Question

$18$-carat gold contains $18$ parts of gold and $6$ parts of copper by mass. Calculate the concentration of gold and copper in $18$-carat gold in $\%(\mathrm{m}/\mathrm{m})$.

Answer 1

$18$-carat gold represents a mixture of $18$-parts Gold $\left(\mathrm{Au}\right)$ and $6$-parts Copper $\left(\mathrm{Cu}\right)$.

• So, it can be considered a solid solution.
• Total parts of the solution= $24[18(\mathrm{from}\mathrm{Gold})+6(\mathrm{from}\mathrm{Copper}\left)\right]$

Step-1: Calculate the concentration of Gold in $18$-carat Gold

Formula to be used:

• $\mathrm{Concentration}\mathrm{in}(\mathrm{m}/\mathrm{m})\%=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{solute}\left(\mathrm{g}\right)}{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{solution}\left(\mathrm{g}\right)}\times 100$

This formula can be modified according to the given scenario as the following by considering gold as the solute first.

• $\mathrm{Concentration}\mathrm{of}\mathrm{Gold}\mathrm{in}(\mathrm{m}/\mathrm{m})\%=\frac{\mathrm{Parts}\mathrm{of}\mathrm{Gold}\mathrm{in}18-\mathrm{carat}\mathrm{gold}}{\mathrm{Parts}\mathrm{of}(\mathrm{Gold}+\mathrm{Copper})}\times 100$
• $\mathrm{Concentration}\mathrm{of}\mathrm{Gold}\mathrm{in}(\mathrm{m}/\mathrm{m})\%=\frac{18}{24}\times 100=\frac{1800}{24}=75\%(\mathrm{m}/\mathrm{m})$
• $\mathrm{Concentration}\mathrm{of}\mathrm{Gold}\mathrm{in}(\mathrm{m}/\mathrm{m})\%=75\%(\mathrm{m}/\mathrm{m})$

Step-2: Calculate the concentration of Copper in $18$-carat Gold

Now considering Copper as the solute,

$\mathrm{Concentration}\mathrm{of}\mathrm{Copper}\mathrm{in}(\mathrm{m}/\mathrm{m})\%=\frac{\mathrm{Parts}\mathrm{of}\mathrm{Copper}\mathrm{in}18-\mathrm{carat}\mathrm{gold}}{\mathrm{Parts}\mathrm{of}(\mathrm{Gold}+\mathrm{Copper})}\times 100$

$\mathrm{Concentration}\mathrm{of}\mathrm{Copper}\mathrm{in}(\mathrm{m}/\mathrm{m})\%=\frac{6}{24}\times 100=\frac{600}{24}=25\%$

$\mathrm{Concentration}\mathrm{of}\mathrm{Copper}\mathrm{in}(\mathrm{m}/\mathrm{m})\%=25\%(\mathrm{m}/\mathrm{m})$