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 \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\)

To prepare a 2.0 molar (M) solution of KCl in water, we need to first determine the volume of water needed to dissolve KCl. Since we need at least 100 mL of the solution, we can use the molar concentration formula: \[C = \frac{n}{V}\] Here, \(C\) is the concentration (2.0 M), \(n\) is the moles of KCl, and \(V\) is the volume of the solution (100 mL in this case). Rearranging the formula to solve for \(n\): \[n = C \times V\] Next, plug in the values and solve for \(n\): \[n = 2.0 \, \text{M} \times 100 \, \text{mL} = 0.2 \, \text{moles KCl}\] Now, we need to convert moles of KCl to mass. Use the molar mass of KCl: \[m = n \times M_{\text{molar}}\] Here, \(m\) is the mass of KCl, and \(M_{\text{molar}}\) is the molar mass of KCl (74.55 g/mol). Plug in the values and solve for the mass: \[m = 0.2 \, \text{moles KCl} \times 74.55 \, \text{g/mol} = 14.91 \, \text{g KCl}\] Finally, weigh out 14.91g of KCl, dissolve it in water, and add water until the total volume of the solution reaches 100 mL.

To prepare 100 mL of a 15% NaOH by mass in water, we need to calculate the mass of the NaOH to be used. Since the density of this solution is 1.00 g/cm³, we can convert the volume to mass using the formula: \[m = V \times d\] Here, \(V\) is the volume of the solution (100 mL) and \(d\) is the density (1.00 g/cm³). Convert the volume to mass: \[m = 100 \, \text{mL} \times 1.00 \, \text{g/cm}^3 = 100 \, \text{g}\] To calculate the mass of NaOH required, multiply this mass by the percentage (15%): \[m_{NaOH} = 100 \, \text{g} \times 0.15 = 15 \, \text{g}\] Weigh out 15 g of NaOH, dissolve it in water, and add water until the total mass of the solution reaches 100 g.

To prepare 100 mL of a 25% NaOH by mass in CH3OH, we will follow a similar approach as part (b). First, convert the volume of the solution to mass using the formula \(m = V \times d\). In this case, the density is 0.79 g/cm³: \[m = 100 \, \text{mL} \times 0.79 \, \text{g/cm}^3 = 79 \, \text{g}\] Next, calculate the mass of NaOH required by multiplying this mass by the percentage (25%): \[m_{NaOH} = 79 \, \text{g} \times 0.25 = 19.75 \, \text{g}\] Weigh out 19.75 g of NaOH, dissolve it in CH3OH, and add CH3OH until the total mass of the solution reaches 79 g.

To prepare a solution containing 0.10 mole fraction of C6H12O6 (glucose) in water, we first need to calculate the moles of glucose and water: Let \(n_G\) and \(n_W\) be the moles of glucose and water, respectively. Based on the mole fraction formula, we have: \[\chi_G = \frac{n_G}{n_G + n_W}\] Rearranging the above equation, we get: \[n_W = \frac{n_G (1 - \chi_G)}{\chi_G}\] Given that the mole fraction of glucose (\(\chi_G\)) is 0.10, we get: \[n_W = \frac{n_G (1 - 0.10)}{0.10}\] Now, we can set an arbitrary number of moles for glucose (let's say, 0.1 mole) and use the above equation to calculate the moles of water: \[\begin{aligned} n_W & = \frac{0.1 \, \text{mole} \times (1 - 0.10)}{0.10} \\ & = 0.9 \, \text{mole} \end{aligned}\] Now, convert moles of glucose and water to mass using their molar masses: \[\begin{aligned} m_G & = n_G \times M_{G} \\ m_W & = n_W \times M_{W} \end{aligned}\] Here, \(M_G\) and \(M_W\) are the molar masses of glucose (180.16 g/mol) and water (18.02 g/mol), respectively. Calculate the masses: \[\begin{aligned} m_G & = 0.1 \, \text{mole} \times 180.16 \, \text{g/mol} = 18.02 \, \text{g} \\ m_W & = 0.9 \, \text{mole} \times 18.02 \, \text{g/mol} = 16.22 \, \text{g} \end{aligned}\] Finally, weigh out 18.02 g of glucose, dissolve it in 16.22 g of water, and add water until the total mass of the solution reaches the combined mass of glucose and water.