Naturally occuring boron is composed of only two isoptopes 10/5B and 11/B with isotopic masses of 10.0129 and 11.00931 respectively. To account for the atomic mass of 10.811 what must be the percentage abundance of each isotope?

Answer 1

Let 10B abundance = x
Then 11B abundance = 1-x
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x(10.0129) + (1-x)(11.00931) = 10.811.
Solve for x and 1-x, then convert to percent.