Question

A sugar solution is prepared in such a way that it contains \(10 . \%\) dextrose by mass. What quantity of this solution do we need to obtain \(25 \mathrm{g}\) of dextrose?

Short answer

To obtain \(25\mathrm{g}\) of dextrose from a \(10\%\) dextrose sugar solution, we need \(250\mathrm{g}\) of the solution.

Step-by-step solution

Set up the proportion equation

We are given that the sugar solution contains \(10\%\) dextrose by mass. This means that for every \(100 \mathrm{g}\) of the solution, there are \(10 \mathrm{g}\) of dextrose. Therefore, the proportion between the dextrose mass (D) and the sugar solution mass (M) is: \(\frac{D}{M} = \frac{10}{100}\) Since we need to find the quantity of sugar solution that contains \(25 \mathrm{g}\) of dextrose, we will set up this proportion: \(\frac{25 \mathrm{g}}{x\mathrm{g}} = \frac{10\mathrm{g}}{100 \mathrm{g}}\)

Solve the equation for the unknown quantity of sugar solution

Now, we will solve the proportion equation to find the mass of sugar solution (x) needed to obtain \(25\mathrm{g}\) of dextrose: \(\frac{25}{x} = \frac{10}{100}\) To solve for x, we will first cross-multiply: \(25 \cdot 100 = 10 \cdot x\) This gives us: \(2500 = 10x\) Now, we will divide both sides by 10: \(x = \frac{2500}{10}\) So, the mass of the sugar solution needed to obtain \(25\mathrm{g}\) of dextrose is: \(x = 250 \mathrm{g}\) Therefore, we need \(250\mathrm{g}\) of the \(10\%\) dextrose sugar solution to obtain \(25\mathrm{g}\) of dextrose.

Key concepts

Understanding Dextrose

Dextrose, often recognized as a simple sugar, is chemically the same as glucose. It is widely used in the food industry and medical treatments due to its rapid energy release. In solutions, dextrose behaves as an essential component that can be easily measured by mass. When we talk about dextrose concentration, especially in this exercise, it is about identifying how much of the entire solution is made up of dextrose.

Knowing the percentage of dextrose, like the 10% in the exercise, means understanding that there are 10 grams of dextrose in every 100 grams of total solution. Such a measurement helps in various applications, including food preparation and pharmaceuticals, ensuring precise nutritional and caloric intake.

• Dextrose = Glucose
• Used for rapid energy
• Measured by mass in solutions

Understanding the concept of dextrose and its role in solutions is critical in solving problems related to its concentration and requirements in different fields.

The Concept of Mass Percentage

Mass percentage is a straightforward method used to describe the concentration of a component in a mixture. It expresses the mass of a particular substance as a percentage of the total mass of the mixture. In our case, the solution features a 10% mass percentage of dextrose.

Calculating mass percentage involves dividing the mass of the component by the total mass of the mixture and then multiplying the result by 100 to convert it to a percentage. This helps in quantifying precise amounts needed in formulations and mixtures.

• Mass Percentage = \(\frac{\text{mass of component}}{\text{total mass}} \times 100\)
• Helpful in precise computations
• Useful in food and chemical solutions

Mass percentage calculation is particularly useful in fields that require meticulous measurements, like pharmacology and chemistry.

Solving with a Proportion Equation

A proportion equation is a mathematical tool that allows us to compare ratios or relationships between two quantities. In this exercise, we used it to determine how much of the sugar solution is needed to reach a specific quantity of dextrose. This is done by setting up two pairs of equivalent ratios and solving for the unknown.

The proportion equation here is set as \( \frac{25}{x} = \frac{10}{100} \). This setup allows us to solve for \( x \), which represents the total mass of the sugar solution required.

• Useful for solving equivalent ratios
• Helps find unknown quantities
• Applied in real-world problems like recipe adjustments

By simplifying these relationships through cross-multiplication, we unlock the ability to derive concrete figures from abstract proportions, as demonstrated with the solution requiring 250g of the mixture to obtain the necessary dextrose.