Nuturally occurring boron consists of two isotopes whese atomic weight are `10.01` and `11.01`The atomic weight of nurons is `10.81` Calculate the percentage of each isotope is natural boron

Explanation

Let the percentage of isotope with atomic weight `10.01` be x .
So the percentage of isotope with atomic weight `11.01` will be `100 - x`.
Average atomic weight = `(m_(1)x_(1) + m_(2) x_(2))/(s_(1) + s_(2))`
`= (s xx 10.01 + (100 - s) xx 11.01)/(100)`
`10.81 = (s xx 10.01 + (100 - s) xx 11.01)/(100) = 20`
`%` of isotope with atomic weight `10.01 = 20`
`%` of isotope with atomic weight `11.01 = 100 - x = 80`