. Number of neutrons = Atomic mass - Number of protons = 4 - 2 = 2 neutrons.

Teacher's Note:
a) Mass number is the sum of protons and neutrons.
b) Helium nucleus (alpha particle) contains 2 protons and 2 neutrons.

20. An ion \(\text{X}^{2+}\) contains 18 electrons and 20 neutrons. Calculate the atomic number and mass no. of element X. Name the element X. [3 Marks]

Answer:
1. Number of electrons in the ion \(\text{X}^{2+}\) = 18. Since it has a 2+ charge, the neutral atom X has 18 + 2 = 20 electrons.
2. Atomic number of element X = Number of protons = 20.
3. Mass number = Number of protons + Number of neutrons = 20 + 20 = 40.
4. The element with atomic number 20 is Calcium (Ca).

Teacher's Note:
a) Positive charges indicate loss of electrons from the neutral atom.
b) Element identification is strictly based on the atomic number.

21. Give one Achievement and one limitation of J.J Thomson’s model of atom? [2 Marks]

Answer:
1. Achievement: It proposed that an atom is electrically neutral as positive and negative charges balance each other.
2. Limitation: It failed to explain how positive and negative charges could remain stable together or account for Rutherford's scattering experiment.

Teacher's Note:
a) Thomson's model successfully introduced the concept of subatomic charge balance.
b) It was subsequently replaced due to lack of experimental validation for scattering.

22. In a given electric field, beta- particles are deflected more than alpha- particles inspite of the fact that alpha- particles have larger charge, why? [3 Marks]

Answer:
1. Beta- particles are fast-moving electrons, which have a very small mass compared to alpha- particles.
2. An alpha- particle is a doubly ionized helium nucleus, which is much heavier than an electron.
3. Due to the extremely small mass of beta- particles, the deflection experienced by them in an electric field is much greater than that of alpha- particles, even though alpha- particles possess a higher charge.

Teacher's Note:
a) Deflection in an electric field depends inversely on the mass of the particle.
b) Students often mistakenly think charge is the only factor determining deflection.
c) Remember that mass plays a dominant role in acceleration under an electric force.

23. What are valence electrons? What is their significance? [3 Marks]

Answer:
1. Valence electrons are the electrons present in the outermost shell of an atom.
2. They determine the chemical properties and reactivity of an element.
3. They are responsible for combining capacity or valency of an atom during chemical bonding.

Teacher's Note:
a) Outermost shell electrons dictate how an element interacts with others.
b) Do not confuse core electrons with valence electrons.
c) Elements with similar valence electrons show similar chemical behavior.

24. What would be the observation if the alpha- particle scattering experiment is carried out using a foil of a metal other than gold? [2 Marks]

Answer:
1. The overall observations would remain nearly the same because the structure of an atom remains identical across different elements.
2. However, if a lighter metal foil is used, fewer alpha- particles might rebound strongly due to a lower nuclear charge and lighter mass of the atomic nuclei.

Teacher's Note:
a) Nuclear model of the atom is universal for all elements.
b) Choice of gold was due to its high malleability to form an extremely thin foil.

25. Electronic configuration of Potassium is 2,8,8,1 and Calcium 2,8,8,2, when M shell can have maximum of 18 electrons then why next element Scandium has electronic configuration 2,8,9,2 and not 2,8,8,3 ? [3 Marks]

Answer:
1. According to the Bohr-Bury scheme, a new shell can be accommodated before the previous shell is completely filled to its maximum capacity.
2. The outermost shell cannot hold more than 8 electrons if the second-to-last shell (penultimate shell) is unstable with less than its maximum capacity.
3. Once the N shell attains 2 electrons in Calcium, further incoming electrons enter the M shell to build up its capacity towards 18 before filling the N shell further.

Teacher's Note:
a) Filling of electrons follows the rule of octet stability for the outermost active shell.
b) Transition elements involve filling inner d-orbitals after outer s-orbitals begin.
c) This rule prevents outer shells from becoming overcrowded prematurely.

26. What are isotopes and Isobars? What are two isotopes of chlorine? Calculate the average atomic mass of a chlorine atom? [5 Marks]

Answer:
1. Isotopes are atoms of the same element having the same atomic number but different mass numbers.
2. Isobars are atoms of different elements having the same mass number but different atomic numbers.
3. The two main isotopes of chlorine are \(^{35}\text{Cl}\) (75%) and \(^{37}\text{Cl}\) (25%).
4. Average atomic mass = \(\left(35 \times \frac{75}{100}\right) + \left(37 \times \frac{25}{100}\right)\)
5. Calculation = 26.25 + 9.25 = 35.5 u.

