Bronze is a solid solution of \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Sn}(\mathrm{s})\) ; 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) \(\mathrm{A} 100.0\) -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.

In a mixture, the component with the larger proportion is the solvent while the other is the solute. It's given that the alloy is 90.0% copper and 10.0% tin by mass. Since the proportion of copper is higher, copper is the solvent and tin is the solute in this alloy.

To calculate the concentration of tin, the solute, in this bronze alloy in molarity, we need to follow these steps: 1. Determine the mass of tin in the sample. 2. Convert the mass of tin to moles, using its molar mass. 3. Determine the volume of the sample. 4. Calculate the concentration of tin in the alloy in molarity. 1. Mass of tin: Given that the alloy sample is 100.0g and 10.0% of it is tin, the mass of tin in the sample is: \(10.0 \% \times 100.0 \mathrm{g} = 10 \mathrm{g}\) 2. Moles of tin: The molar mass of tin (\(\mathrm{Sn}\)) is approximately 118.71 g/mol. So, the moles of tin in the sample are: \(\frac{10 \mathrm{g}}{118.71 \mathrm{g/mol}} \approx 0.0842 \mathrm{mol}\) 3. Volume of the sample: Given that the density of the alloy is 7.9g/cm³, and the mass of the sample is 100g, the volume of the sample is: \(\frac{100 \mathrm{g}}{7.9 \mathrm{g/cm³}} \approx 12.66 \mathrm{cm³}\) 4. Concentration of tin in the alloy in molarity: Converting the volume of the sample from cm³ to L, we get: \(\frac{12.66 \mathrm{cm³}}{1000} = 0.01266 \mathrm{L}\) Now, we can calculate the concentration of tin in the alloy: \(\frac{0.0842 \mathrm{mol}}{0.01266 \mathrm{L}} \approx 6.65 \mathrm{M}\) The concentration of tin in the alloy is approximately 6.65M.

One possible reaction to remove tin from the bronze alloy is to treat it with nitric acid, which can selectively dissolve tin to produce soluble tin nitrate while leaving copper behind. The reaction is as follows: \(\mathrm{Sn(s) + 4HNO_3(aq) \rightarrow Sn(NO_3)_4(aq) + 2NO_2(g) + 2H_2O(l)}\) The reason for choosing nitric acid is that it reacts with tin to form a soluble compound, tin nitrate, which can be easily separated from the copper. Copper, on the other hand, is relatively unreactive under the same conditions, especially if it forms a protective oxide layer on its surface. After removing the tin from the alloy, the remaining copper can be purified further through an electrorefining process.