Problem 5 - Chemistry for Allied Health

Question

What mass of glucose is in \(250.0 \mathrm{~mL}\) of solution that is \(5.00 \%\) by mass? Assume the density of the solution is \(1.00 \mathrm{~g} / \mathrm{mL}\).

Step-by-Step Solution

Answer

The mass of glucose is 12.5 g.

Step 1: Understand the Problem

We need to find the mass of glucose in a solution. We are given that the solution is 5% glucose by mass and the density of the solution is 1.00 g/mL. Additionally, the volume of the solution is 250.0 mL. We need to convert this information into a mass of glucose.

Step 2: Find the Mass of the Solution

The density of the solution is given as 1.00 g/mL, which means that 1 mL of the solution has a mass of 1 g. Therefore, the mass of 250.0 mL of the solution is calculated by multiplying the volume by the density:\[ \text{Mass of Solution} = \text{Volume} \times \text{Density} = 250.0 \, \text{mL} \times 1.00 \, \text{g/mL} = 250.0 \, \text{g} \]

Step 3: Calculate the Mass of Glucose

Since the solution is 5.00% glucose by mass, this means that 5% of the total mass of the solution is glucose. To find the mass of glucose, multiply the total mass of the solution by 5%:\[ \text{Mass of Glucose} = \frac{5}{100} \times 250.0 \, \text{g} = 12.5 \, \text{g} \]

Step 4: Review and Conclude

We have determined that the mass of glucose in the 250.0 mL solution is 12.5 g. This is based on the calculations using the given percentage and the density of the solution.

Key Concepts

Mass Percent

Mass percent is a way of expressing the concentration of a component in a mixture. Imagine you have a solution made up of various components. Mass percent specifically tells you how much of the total mass of the solution is made up of the component of interest. For example, if a solution is 5% glucose by mass, this means that out of every 100 grams of the solution, 5 grams are glucose.
To calculate mass percent, you can use the formula:

• Mass Percent = (Mass of Component / Total Mass of Solution) × 100%

This equation allows us to express the concentration of any component in the solution as a percentage of the total solution mass.

Density

Density is a physical property that is crucial in finding the relationship between the mass and volume of a substance. It tells you how compact a substance is. Density is defined as the mass per unit volume and is expressed in units such as grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³).
The formula for density is:

• Density = Mass / Volume

In this exercise, we're told that the solution has a density of 1.00 g/mL. This indicates that every 1 mL of solution weighs 1 gram. Utilizing this information, we can determine the mass of the solution if the volume is known, by rearranging the formula to find mass: Mass = Density × Volume.

Mass Calculation

Mass calculation involves determining the total mass of a substance or component using its volume and density. Once density and volume are known, you can easily calculate the mass of a solution by multiplying these two values. For example, if you have a 250.0 mL solution with a density of 1.00 g/mL, the mass of the solution can be calculated as follows:

• Mass of Solution = 250.0 mL × 1.00 g/mL = 250.0 g

Such calculations are essential when working with solutions, as they help in determining how much of each component, like glucose, is present within a given volume.

Glucose Concentration

Glucose concentration in this context refers to the amount of glucose present in a given volume or mass of solution. It reflects the proportion of glucose relative to the entire solution based on its mass percentage. In the given exercise, the solution is 5.00% glucose by mass. This means that 5% of the total mass of the solution is glucose.
To calculate the mass of glucose, we can use the known mass percent of 5% and the total mass of the solution (250.0 g, as calculated previously):

• Mass of Glucose = (5/100) × 250.0 g = 12.5 g

This calculation helps answer how much glucose is actually within the 250.0 mL of solution, ensuring that the concentration is understood in practical terms.