Chapter 2 - Exercises - Page 82: 116

\[\underline{20\,\text{percent}}\] and \[80\,\text{percent}\]

Calculate the fraction of boron-10 as follows: \[\text{Fraction of isotope boron-}10=\frac{\text{Relative abundance of boron-}10}{\text{100}}\] Calculate the fraction of boron-11 as follows: \[\text{Fraction of isotope boron-}11=\frac{\text{Relative abundance of boron-}11}{\text{100}}\] Atomic mass is the average of the mass of isotopes. So, the atomic mass of boron is \[10.81\text{ amu}\]. Calculate the relative abundance as follows: \[\begin{align} & \text{Atomic mass}=\left( \text{Fraction of boron-}10\times \text{Mass of boron-}10 \right)+ \\ & \left( \text{Fraction of boron-}11\times \text{Mass of boron-}11 \right) \\ & 10.81=\left( \frac{\text{Relative abundance of boron-}10}{\text{100}}\times 10.01294\text{ amu} \right) \\ & +\left( \frac{\text{Relative abundance of boron-}11}{\text{100}}\times 11.00931\text{ amu} \right) \end{align}\] Substitute relative abundance of boron-11 as \[100-\text{relative abundance of boron-}10\] in the above expression as follows: \[\begin{align} & 10.81=\left( \frac{\text{Relative abundance of boron-}10}{\text{100}}\times 10.01294\text{ amu} \right)+ \\ & \left( \frac{\text{100}-\text{Relative abundance of boron-}10}{\text{100}}\times 11.00931\text{ amu} \right) \end{align}\] \[\text{Relative abundance of boron}-10=20\,\text{percent}\] Calculate the relative abundance of boron-11 as follows: \[\begin{align} & \text{Relative abundance of boron-}11=100\,\text{percent}-\underline{20\,\text{percent}} \\ & =80\,\underline{\text{percent}} \end{align}\] The relative abundance of boron-10 is \[\underline{20\,\text{percent}}\] and relative abundance of boron-11 is \[80\,\text{percent}\].