Question

# 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.

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Solution

## Step 1: Finding the number of atoms of Gold and Silver:As, Mass of Silver = $\mathrm{m}\mathrm{g}$And mass of Gold = $\frac{\mathrm{m}}{100\mathrm{g}}$Because, $\mathrm{Number}\mathrm{of}\mathrm{atoms}=\frac{\mathrm{Mass}}{\mathrm{Atomic}\mathrm{mass}}×{\mathrm{N}}_{\mathrm{A}}$So, $\mathrm{Number}\mathrm{of}\mathrm{atoms}\mathrm{of}\mathrm{Silver}=\left(\frac{\mathrm{m}}{108}\right)×{\mathrm{N}}_{\mathrm{A}}$and $\mathrm{Number}\mathrm{of}\mathrm{atoms}\mathrm{of}\mathrm{Gold}=\left(\frac{\mathrm{m}}{\left(100×197\right)}\right)×{\mathrm{N}}_{\mathrm{A}}$.Step 2: Finding the ratio of number of atoms of Gold to Silver:As, the ratio of number of atoms from Gold to Silver would be = $\mathrm{Au}:\mathrm{Ag}$So, the ratio = $\frac{\mathrm{m}}{\left(100×197\right)}×{\mathrm{N}}_{\mathrm{A}}:\frac{\mathrm{m}}{108}×{\mathrm{N}}_{\mathrm{A}}$Ratio would be: $108:100×197$Therefore, the ratio of number of atoms of Gold to Silver is : $1:182.41$

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