Boron occurs in nature in the form of two isotopes having atomic masses 10 and 11. What are the percentage abundances of these isotopes in a simple a boron having average atomic mass of 10.8 ?

Answer

`B - 10 = 20,B - 11= 80`

Explanation

Let the percentage abundance of B-10 isotope=x
`:.` The percentage abundance of B-11 isotope = (100-x)
From the available data :
`(x xx10)/(100)+((100-x)xx11)/(100)=10.8`
or `10x + 1100 - 11x = 1080or x = 20`
Percentage abundance of B - 10 isotope = 20 %
Percentage abundance of B - 11 isotope = 80 %.