A silver ornament of mass ' $m$ ' gram is polished with gold equivalent to $1 \%$ of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
Mass of silver $=m \mathrm{~g}$
Mass of gold deposited $=\frac{m \times 1}{100}=\frac{m}{100} \mathrm{~g}$
No. of atoms of silver $=\frac{\text { Mass }}{\text { Atomic mass }} \times \mathrm{N}_{\mathrm{A}}=\frac{m}{108} \times \mathrm{N}_{\mathrm{A}}$
No. of atoms of gold $=\frac{\text { Mass }}{\text { Atomic mass }} \times N_{A}=\frac{m}{100 \times 197} \times N_{A}$
Ratio of the number of atoms of gold to silver $=\frac{m}{100 \times 197} \times \mathrm{N}_{\mathrm{A}}: \frac{m}{108} \times \mathrm{N}_{\mathrm{A}}$
$108: 100 \times 197=108: 19700=1: 182.41 .$
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