n lab you need to prepare at least 100 mL of each of the following solutions. Explain how you would proceed using the given information. a. 2.0 \(\mathrm{mKCl}\) 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 \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\)

Firstly, we must define that a 2.0 m (molal) KCl solution means that there are 2.0 moles of KCl in 1 kg of solvent (water). Now, let's assume that we have to prepare 100 mL of this solution. Since the density of water is 1.00 g/cm³, the mass of 100 mL of water is: Mass of water = Density × Volume = 1.00 g/cm³ × 100 mL = 100 g In 1 kg of water, there are 2.0 moles of KCl, so in 100 g of water, there will be: 2.0 moles × 100 g/1000 g = 0.2 moles of KCl Now, we need to convert moles to mass. The molar mass of KCl is approximately 39.10 g/mol (K) + 35.45 g/mol (Cl) = 74.55 g/mol. Thus, the mass of KCl required is: Mass = Moles × Molar mass= 0.2 moles × 74.55 g/mol = 14.91 g

To prepare the 2.0 m KCl solution: 1. Weigh 14.91 g of KCl using a balance. 2. Add the KCl to a beaker containing approximately 80 mL of water. 3. Stir until the KCl is completely dissolved. 4. Transfer the solution to a 100 mL volumetric flask using a funnel, rinse the beaker with water, and pour the rinsing into the flask. 5. Add water to the volumetric flask until the 100 mL mark is reached. 6. Stir or invert the flask to ensure uniform mixing. b. 15% NaOH by mass in water

Let's assume we need to prepare 100 mL of the 15% NaOH solution. The mass of 100 mL of water is 100 g. A 15% NaOH by mass solution means that there are 15 g of NaOH in 100 g of solution. Therefore, the mass of NaOH required is 15 g.

To prepare the 15% NaOH solution: 1. Weigh 15 g of NaOH using a balance. 2. Add the NaOH to a beaker containing approximately 80 mL of water. 3. Stir until the NaOH is completely dissolved. 4. Transfer the solution to a 100 mL volumetric flask using a funnel, rinse the beaker with water, and pour the rinsing into the flask. 5. Add water to the volumetric flask until the 100 mL mark is reached. 6. Stir or invert the flask to ensure uniform mixing. c. 25% NaOH by mass in CH₃OH

For a 25% NaOH by mass solution, there are 25 g of NaOH in every 100 g of solution. Firstly, we need to determine the mass of the CH₃OH required to obtain 100 mL of the final solution. The density of CH₃OH is 0.79 g/cm³: Mass of CH₃OH = Density × Volume = 0.79 g/cm³ × 100 mL = 79 g Now, we must calculate the total mass of the NaOH and CH₃OH in the solution: Total mass = mass of NaOH + mass of CH₃OH = 25 g + 79 g = 104 g

To prepare the 25% NaOH in CH₃OH solution: 1. Weigh 25 g of NaOH using a balance. 2. Add the NaOH to a beaker containing approximately 60 mL of CH₃OH. 3. Stir until the NaOH is completely dissolved. 4. Add more CH₃OH to the beaker until the total mass reaches 104 g using a balance. 5. Stir or invert the beaker to ensure uniform mixing. d. 0.10 mole fraction of C₆H₁₂O₆ in water

Let's assume we are preparing 100 mL of a solution of C₆H₁₂O₆ in water with a 0.10 mole fraction. Firstly, we will find the moles of C₆H₁₂O₆ and water in the solution: When X(C₆H₁₂O₆) = 0.10, X(H₂O) = 1 - X(C₆H₁₂O₆) = 0.90 Using mole fraction equation: n(C₆H₁₂O₆) = X(C₆H₁₂O₆) × n(total) = 0.10 × n(total); n(H₂O) = X(H₂O) × n(total) = 0.90 × n(total) Total mass of 100 mL of water is 100 g. Since we are given the solution's density as 1.00 g/cm³, the volume of the solution is also 100 mL. To calculate the moles of H₂O in 100 g, using its molar mass (18 g/mol): n(H₂O) = mass(H₂O) / Molar mass(H₂O) = 100 g / 18 g/mol ≈ 5.56 moles Now, we can find n(total): n(total) = n(H₂O) / X(H₂O) = 5.56 moles / 0.90 ≈ 6.18 moles And, we calculate the moles of C₆H₁₂O₆: n(C₆H₁₂O₆) = X(C₆H₁₂O₆) × n(total) = 0.10 × 6.18 moles ≈ 0.618 moles Now we can find the mass of C₆H₁₂O₆, using its molar mass (180 g/mol): Mass(C₆H₁₂O₆) = n(C₆H₁₂O₆) × Molar mass(C₆H₁₂O₆) = 0.618 moles × 180 g/mol ≈ 111.24 g

To prepare the 0.10 mole fraction C₆H₁₂O₆ solution: 1. Weigh 111.24 g of C₆H₁₂O₆ using a balance. 2. Add the C₆H₁₂O₆ to a beaker containing approximately 80 mL of water. 3. Stir until the C₆H₁₂O₆ is completely dissolved. 4. Transfer the solution to a 100 mL volumetric flask using a funnel, rinse the beaker with water, and pour the rinsing into the flask. 5. Add water to the volumetric flask until the 100 mL mark is reached. 6. Stir or invert the flask to ensure uniform mixing.