Calculating the Amount of Alcohol Evaporated

Oct 13, 2025 by vinooalluri.com 86 views

Introduction

In this article, we will delve into a fascinating mathematical problem involving the evaporation of a 60% alcohol solution and its subsequent dilution with distilled water. The problem presents a classic example of a mixture problem, where we need to determine the amount of water added to achieve a specific concentration of alcohol in the resulting solution.

The Problem

We start with 40 liters of a 60% alcohol solution. When heated, some of the alcohol evaporates, but none of the water does. As a result, the concentration of alcohol in the remaining solution increases. Let's assume that x liters of alcohol evaporate. The amount of water in the original solution remains unchanged at 16 liters (40% of 40 liters). After evaporation, the remaining solution has a concentration of 80% alcohol (60% + 20% due to evaporation).

Calculating the Amount of Alcohol Evaporated

Let's calculate the amount of alcohol evaporated, denoted by x. We know that the initial amount of alcohol is 24 liters (60% of 40 liters). After evaporation, the amount of alcohol remaining is 24 - x liters. Since the concentration of alcohol in the remaining solution is 80%, we can set up the following equation:

(24 - x) / (40 - x) = 0.8

Solving for x

To solve for x, we can start by multiplying both sides of the equation by (40 - x) to eliminate the fraction:

24 - x = 0.8(40 - x)

Expanding the right-hand side of the equation, we get:

24 - x = 32 - 0.8x

Now, let's isolate x by moving all terms involving x to one side of the equation:

x + 0.8x = 32 - 24

Combine like terms:

0.2x = 8

Finally, divide both sides by 0.2 to solve for x:

x = 40

The Amount of Water Added

Now that we know the amount of alcohol evaporated, we can determine the amount of water added to achieve a 45% alcohol solution. Let's denote the amount of water added as y. The total volume of the resulting solution is still 40 liters. The amount of alcohol in the resulting solution is 18 liters (45% of 40 liters). Since 40 liters of the original solution had 24 liters of alcohol, the amount of alcohol lost due to evaporation is 24 - 18 = 6 liters.

Setting Up the Equation

We know that the amount of water added is y liters. The total volume of the resulting solution is 40 liters, and the amount of alcohol in the resulting solution is 18 liters. We can set up the following equation:

(24 - 6) + 18 = 40 - y

Simplifying the equation, we get:

18 + 18 = 40 - y

Combine like terms:

36 = 40 - y

Solving for y

Now, let's solve for y by isolating it on one side of the equation:

y = 40 - 36

y = 4

Conclusion

In this article, we explored a mathematical problem involving the evaporation of a 60% alcohol solution and its subsequent dilution with distilled water. We calculated the amount of alcohol evaporated and determined the amount of water added to achieve a 45% alcohol solution. The problem presented a classic example of a mixture problem, where we needed to determine the amount of water added to achieve a specific concentration of alcohol in the resulting solution.

Final Answer

The amount of water added to achieve a 45% alcohol solution is 4 liters.

Discussion

This problem is a great example of how mathematical concepts can be applied to real-world scenarios. The problem requires a deep understanding of mixture problems and the ability to set up and solve equations to determine the unknown quantities. The problem also highlights the importance of careful calculation and attention to detail in mathematical problem-solving.

Solutions to Related Problems

A 20% salt solution is mixed with a 50% salt solution to create a 30% salt solution. If the resulting solution has a volume of 120 liters, how many liters of the 20% salt solution were used? Let's denote the amount of the 20% salt solution used as x liters. The amount of the 50% salt solution used is 120 - x liters. The total amount of salt in the resulting solution is 0.3(120) = 36 liters. The amount of salt from the 20% salt solution is 0.2x liters, and the amount of salt from the 50% salt solution is 0.5(120 - x) liters. We can set up the following equation:

0.2x + 0.5(120 - x) = 36

Simplifying the equation, we get:

0.2x + 60 - 0.5x = 36

Combine like terms:

-0.3x = -24

Divide both sides by -0.3:

x = 80

Therefore, 80 liters of the 20% salt solution were used.

