7 Solutions

By Jennifer Elmo

Provided by Kenan Fellows Program.

Systems of equations can be used to solve many different kinds of problems. One application is in making solutions such as is done in chemistry.

Learning outcomes

Students will be able to:

• Set up a system of equations to determine the volume of solution
• Connect this activity to the way scientists make solutions
• Set up and solve a system of equations from the information given in the problems which describe the solutions provided and the one needed to be made

Teacher planning

Time required

45 minutes

Materials needed

No special materials are required for this lesson.

Student handouts

Pre-activities

Before beginning this lesson, students should be familiar with:

• Setting up and solving a system of equations
• How a solution is made

Activities

1. Pass out the Solutions worksheet.
2. Lead a class discussion on solutions. Make sure to explain that a solution is made up of two parts: a solute and a solvent. The solute is the substance that is dissolved into the solvent, usually water. The solute is generally the smallest portion of a solution. Solutions are quantitatively described by concentration. Concentration is the amount of solute per give amount of solvent or solution. Concentration is generally described as a percentage or in molarity. Molarity is a chemistry term that describes the concentration in terms of number of moles of solute per liter of solution. Students will focus on molarity more in chemistry. For this lesson we will focus on concentration in terms of percentages.
3. Continue on to explain to students that if they have a 10 percent alcohol solution, then 10 percent is alcohol and 90 percent is water. Likewise, if they have a 10 M alcohol solution, then 10 moles of alcohol is in 1 liter of solution. Volumes are not always additive, but for our purposes you may assume they are. This would mean a 10 percent alcohol solution will have 10 milliliters of alcohol and 90 milliliters of water.
4. Tell students that making solutions can be a little more complicated if you want to use solutions you have already made instead of starting from scratch. Demonstrate the following example on the board or overhead.

   You have a 10% and a 5% solution of alcohol and you want to make 100 mL of a 7% solution. What volume of each solution do you need to make the desired concentration?

   In order to solve this problem, you will need a system of equations. One equation will represent the volumes of each solution and the other will represent the combining of solutions.

   Since you know you need a total of 100 ml, then you know the volume of the 10% (X) and the volume of the 5% (Y) solution must add up to 100, so you have the following equation:

   X + Y = 100

   Since you know combining some amount of the two solutions will make 100 mL of a 7% solution you can set up the following equation:

   10X + 5Y = 7(100)

   Solve one equation for X and substitute it for X in the other equation.

   X = 100-Y
   10(100-Y) + 5Y = 700
   1000 – 10Y + 5Y = 700
   -5Y = -300
   Y = 60
   X + Y =100
   X + 60 = 100
   X = 40

   Solution: You need to mix 40 mL of the 10% solution with 60 mL of the 5% solution in order to get 100 mL of a 7% solution.

5. Solve the following problems as a class:
   • Alex has added 3 liters of alcohol to 7 liters of a 22% sugar in alcohol solution. What is the sugar concentration of the resulting solution?
   • Stacy mixed 25 grams of salt with 400 grams of a 13% salt solution in water. What is the percentage of salt in his new solution?
   • What is the concentration of salt in the solution obtained by mixing 75 ml of water with 280 ml of a 33% salt solution in water?
   • Kasey has added 2.5 quarts of an alcohol in water solution to 3 quarts of water and her ending product is a 55% alcohol in water solution. What was the concentration of the 4 quarts of alcohol solution that was added?
   • Megan wishes to increase the percent of salt in 125 ml of a 35% acid solution in water to 65% acid, how much salt must she add?
   • Victor has 34 kg of pinto beans which cost $1.50 per kilogram. How many kilograms of kidney beans at $1.09 per kilogram should be added for the complete mixture to have a value of $95.00?
   • Mark’s teacher has asked him to add acid to 18 gallons of a 12% acid in water solution to get a solution which is 75% acid. How much pure acid should he add?
   • How many ounces of water should be evaporated from 75 ounces of a 5% salt solution in order to make a 25% salt solution?
6. Have students complete the following problems on their own.
   • 5 liters of alcohol are added to 12 liters of a 58% sugar in alcohol solution. What is the sugar concentration of the resulting solution?
   • 45 grams of salt with 330 grams of a 15% salt solution in water. What is the percent of salt in this new solution?
   • What is the concentration of sugar in the solution obtained by mixing 225 ml of water with 425 ml of a 40% salt solution in water?
   • Victor has 60 kg of pinto beans which cost $0.69 per kilogram. How many kilograms of kidney beans at $0.89 per kilogram should be added for the complete mixture to have a value
     of $101.00?
   • How many ounces of water should be evaporated from 240 ounces of a 34% salt solution in order to make a 65% salt solution?

Assessment

Evaluate student responses on the worksheet.

Critical vocabulary

solution

a homogeneous mixture of two or more substances, which may be solids, liquids, gases, or a combination of these

solute

a substance dissolved in another substance, usually the component of a solution present in the lesser amount

solvent

a substance in which another substance is dissolved, forming a solution

dilute

to make thinner or less concentrated by adding a liquid such as water

concentration

the amount of a specified substance in a unit amount of another substance

molarity

the molar concentration of a solution, usually expressed as the number of moles of solute per liter of solution

percent by mass (volume)

the mass of the solute divided by the mass of the solution (mass of the solute plus mass of the solvent), multiplied by 100

Comments

This activity can be taught by the Algebra II teacher during the unit on systems of equations even if students have not yet studied solutions in chemistry.

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