Question

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

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

Answer 1

The correct option is C $20:80$

Let us consider in boron, $B^{10}$ and $B^{11}$ isotopes are present in the ratio of $20:80$.

$B^{10}$ has a mass of $10.013amu$. Thus $20$% of it will have a weight of $\frac{20}{100}$$\times$$10.013$, that is $2.0026$.

$B^{11}$ has a mass of $11.0093amu$. Thus $80$% of it will have a mass of $\frac{80}{100}$$\times$$11.0093$, that is $8.8074$

Thus, total weight of boron which consists of $B^{10}$ and $B^{11}$ isotopes is $2.0026+8.8074$, which is equal to $10.81$, which matches with our given question.

Thus, by back calculation, we find that $20:80$ gives the correct ratio of isotopes.

Option C is the correct ratio.