Thank you for reporting, we will resolve it shortly
Q.
The natural boron of atomic weight $ 10.81 $ is found to have two isotopes $ {{B}^{10}} $ and $ {{B}^{11}} $ . The ratio of abundance of isotopes in natural boron should be
Let abundance of $ {{B}^{10}} $ be m% .
So, abundance of $ {{B}^{11}}=(100-m)% $
$ \therefore $ $ 10.81=\frac{(10\times m)+11(100-m)}{100} $
Or $ 1081=10m+1100-11m $
Or $ m=19 $
$ \therefore $ Ratio of abundances
$ =\frac{19}{100-19}=\frac{19}{81} $