Bronze is a solid solution of \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Sn}(s) ;\) solutions of metals like this that are solids are called alloys. There is a range of compositions over which the solution is considered a bronze. Bronzes are stronger and harder than either copper or tin alone. (a) A \(100.0-\mathrm{g}\) sample of a certain bronze is \(90.0 \%\) copper by mass and \(10.0 \%\) tin. Which metal can be called the solvent, and which the solute? (b) Based on part (a), calculate the concentration of the solute metal in the alloy in units of molarity, assuming a density of $7.9 \mathrm{~g} / \mathrm{cm}^{3} .$ (c) Suggest a reaction that you could do to remove all the tin from this bronze to leave a pure copper sample. Justify your reasoning.

To identify the solvent and solute in this case, we need to determine which metal is present in a greater proportion. Since the bronze alloy is composed of \(90.0\%\) copper and \(10.0\%\) tin, copper is clearly the larger component. Thus, copper (Cu) can be called the solvent and tin (Sn) the solute.

To calculate the molarity of the solute (tin) in the alloy, follow these steps: 1. Find the mass of tin in the bronze sample: \(10.0\%\) of \(100.0\mathrm{g}\) gives us \(10.0\mathrm{g}\) of tin. 2. Convert the mass of tin to moles by dividing the mass by tin's molar mass (approximately \(118.71\mathrm{g/mol}\)): \(\frac{10.0\mathrm{g}}{118.71\mathrm{g/mol}} \approx 0.0842\mathrm{mol}\) of tin. 3. Determine the volume of the bronze sample using its density: \(density = \frac{mass}{volume}\), so \(volume = \frac{mass}{density} = \frac{100.0 \mathrm{g}}{7.9 \mathrm{g/cm^{3}}} \approx 12.66\mathrm{cm^{3}}\) 4. Convert the volume to liters: \(12.66\mathrm{cm^{3}} \times \frac{1 \mathrm{L}}{1000 \mathrm{cm^{3}}} \approx 0.01266\mathrm{L}\) 5. Finally, calculate the molarity of tin by dividing its number of moles by the volume of the bronze sample in liters: \[Molarity = \frac{moles}{volume} = \frac{0.0842\mathrm{mol}}{0.01266\mathrm{L}} \approx 6.65\mathrm{M}\] Hence, the concentration of the solute metal (tin) in the bronze alloy is approximately \(6.65\mathrm{M}\).

A possible reaction to remove the tin from the bronze, leaving a pure copper sample, is an acid for which tin will be more reactive than copper (e.g., hydrochloric acid). Here, tin will react with the acid, forming tin(II) chloride and hydrogen gas. The unreacted copper can then be separated from the tin-containing solution. Reaction: \(\mathrm{Sn}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{SnCl_2}(aq) + \mathrm{H_2}(g)\) This reaction is feasible because tin is more reactive with hydrochloric acid than copper, which will not react with the acid and can be easily separated from the solution afterward. This process should effectively remove all the tin from the bronze, leaving a pure copper sample.