Question

Atomic weight of Boron is 10.81 and it has two isotopes $_5B^{10}$ and $_5B^{11}$ Then the ratio of $_5B^{10} :_5B^{11}$ in nature would be

• A. 0.636111111
• B. 0.424305556
• C. 0.847916667
• D. 3.388194444

Answer 1

Answer: (C)

0.847916667

Answer 2

Let $_5B^{10}$ be present as x% and percentage of $_5B^{10}$ = (100 - x) $\therefore \,$ Average atomic coefficient $=\frac{10x+11(100-x)}{100}$ $=10.81 \, \Rightarrow \, \, x=19$ $\therefore$ % of $_5B^{11}$ is 100- 19 = 81. Ratio is 19 : 81