Ordinary chlorine has two isotopes $_{17}^{35} C l$ and $_{17}^{37} C l$ in the ration of 3:1 Calculate the relative atomic mass of chlorine.

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Solution

1. Given that:
Isotope one = $_{17}^{35} C l$
Isotope two = $_{17}^{37} C l$
2. The percentage of the abundance of $_{17}^{37} C l$ is 25%
3. The percentage of the abundance of $_{17}^{35} C l$ is 75%
4. Now, the average atomic mass of chlorine can be calculated as:
Average atomic mass = relative abundance of $_{17}^{35} C l$ isotope $\times$ atomic mass of $_{17}^{35} C l$ $+$ relative abundance of $_{17}^{37} C l$ $\times$ atomic mass of $_{17}^{37} C l$ isotope
$= (37 \times 0.25) + (35 \times 0.75)$
$= 9.25 + 25.25$
$= 35.5 u$
Hence, The relative atomic mass of chlorine is 35.5 u.

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