Question

What amounts of solute or solvent are needed to prepare the following solutions? (a) Mass of glucose needed to prepare \(125.0 \mathrm{~mL}\) of \(16 \%(\mathrm{~m} / \mathrm{v})\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\). (b) Volume of water needed to prepare a \(2.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{KCl}\) solution using \(1.20 \mathrm{~g} \mathrm{KCl}\).

Short answer

(a) 20 g of glucose is needed. (b) About 58.8 mL of water is needed.

Step-by-step solution

Understanding the problem for part (a)

We need to find the mass of glucose required to make a 16% (m/v) solution with a volume of 125.0 mL. This percentage means 16 grams of glucose per 100 mL of solution.

Calculating mass of glucose for part (a)

Using the percentage concentration, we set up the equation: \( \frac{16 ext{ g}}{100 ext{ mL}} = \frac{x ext{ g}}{125.0 ext{ mL}} \), where \( x \) is the mass of glucose needed. We solve for \( x \) to find that \( x = \frac{16 imes 125}{100} = 20 \text{ g}\).

Understanding the problem for part (b)

We need the volume of water required to make a 2.0% (m/v) KCl solution using 1.20 g of KCl. The percentage indicates 2 grams of KCl per 100 mL of solution.

Calculating volume of water for part (b)

Using the percentage concentration, we set the equation: \( \frac{2 ext{ g}}{100 ext{ mL}} = \frac{1.20 ext{ g}}{y ext{ mL}} \), where \( y \) is the total volume of the solution. Solving for \( y \), we find \( y = \frac{100 imes 1.20}{2} = 60 ext{ mL} \). Therefore, the volume of water needed (after subtracting the volume occupied by KCl) is approximately 58.8 mL.

Key concepts

Solute Mass Calculation

Calculating the mass of a solute required for a solution is a common task in chemistry. To determine this, you need to know the concentration and the volume of the solution you wish to prepare. For instance, if you are asked to prepare a certain percentage concentration solution, like a 16% (m/v) glucose solution in 125 mL of water:

• Identify the percentage concentration: this is 16% (m/v), meaning 16 grams of glucose per 100 mL of solution.
• Set up a proportion using the formula: \( \frac{16 \, \text{g}}{100 \, \text{mL}} = \frac{x \, \text{g}}{125 \, \text{mL}} \).
• Solve for \(x\) to find that \(x = \frac{16 \times 125}{100} = 20\) grams.

This calculation helps you find out how much solute is needed given the desired concentration and volume.

Volume Percent Concentration

Volume percent concentration is a way to express the concentration of a component in a solution. It's used primarily for liquid-in-liquid solutions but can also apply to solids like glucose in water. The formula for volume percent concentration is:

• Concentration = \( \frac{\text{Volume of solute}}{\text{Total volume of solution}} \times 100 \% \).

In our exercise, we applied mass/volume percent instead, which is similar:

• For glucose: 16% (m/v) means 16 g of glucose in every 100 mL of solution.
• Concentration tells us the required mass of solute relative to volume.

This concept helps understand how much of a solute you need to mix with a solvent to achieve a specific concentration.

Mass/Volume Percent Solution

Mass/volume percent solutions are easy to use and understand. The formula generally used for calculations involving mass/volume percent (m/v) includes:

• \( \text{Mass/Volume %} = \frac{\text{g of solute}}{\text{mL of solution}} \times 100 \% \).

It's especially useful in laboratory settings where liquids often dilute.

• For example: In part (a), a 16% (m/v) glucose solution requires 16 g of glucose per 100 mL of total solution.
• This helps you know exactly the mass of solute you need for any volume of solution.

This calculation is crucial for applications needing precise solute concentrations.

Density of Solutions

Density in solutions is a factor sometimes required for conversions between mass and volume, although not explicitly needed in the original problem. Density offers insight into how packed or concentrated components are in a solution.

• Density = \( \frac{\text{mass}}{\text{volume}} \).
• Understanding density helps if a concentration by mass is needed.

In the context of the exercise, density isn't directly involved, but understanding how density influences other properties can be valuable. This helps grasp the importance of concentration differences and how to calculate mass when density is a factor.