Question

If the relative atomic mass of boron is $10.8$, then calculate the percentage of its isotopes ${}_5{}^{10}\mathrm{B}$ and ${}_5{}^{11}\mathrm{B}$ occurring in nature.

Answer 1

Given:

Atomic mass of Boron$=10.8$

For isotope ${}_5{}^{10}\mathrm{B}$: Atomic mass$=10$

For isotope ${}_5{}^{11}\mathrm{B}$: Atomic mass$=11$

Formula used:

Atomic mass of an element$=\frac{\%\mathrm{of}\mathrm{isotope}-1}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-1+\frac{\%\mathrm{of}\mathrm{isotope}-2}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-2$

Calculate the percentage of ${}_5{}^{10}\mathrm{B}$:

Let the percentage of ${}_5{}^{10}\mathrm{B}$ be $\mathrm{x}$.

Then the percentage of ${}_5{}^{11}\mathrm{B}$ will be $100-\mathrm{x}$.

Thus,

Atomic mass of an element$=\frac{\%\mathrm{of}\mathrm{isotope}-1}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-1+\frac{\%\mathrm{of}\mathrm{isotope}-2}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-2$

Substituting the values,

$$\begin{gathered} 10.88=\frac{\mathrm{x}}{100}\times 10+\frac{100-\mathrm{x}}{100}\times 11 \\ 10.88=\frac{10\mathrm{x}}{100}+\frac{1100-11\mathrm{x}}{100} \\ 1088=1100-\mathrm{x} \\ \mathrm{x}=12 \end{gathered}$$

Thus, the percentage of ${}_5{}^{10}\mathrm{B}$ is $12\%$.

Calculate the percentage of ${}_5{}^{11}\mathrm{B}$:

The percentage of ${}_5{}^{11}\mathrm{B}$ $=100-\mathrm{x}$

Thus,

The percentage of ${}_5{}^{11}\mathrm{B}$ $=100-12=88$

Thus, percentage of isotopes of Boron ${}_5{}^{10}\mathrm{B}$ and ${}_5{}^{11}\mathrm{B}$ occurring in nature is $12\%$ and $88\%$ respectively.