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

Explanation

Let the percentage of isotope with atomic weight `10.01 = X`
Let the percentage of isotope with atomic weight `11.01=(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%`
`therefore` The percentage of isotope with A. wt. `10.01 = 20%`
The percentage of isotope with A. wt. `11.01=80.0%`