Teacher's Note:
a) Isotopes differ in number of neutrons.
b) Fractional atomic masses arise due to the percentage abundance of isotopes in nature.
c) Always weigh the percentage contribution correctly during average mass calculations.
d) Isobars have different chemical properties because they are different elements.

27. What is present concept of an atom? Explain in detail? Why this model is considered to be the most appropriate model? [4 Marks]

Answer:
1. The present concept of an atom is based on the quantum mechanical model, where an atom consists of a central nucleus containing protons and neutrons.
2. Electrons do not revolve in fixed circular orbits; instead, they exist in probability regions called orbitals around the nucleus.
3. The model incorporates wave-particle duality and the Heisenberg uncertainty principle.
4. It is considered most appropriate because it accurately explains atomic spectra, chemical bonding, and the stability of atoms which older models failed to do.

Teacher's Note:
a) Modern quantum model supersedes Rutherford and Bohr models.
b) Orbitals represent regions of highest electron probability.
c) Explains stability by avoiding continuous energy loss as predicted by classical physics.
d) Essential foundation for modern chemistry and physics.

28. Explain the Rutherford’s alpha particle scattering experiment. What were the main conclusions drawn from this experiment? [4 Marks]

Answer:
1. Fast-moving alpha- particles were made to bombard a very thin gold foil in a vacuum chamber.
2. Most alpha- particles passed straight through the foil undeflected.
3. A small fraction of particles was deflected at small angles, and very few (1 in 12,000) rebounded back completely.
4. Conclusions: Most of the space inside the atom is empty; the positive charge and mass are concentrated in a very small volume called the nucleus at the center.

Teacher's Note:
a) Experiment disproved Thomson's plum pudding model.
b) Highlights the tiny size of the nucleus relative to the atom.
c) Large deflections proved electrostatic repulsion from a dense positive center.
d) Foundation stone for nuclear physics.

29. A naturally occurring sample of :-
(i) 69.2% of ⁶³Cu & 30.8% of ⁶⁵Cu. Find the average atomic mass of a naturally occuring sample of copper.
(ii) 7.42% of ⁶Li & 92.58% of ⁷Li.Find the average atomic mass of a naturally occuring sample of Lithium. [4 Marks]

Answer:
1. For copper (i): Average atomic mass = \(\left(63 \times \frac{69.2}{100}\right) + \left(65 \times \frac{30.8}{100}\right)\)
2. Calculation (i) = 43.596 + 20.02 = 63.616 u.
3. For lithium (ii): Average atomic mass = \(\left(6 \times \frac{7.42}{100}\right) + \left(7 \times \frac{92.58}{100}\right)\)
4. Calculation (ii) = 0.4452 + 6.4806 = 6.9258 u.

Teacher's Note:
a) Use percentage abundance directly multiplied by isotope mass.
b) Check decimal placements carefully during calculation.
c) Units must be written as u (atomic mass unit).
d) Note that the question paper had a slight typographical error writing 53Cu instead of 63Cu in part (i).

30. Calculate the no. of atoms of each element present in 9.8 g of sulphuric acid, H₂SO₄. (H=1, S=32, O=16) [4 Marks]

Answer:
1. Molar mass of H₂SO₄ = \((2 \times 1) + 32 + (4 \times 16)\) = 2 + 32 + 64 = 98 g/mol.
2. Number of moles of H₂SO₄ = \(\frac{9.8}{98}\) = 0.1 mol.
3. Number of molecules = 0.1 × \(6.022 \times 10^{23}\) = \(6.022 \times 10^{22}\) molecules.
4. Number of atoms of H = 2 × \(6.022 \times 10^{22}\) = \(1.2044 \times 10^{23}\); S = \(6.022 \times 10^{22}\); O = 4 × \(6.022 \times 10^{22}\) = \(2.4088 \times 10^{23}\).

Teacher's Note:
a) Always calculate molar mass first from given atomic masses.
b) Use Avogadro number to find total molecules before finding individual atoms.
c) Multiply molecules by the respective atomicity of each element in the formula.
d) Maintain correct exponent notation for scientific values.

