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Question #a535f

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Nov 10, 2016

Atomic mass of an element is defined as the weighted average mass of atoms of an element, calculated using the relative abundance of nuclides in a naturally-occurring mass of that element.

In nature the in a mass of CHlORINE two types of nuclides are present $-_{17}^{35} C l and_{17}^{37} C l$ $-""_17^35Cl and ""_17^37Cl$

If percent abundance of $_{17}^{35} C l$ $""_17^35Cl$ be $x %$ $x%$ then percent abundance of $_{17}^{37} C l$ $""_17^37Cl$ will be $(100 - x) %$ $(100-x)%$

The atomic mass of $C l \to 35.453 g/mol$ $Cl->35.453" g/mol"$

So weighted average atomic mass

$\frac{\times x 35 + (100 - x) 37}{100} = 35.453$ $(xxx35+(100-x)37)/100=35.453$

$\Rightarrow 35 x - 37 x = 3545.3 - 3700 = - 154.7$ $=>35x-37x=3545.3-3700=-154.7$

$x = \frac{154.7}{2} = 77.35$ $x=154.7/2=77.35$

So Abundance

$_{17}^{35} C l \to 77.35 %$ $""_17^35Cl->77.35%$

and

$_{17}^{37} C l \to 22.65 %$ $""_17^37Cl->22.65%$

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