Some Basic Concepts of Chemistry 1 Question 41
61. Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01 . The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron.
$(1978,2 M)$
Show Answer
Solution:
- Average atomic weight
$\begin{aligned} & =\frac{\Sigma \text { Percentage of an isotope } \times \text { Atomic weight }}{100} \ \Rightarrow 10.81 & =\frac{10.01 x+11.01(100-x)}{100} \Rightarrow x=20 %\end{aligned}$
Therefore, natural boron contains 20% (10.01) isotope and 80% other isotope.
