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(a) Two stable isotopes of lithium ${ }_3^6 \mathrm{Li}$ and ${ }_3^7 \mathrm{Li}$ have respective abundances of $7.5 \%$ and $92.5 \%$. These isotopes have masses $6.01512 \mathrm{u}$ and $7.01600 \mathrm{u}$ respectively. Find the atomic mass of lithium.
(b) Boron has two stable isotopes, ${ }_5^{10} \mathrm{~B}$ and ${ }_5^{11} \mathrm{~B}$. Their respective masses are $10.01294 \mathrm{u}$ and $11.00931 \mathrm{u}$, and the atomic mass of boron is $10.811 \mathrm{u}$. Find the abundances of ${ }_5^{10} \mathrm{~B}$ and ${ }_5^{11} \mathrm{~B}$.
(b) Boron has two stable isotopes, ${ }_5^{10} \mathrm{~B}$ and ${ }_5^{11} \mathrm{~B}$. Their respective masses are $10.01294 \mathrm{u}$ and $11.00931 \mathrm{u}$, and the atomic mass of boron is $10.811 \mathrm{u}$. Find the abundances of ${ }_5^{10} \mathrm{~B}$ and ${ }_5^{11} \mathrm{~B}$.
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(a) The isotope ${ }_3^6 \mathrm{Li}$ has abundance $7.5 \%$ and ${ }_3^7 \mathrm{Li}$ has $92.5 \% . \mathrm{m}\left({ }_3^6 \mathrm{Li}\right)=6.01512 \mathrm{u} \mathrm{m}\left({ }_3^7 \mathrm{Li}\right)=7.01600 \mathrm{u}$ Average mass of Li $=\frac{6.01512 \times 7.5+7.01600 \times 92.5}{7.5+92.5}$ $=\frac{45.1140+648.98}{7.5+92.5}$ $=\frac{694.094}{100}=6.94000 \mathrm{u}$.
(b) $\mathrm{m}\left({ }_5^{10} \mathrm{~B}\right)=10.01294 \mathrm{u} ; \mathrm{m}\left({ }_5^{11} \mathrm{~B}\right)=11.00931 \mathrm{u} ; \mathrm{m}$ (B) $=10.811 \mathrm{u}$.
Let abundance of ${ }_5^{10} \mathrm{~B}$ be $\mathrm{x} \%$ thus abundance of ${ }_5^{11} \mathrm{~B}$ will be $(100-\mathrm{x}) \%$
Average mass of Boron
$$
\begin{aligned}
&=\frac{10.01294 \mathrm{x}+11.00931(100-\mathrm{x})}{100} \\
&=10.811 \mathrm{u}
\end{aligned}
$$
$\Rightarrow 10.01294 \mathrm{x}+1100.931-11.00931 \mathrm{x}=1081.1$
$\Rightarrow \quad 0.996 \mathrm{x}=19.831$
$\Rightarrow \mathrm{x}=19.91 \%=$ abundance of ${ }_5^{10} \mathrm{~B}$
$\quad$ and $(100-\mathrm{x})$
$\quad=(100-19.91)=80.1 \%$ abundance of ${ }_5^{11} \mathrm{~B}$
(b) $\mathrm{m}\left({ }_5^{10} \mathrm{~B}\right)=10.01294 \mathrm{u} ; \mathrm{m}\left({ }_5^{11} \mathrm{~B}\right)=11.00931 \mathrm{u} ; \mathrm{m}$ (B) $=10.811 \mathrm{u}$.
Let abundance of ${ }_5^{10} \mathrm{~B}$ be $\mathrm{x} \%$ thus abundance of ${ }_5^{11} \mathrm{~B}$ will be $(100-\mathrm{x}) \%$
Average mass of Boron
$$
\begin{aligned}
&=\frac{10.01294 \mathrm{x}+11.00931(100-\mathrm{x})}{100} \\
&=10.811 \mathrm{u}
\end{aligned}
$$
$\Rightarrow 10.01294 \mathrm{x}+1100.931-11.00931 \mathrm{x}=1081.1$
$\Rightarrow \quad 0.996 \mathrm{x}=19.831$
$\Rightarrow \mathrm{x}=19.91 \%=$ abundance of ${ }_5^{10} \mathrm{~B}$
$\quad$ and $(100-\mathrm{x})$
$\quad=(100-19.91)=80.1 \%$ abundance of ${ }_5^{11} \mathrm{~B}$
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