Question

Boron Occurs in nature in the form of two isotopes having atomic masses 10 and 11.Calculate the percentage abundances of two isotopes in a sample having atomic masses 10.80.

Answer 1

• Let the percentage abundance of the 1st Isotope be $x$
• Hence, the percentage abundance of the second isotope becomes $100-x$.
• Using the Formula of Average atomic mass,

$\mathrm{Average}\mathrm{atomic}\mathrm{mass}=\frac{\mathrm{Mass}\mathrm{of}1\mathrm{st}\mathrm{Isotope}\times \mathrm{Percentage}\mathrm{abundance}\mathrm{of}1\mathrm{st}\mathrm{Isotope}+\hspace{0.17em}\mathrm{mass}\mathrm{of}2\mathrm{nd}\mathrm{Isotope}\times \mathrm{Pefcentage}\mathrm{abundance}\mathrm{of}2\mathrm{nd}\mathrm{Isotope}}{100}$

• Substituting the given values,

$10.80=\frac{10\times x+11\times (100-x)}{100}$

$1080=10x+1100-11x$

$1080=1100-x$

$x=20$.

• Hence, the Percentage abundance of the First element ( ${}_{}{}^{10}\mathrm{B}$ ) is 20%
• and the percentage abundance of the Second Element ( ${}_{}{}^{11}\mathrm{B}$ ) is 80%.