Question 9 of 36intermediate🔧 ApplyNumerical3 marks

Calculate the atomic mass (average) of chlorine using the following data:

% Natural Abundance Molar Mass ³⁵Cl 75.77 34.9689 ³⁷Cl 24.23 36.9659

Correct Answer

The average atomic mass of an element is calculated using the natural abundance and isotopic masses of its isotopes. For chlorine, we use the given percentage abundances and molar masses of 35Cl^{35}\text{Cl} and 37Cl^{37}\text{Cl}.

The formula is: Average atomic mass = (Abundance of 35Cl×Mass of 35Cl)+(Abundance of 37Cl×Mass of 37Cl)100\frac{(\text{Abundance of } ^{35}\text{Cl} \times \text{Mass of } ^{35}\text{Cl}) + (\text{Abundance of } ^{37}\text{Cl} \times \text{Mass of } ^{37}\text{Cl})}{100}

Substituting the values: Average atomic mass = (75.77×34.9689)+(24.23×36.9659)100=2649.17+895.57100=35.45 u\frac{(75.77 \times 34.9689) + (24.23 \times 36.9659)}{100} = \frac{2649.17 + 895.57}{100} = 35.45 \text{ u}.

Exercise: EXERCISES | Q: 1.9 | (Page 26)
For More Understanding

Explanation

The average atomic mass of an element is the weighted average of the atomic masses of its naturally occurring isotopes. It is calculated by multiplying the fractional abundance of each isotope by its atomic mass and summing the results.

Solution Steps

  1. Step 1: Write the formula for average atomic mass: Average atomic mass = ∑(Natural Abundance×Isotopic Mass)100\frac{\sum (\text{Natural Abundance} \times \text{Isotopic Mass})}{100}

  2. Step 2: Substitute the given values for both isotopes of chlorine: Average atomic mass = (75.77×34.9689)+(24.23×36.9659)100\frac{(75.77 \times 34.9689) + (24.23 \times 36.9659)}{100}

  3. Step 3: Calculate the individual products: (75.77×34.9689)=2649.17(75.77 \times 34.9689) = 2649.17 and (24.23×36.9659)=895.57(24.23 \times 36.9659) = 895.57

  4. Step 4: Add the products and divide by 100: Average atomic mass = 2649.17+895.57100=3544.74100=35.45 u\frac{2649.17 + 895.57}{100} = \frac{3544.74}{100} = 35.45 \text{ u}

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