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 $250 \mathrm{~mL}$ of a $0.250 \mathrm{M}$ sucrose solution, measure $21.39 \mathrm{~g}$ of solid sucrose, dissolve it in a small amount of distilled water, and transfer it to a $250 \mathrm{~mL}$ volumetric flask. Add more distilled water to the flask until the total volume reaches the $250 \mathrm{~mL}$ mark and mix well. (b) To prepare $350.0 \mathrm{~mL}$ of $0.100 \mathrm{M}$ sucrose solution starting with $1.50 \mathrm{M}$ sucrose solution, measure $23.3 \mathrm{~mL}$ of the $1.50 \mathrm{M}$ sucrose solution and transfer it to a $350 \mathrm{~mL}$ volumetric flask. Add distilled water to the flask until the total volume reaches the $350 \mathrm{~mL}$ mark and mix well.

Step 1: Calculate the Moles of Sucrose Required

To prepare a 0.250 M sucrose solution, first calculate the moles of sucrose needed using the equation: Molarity (M) = Moles of solute / Volume of solution (L) Rearrange the equation and solve for moles: Moles of solute = Molarity * Volume of solution (L) Remember to convert milliliters to liters: 250 mL = 0.250 L Moles of sucrose = (0.250 M) * (0.250 L) = 0.0625 moles

Step 2: Calculate the Mass of Sucrose Required

Now, calculate the mass of sucrose needed using the molecular weight of sucrose, which is 342.3 g/mol: Mass of sucrose = Moles of sucrose * Molecular weight of sucrose Mass of sucrose = 0.0625 moles * 342.3 g/mol = 21.39 g

Step 3: Prepare the Sucrose Solution

To prepare the 0.250 M sucrose solution, follow these steps: 1. Measure 21.39 g of solid sucrose using a balance. 2. Dissolve the measured solid sucrose in a small amount of distilled water. 3. Pour the dissolved sucrose into a 250 mL volumetric flask. 4. Add more distilled water to the flask until the total volume of the solution reaches the 250 mL mark. 5. Cap the flask and mix the solution well. You now have a 0.250 M sucrose solution in 250 mL of water. (b) Preparation of \(0.100 \mathrm{M}\) Sucrose Solution using \(1.50 \mathrm{M}\) Sucrose Solution:

Step 4: Calculate the Volume of the Concentrated Sucrose Solution Required

To prepare a 0.100 M sucrose solution using a 1.50 M sucrose solution, use the dilution formula: M₁* V₁ = M₂ * V₂ where M₁ and V₁ are the molarity and volume of the concentrated solution, and M₂ and V₂ are the molarity and volume of the diluted solution. Solve for V₁: V₁ = (M₂ * V₂) / M₁ Convert milliliters to liters: 350 mL = 0.350 L V₁ = (0.100 M * 0.350 L) / 1.50 M = 0.0233 L

Step 5: Prepare the Diluted Sucrose Solution

To prepare the 0.100 M sucrose solution, follow these steps: 1. Measure 0.0233 L (23.3 mL) of the 1.50 M sucrose solution using a graduated cylinder or pipette. 2. Transfer the measured volume to a 350 mL volumetric flask. 3. Add distilled water to the flask until the total volume of the solution reaches the 350 mL mark. 4. Cap the flask and mix the solution well. You now have a 0.100 M sucrose solution in 350 mL of water.

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry that allows scientists and students to understand the concentration of a solution. By definition, it is the number of moles of a solute present in one liter of solution. Expressed as the unit molarity (M), it forms the basis for many calculations in chemistry, particularly when preparing solutions.

In our exercise, for instance, we calculate the molarity of a sucrose solution by dividing the number of moles of sucrose by the volume of the solution in liters. This foundational measure helps to ensure that the prepared solution has the precise chemical composition required for a particular experiment or application. To grasp molarity fully, remember this relation:
\[\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}\].

When dealing with exercises like the one above, it's necessary to be comfortable with converting milliliters to liters — since molarity is specifically moles per liter — and using the molarity formula to calculate either the number of moles needed for a particular volume or figuring out what volume an existing number of moles will occupy at a desired molarity.

Solution Dilution

Understanding solution dilution is essential for performing a range of chemistry tasks, from experimental design to practical applications like medicine formulation. Dilution involves reducing the concentration of a solute in a solution, usually by adding more solvent. In the context of our sucrose solution, it's about how to properly dilute a more concentrated solution to achieve a lower molarity.

The dilution formula connects the volume and concentration (molarity) of the original solution (concentrated) to the volume and concentration of the resulting solution (diluted):
\[M_1V_1 = M_2V_2\].

Where \(M_1\) and \(V_1\) are the molarity and volume of the concentrated solution, while \(M_2\) and \(V_2\) are for the diluted solution. To dilute a solution, you must first calculate the volume needed from the concentrated stock using this formula and then simply add enough solvent to reach the intended final volume. As seen in the exercise, starting with a high molarity solution and using these calculations leads to the correct volume of a lower molarity solution.

Mole Concept

Lastly, let's delve into the mole concept, which is a pillar in the study of chemistry. A mole is a unit that represents a specific number of particles, specifically Avogadro's number (\(6.022 \times 10^{23}\)) of particles. This could be atoms, molecules, ions, or electrons – in our case, it's sucrose molecules. The mole concept is central to the quantitative description of chemical reactions and stoichiometry.

When preparing a sucrose solution, as in our problem, we first need to figure out the number of moles of sucrose to achieve the desired molarity. This is done by utilizing the formula:
\[\text{Moles of solute} = \text{Molarity} \times \text{Volume of solution (L)}\].

With these moles, we can then calculate the corresponding mass of sucrose needed by multiplying the number of moles by sucrose's molar mass. The mole concept helps bridge the gap between the microscopic scale of atoms and molecules and the macroscopic world we can see and measure in the lab. By grasping this concept, students can securely navigate through a multitude of chemistry problems, confident in their ability to relate quantities of substances in terms of particles and their masses.