Naturally occurring boron has 20% _5B^{10} an | vedclass

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(C) The average atomic weight is calculated by the weighted average of the isotopes: $	ext{Average atomic weight} = \frac{(	ext{Mass of isotope } 1

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$10.80$

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(C) The average atomic weight is calculated by the weighted average of the isotopes: $	ext{Average atomic weight} = \frac{(	ext{Mass of isotope } 1 	imes 	ext{Percentage } 1) + (	ext{Mass of isotope } 2 	imes 	ext{Percentage } 2)}{100}$ $	ext{Average atomic weight} = \frac{(10 	imes 20) + (11 	imes 80)}{100}$ $	ext{Average atomic weight} = \frac{200 + 880}{100} = \frac{1080}{100} = 10.80$

Naturally occurring boron has $20\%$ $_5B^{10}$ and $80\%$ $_5B^{11}$. The average atomic weight of boron is

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