You make \(1.000 \mathrm{~L}\) of an aqueous solution that contains $35.0 \mathrm{~g}\( of sucrose \)\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) .$ (a) What is the molarity of sucrose in this solution? (b) How many liters of water would you have to add to this solution to reduce the molarity you calculated in part (a) by a factor of two?

To find the molarity of the sucrose solution, we need to calculate the number of moles of sucrose in the solution. We can do this using the given mass of sucrose and its molar mass. After that, we will use the molarity formula to calculate the molarity. 1. Calculate the molar mass of sucrose: The molar mass of sucrose (\(C_{12}H_{22}O_{11}\)) can be calculated by adding the molar masses of its individual atoms: Molar mass of sucrose = (12 × molar mass of C) + (22 × molar mass of H) + (11 × molar mass of O) Molar mass of sucrose = (12 × 12.01 g/mol) + (22 × 1.008 g/mol) + (11 × 16.00 g/mol) ≈ 342.30 g/mol 2. Calculate the number of moles of sucrose: Number of moles = (mass of sucrose) / (molar mass of sucrose) Number of moles = (35.0 g) / (342.30 g/mol) ≈ 0.102 mol 3. Calculate the molarity of the sucrose solution: Molarity = (number of moles of solute) / (volume of solution in liters) Molarity ≈ 0.102 mol / 1.000 L ≈ 0.102 M So, the molarity of the sucrose solution is approximately 0.102 M.

Now, we need to find the volume of water that needs to be added to the solution in order to reduce the molarity by a factor of two. To do this, we will use the dilution formula: \(C_1 V_1 = C_2 V_2\) Where: - \(C_1\) is the initial concentration (0.102 M) - \(V_1\) is the initial volume (1.000 L) - \(C_2\) is the final concentration (half of the initial concentration, or 0.051 M) - \(V_2\) is the final volume (unknown) 1. Rearrange the dilution formula to solve for \(V_2\): \(V_2 = \frac{C_1 V_1}{C_2}\) 2. Insert the values and calculate \(V_2\): \(V_2 = \frac{0.102 M * 1.000 L}{0.051 M} ≈ 2.000 L\) 3. Calculate the volume of water to be added: Volume of water to be added = final volume - initial volume Volume of water to be added = 2.000 L - 1.000 L = 1.000 L So, 1.000 L of water needs to be added to the solution in order to reduce the molarity of sucrose by a factor of two.