Question

Boron has two isotopes, ${}_5{B}^{10}$ and ${}_5{B}^{11}$ and the atomic weight of boron is $10.81$.

The ratio of ${}_5{B}^{10}:{}_5{B}^{11}$ in nature would be:

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

Answer 1

The correct option is A $19:81$

Let, $100$ mol of sample contains $x$ mol of ${}_5{B}^{10}$ and $100-x$ mol of ${}_5{B}^{11}$.
The average atomic weight of boron is $10.81$.

Hence, $10.81=\frac{x\times 10+(100-x)\times 11}{100}$

$1081=10x+1100-11x$

$19=x$

$100-x=100-19=81$

Thus, $100$ mol of sample contains $19$ mol of ${}_5{B}^{10}$ and $81$ mol of ${}_5{B}^{11}$.

Thus, the ratio is $19:81$.