31. How to calculate the atomicity and the atomic mass of an atom? [2 Marks]

Answer:
1. Atomicity is calculated by counting the total number of atoms present in a single molecule of an element or compound.
2. Atomic mass is determined relative to one-twelfth the mass of an unbound carbon-12 atom at rest.

Teacher's Note:
a) Atomicity is a simple count per molecule (e.g., O₂ is 2).
b) Atomic mass is a standardized relative scale value.

32. To weight BaCl₂ or Na₂SO₄
a) use a polythene bags and spring balance
b) use a watch glass and spring balance
c) use a polythene bags and physical balance
d) use a watch glass and physical balance [1 Mark]

Answer: (d) use a watch glass and physical balance

1. Salts like barium chloride and sodium sulphate are weighed accurately in chemistry laboratories using a watch glass to hold the chemical and a physical balance for precise measurement.
2. Hence, the correct option is (d).

Teacher's Note:
a) Physical balances provide the high precision needed for chemical quantities.

33. To weight sodium sulphate or barium chloride it should be in form of
a) saturated solution b) large crystals c) small crystals d) fine powder [1 Mark]

Answer: (d) fine powder

1. Salts are preferably taken in the form of fine powder for quick dissolution and uniform sampling during experiments.
2. Hence, the correct option is (d).

Teacher's Note:
a) Fine powder increases surface area for faster reactions or dissolution.

34. What is the distribution of electrons in an atom of Phosphorus and how can it have two valencies [3 Marks]

Answer:
1. The atomic number of Phosphorus is 15, so its electronic configuration is 2, 8, 5.
2. It can gain 3 electrons to complete its octet, giving it a valency of 3.
3. It can also share or lose 5 valence electrons in certain compounds (like PCl₅), exhibiting a higher valency of 5.

Teacher's Note:
a) Phosphorus shows variable valency of 3 and 5.
b) Configuration is filled strictly as per 2n² rule per shell.
c) Elements with multiple oxidation states often display multiple valencies.

35. The maximum no. of electrons present in shell is given by the formula 2n² and the maximum no. of electrons filled in M shell is 18. But in the element calcium we only fill 8 electrons in the M shell and move on to the N shell. why ? [3 Marks]

Answer:
1. According to the Bohr-Bury scheme, the outermost shell of an atom cannot accommodate more than 8 electrons, even if its capacity is higher.
2. Calcium has an atomic number of 20 with configuration 2, 8, 8, 2.
3. Once the M shell has 8 electrons, it temporarily acts as a stable outer configuration before further inner d-orbital filling takes place in heavier elements.

Teacher's Note:
a) Octet rule takes precedence over maximum theoretical capacity for outer shells.
b) Calcium stops filling M at 8 to maintain stability.
c) Helps explain the periodic recurrence of chemical properties.

36. What are alpha particles? [2 Marks]

Answer:
1. Alpha- particles are high-energy, positively charged particles consisting of two protons and two neutrons.
2. They are essentially identical to a doubly ionized helium nucleus (He²⁺).

Teacher's Note:
a) Symbolized as \(^4_2\text{He}\).
b) Heavy and relatively slow compared to beta particles.

37. Write difference between atomic mass and mass number? [3 Marks]

Answer:
1. Atomic mass is the average relative mass of an atom of an element compared to carbon-12, whereas mass number is the total count of protons and neutrons in the nucleus.
2. Atomic mass can be a fractional value due to isotopes, while mass number is always a whole number.
3. Atomic mass is expressed in atomic mass units (u), whereas mass number is a simple dimensionless count.

Teacher's Note:
a) Clear distinction between average relative mass and nucleon count.
b) Mass number is always an integer.
c) Frequently asked conceptual question in board exams.

38. An ion (M²⁺) contain 10 electrons and 12 neutrons what is the atomic number and mass number of the element M. [3 Marks]

Answer:
1. The ion M²⁺ has lost 2 electrons, meaning its neutral atom has 10 + 2 = 12 electrons.
2. Atomic number of element M = number of electrons in neutral atom = 12.
3. Mass number = number of protons + number of neutrons = 12 + 12 = 24.

Teacher's Note:

a) Positive charge means loss of electrons; negative charge means gain.
b) Proton count equals atomic number and remains unchanged in ions.
c) Mass number is the sum of protons and neutrons.

