Problem 75 - Chemistry: The Central Science

Question

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with 3.00 \(\mathrm{L}\) of \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

Step-by-Step Solution

Answer

(a) To prepare 250 mL of a 0.250 M sucrose solution: 1. Calculate the mass of sucrose required: 0.0625 mol * 342.30 g/mol = 21.39 g. 2. Weigh 21.39 g of solid sucrose and transfer it to a 250 mL volumetric flask. 3. Add distilled water to dissolve the sucrose, and then fill the flask to the 250 mL mark. 4. Mix the solution thoroughly. (b) To prepare 350 mL of a 0.100 M sucrose solution from a 1.50 M sucrose solution: 1. Use the dilution formula (C1V1 = C2V2) to find the initial volume needed: V1 = (0.100 M * 350 mL) / 1.50 M = 23.33 mL. 2. Measure 23.33 mL of the 1.50 M sucrose solution and transfer it to a 350 mL volumetric flask. 3. Add distilled water to fill the flask to the 350 mL mark. 4. Mix the solution thoroughly.

Step 1: Calculate the amount of sucrose for the desired solution

Firstly, we need to determine the number of moles of sucrose required to prepare the solution. Using the equation, Molarity (M) = Moles of solute / Volume of solution (L) 0.250 M = Moles of sucrose / (250 mL * (1 L / 1000 mL)) Moles of sucrose = 0.250 M * 0.250 L = 0.0625 mol Now, we need to determine the mass of sucrose required. The molar mass of sucrose is given by the sum of atomic masses of its elements in the formula: Molar mass of sucrose = \(12 * 12.01 + 22 * 1.01 + 11 * 16.00\) g/mol ≈ 342.30 g/mol. Mass of sucrose = Moles * Molar mass = 0.0625 mol * 342.30 g/mol = 21.39 g We need 21.39 g of solid sucrose to make the solution.

Step 2: Prepare the solution

To prepare the 0.250 M sucrose solution: 1. Weigh 21.39 g of solid sucrose on a balance. 2. Transfer the sucrose to a 250 mL volumetric flask. 3. Add distilled water to the volumetric flask to dissolve the sucrose. Stir or swirl the solution to aid in the dissolution process. 4. Once the sucrose is completely dissolved, carefully add more distilled water to bring the volume of the solution up to the 250 mL mark on the volumetric flask. 5. Mix the solution thoroughly, and it is ready to use. (b) Preparing 350 mL of a 0.100 M sucrose solution starting with 3.00 L of 1.50 M sucrose solution

Step 1: Calculate the volume of the initial solution needed

In this case, we are diluting an existing solution instead of starting with a solid solute. To find the required volume of the 1.50 M solution to use in preparing the 0.100 M solution, we will use the dilution formula: C1V1 = C2V2 Where C1 is the initial concentration (1.50 M), V1 is the initial volume, C2 is the final concentration (0.100 M), and V2 is the final volume (350 mL). V1 = (C2 * V2) / C1 V1 = (0.100 M * 350 mL) / 1.50 M = 23.33 mL We need 23.33 mL of the 1.50 M sucrose solution to make the new 0.100 M sucrose solution.

Step 2: Dilute the initial solution

To prepare the 0.100 M sucrose solution: 1. Measure 23.33 mL of the 1.50 M sucrose solution using a graduated cylinder or pipette. 2. Transfer the measured solution to a 350 mL volumetric flask. 3. Add distilled water to the volumetric flask to bring the volume of the solution up to the 350 mL mark. 4. Mix the solution thoroughly, and it is ready to use.

Key Concepts

Molarity Calculation

Understanding molarity is fundamental in chemistry, especially when preparing solutions. Molarity, denoted by M, is the measure of the concentration of a solute in a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The formula for calculating molarity is:\[ M = \frac{\text{moles of solute}}{\text{Volume of solution in liters}} \]

When preparing a sucrose solution, like in our exercise, the first step involves calculating how many moles you need to achieve the desired concentration. For instance, to prepare 250 mL of a 0.250 M sucrose solution, you would perform the following calculation:\[ 0.250\,\text{M} = \frac{\text{moles of sucrose}}{0.250\,\text{L}} \]This equation can be rearranged to solve for moles of sucrose:\[ \text{Moles of sucrose} = 0.250\,\text{M} \times 0.250\,\text{L} = 0.0625\,\text{mol} \]

Once you know the moles of the solute required, the next step (discussed in molar mass section) is to calculate the mass of the solute needed by using its molar mass.

Solution Dilution

Diluting a solution is another critical laboratory skill. We often need to change the concentration of a solution to suit a particular test or reaction. To do this without altering the chemical composition, we use the dilution formula:\[ C_1V_1 = C_2V_2 \]where:\

• \( C_1 \) is the initial concentration of the solution,
• \( V_1 \) is the volume of the solution we need to use,
• \( C_2 \) is the final concentration we want to achieve,
• \( V_2 \) is the final volume of the diluted solution.

Using our example, to dilute 3.00 L of a 1.50 M sucrose solution to 350.0 mL of 0.100 M, you use the volumes in liters for accurate calculations:\[ 23.33\,\text{mL} = \frac{0.100\,\text{M} \times 0.350\,\text{L}}{1.50\,\text{M}} \]After measuring the required volume of the initial concentrated solution, you add sufficient water to reach the final desired volume. It's important to do this addition carefully to avoid exceeding the volume and diluting the solution too much.

Molar Mass of Sucrose

The molar mass of a substance is the weight in grams of one mole of that substance. The unit is grams per mole (g/mol). For sucrose (C12H22O11), this is particularly important because precise measurements are necessary to prepare solutions with the right concentration.

To calculate the molar mass of sucrose, sum the atomic masses of all the atoms in the molecule:\[ 12 \times 12.01\,(\text{C}) + 22 \times 1.01\,(\text{H}) + 11 \times 16.00\,(\text{O}) \approx 342.30\,\text{g/mol} \]

Knowing the molar mass lets you convert between moles of sucrose and mass which is practical since we measure mass (in grams) more easily than moles in a lab. For instance, for our 0.250 M solution, we need:\[ \text{Mass of sucrose} = 0.0625\,\text{mol} \times 342.30\,\text{g/mol} = 21.39\,\text{g} \]
You’ll weigh out this amount of sucrose and proceed with the steps to prepare your sucrose solution.