Problem 17 - Chemistry The Science in Context

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

Answer: The required masses of solute are 1) 11.69 g of NaCl, 2) 4.98 g of CuSO4, and 3) 6.41 g of CH3OH.

Step 1: Finding the mass of solute for 1.000 L of 0.200 M NaCl

Firstly, we need to find the moles of NaCl in the 1.000 L solution. Using the formula: Moles of NaCl = Molarity * Volume in Liters Moles of NaCl = 0.200 M * 1.000 L Moles of NaCl = 0.200 moles Now, we find the mass of NaCl using its molar mass (58.44 g/mol): Mass of NaCl = Moles of NaCl * Molar Mass Mass of NaCl = 0.200 moles * 58.44 g/mol Mass of NaCl = 11.69 g

Step 2: Finding the mass of solute for 250.0 mL of 0.125 M CuSO4

Firstly, we need to find the moles of CuSO4 in the 250.0 mL solution. Convert volume in mL to L: Volume in Liters = 250.0 mL * (1 L / 1000 mL) Volume in Liters = 0.250 L Now, find the moles of CuSO4 using the formula: Moles of CuSO4 = Molarity * Volume in Liters Moles of CuSO4 = 0.125 M * 0.250 L Moles of CuSO4 = 0.03125 moles Now, find the mass of CuSO4 using its molar mass (159.61 g/mol): Mass of CuSO4 = Moles of CuSO4 * Molar Mass Mass of CuSO4 = 0.03125 moles * 159.61 g/mol Mass of CuSO4 = 4.98 g

Step 3: Finding the mass of solute for 500.0 mL of 0.400 M CH3OH

Firstly, we need to find the moles of CH3OH in the 500.0 mL solution. Convert volume in mL to L: Volume in Liters = 500.0 mL * (1 L / 1000 mL) Volume in Liters = 0.500 L Now, find the moles of CH3OH using the formula: Moles of CH3OH = Molarity * Volume in Liters Moles of CH3OH = 0.400 M * 0.500 L Moles of CH3OH = 0.200 moles Now, find the mass of CH3OH using its molar mass (32.04 g/mol): Mass of CH3OH = Moles of CH3OH * Molar Mass Mass of CH3OH = 0.200 moles * 32.04 g/mol Mass of CH3OH = 6.41 g Thus, the required masses of solute are: a. 11.69 g of NaCl b. 4.98 g of CuSO4 c. 6.41 g of CH3OH

Key Concepts

Molarity

Molarity is a way to express the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The unit of molarity is noted as 'M', which stands for moles per liter. To calculate molarity, you can use the formula:

• Molarity (M) = \( \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}} \)

This concept is essential when preparing chemical solutions because it allows us to know the exact amount of solute needed to achieve the desired concentration. When you understand molarity, you can prepare solutions with precise chemical proportions, which is crucial for experiments and reactions.
To apply this concept practically, think about trying to prepare a saltwater solution for an aquarium. You would need to know the exact amount of salt to achieve the right conditions for aquatic life. Similarly, in a lab, preparing a 1 M solution means having 1 mole of the solute dissolved in 1 liter of total solution.

Mass Calculation

Mass calculation of a solute is vital for any solution preparation. Once you have determined the number of moles of a solute required, you can find its mass using the molar mass of the solute. The process involves multiplying the number of moles by the molar mass (which is the mass of one mole of that substance in grams). The formula is:

• Mass = Moles \( \times \) Molar Mass

For example, if you have determined that you need 0.200 moles of NaCl, and the molar mass of NaCl is 58.44 g/mol, the mass of NaCl required would be \(0.200 \times 58.44 = 11.69 \text{ g} \).
The concept of mass calculation helps in ensuring that the correct amount of solute is used, thereby preventing errors in solution concentration which can affect experimental results. Calculating mass accurately is also important in industries like pharmaceuticals, where the precision of the active ingredient's mass can determine the effectiveness of a product.

Moles

Understanding moles is crucial in chemistry as it allows us to count particles by weighing them. A mole is a unit that represents \(6.022 \times 10^{23}\) entities of a substance, which is Avogadro's number. In simple terms, it's a way to count large quantities of atoms, molecules, or other chemical units in a sample.
To determine the number of moles, you can use the formula:

• Moles = \( \frac{\text{Mass of Substance}}{\text{Molar Mass}} \)

For instance, if you have 100 grams of a substance with a molar mass of 50 g/mol, then you have \( \frac{100}{50} = 2 \text{ moles}\).
Moles provide a bridge between the atomic scale and the laboratory scale. When you calculate moles, you can convert between the masses of substances and the volumes of gases, as well as predict the amounts of substances consumed or produced in chemical reactions. This concept is foundational not only in carrying out calculations but also in understanding chemical formulas and equations.