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Chlorine has two naturally occurring isotopes, ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$. If the atomic mass of $\mathrm{Cl}$ is $35.45$ the ratio of natural abundance of ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$ is closest to
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$3: 1$
Mav $=\frac{\mathrm{M}_{1} \mathrm{n}_{1}+\mathrm{M}_{2} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$35.45=\frac{35 \mathrm{n}_{1}+37 \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$\therefore \mathrm{n}_{1}: \mathrm{n}_{2}=3: 1$
$35.45=\frac{35 \mathrm{n}_{1}+37 \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$\therefore \mathrm{n}_{1}: \mathrm{n}_{2}=3: 1$
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