Question

In lab you need to prepare at least \(100 \mathrm{~mL}\) of each of the following solutions. Explain how you would proceed using the given information. a. \(2.0 \mathrm{~m} \mathrm{KCl}\) in water (density of \(\mathrm{H}_{2} \mathrm{O}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\) ) b. \(15 \% \mathrm{NaOH}\) by mass in water \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\) c. \(25 \% \mathrm{NaOH}\) by mass in \(\mathrm{CH}_{3} \mathrm{OH}\left(d=0.79 \mathrm{~g} / \mathrm{cm}^{3}\right)\) d. \(0.10\) mole fraction of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\)

Short answer

To prepare the given solutions, follow these steps for each: a. Weigh 14.91 g of KCl, dissolve it in approximately 80 mL of distilled water, and then top up with water to the 100 mL mark. b. Weigh 15 g of NaOH, dissolve it in approximately 80 mL of distilled water, and then top up with water to the 100 mL mark. c. Weigh 25 g of NaOH, dissolve it in approximately 40 mL of methanol, and then top up with methanol to the 100 mL mark. d. Weigh 11.17 g of C₆H₁₂O₆, dissolve it in approximately 80 mL of distilled water, and then top up with water to the 100 mL mark.

Step-by-step solution

Calculate the required mass of KCl

First, we need to find the molarity of the solution, which is \(2.0 M\). The molar mass of \(\mathrm{KCl} = \mathrm{(39.10+35.45)~g/mol} = 74.55 g/mol\). For 100 mL (0.1 L) of 2.0 M KCl solution, we can calculate the required mass as follows: \(mass = molarity * volume * molar~mass = 2.0 * 0.1 * 74.55 = 14.91 g\).

Prepare the solution

Weigh 14.91 g of KCl and dissolve it in a 100 mL graduated cylinder containing approximately 80 mL of distilled water. Stir the solution until all KCl has dissolved, then top up with water to the 100 mL mark. Solution b: Preparing 15% NaOH in water:

Calculate the required mass of NaOH

To prepare 100 mL of a 15% NaOH solution (by mass), we need 15 g of NaOH in every 100 g of the solution. Since the density of the solution is 1.00 g/cm³, the solution will weigh the same as 100 mL of water. We need to add 15 g of NaOH to distilled water to create this solution.

Prepare the solution

Weigh 15 g of NaOH and dissolve it in a 100 mL graduated cylinder containing approximately 80 mL of distilled water. Stir the solution until all NaOH has dissolved, then carefully top up with water to the 100 mL mark. Solution c: Preparing 25% NaOH in CH₃OH:

Calculate the required mass of NaOH

To prepare 100 mL of a 25% NaOH solution (by mass) in methanol, we need 25 g of NaOH in every 100 g of the solution. Since the density of methanol is 0.79 g/cm³, the methanol volume in the solution will be \(100 \times (1-0.25) \times 0.79 = 59.25 \mathrm{~mL}\).

Prepare the solution

Weigh 25 g of NaOH and dissolve it in a 100 mL graduated cylinder containing approximately 40 mL of methanol. Stir the solution until all NaOH has dissolved, then carefully top up with methanol to the 100 mL mark. Solution d: Preparing 0.10 mole fraction of C₆H₁₂O₆ in water:

Calculate the required mass of C6H12O6

First, we need to find the molecular weights of \(\mathrm{C_6H_{12}O_{6}}\) and \(\mathrm{H_2O}\). The molecular weight of glucose is \(\mathrm{(6\times 12.01) + (12\times 1.01) + (6\times 16.00) = 180.18 ~g/mol}\) and water is \(\mathrm{(2\times 1.01) + 16.00 = 18.02 ~g/mol}\). For a solution with a 0.10 mole fraction, the ratio of moles of glucose to water must be \(0.10\). Therefore, the mass ratio of glucose to water will be \(\frac{0.10 * 180.18}{(1-0.10) * 18.02}\) To prepare 100 mL of this solution, the total mass of the solution will be \(100 \mathrm{~mL} \times 1.00 \mathrm{~g/cm^3} = 100 \mathrm{~g}\). Now we can calculate the required mass of glucose to add to the solution: \[mass_{C_6H_{12}O_{6}} = \frac{0.10 \times 180.18}{(1-0.10) \times 18.02} \times 100 \mathrm{~g} = 11.17 \mathrm{~g}\]

Prepare the solution

Weigh 11.17 g of C₆H₁₂O₆ and dissolve it in a 100 mL graduated cylinder containing approximately 80 mL of distilled water. Stir the solution until all glucose has dissolved, then carefully top up with water to the 100 mL mark.

Key concepts

Molarity and Solution Preparation

Understanding molarity is essential when preparing laboratory solutions. Molarity, represented as 'M', is the number of moles of solute per liter of solution. To make a solution of specific molarity, you need to calculate the required mass of the solute based on its molar mass and then dissolve that solute in a solvent to a particular volume.

For example, to prepare a 2.0 M solution of potassium chloride (KCl), calculate the molar mass of KCl, which is the sum of the atomic masses of potassium (K) and chlorine (Cl). If the molarity of the desired solution is 2.0 M and the required volume is 100 mL, you apply the formula:
\[mass = molarity \times volume \times molar~mass\].
You must then dissolve the calculated mass of KCl in water and adjust the volume to the exact 100 mL mark in a graduated cylinder. Ensuring complete dissolution and accurate volume measurement is crucial for precision.

Percent by Mass Solution

Percent by mass (weight percent) is another way to express concentration, referring to the mass of solute divided by the total mass of the solution, multiplied by 100. When a solution calls for a certain percent by mass, such as a 15% NaOH solution, this means 15 grams of NaOH is required for every 100 grams of the final solution.

Given the density of water (1.00 g/cm³), a 100 mL volume translates to 100 grams. Therefore, to make a 15% NaOH solution, you would weigh 15 grams of NaOH and add enough water to obtain a final mass of 100 grams. For a solution with a different solvent or density, such as methanol, adjustments in volume are needed based on methanol's density (0.79 g/cm³) to ensure the final mass corresponds to the desired percent by mass.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a solution. It is the ratio of the number of moles of that component to the total number of moles of all substances in the mixture. The mole fraction is dimensionless and does not change with temperature, because it is a ratio.

For example, when preparing a solution with a mole fraction of 0.10 for glucose (C₆H₁₂O₆) in water, calculate the number of moles of glucose relative to water, which can be determined using their molar masses. Then, use the mole fraction to find the mass of glucose needed per mass of the solution. Convert the calculation into grams for practical measurement and proceed to prepare the solution accordingly, giving attention to maintaining the ratio of glucose to water as calculated.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule. For instance, the molar mass of glucose (C₆H₁₂O₆) is calculated by adding six times the atomic mass of carbon, twelve times the atomic mass of hydrogen, and six times the atomic mass of oxygen.

The accurate molar mass calculation is critical in preparing solutions with specific molarities or mole fractions. It provides the conversion factor between the number of moles and the mass in grams, enabling the precise weighing of the substance for solution preparation. Always ensure to use the latest atomic mass values from the periodic table for your calculations to ensure the accuracy of your molar mass determinations.