Question

Boron found in nature has an atomic weight of 10.811 and is made up of the isotopes $B^{10}$ (mass 10.013 amu) and $B^{11}$ (mass 11.0093). What percentage of naturally occurring boron is made up of $B^{10}$ and $B^{11}$, respectively?

• A. 30 : 70
• B. 25 : 75
• C. 20 : 80
• D. 15 : 85

Answer 1

The correct option is C 20 : 80

Let the abundance of $B^{11}$ be $x$% and $B^{10}$ be $(100-x)$%

Average atomic mass= [Atomic mass of $B^{11}\times$ abundance + Atomic mass of $B^{10}\times$ abundance]$/100$

$\Rightarrow 10.811=\frac{11.0093\times x\left(percent\right)+10.013\times (100-x)\left(percent\right)}{100}$

$\Rightarrow 10.811\times 100=11.0093x$%$+1001.3-10.013x$%

$\Rightarrow 1081.1=0.9963x$%$+1001.3$

$\Rightarrow 1081.1-1001.3=0.9963x$%

$\Rightarrow 79.8=0.9963x$%

$\Rightarrow x$%=$\frac{79.8}{0.9963}$

$\therefore x$%=$80$

$(100-x)$%=$20$

$\therefore$ Natural abundance of $B^{10}=20$

Natural abundance of $B^{11}=80$

Ratio= $20:80$