Chlorine has two isotopes \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\) with average atomic mass of 35.5 . What is the ratio of their relative abundance respectively?

Average atomic mass
(atomic mass of \({ }^{35} \mathrm{Cl} \times\) percentage)
\(=\frac{+ \text { atomic mass of }{ }^{37} \mathrm{Cl} \times \text { percentage }}{100}\)
Let the \(\%\) abundance of \({ }^{35} \mathrm{Cl}\) isotope \(=x\). \(\%\) abundance of \({ }^{37} \mathrm{Cl}\) isotope \(=100-x\).
Average atomic mass \(=35.5\)
From formula, average atomic mass
\(=\frac{35 \times x+37 \times(100-x)}{100}=35.5\)
\(\begin{array}{ll}
\therefore & 35 x+3700-37 x=35.5 \times 100 \\
\therefore & -2 x=-3700+3550 \\
\therefore & -2 x=-150 \\
\therefore \quad & x=75 \% \text { and }(100-x)=25 \%
\end{array}\)
Ratio of relative abundance
\(=\frac{\% \text { abundance of }{ }^{35} \mathrm{Cl}}{\% \text { abundance of }{ }^{37} \mathrm{Cl}}=\frac{75}{25}=3: 1\)