Question

Naturally occurring $B$ consists of two isotopes whose atomic masses are $10.01$ and $11.01$. The atomic mass of natural boron is $10.81$. What is the $\%$ of each isotope in natural boron?

• A. ${}^{10}B=20$% ,${}^{11}B=80$%
• B. ${}^{10}B=80$% ,${}^{11}B=20$% `
• C. ${}^{10}B=40$% ,${}^{11}B=60$%
• D. ${}^{10}B=60$% ,${}^{11}B=40$%

Answer 1

The correct option is A ${}^{10}B=20$% ,${}^{11}B=80$%

As we know,
Average atomic mass =$\frac{\Sigma AX}{100}$
Let $\%$ of isotope of mass $10.01$ be $a$.
$\therefore 10.81=\frac{10.01\times a+11.01(100-a)}{100}$
$\therefore a=20$
$\therefore$ $\%$ of isotope of mass $10.01$ $=20\%$
$\therefore$ $\%$ of isotope of mass $11.01$ $=80\%$