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SS1: CHEMISTRY - 1ST TERM

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  1. Introduction to Chemistry and Laboratory Apparatus | Week 1
    5Topics
    |
    1 Quiz
  2. Nature of Matter | Week 2
    3Topics
    |
    1 Quiz
  3. Separation Techniques I | Week 3
    1Topic
    |
    1 Quiz
  4. Separation Techniques II | Week 4
    5Topics
    |
    1 Quiz
  5. Particulate Nature of Matter I | Week 5
    5Topics
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    1 Quiz
  6. Particulate Nature of Matter II | Week 6
    9Topics
    |
    1 Quiz
  7. Symbols, Formulae & Oxidation Number | Week 7
    7Topics
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    1 Quiz
  8. Laws of Chemical Combination | Week 8
    4Topics
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    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I | Week 9
    4Topics
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    1 Quiz
  10. Chemical Combination (Chemical Bonding) II | Week 10
    4Topics
    |
    1 Quiz
  11. Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules | Week 11
    3Topics
    |
    1 Quiz
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The relative Atomic mass of an element is the number of times the average mass of one atom of that element is heavier than one-twelfth the mass of one atom of carbon-12.

\(\scriptsize Relative \: Atomic\: Mass = \normalsize \frac{mass \: of \: 1 \: atom \: of \: the\: element }{\frac{1}{12} \: \times \: mass \: of \:1\: atom \: of \: Carbon-12 } \)

From the definition,

1 carbon atom has a mass of 12

1 oxygen atom has a mass of 16

1 hydrogen atom has mass of 1

The relative atomic mass of an element which exhibits Isotopy is the average mass of its various isotopes.

Relative atomic mass is not a whole number because of the existence of an isotope.

Example 1

A natural occurring Chlorine contains 75% of \( \scriptsize _{17} ^{35} \textrm {Cl} \) and 25% of \( \scriptsize _{17} ^{37} \textrm {Cl} \)

Calculate the relative atomic mass of Chlorine

Solution:

:- \( \left( \frac{75}{100} \: \times \: \frac{35}{1} \right) \: +\: \left( \frac{25}{100} \: \times \: \frac{37}{1} \right)\\ \scriptsize 26.25 \: + \: 9.25 = 35.5 \)

Example 2

An element x has two Isotopes of \( \scriptsize _{10} ^{20} \textrm {X} \) and \( \scriptsize _{10} ^{22} \textrm {X} \) in the ratio of 1:3. 

What is the relative atomic mass of X?

Solution:

Add ratio (1:3), 1 + 3 = 4

:- \( \left( \frac{1}{4} \: \times \: \frac{20}{1} \right) \:+\: \left( \frac{3}{4} \: \times \: \frac{22}{1} \right)\\ \scriptsize 5 \: + \: 16.5 = 21.5 \)

Example 3:

Natural occurring Lithium consists of two 

Isotopes of 7.4% of \( \scriptsize _{33} ^{6} \textrm {Li} \)and 92.6% of \( \scriptsize _{33} ^{7} \textrm {Li} \)

Solution:

Determine the relative atomic mass of Lithium

:- \( \left( \frac{7.4}{100} \; \times \; \frac{6}{1} \right) \;+\; \left( \frac{92.6}{100} \; \times \; \frac{7}{1} \right)\\ \scriptsize 0.444 \; + \; 6.483 = 6.926 \)

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