The solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) is \(70 \mathrm{~g}\) per \(100 \mathrm{~mL}\) of water. (a) Is a \(1.22 \mathrm{M}\) solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) saturated, supersaturated, or unsaturated? (b) Given a solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

First, we need to determine the solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in moles per liter. We know that \(70\,\mathrm{g}\) of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) dissolve in \(100\,\mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\). To convert this to moles per liter, we will first need to find the molecular weight of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\): Molecular weight = (Mass of Mn) + (Mass of S) + (4 * Mass of O) + (2 * Mass of H) + (Mass of O) = \(54.94 + 32.07 + 4\times 16 + 2\times 1.01 + 16\) Now, we can convert grams to moles: Moles of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) = \(\frac{70\,\mathrm{g}}{\text{Molecular weight}}\) Finally, since the amount of solute is given per \(100\,\mathrm{mL}\), we need to convert the volume to liters. To do so, divide the volume by \(1000\): Volume = \(\frac{100\,\mathrm{mL}}{1000}\,\mathrm{L}\) Now, we can find the solubility in moles per liter: Solubility = \(\frac{\text{Moles of }\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}}{\text{Volume in L}}\)

Now, we will compare the solubility we calculated in moles per liter to the given concentration of \(1.22\,\mathrm{M}\) to determine whether the solution is saturated, supersaturated, or unsaturated. - If the solubility is equal to the concentration, the solution is saturated. - If the solubility is greater than the concentration, the solution is unsaturated. - If the solubility is less than the concentration, the solution is supersaturated.

For a solution of unknown concentration, you can perform the following experiment to determine if the solution is saturated, supersaturated, or unsaturated: 1. Add a small amount of solid \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) to the solution and stir. 2. Observe whether the solid dissolves completely, partially, or not at all. - If the solid dissolves completely, the solution is unsaturated, as it can still dissolve more solute. - If the solid does not dissolve at all, the solution is saturated, as it cannot dissolve any more solute. - If the solid dissolves and then begins to form crystals again, the solution is supersaturated, as it is unstable and can only hold the excess solute temporarily.