Naturally occurring boron consists of two isotopes whose atomic weight are `10.01 and 11.01 `. The atomic weight of the natural boron is `10.81 `. Calculate the percentage of each isotopes in natural boron.

Explanation

Let the percentage of isotope with atomic weight `10.01 = x`
The percentage of isotope with atomic weight `11.0 = (100 -x)`
Avg. Awt `=(m_(1)x_(1) +m_(2)x_(2))/(x_(1)+x_(2))`
`10.81 = (x +10.01 +(100-x)11)/(100)`
`x = 20%`
`:.` The % of isotope with At.wt `10.01 = 20%`
The % of isotope with At.wt `11.01 = 80.0%`