39. An ion M³⁺ contains 10 electrons and 14 neutrons. What are the atomic mass and mass number of the element M ? Name the element. [3 Marks]

Answer:
1. Number of electrons in ion M³⁺ = 10
2. Number of protons in neutral atom M = Number of electrons in M³⁺ + 3 = 10 + 3 = 13
3. Mass number = Number of protons + Number of neutrons = 13 + 14 = 27
4. Atomic mass is approximately equal to the mass number, which is 27 u.
5. The element with atomic number 13 is Aluminium (Al).

Teacher's Note:
a) A positive ion is formed by the loss of electrons, so protons exceed electrons by the charge magnitude.
b) Students often forget to add back the lost electrons to find the atomic number.
c) Atomic number determines the identity of the element in the periodic table.

40. 10 gm of silver nitrate solution is added to 10 gm of sodium chloride solution. What change in mass do you expect after the reaction and why? [2 Marks]

Answer:
1. There will be no change in the total mass after the reaction.
2. This is in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed in a chemical reaction.

Teacher's Note:
a) The total mass of reactants equals the total mass of products.
b) Precipitation reactions in a closed system do not alter the overall mass.

41. Write the atomicity of the following molecules: (i) H₂SO₄ (ii) CCl₄ [2 Marks]

Answer:
1. (i) In H₂SO₄, total atoms = 2 (Hydrogen) + 1 (Sulphur) + 4 (Oxygen) = 7. Atomicity = 7.
2. (ii) In CCl₄, total atoms = 1 (Carbon) + 4 (Chlorine) = 5. Atomicity = 5.

Teacher's Note:
a) Atomicity is the total number of atoms constituting a molecule.
b) Count each element subscript carefully to avoid summation errors.

42. Define the term mole. [2 Marks]

Answer:
1. A mole is the amount of a substance that contains as many particles (atoms, molecules, or ions) as there are atoms in exactly 12 g of carbon-12.
2. One mole of any substance contains exactly \(6.022 \times 10^{23}\) representative particles.

Teacher's Note:
a) The mole is the SI unit for the amount of substance.
b) Emphasize both the mass definition and the Avogadro number equivalence.

43. What is the law of constant proportions? [2 Marks]

Answer:
1. The law of constant proportions (or law of definite proportions) states that in a chemical substance, the elements are always present in definite proportions by mass.
2. For example, in water (H₂O), the ratio of the mass of hydrogen to the mass of oxygen is always 1:8, regardless of the source of water.

Teacher's Note:
a) Formulated by Joseph Proust.
b) Compounds have a fixed elemental composition by weight.

44. What is molar mass? What are its units? [2 Marks]

Answer:
1. The molar mass of a substance is the mass of one mole of that substance.
2. Its SI unit is grams per mole (g/mol).

Teacher's Note:
a) Molar mass is numerically equal to atomic or molecular mass expressed in grams.
b) Always include units when reporting molar mass.

45. Define atomicity. [1 Mark]

Answer:
Atomicity is defined as the number of atoms present in one molecule of an element or compound.

Teacher's Note:
a) Monatomic, diatomic, and polyatomic are classifications based on atomicity.

46. Calculate the formula unit mass of Na₂CO₃.
[Atomic mass of Na = 23 u, C = 12 u, O = 16 u] [2 Marks]

Answer:
1. Formula unit mass of Na₂CO₃ = (2 × Atomic mass of Na) + (1 × Atomic mass of C) + (3 × Atomic mass of O)
2. = (2 × 23) + (1 × 12) + (3 × 16) = 46 + 12 + 48 = 106 u

Teacher's Note:
a) Sum the atomic masses of all atoms in the ionic formula unit.
b) Use unified atomic mass unit (u) as the unit.

47. Give the definition of a cation and an anion. [2 Marks]

Answer:
1. Cation: A positively charged ion formed by the loss of one or more electrons from an atom.
2. Anion: A negatively charged ion formed by the gain of one or more electrons by an atom.

Teacher's Note:
a) Metals generally form cations, while non-metals generally form anions.
b) Cations are attracted to the cathode, anions to the anode.

48. An element X has a valency 3. Write the formula of its oxide. [2 Marks]

Answer:
1. Valency of element X = 3
2. Valency of oxygen = 2
3. By criss-cross method of valencies, the formula is X₂O₃.

Teacher's Note:
a) Apply chemical formula writing rules by balancing valencies.
b) Oxygen always exhibits a valency of 2 in simple oxides.

