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.

Answer 1

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, $\mathrm{Number}\mathrm{of}\mathrm{atoms}=\frac{\mathrm{Mass}}{\mathrm{Atomic}\mathrm{mass}}\times {\mathrm{N}}_{\mathrm{A}}$

So, $\mathrm{Number}\mathrm{of}\mathrm{atoms}\mathrm{of}\mathrm{Silver}=\left(\frac{\mathrm{m}}{108}\right)\times {\mathrm{N}}_{\mathrm{A}}$

and $\mathrm{Number}\mathrm{of}\mathrm{atoms}\mathrm{of}\mathrm{Gold}=\left(\frac{\mathrm{m}}{(100\times 197)}\right)\times {\mathrm{N}}_{\mathrm{A}}$.

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 = $\frac{\mathrm{m}}{(100\times 197)}\times {\mathrm{N}}_{\mathrm{A}}:\frac{\mathrm{m}}{108}\times {\mathrm{N}}_{\mathrm{A}}$

Ratio would be: $108:100\times 197$

Therefore, the ratio of number of atoms of Gold to Silver is : $1:182.41$