Question

Naturally occurring boron consists of two isotopes, whose atomic masses are 10.01 and 11.01. The atomic mass of natural boron is 10.81. Calculate the percentage of each isotope in natural boron.

Answer 1

% of isotope with atomic mass 10.01 = 20; % of isotope with atomic mass 11.01= 80
Let x be the percentage of the isotope with atomic mass 10.01.
$\frac{10.01\times x}{100}+\frac{10.01(100-x)}{100}=10.81$ or $x=20$