Question

Atomic weight of Boron is 10.81 and it has two isotopes ${}_5{B}^{10}$ and ${}_5{B}^{11}$ Then the ratio of ${}_5{B}^{10}{:}_5{B}^{11}$ in nature would be

• A. 15:16
• B. 10:11
• C. 19:81
• D. 81:19

Answer 1

Answer: (C)

Let ${}_5{B}^{10}$ be present as $x\%$
and percentage of ${}_5{B}^{10}=(100-x)$
$\therefore$ Average atomic coefficient $=\frac{10x+11(100-x)}{100}$
$=10.81\Rightarrow x=19$
$\therefore \%$ of ${}_5{B}^{11}$ is $100-19=81$.
Ratio is $19:81$