1. The formula for copper(II) sulfate pentahydrate is CuSO4·5H2O. The molar mass of CuSO4 is 159.61 g/mol and the molar mass of 5H2O is 90.10 g/mol. Therefore, the molar mass of CuSO4·5H2O is 249.71 g/mol. The mass of water in one mole of CuSO4·5H2O is 5 x 18.02 g = 90.10 g. The mass of CuSO4 in one mole of CuSO4·5H2O is 159.61 g. The percent water in CuSO4·5H2O is: (90.10 g water / 249.71 g CuSO4·5H2O) x 100% = 36.06% Therefore, the percent water in CuSO4·5H2O to three significant figures is 36.1%. 2. The moles of Fe(NO3)3 needed to prepare 1.00 L of 0.250 M solution can be calculated using the formula: moles = concentration x volume moles = 0.250 M x 1.00 L = 0.250 moles The initial volume of the 1.00 M Fe(NO3)3 solution needed can be calculated using the same formula: moles = concentration x volume 0.250 moles = 1.00 M x volume volume = 0.250 moles / 1.00 M = 0.250 L Therefore, 0.250 L or 250 mL of the 1.00 M Fe(NO3)3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M. 3. A. The molar mass of cholesterol (C27H46O) is 386.65 g/mol. The moles of cholesterol in 0.195 g can be calculated using the formula: moles = mass / molar mass moles = 0.195 g / 386.65 g/mol = 0.000504 moles The molarity of cholesterol in 0.100 L of serum can be calculated using the formula: molarity = moles / volume molarity = 0.000504 moles / 0.100 L = 0.00504 M or 5.04 mM (millimolar) B. The molar mass of NH3 is 17.03 g/mol. The moles of NH3 in 4.25 g can be calculated using the formula: moles = mass / molar mass moles = 4.25 g / 17.03 g/mol = 0.249 moles The molarity of NH3 in 0.500 L of solution can be calculated using the formula: molarity = moles / volume molarity = 0.249 moles / 0.500 L = 0.498 M or 498 mM (millimolar)