How many moles of each substance are needed to prepare the following solutions?
a. 50.0 mL of 8.0% (m/v) KCl (MW = 74.55 g/mol)
b. 200.0 mL of 7.5% (m/v) acetic acid (MW = 60.05 g/mol)

Verified step by step guidance

1. Step 1: Understand the problem. The (m/v) percentage indicates the mass of solute (in grams) per 100 mL of solution. To find the moles of solute, we need to calculate the mass of solute first using the given volume and percentage, then convert the mass to moles using the molar mass (MW).
2. Step 2: For part (a), calculate the mass of KCl required. Use the formula: \( \text{Mass of solute (g)} = \text{(m/v percentage)} \times \frac{\text{Volume of solution (mL)}}{100} \). Substitute \( 8.0\% \) and \( 50.0 \ \text{mL} \) into the formula.
3. Step 3: Convert the mass of KCl obtained in Step 2 to moles using the molar mass of KCl. Use the formula: \( \text{Moles of KCl} = \frac{\text{Mass of KCl (g)}}{\text{Molar mass of KCl (g/mol)}} \). Substitute the calculated mass and \( 74.55 \ \text{g/mol} \).
4. Step 4: For part (b), calculate the mass of acetic acid required. Use the same formula as in Step 2: \( \text{Mass of solute (g)} = \text{(m/v percentage)} \times \frac{\text{Volume of solution (mL)}}{100} \). Substitute \( 7.5\% \) and \( 200.0 \ \text{mL} \) into the formula.
5. Step 5: Convert the mass of acetic acid obtained in Step 4 to moles using the molar mass of acetic acid. Use the formula: \( \text{Moles of acetic acid} = \frac{\text{Mass of acetic acid (g)}}{\text{Molar mass of acetic acid (g/mol)}} \). Substitute the calculated mass and \( 60.05 \ \text{g/mol} \).

Key Concepts

Molarity and Moles

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. To find the number of moles, one can use the formula: moles = mass (g) / molar mass (g/mol). Understanding how to convert between mass and moles is essential for preparing solutions accurately.

Percent by Mass/Volume (m/v)

Percent by mass/volume (m/v) is a way to express the concentration of a solution, indicating the grams of solute per 100 mL of solution. For example, an 8.0% (m/v) KCl solution contains 8.0 grams of KCl in 100 mL of solution. This concept is crucial for calculating the required mass of solute for a given volume of solution.

Dilution and Solution Preparation

Dilution involves reducing the concentration of a solute in a solution, typically by adding more solvent. When preparing solutions, it is important to calculate the correct amount of solute needed based on the desired concentration and volume. This ensures that the final solution meets the specified concentration requirements.