Problem 59 - Algebra for College Students

Question

Solve each problem. In a chemistry class, 12 L of a \(12 \%\) alcohol solution must be mixed with a \(20 \%\) solution to obtain a \(14 \%\) solution. How many liters of the \(20 \%\) solution are needed?

Step-by-Step Solution

Answer

4 liters

Step 1 - Define Variables

Let \( x \) be the number of liters of the 20% solution needed. We need to express the total volume and concentration of the resulting mixture in terms of \( x \).

Step 2 - Write the Equation for Alcohol Content

The amount of pure alcohol in the 12 L of 12% solution is \( 0.12 \times 12 = 1.44 \) liters. The amount of pure alcohol in \( x \) liters of 20% solution is \( 0.20 \times x \) liters. The amount of pure alcohol in the final mixture of \( 12 + x \) liters of 14% solution is \( 0.14 \times (12 + x) \) liters.

Step 3 - Set up the Equation

Set the sum of the alcohol from both solutions equal to the alcohol in the final mixture: \[ 1.44 + 0.20x = 0.14(12 + x) \]

Step 4 - Solve the Equation

Distribute and simplify the equation: \[ 1.44 + 0.20x = 1.68 + 0.14x \] Subtract \( 0.14x \) from both sides: \[ 1.44 + 0.06x = 1.68 \] Subtract 1.44 from both sides: \[ 0.06x = 0.24 \] Divide both sides by 0.06: \[ x = \frac{0.24}{0.06} = 4 \]

Step 5 - Interpret the Result

The solution indicates that 4 liters of the 20% alcohol solution are needed to achieve the desired 14% alcohol mixture.

Key Concepts

Algebraic Equations

In the context of mixture problems, algebraic equations are used to create mathematical models that represent real-world scenarios. Let's break this down. Algebra allows us to represent unknown quantities with variables, such as using 'x' to signify the amount of a solution needed. By setting up an equation that represents the problem, we can solve for the unknown variable.

In our given problem, we defined 'x' as the liters of 20% solution needed. We then wrote an equation based on the amount of pure alcohol in all solutions involved. This equation, represents the key relationships between given amounts. Understanding this foundation is critical for solving algebraic problems in chemistry.

Always start any algebraic problem by identifying your unknowns, establishing relationships between knowns and unknowns, and setting up an equation that models the scenario.

Solution Concentration

Solution concentration is a way of expressing how much solute is present in a given amount of solution. In percentage terms, it tells us what portion of the solution consists of the solute. For instance, a 12% alcohol solution means that for every 100 liters of solution, there are 12 liters of alcohol.

In this exercise, we work with three different concentrations: 12%, 20%, and 14%. We calculate the amount of pure alcohol by multiplying the concentration percentage by the volume of solution. Example:

• 12 L of 12% solution: 0.12 × 12 = 1.44 L of alcohol
• x L of 20% solution: 0.20 × x = 0.20x L of alcohol
• (12 + x) L of 14% solution: 0.14 × (12 + x) = 1.68 + 0.14x L of alcohol

By understanding solution concentrations, you can solve to find out how much of one solution is needed to achieve a desired concentration when mixed with another.

Chemistry Applications in Algebra

Mixing solutions with different concentrations lets chemistry and algebra intersect in interesting ways. Problems like these are practical applications of algebra in real life.

In classrooms and laboratories, different solutions are often mixed to achieve a desired concentration. The algebra behind this process includes setting up equations to balance the amounts of solute. For example, when mixing two solutions, you will consider:

• The volume of each solution
• The concentration of solute in each solution
• The desired concentration for the final mixture

In our specific problem, we began by defining the variables and setting up an equation that relates the concentrations. We solved this equation to find that 4 liters of a 20% alcohol solution must be added to 12 liters of a 12% alcohol solution to achieve a 14% alcohol solution.

This emphasizes how algebra provides the tools necessary to solve practical problems in chemistry and other fields.