Question

Naturally occurring B consists of two isotopes, whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. The % of each isotope in natural boron is :

• A. % of isotope of mass 10.01 = 20, % of isotope of mass 11.01 = 80
• B. % of isotope of mass 10.01 = 30, % of isotope of mass 11.01 = 70
• C. % of isotope of mass 10.01 = 50, % of isotope of mass 11.01 = 50
• D. none of these

Answer 1

The correct option is A % of isotope of mass 10.01 = 20, % of isotope of mass 11.01 = 80

Average atomic weight =$\frac{\text{relative atomic mass of type 1 X \% abundance of type 1}+\text{relative mass of type 2 X \% abundance of type 2}}{100}$

Let % of isotope of mass 10.01 be $a$.

$10.81=\frac{10.01\times a+11.01(100-a)}{100}$

$a=20$

% of isotope of mass 10.01 = 20

% of isotope of mass 11.01 = 80