49. Calculate the number of moles in 52 g of He (Helium) [2 Marks]

Answer:
1. Given mass of Helium = 52 g
2. Molar mass of He = 4 g/mol
3. Number of moles = \(\frac{\text{Given mass}}{\text{Molar mass}}\) = \(\frac{52}{4}\) = 13 moles

Teacher's Note:
a) Use the mole formula relating mass and molar mass.
b) Helium is a monoatomic gas with atomic mass 4 u.

50. If 12 gm of carbon is burnt in the presence of 32 gm of oxygen, how much carbon dioxide will be formed ? [3 Marks]

Answer:
1. Chemical equation: C + O₂ → CO₂
2. Masses: 12 g of carbon reacts with 32 g of oxygen to form 44 g of carbon dioxide.
3. Since exactly 12 g of carbon and 32 g of oxygen are taken, there is no limiting reagent.
4. Total carbon dioxide formed = 44 g.

Teacher's Note:
a) Verify stoichiometric proportions before calculating product mass.
b) This illustrates the law of definite proportions and conservation of mass.

51. Calculate the number of moles in 17 gm of H₂O₂. (Atomic weight of H = 1 u, O = 16 u). [2 Marks]

Answer:
1. Molar mass of H₂O₂ = (2 × 1) + (2 × 16) = 2 + 32 = 34 g/mol
2. Given mass = 17 g
3. Number of moles = \(\frac{17}{34}\) = 0.5 moles

Teacher's Note:
a) Calculate accurate molar mass for hydrogen peroxide.
b) Half a mole corresponds to 17 g of H₂O₂.

52. Define and explain atomic mass of an element. [2 Marks]

Answer:
1. Atomic mass of an element is the average mass of an atom of the element compared to one-twelfth (1/12th) the mass of one carbon-12 atom.
2. It is expressed in atomic mass units (u), where 1 u is equal to \(\frac{1}{12}\)th of the mass of a carbon-12 atom.

Teacher's Note:
a) Carbon-12 is the international standard reference isotope.
b) Relative atomic mass takes into account isotopic abundances.

53. If one mole of carbon atoms weighs 12 g. What is the mass of 1 atom of carbon ? [2 Marks]

Answer:
1. Mass of 1 mole of carbon atoms (6.022 × 10²³ atoms) = 12 g
2. Mass of 1 atom of carbon = \(\frac{12}{6.022 \times 10^{23}}\) = \(1.993 \times 10^{-23}\) g

Teacher's Note:
a) Divide the molar mass by the Avogadro constant.
b) Express the final very small value in standard scientific notation.

54. Calculate the mass of 1 molecule of oxygen gas. [2 Marks]

Answer:
1. Molar mass of oxygen gas (O₂) = 32 g/mol
2. Mass of 1 molecule of O₂ = \(\frac{\text{Molar mass}}{\text{Avogadro constant}}\)
3. = \(\frac{32}{6.022 \times 10^{23}}\) = \(5.31 \times 10^{-23}\) g

Teacher's Note:
a) Remember oxygen gas exists as a diatomic molecule (O₂).
b) Apply Avogadro's number to convert molar mass to single molecular mass.

55. The mass of single atom of an element is 2.65 × 10 ^{- 23} g. Calculate its atomic mass. [N_A = 6.022 × 10²³ mol^-1 ] [3 Marks]

Answer:
1. Mass of 1 atom = \(2.65 \times 10^{-23}\) g
2. Molar mass = Mass of 1 atom × Avogadro constant (N_A)
3. = \(2.65 \times 10^{-23} \times 6.022 \times 10^{23}\)
4. = 15.958 g/mol ≈ 16 u (The element is Oxygen).

Teacher's Note:
a) Multiply single atom mass by Avogadro constant to find molar mass.
b) Atomic mass in u is numerically equal to molar mass in g/mol.

56. Calculate percentage composition of glucose (C₆H₁₂O₆). [3 Marks]

Answer:
1. Molar mass of C₆H₁₂O₆ = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g/mol
2. Percentage of Carbon = \(\frac{72}{180} \times 100\)% = 40%
3. Percentage of Hydrogen = \(\frac{12}{180} \times 100\)% = 6.67%
4. Percentage of Oxygen = \(\frac{96}{180} \times 100\)% = 53.33%

Teacher's Note:
a) Total percentage sum must equal 100%.
b) Use standard atomic masses: C = 12, H = 1, O = 16.

