Naturally occurring Boron is made up of 19.9% \(^{10}\)B and 80.1% \(^{11}\)B. The relative atomic mass of Boron is
Relative Atomic Mass of Boron = [19.9% \(^{10}\)B} + [80.1% \(^{11}\)B]
= [ 19.9% of 10] + [80.1% of 11]
= [ \(\frac{19.9}{100}\times{10}\) ] + [ \(\frac{80.1}{100}\times{11}\)]
= 1.99 + 8.811
= 10.801
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