Class 9 Chemistry |New Book Chapter 2 Lecture 13| Calculation of Relative Atomic Mass🧮| Punjab Board

Complete Lecture Series: Class 9 Chemistry | Complete New Book (Chapters 1-13) | Punjab Board 🔥📘    • Class 9 Chemistry | Complete New Book (Cha...   🚀 Welcome to Chapter 2 of Class 9 Chemistry – Calculation of Relative Atomic Mass from Isotopic Abundance! In this lecture, we delve into the method of calculating the Relative Atomic Mass (Ar) of an element based on the abundance of its isotopes. This concept is crucial for understanding the atomic structure and properties of elements, aligning with the Punjab Board curriculum. 📘 Topics Covered in This Lecture: 🔹 Understanding Isotopes:     • Definition of isotopes as atoms of the same element with different numbers of neutrons.     • Examples of common isotopes and their significance. 🔹 Relative Atomic Mass (Ar):     • Explanation of relative atomic mass as the weighted average mass of an element's isotopes compared to one-twelfth the mass of a carbon-12 atom.     • Importance of Ar in chemical calculations and reactions. 🔹 Calculating Relative Atomic Mass:     • Step-by-step method to calculate Ar using isotopic masses and their relative abundances.     • Worked examples to illustrate the calculation process. Example 1: Calculation for Chlorine Chlorine has two stable isotopes: Isotope ^35Cl: Atomic mass = 34.9688527 u, Abundance = 75.78% Isotope ^37Cl: Atomic mass = 36.9659026 u, Abundance = 24.22% To calculate the relative atomic mass of chlorine: Convert percentages to decimals: 75.78% = 0.7578 24.22% = 0.2422 Multiply each isotope's mass by its relative abundance: (34.9688527 u × 0.7578) = 26.50 u (36.9659026 u × 0.2422) = 8.95 u Add the results: 26.50 u + 8.95 u = 35.45 u Therefore, the relative atomic mass of chlorine is approximately 35.45 u. Example 2: Calculation for Krypton Krypton has several stable isotopes, with the following abundances: Isotope ^80Kr: Abundance = 2.28% Isotope ^82Kr: Abundance = 11.58% Isotope ^83Kr: Abundance = 11.50% Isotope ^84Kr: Abundance = 57.00% Isotope ^86Kr: Abundance = 17.30% To calculate the relative atomic mass of krypton: Convert percentages to decimals: 2.28% = 0.0228 11.58% = 0.1158 11.50% = 0.1150 57.00% = 0.5700 17.30% = 0.1730 Multiply each isotope's mass by its relative abundance: (79.916 u × 0.0228) = 1.82 u (81.913 u × 0.1158) = 9.48 u (82.914 u × 0.1150) = 9.53 u (83.911 u × 0.5700) = 47.83 u (85.910 u × 0.1730) = 14.87 u Add the results: 1.82 u + 9.48 u + 9.53 u + 47.83 u + 14.87 u = 83.53 u Therefore, the relative atomic mass of krypton is approximately 83.53 u. 🔹 Practice Problems:     • Exercises for students to apply the calculation method.     • Solutions and explanations to reinforce learning. This lecture is specially designed for Class 9 students under the Punjab Board system, encompassing: Board of Intermediate and Secondary Education, Lahore Board of Intermediate and Secondary Education, Faisalabad Board of Intermediate and Secondary Education, Multan Board of Intermediate and Secondary Education, Rawalpindi Board of Intermediate and Secondary Education, Sargodha Board of Intermediate and Secondary Education, Dera Ghazi Khan Board of Intermediate and Secondary Education, Bahawalpur Board of Intermediate and Secondary Education, Gujranwala Board of Intermediate and Secondary Education, Sahiwal ✅ Watch till the end for comprehensive explanations and insights that will aid in your exam preparation and deepen your understanding of atomic structures and masses. 📌 Stay Connected: 👉 Like, share, and subscribe for more educational Chemistry lectures and complete coverage of the Punjab Board syllabus! #Class9Chemistry #RelativeAtomicMass #IsotopicAbundance #PunjabBoard #NewBook2025 #ChemistryLecture #StudyWithMe #LearnChemistry #LahoreBoard #FaisalabadBoard #MultanBoard #RawalpindiBoard #SargodhaBoard #GujranwalaBoard #SahiwalBoard #ExamPreparation #EducationalContent #ScienceForPakistan #pakistan