57. John placed 10 moles of sulphur molecules (S₈) and 5 moles of glucose (C₆H₁₂O₆) in the two different pans of a physical balance. Find which pan of the balance would be heavier, support your answer with calculations.
(atomic mass : O = 16 u, C = 12 u, H = 1 u, S = 32 u) [4 Marks]

Answer:
1. Mass of 10 moles of S₈ molecules:
- Molar mass of S₈ = 8 × 32 = 256 g/mol
- Mass of 10 moles of S₈ = 10 × 256 = 2560 g
2. Mass of 5 moles of glucose (C₆H₁₂O₆):
- Molar mass of C₆H₁₂O₆ = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
- Mass of 5 moles of glucose = 5 × 180 = 900 g
3. Comparison: 2560 g (S₈ pan) > 900 g (glucose pan).
4. Conclusion: The pan containing 10 moles of sulphur molecules (S₈) would be heavier.

Teacher's Note:
a) Calculate individual molar masses first before multiplying by mole quantities.
b) Pay close attention to molecular formulas, noting S₈ contains 8 sulphur atoms per molecule.

58. What is Avogadro constant ? [1 Mark]

Answer:
The Avogadro constant is the number of particles (atoms, molecules, or ions) present in one mole of any substance, having the numerical value of \(6.022 \times 10^{23} \text{ mol}^{-1}\).

Teacher's Note:
a) Denoted by N_A, it bridges the macroscopic and microscopic worlds.

59. Calculate the number of particles present in 56 gm of N₂ molecule. [3 Marks]

Answer:
1. Molar mass of N₂ = 2 × 14 = 28 g/mol
2. Number of moles in 56 g of N₂ = \(\frac{56}{28}\) = 2 moles
3. Number of molecules = Number of moles × Avogadro constant = 2 × \(6.022 \times 10^{23}\) = \(1.2044 \times 10^{24}\) molecules.

Teacher's Note:
a) Convert given mass to moles first.
b) Multiply moles by Avogadro number to get total particle count.

60. Calculate the number of moles present in (a) 60 g of calcium. (b) 3.011 × 10²³ number of oxygen atoms.
[Given that Ca = 40u; Avogadro number, N_A = 6.022 × 10²³ per mole] [3 Marks]

Answer:
1. (a) Moles in 60 g of Calcium = \(\frac{\text{Given mass}}{\text{Atomic mass}}\) = \(\frac{60}{40}\) = 1.5 moles.
2. (b) Moles of oxygen atoms = \(\frac{\text{Given number of atoms}}{\text{Avogadro number}}\)
3. = \(\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}}\) = 0.5 moles.

Teacher's Note:
a) Apply mass-to-mole relation for part (a).
b) Apply particle-to-mole relation for part (b).

61. Give an account of the ‘mole concept’. [3 Marks]

Answer:
1. The mole concept provides a bridge between the macroscopic mass of substances and the number of atoms or molecules contained in them.
2. One mole represents a definite collection of \(6.022 \times 10^{23}\) particles, known as the Avogadro number.
3. It allows chemists to weigh out macroscopic amounts of chemicals and know the exact count of reacting atoms or molecules.

Teacher's Note:
a) Central concept in stoichiometry and quantitative chemistry.
b) Connects mass, molar mass, volume of gases, and particle count.

62. Calculate the ratio by number of atoms for Magnesium sulphide. [2 Marks]

Answer:
1. Chemical formula of Magnesium sulphide is MgS.
2. It contains 1 atom of Magnesium (Mg) for every 1 atom of Sulphur (S).
3. The ratio by number of atoms is 1:1.

Teacher's Note:
a) Determine the chemical formula using valencies (Mg²⁺ and S^2-).
b) Simplify atomic counts to find the final atom ratio.

63. (a) The average atomic mass of a sample of an element X is 35.5u. What are the percentages of isotopes \(^{37}_{17}\text{X}\) and \(^{35}_{17}\text{X}\) in the sample ?
(b) Write any two applications of isotopes. [4 Marks]

Answer:
1. (a) Let percentage of \(^{35}_{17}\text{X}\) be \(x\)% and \(^{37}_{17}\text{X}\) be \((100 - x)\)%.
2. Average atomic mass equation: \(\frac{35x + 37(100 - x)}{100} = 35.5\)
3. Solving: \(35x + 3700 - 37x = 3550 \implies -2x = -150 \implies x = 75\).
4. Thus, percentage of \(^{35}_{17}\text{X}\) = 75% and \(^{37}_{17}\text{X}\) = 25%.
5. (b) Applications of isotopes: (1) An isotope of cobalt (Co-60) is used in the treatment of cancer. (2) An isotope of iodine (I-131) is used in the treatment of goitre.

