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9th Class Chemistry · Punjab Board · Chapter 2

Chapter 2: Atomic StructureNumericals Solved Step by Step

All 9 numericals of 9th Class Chemistry Chapter 2, Atomic Structure, solved step by step: Given Data, To Find, Formula, Calculation and Result, the way you write them in the exam. Punjab Board new syllabus.

Chapter 2· Atomic Structure

Numericals Solved Step by Step

Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.

Sample Example (neutrons)

Calculate the number of neutrons, protons and electrons in barium, ¹³⁷₅₆Ba.

Given Data

  • Mass number A = 137
  • Atomic number Z = 56

To Find

  • Protons, electrons and neutrons

Formulas

protons = electrons = Z
N = A − Z

Calculation

  1. Protons = 56; electrons = 56 (neutral atom).
  2. N = 137 − 56 = 81.
Result

56 protons, 56 electrons, 81 neutrons

Exercise (in text)

Calculate the number of neutrons, protons and electrons in ¹⁹⁵₇₈Pt, ⁵⁵₂₅Mn and ¹²⁷₅₃I.

Given Data

  • Pt: A = 195, Z = 78
  • Mn: A = 55, Z = 25
  • I: A = 127, Z = 53

To Find

  • p, e and n for each atom

Formulas

p = e = Z
N = A − Z

Calculation

  1. Pt: p = e = 78; N = 195 − 78 = 117.
  2. Mn: p = e = 25; N = 55 − 25 = 30.
  3. I: p = e = 53; N = 127 − 53 = 74.
Result

Pt 78 p, 78 e, 117 n; Mn 25 p, 25 e, 30 n; I 53 p, 53 e, 74 n

Short Q 2.iii

How many neutrons are present in ²¹⁰₈₃Bi?

Given Data

  • A = 210
  • Z = 83

To Find

  • Neutrons N

Formula

N = A − Z

Calculation

  1. N = 210 − 83 = 127.
Result

127 neutrons

Sample Example 1 (krypton)

Krypton has five isotopes: ⁸⁰Kr 2.0 %, ⁸²Kr 12.0 %, ⁸³Kr 12.0 %, ⁸⁴Kr 57.0 % and ⁸⁵Kr 17.0 %. Calculate its relative atomic mass.

Given Data

  • masses 80, 82, 83, 84, 85
  • abundances 2.0, 12.0, 12.0, 57.0, 17.0 %

To Find

  • Relative atomic mass of Kr

Formula

RAM = (m₁p₁ + m₂p₂ + …) ÷ 100

Calculation

  1. 80 × 2.0 = 160; 82 × 12.0 = 984; 83 × 12.0 = 996; 84 × 57.0 = 4788; 85 × 17.0 = 1445.
  2. Sum = 8373.
  3. RAM = 8373 ÷ 100 = 83.73.
Result

RAM of krypton ≈ 83.7

Note: The book's working prints "85 × 57.0" for the last term; it must be 17.0 (from the table) to give 83.7.

Sample Example 2 (chlorine)

Calculate the relative mass of the light isotope of chlorine when the relative atomic mass of chlorine is 35.45, the abundance of Cl-37 is 24.23 % and that of the light isotope is 75.77 %.

Given Data

  • RAM = 35.45
  • ³⁷Cl: 24.23 %
  • light isotope: 75.77 %, mass m

To Find

  • Mass m of the light isotope

Formula

RAM = (m × 75.77 + 37 × 24.23) ÷ 100

Calculation

  1. 35.45 × 100 = 75.77 m + 37 × 24.23.
  2. 3545 = 75.77 m + 896.51.
  3. 75.77 m = 2648.49, so m = 2648.49 ÷ 75.77 = 34.95.
Result

Mass of the light isotope ≈ 34.95 (chlorine-35)

Exercise (lead)

Calculate the relative atomic mass of lead (Pb). The isotopic abundances of ²⁰⁴Pb, ²⁰⁶Pb, ²⁰⁷Pb and ²⁰⁸Pb are 2.0, 24.0, 22.0 and 52.0 % respectively.

Given Data

  • masses 204, 206, 207, 208
  • abundances 2.0, 24.0, 22.0, 52.0 %

To Find

  • Relative atomic mass of Pb

Formula

RAM = Σ(m × p) ÷ 100

Calculation

  1. 204 × 2 = 408; 206 × 24 = 4944; 207 × 22 = 4554; 208 × 52 = 10 816.
  2. Sum = 20 722.
  3. RAM = 20 722 ÷ 100 = 207.22.
Result

RAM of lead ≈ 207.2

Descriptive Q 4.iv (mercury)

Find the relative atomic mass of mercury: ¹⁹⁶Hg 0.0146 %, ¹⁹⁸Hg 10.02 %, ¹⁹⁹Hg 16.34 %, ²⁰⁰Hg 23.13 %, ²⁰¹Hg 13.22 %, ²⁰²Hg 29.80 %, ²⁰⁴Hg 6.85 %.

Given Data

  • seven isotopes, masses 196-204
  • abundances as listed (total 99.37 %)

To Find

  • Relative atomic mass of Hg

Formula

RAM = Σ(m × p) ÷ Σp

Calculation

  1. Σ(m × p) = 2.86 + 1983.96 + 3251.66 + 4626.00 + 2657.22 + 6019.60 + 1397.40 = 19 938.70.
  2. Σp = 99.37.
  3. RAM = 19 938.70 ÷ 99.37 = 200.6.
Result

RAM of mercury ≈ 200.6

Note: The book's abundances total 99.37 %, so dividing by 100 gives 199.4. Dividing by the real total gives 200.6, the accepted value (200.59).

MCQ 1(vi) (nitrogen)

Calculate the relative atomic mass of nitrogen, given the abundances of its isotopes ¹⁴N and ¹⁵N are 99.64 % and 0.35 %.

Given Data

  • ¹⁴N: 99.64 %
  • ¹⁵N: 0.35 %

To Find

  • Relative atomic mass of N

Formula

RAM = (m₁p₁ + m₂p₂) ÷ 100

Calculation

  1. 14 × 99.64 = 1394.96; 15 × 0.35 = 5.25.
  2. Sum = 1400.21; ÷ 100 = 14.0021.
Result

RAM of nitrogen ≈ 14.0021

MCQ 1(x) (rubidium)

Rubidium has two isotopes, ⁸⁵Rb (72.2 %) and ⁸⁷Rb. Find the percentage of ⁸⁷Rb and check the atomic mass, 85.47.

Given Data

  • ⁸⁵Rb: 72.2 %
  • two isotopes only

To Find

  • Abundance of ⁸⁷Rb
  • Relative atomic mass

Formulas

p₁ + p₂ = 100
RAM = (m₁p₁ + m₂p₂) ÷ 100

Calculation

  1. p(⁸⁷Rb) = 100 − 72.2 = 27.8 %.
  2. RAM = (85 × 72.2 + 87 × 27.8) ÷ 100 = (6137 + 2418.6) ÷ 100 = 85.56.
Result

⁸⁷Rb = 27.8 %; RAM ≈ 85.6

Note: Using mass numbers gives 85.56; the exact isotopic masses (84.91 and 86.91) give the stated 85.47.

End of Chapter 2 NumericalsNext: Chapter 2 MCQs →
2Chapter
Full guide: Atomic Structure (notes, diagrams, all questions, self-test)
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States of Matter and Phase Changes
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Chemical Bonding
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Periodic Table and Periodicity
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