Atomic Structure Question 472
Question: 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.
Options:
A) 20, 80
B) 30, 70
C) 10, 90
D) 15, 85
Show Answer
Answer:
Correct Answer: A
Solution:
- Let the percentage of isotope with atomic wt. 10.01 = x Percentage of isotope with atomic wt. Average atomic wt. $ =\frac{m_1x_1+m_2x_2}{x_1+x_2} $ or Average atomic wt. $ =\frac{x\times 10.01+(100-x)\times 11.01}{100} $ $ 10.81=\frac{x\times 10.01+(100-x)\times 11.01}{100}\Rightarrow x=20 $
$ \therefore $ % of isotope with atomic wt. 10.01 = 20 % of isotope with atomic wt. 11.01 = 100 - x = 80.
