Problem 15 - Chemistry: An Atoms-Focused Approach

Question

How many grams of solute are needed to prepare each of the following solutions? a. \(1.000 \mathrm{L}\) of \(0.200 M \mathrm{NaCl}\) b. \(250.0 \mathrm{mL}\) of \(0.125 M \mathrm{CuSO}_{4}\) c. \(500.0 \mathrm{mL}\) of \(0.400 M \mathrm{CH}_{3} \mathrm{OH}\)

Step-by-Step Solution

Answer

a. 1.000 L of 0.200 M NaCl solution b. 250.0 mL of 0.125 M CuSO4 solution c. 500.0 mL of 0.400 M CH3OH solution Answer: a. 11.69 g of NaCl b. 4.99 g of CuSO4 c. 6.41 g of CH3OH

Step 1: Calculate moles of NaCl

To calculate the moles of sodium chloride (NaCl) needed, multiply the volume of the solution in liters by its molarity: moles of NaCl = volume × molarity = 1.000 L × 0.200 M = 0.200 mol

Step 2: Calculate mass of NaCl

Now, we need to calculate the mass of 0.200 moles of NaCl. To do this, we multiply the moles of NaCl by its molar mass (58.44 g/mol). mass of NaCl = moles × molar mass = 0.200 mol × 58.44 g/mol = 11.69 g Thus, 11.69 grams of NaCl is needed. #b)

Step 3: Calculate moles of CuSO4

Convert the volume of the solution to liters and then multiply it by the molarity to find the moles of copper(II) sulfate (CuSO4) required: volume = 250.0 mL ÷ 1000 = 0.250 L moles of CuSO4 = volume × molarity = 0.250 L × 0.125 M = 0.03125 mol

Step 4: Calculate mass of CuSO4

To calculate the mass of 0.03125 moles of CuSO4, multiply it by its molar mass (159.62 g/mol): mass of CuSO4 = moles × molar mass = 0.03125 mol × 159.62 g/mol = 4.99 g Thus, 4.99 grams of CuSO4 is needed. #c)

Step 5: Calculate moles of CH3OH

Convert the volume of the solution to liters and then multiply it by the molarity to find the moles of methanol (CH3OH) required: volume = 500.0 mL ÷ 1000 = 0.500 L moles of CH3OH = volume × molarity = 0.500 L × 0.400 M = 0.200 mol

Step 6: Calculate mass of CH3OH

To calculate the mass of 0.200 moles of CH3OH, multiply it by its molar mass (32.04 g/mol): mass of CH3OH = moles × molar mass = 0.200 mol × 32.04 g/mol = 6.41 g Thus, 6.41 grams of CH3OH is needed. So, the required grams of solute for each solution are: a. 11.69 g of NaCl b. 4.99 g of CuSO4 c. 6.41 g of CH3OH