A 60% sugar solution is mixed with a 40% sugar solution to create a 50% sugar solution. If the resulting solution has a volume of 80 liters, how many liters of the 60% sugar solution were used? Let's denote the amount of the 60% sugar solution used as x liters. The amount of the 40% sugar solution used is 80 - x liters. The total amount of sugar in the resulting solution is 0.5(80) = 40 liters. The amount of sugar from the 60% sugar solution is 0.6x liters, and the amount of sugar from the 40% sugar solution is 0.4(80 - x) liters. We can set up the following equation:

0.6x + 0.4(80 - x) = 40

Simplifying the equation, we get:

0.6x + 32 - 0.4x = 40

Combine like terms:

0.2x = 8

Divide both sides by 0.2:

x = 40

Therefore, 40 liters of the 60% sugar solution were used.

Q: What is the initial concentration of the alcohol solution?

A: The initial concentration of the alcohol solution is 60%.

Q: How much alcohol evaporates from the solution?

A: 6 liters of alcohol evaporate from the solution.

Q: What is the concentration of the remaining solution after evaporation?

A: The concentration of the remaining solution after evaporation is 80%.

Q: How much water is added to the solution to achieve a 45% alcohol concentration?

A: 4 liters of water are added to the solution to achieve a 45% alcohol concentration.

Q: What is the total volume of the resulting solution?

A: The total volume of the resulting solution is 40 liters.

Q: How much alcohol is present in the resulting solution?

A: 18 liters of alcohol are present in the resulting solution.

Q: What is the percentage of alcohol in the resulting solution?

A: The percentage of alcohol in the resulting solution is 45%.

Q: Can you explain the concept of dilution in this problem?

A: Yes, dilution is the process of adding a solvent (in this case, water) to a solution to decrease its concentration. In this problem, we add 4 liters of water to the solution to decrease its concentration from 80% to 45%.

Q: How do you calculate the amount of water added to achieve a specific concentration?

A: To calculate the amount of water added, you need to set up an equation based on the initial and final concentrations of the solution. In this problem, we set up an equation using the amount of alcohol evaporated and the total volume of the resulting solution.

Q: What are some real-world applications of this problem?

A: This problem has real-world applications in fields such as chemistry, biology, and engineering, where solutions with specific concentrations are often required. For example, in pharmaceutical manufacturing, solutions with precise concentrations of active ingredients are crucial for the production of safe and effective medications.

Q: Can you provide more examples of mixture problems like this one?

A: Yes, here are a few more examples:

A 20% salt solution is mixed with a 50% salt solution to create a 30% salt solution. If the resulting solution has a volume of 120 liters, how many liters of the 20% salt solution were used?
A 60% sugar solution is mixed with a 40% sugar solution to create a 50% sugar solution. If the resulting solution has a volume of 80 liters, how many liters of the 60% sugar solution were used?

Q: How do you solve mixture problems like this one?

A: To solve mixture problems, you need to set up an equation based on the initial and final concentrations of the solution. You can use the concept of dilution to determine the amount of solvent (water) added to achieve a specific concentration.

Q: What are some common mistakes to avoid when solving mixture problems?

A: Some common mistakes to avoid when solving mixture problems include:

Failing to set up the correct equation
Not considering the initial and final concentrations of the solution
Not taking into account the amount of solvent (water) added to achieve a specific concentration
Not checking the units of the solution (e.g., liters, grams, etc.)

Q: Can you provide more resources for learning about mixture problems?

A: Yes, here are some additional resources:

Online tutorials and videos
Textbooks and study guides
Practice problems and worksheets
Online forums and discussion groups

Q: How can I practice solving mixture problems?

A: You can practice solving mixture problems by:

Working through practice problems and worksheets
Using online resources and tutorials
Joining online forums and discussion groups
Asking a teacher or tutor for help
Solving real-world problems and applications of mixture problems.