Teacher's Note:
a) Set up a linear equation assuming percentage sums to 100.
b) This calculation explains the fractional atomic mass of Chlorine (Cl = 35.5 u).

64. Nitu presented a silver lamp to her mother on her birthday. The lamp contained 3.011 × 10²³ atoms of silver in it. What is the mass of silver lamp and the cost of it if 1 gm silver costs Rs 60. Atomic mass of Ag = 108 u, N₀ = 6.022 × 10²³ per mole. [4 Marks]

Answer:
1. Given number of silver atoms = 3.011 × 10²³
2. Number of moles of silver (n) = \(\frac{\text{Given atoms}}{\text{Avogadro number}}\) = \(\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}}\) = 0.5 moles
3. Mass of silver lamp = Number of moles × Atomic mass of Ag = 0.5 × 108 = 54 g
4. Cost of the silver lamp = Mass in grams × Cost per gram = 54 × 60 = Rs. 3,240

Teacher's Note:
a) Use the mole formula relating the number of particles to Avogadro constant.
b) Remember that atomic mass in grams represents the molar mass of an element.
c) Always multiply the total mass by the given rate per unit weight to find the cost.
d) Check the arithmetic carefully during division of exponents of 10.

65. Verify by calculating that 5 mole of CO₂ and 5 mole of H₂S do not have the same mass. (Atomic mass of C = 12 u, O = 16 u, H = 1 u, S = 32 u) [3 Marks]

Answer:
1. Molar mass of CO₂ = 12 + (2 × 16) = 12 + 32 = 44 g/mol
2. Mass of 5 moles of CO₂ = Number of moles × Molar mass = 5 × 44 = 220 g
3. Molar mass of H₂S = (2 × 1) + 32 = 2 + 32 = 34 g/mol
4. Mass of 5 moles of H₂S = Number of moles × Molar mass = 5 × 34 = 170 g
5. Since 220 g is not equal to 170 g, 5 moles of CO₂ and H₂S do not have the same mass.

Teacher's Note:
a) The mass of a substance depends directly on its molar mass.
b) Calculate each molar mass independently by summing the atomic masses of constituent atoms.
c) Conclude clearly by comparing the final calculated values for verification.

66. A solution is made by dissolving sodium chloride in water and its concentration is expressed as 0.9% by mass. Calculate : (i) the number of moles, and (ii) number of molecules present in NaCl for this solution. [Given : mass % = (Mass of solute × 100) / Mass of solution] (Atomic mass of Na = 23.0 u, Cl = 35.5 u) [4 Marks]

Answer:
1. Assuming 100 g of the solution, mass of solute (NaCl) = 0.9 g
2. Molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol
3. (i) Number of moles of NaCl = \(\frac{\text{Given mass}}{\text{Molar mass}}\) = \(\frac{0.9}{58.5}\) = 0.0154 moles
4. (ii) Number of molecules of NaCl = Number of moles × Avogadro number = 0.0154 × 6.022 × 10²³ = 9.27 × 10²¹ molecules

Teacher's Note:
a) Percentage by mass lets you assume a convenient total mass like 100 grams for calculation.
b) Pay close attention to unit conversions between grams, moles, and molecules.
c) Use Avogadro number to convert moles into actual molecular entities.
d) Maintain proper significant figures in division and multiplication steps.

67. (i) What do the following symbols / formulae stand for :
(a) 2O (b) O₂ (c) O₃ (d) H₂O [4 Marks]
(ii) Give the chemical formula of the following compounds :
(a) Potassium carbonate (b) Calcium chloride
(iii) Calculate the formula unit mass of Al₂(SO₄)₃.
(Given : Atomic mass of Al = 27 u, S = 32 u, O = 16 u)

Answer:
1. (i) (a) 2O stands for 2 separate atoms of oxygen.
2. (i) (b) O₂ stands for one molecule of oxygen gas consisting of two oxygen atoms.
3. (i) (c) O₃ stands for one molecule of ozone consisting of three oxygen atoms.
4. (i) (d) H₂O stands for one molecule of water consisting of two hydrogen atoms and one oxygen atom.
5. (ii) (a) Potassium carbonate: K₂CO₃
6. (ii) (b) Calcium chloride: CaCl₂
7. (iii) Formula unit mass of Al₂(SO₄)₃ = (2 × 27) + 3 × [32 + (4 × 16)] = 54 + 3 × (32 + 64) = 54 + 3 × 96 = 54 + 288 = 342 u

Teacher's Note:
a) Differentiate carefully between atoms, molecules, and chemical coefficients.
b) Balance valencies properly while writing chemical formulas of ionic compounds.
c) Multiply subscripts correctly inside brackets when calculating formula unit mass.

