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 \(0.100 \mathrm{MC}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\).

Short answer

(a) To prepare a \(0.250\mathrm{M}\) sucrose solution, weigh out \(21.39\mathrm{g}\) of solid sucrose, dissolve it in water, and adjust the volume to \(250\mathrm{mL}\) using a volumetric flask. (b) To prepare a \(0.100\mathrm{M}\) sucrose solution with a volume of \(350.0\mathrm{mL}\) from a \(1.50\mathrm{M}\) stock solution, take \(23.3\mathrm{mL}\) of the stock solution and dilute it with water to make a final volume of \(350.0\mathrm{mL}\).

Step-by-step solution

Calculate the required mass of sucrose.

We need to use the molarity formula considering the volume of the desired solution: \(M = \frac{moles}{volume(L)}\) To find the number of moles required, we can rearrange the equation: \(moles = M \times volume(L)\) Then, we convert moles to mass using the molar mass of sucrose: \(mass = moles \times molar\ mass\) For sucrose, the molar mass is \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} = 342.3\mathrm{g/mol}\).

Calculate the number of moles for the desired solution.

We will plug in the values for M and volume into the equation from step 1: \(moles = 0.250\mathrm{M} \times 0.250\mathrm{L} = 0.0625\mathrm{~mol}\)

Calculate the mass of sucrose needed.

Now we will convert the moles to mass using the molar mass: \(mass = 0.0625\mathrm{~mol} \times 342.3\mathrm{g/mol} = 21.39\mathrm{~g}\)

Prepare the solution.

Weigh out \(21.39\mathrm{g}\) of solid sucrose, dissolve it in the water in a volumetric flask, and add enough water to bring the final volume to \(250\mathrm{mL}\). This will give us a \(0.250\mathrm{M}\) sucrose solution. For part (b):

Use the dilution formula to calculate the required volume of the stock solution.

The dilution formula is as follows: \(C_1V_1 = C_2V_2\) Where: - \(C_1\) is the concentration of the stock solution - \(V_1\) is the volume of the stock solution needed - \(C_2\) is the concentration of the desired diluted solution - \(V_2\) is the volume of the desired diluted solution

Calculate the volume of the stock solution needed.

Plugging in the given values in the dilution formula, we get: \((1.50\mathrm{M})(V_1) = (0.100\mathrm{M})(350.0\mathrm{mL})\) Solving for \(V_1\): \(V_1 = \frac{(0.100\mathrm{M})(350.0\mathrm{mL})}{1.50\mathrm{M}}\) \(V_1 = 23.3\mathrm{mL}\)

Prepare the new solution.

To prepare the new solution, take 23.3 mL of the.\(1.50\mathrm{M}\) stock solution of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) and dilute it with water in a volumetric flask to make the final volume of \(350.0\mathrm{mL}\). This will give us a \(0.100\mathrm{M}\) sucrose solution.

Key concepts

Molarity Formula

The molarity formula is a cornerstone concept in chemistry, used to express the concentration of a solution. Simply defined, molarity (M) is the number of moles of solute (substance being dissolved) per liter of solution. The equation is as follows:

\[\begin{equation}M = \frac{moles}{volume(L)}\end{equation}\]

This equation is crucial when preparing solutions in a lab, as it helps determine how much of a compound is needed to achieve a specific concentration. For instance, if you're asked to prepare a 0.250 M sucrose solution, you would start by calculating the number of moles required using the target molarity and volume in liters of the final solution. Once the moles are calculated, you can use the moles to mass conversion to find the precise mass of sucrose needed.

Moles to Mass Conversion

Converting moles to mass is a fundamental skill in chemistry. After calculating the moles needed for a solution using the molarity formula, the next step is to translate this amount into a measurable weight using the molar mass of the solute. The molar mass is the weight in grams of one mole of a substance, determined by summing the atomic masses of all the atoms in a molecule.

The equation to convert moles to mass is given as:

\[\begin{equation}mass = moles \times molar\ mass\end{equation}\]

For our sucrose example with a molar mass of 342.3 g/mol, if we need 0.0625 moles for a solution, the mass in grams is found by multiplying the two:

\[\begin{equation}mass = 0.0625\mathrm{~mol} \times 342.3\mathrm{g/mol}= 21.39\mathrm{~g}\end{equation}\]

This calculation grants us the exact amount of sucrose we need to weigh out and dissolve to achieve the desired solution concentration.

Solution Dilution

Solution dilution involves reducing the concentration of a solution by adding more solvent. The dilution formula (also known as the dilution equation) provides a simple method to determine the volume of a concentrated stock solution needed to achieve a more dilute solution. The formula is:

\[\begin{equation}C_1V_1 = C_2V_2\end{equation}\]

Here,

• \(C_1\) represents the initial concentration of the stock solution,
• \(V_1\) is the volume of the stock solution required,
• \(C_2\) is the final concentration of the diluted solution, and
• \(V_2\) is the final volume of the diluted solution.

By using this formula, one can calculate how much of the stock solution is needed and then add solvent to reach the desired final volume. In the scenario with sucrose, you would start with a concentrated stock solution and use the formula to find out that 23.3 mL of this solution is required to make a 0.100 M solution in a final volume of 350.0 mL.