Solving a Complex Mixture Problem: A Fun and Challenging Puzzle

Our goal today is to explore a fascinating problem involving the mixture of solutions. Let's dive into the details of a mathematical puzzle that requires a keen understanding of algebraic equations and a solid grasp of chemical solutions. The problem at hand is as follows:

Problem Statement

We are given a 50 salt solution that has a total volume of 40 liters. This solution is then reduced to a 40 salt solution. The question asks us to determine how much of the original salt solution must be drained off and replaced with distilled water so that the final solution contains only a 40 salt concentration.

Breaking Down the Problem

First, let's break down what we know and what we need to find:

We start with 40 liters of a 50 salt solution. After some process, we end up with a 40 salt solution. We need to determine how much of the original solution must be removed and replaced with water to achieve the final solution.

Mathematical Solution

Let's denote the volume of the original solution that needs to be replaced with water as ( x ) liters. The total volume of the solution remains constant at 40 liters.

Initially, the amount of salt in the solution is:

( 40 , text{liters} times 0.5 20 , text{liters of salt} )

After we remove ( x ) liters of the solution, we are left with:

( 40 - x , text{liters of the original solution} )

The amount of salt left in the solution is:

( (40 - x) times 0.5 )

We then add ( x ) liters of distilled water, so the final volume of the solution is 40 liters and the new amount of salt is:

( (40 - x) times 0.5 )

We want the final concentration of salt to be 40, so the equation becomes:

( frac{(40 - x) times 0.5}{40} 0.4 )

Let's solve this equation step by step:

Multiplying both sides by 40: ( (40 - x) times 0.5 16 ) Multiplying both sides by 2: ( 40 - x 32 ) Solving for ( x ): ( x 8 )

Therefore, 8 liters of the original 50 salt solution must be drained off and replaced with distilled water to achieve a 40 salt concentration.

Why Not Solve It Right Away?

While the mathematically inclined might be tempted to solve this problem, there are valid reasons for not providing an outright answer:

Many homework questions, especially those that appear on the internet, are often posted by anonymous sources, potentially even by cheating students. The student seeking help should be encouraged to develop their own problem-solving skills, which is a crucial part of learning. Problems like these are designed to challenge the student's understanding and teach them important concepts.

Conclusion

Understanding how to solve problems involving mixture and concentration is a fundamental skill in both chemistry and mathematics. By breaking down the problem and solving it step by step, we can gain valuable insights into the nature of chemical solutions and the principles that govern them.

Through practice and critical thinking, students can master these types of problems and develop a deeper appreciation for the subject matter.