68. An element ¹⁴₇A 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 its nitrite and nitrate ions. [4 Marks]
(b) How many moles of this gas would contain 12.044 × 10²³ atoms of this element ? (Avogadro’s no. = 6.022 × 10²³)
(c) Calculate the molecular mass of (a) NH₄NO₃ and (b) HNO₃
(Given atomic masses N = 14 u, O = 16 u, H = 1 u)

Answer:
1. (a) The element with atomic number 7 is Nitrogen. The diatomic gas is Nitrogen gas, N₂. Formula of nitrite ion is NO₂⁻ and nitrate ion is NO₃⁻.
2. (b) Given atoms = 12.044 × 10²³. Number of nitrogen atoms in terms of moles = \(\frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\) = 2 moles of nitrogen atoms. Since the gas is diatomic (N₂), moles of N₂ gas = \(\frac{2}{2}\) = 1 mole.
3. (c) (a) Molecular mass of NH₄NO₃ = (2 × 14) + (4 × 1) + (3 × 16) = 28 + 4 + 48 = 80 u
4. (c) (b) Molecular mass of HNO₃ = 1 + 14 + (3 × 16) = 15 + 48 = 63 u

Teacher's Note:
a) Identify elements using their atomic number from the periodic table.
b) Pay attention to whether the question asks for moles of atoms or moles of molecules.
c) Sum up atomic masses accurately according to each element's subscript in the formula.

69. (a) In ammonia, nitrogen and hydrogen are always present in the ratio 14 : 3 by mass. State the law which explains the above statement. [3 Marks]
(b) During the formation of ammonia, what mass of hydrogen gas would be required to react completely with 42 g of nitrogen gas ?

Answer:
1. (a) Law of Definite Proportions (or Law of Constant Composition), which states that a chemical substance is always made up of the same elements combined together in the same fixed proportion by mass.
2. (b) According to the fixed mass ratio, 14 g of nitrogen requires 3 g of hydrogen.
3. Therefore, 42 g of nitrogen gas requires mass of hydrogen = \(\frac{3}{14}\) × 42 = 3 × 3 = 9 g of hydrogen gas.

Teacher's Note:
a) State chemical laws precisely using standard textbook definitions.
b) Use unitary method or proportionality based on given mass ratios to solve numerical parts.
c) Ensure units are written clearly alongside numerical values.

70. (a) Calculate the number of moles in 112 g of iron.
(b) Calculate the mass of 0.5 moles of sugar (C₁₂H₂₂O₁₁).
(c) Define the term molecular mass.
(d) Determine the molecular mass of ZnSO₄. [Atomic mass of Zn = 65 u, S = 32 u, O = 16 u]
(e) Calculate the number of molecules of carbon dioxide, present in 4.4 g of CO₂. [5 Marks]

Answer:
1. (a) Number of moles in 112 g of iron (Atomic mass = 56 u) = \(\frac{112}{56}\) = 2 moles.
2. (b) Molar mass of sugar (C₁₂H₂₂O₁₁) = (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 g/mol. Mass of 0.5 moles = 0.5 × 342 = 171 g.
3. (c) Molecular mass is defined as the sum of the atomic masses of all the atoms in a molecule of the substance, expressed in atomic mass units (u).
4. (d) Molecular mass of ZnSO₄ = 65 + 32 + (4 × 16) = 97 + 64 = 161 u.
5. (e) Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol. Number of moles in 4.4 g of CO₂ = \(\frac{4.4}{44}\) = 0.1 moles. Number of molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules.

Teacher's Note:
a) Apply standard formulas linking mass, moles, and molar mass interchangeably.
b) Write clear definitions using accurate scientific terminology.
c) Perform step-by-step calculations for complex molecular formulas like sugar.
d) Verify each sub-part independently to ensure accurate numerical results.