Mixture and Concentration Problems: Comprehensive Exercises & Solutions

Master mixture and concentration problems: solution dilution, alloy creation, liquid mixing, and chemical concentration calculations through these 5 detailed exercises.

Core Concepts & Formulas
\(\text{Concentration} = \frac{\text{Amount of Solute}}{\text{Total Volume}} \times 100\%\)
Concentration Formula
\(\text{Mixture Equation: } C_1V_1 + C_2V_2 = C_fV_f\)
Mixture Equation
\(\text{Dilution: } C_1V_1 = C_2V_2\)
Dilution Formula
Solute
Substance dissolved
Active ingredient
Solvent
Substance doing dissolving
Usually water
Solution
Solute + Solvent
Mixed result
Key Definitions:

Mixture Problem: A problem involving combining different substances with different concentrations to achieve a desired concentration

Concentration: The amount of solute in a solution, usually expressed as a percentage

Solute: The substance being dissolved (e.g., salt, alcohol)

Solvent: The substance doing the dissolving (usually water)

Dilution: Reducing concentration by adding more solvent

Tip 1: Always keep track of the amount of pure substance (solute) in each solution.
Tip 2: In mixture problems, the total amount of solute before mixing equals the total amount after mixing.
Tip 3: Draw a diagram to visualize the problem with volumes and concentrations.
Tip 4: Check that your answer makes physical sense (concentration should be between the original values).
Mixture & Concentration Exercises 1-3
1 Solution Mixing
Exercise 1
How many liters of a 20% salt solution must be mixed with 15 liters of a 40% salt solution to obtain a 30% salt solution?
Definition:

Solution Mixing: Combining two solutions with different concentrations to create a solution with a specific target concentration

Method:
  1. Define variables for unknown quantities
  2. Set up the mixture equation: C₁V₁ + C₂V₂ = C₃V₃
  3. Solve for the unknown variable
  4. Verify the solution
Unknown Volume
x liters
Equation Setup
0.20x + 0.40(15) = 0.30(x + 15)
Solution
x = 15 liters
Step 1: Define variables

Let x = liters of 20% solution needed

We have 15 liters of 40% solution

Final volume = x + 15 liters

Step 2: Set up the mixture equation

Amount of salt in 20% solution + Amount in 40% solution = Amount in 30% solution

0.20x + 0.40(15) = 0.30(x + 15)

Step 3: Solve the equation

0.20x + 6 = 0.30x + 4.5

6 - 4.5 = 0.30x - 0.20x

1.5 = 0.10x

x = 15 liters

Step 4: Verify the solution

Amount of salt in 20% solution: 0.20 × 15 = 3 liters

Amount of salt in 40% solution: 0.40 × 15 = 6 liters

Total salt: 3 + 6 = 9 liters

Total volume: 15 + 15 = 30 liters

Final concentration: 9/30 = 0.30 = 30% ✓

15 liters of 20% solution needed
Final answer:

15 liters of the 20% salt solution must be mixed with 15 liters of the 40% solution to obtain a 30% solution.

Applied rules:

• Mixture Conservation: Total solute before = Total solute after

• Concentration Formula: Amount of solute = Concentration × Volume

• Algebraic Solution: Solve linear equations systematically

2 Dilution Problem
Exercise 2
A chemist has 500 mL of a 15% alcohol solution. How much water should be added to dilute it to a 10% alcohol solution?
Definition:

Dilution: Process of reducing concentration by adding solvent (water) while keeping the amount of solute constant

Initial Solution
500 mL at 15%
Pure Alcohol
75 mL
Final Volume
750 mL
Step 1: Calculate amount of pure alcohol

Pure alcohol = 500 mL × 0.15 = 75 mL

This amount remains constant during dilution

Step 2: Set up the dilution equation

Final concentration = Pure alcohol / Final volume

0.10 = 75 / (500 + x)

Where x = mL of water to add

Step 3: Solve for x

0.10(500 + x) = 75

50 + 0.10x = 75

0.10x = 25

x = 250 mL

Step 4: Verify the solution

Final volume: 500 + 250 = 750 mL

Final concentration: 75 mL / 750 mL = 0.10 = 10% ✓

250 mL of water needed
Final answer:

The chemist needs to add 250 mL of water to dilute the solution to 10% alcohol.

Applied rules:

• Dilution Principle: Amount of solute remains constant

• Concentration Formula: Concentration = Solute / Total Volume

• Conservation Law: What goes in stays in (for pure substances)

3 Alloy Creation
Exercise 3
A jeweler needs to create 100 grams of 18-karat gold by mixing pure gold (24-karat) with silver. If 18-karat gold is 75% pure gold, how much pure gold and how much silver are needed?
Definition:

Alloy Creation: Creating a mixture of metals with specific proportions to achieve desired properties

Target Alloy
100g at 75% gold
Pure Gold Needed
75g
Silver Needed
25g
Step 1: Understand the karat system

24-karat = 100% pure gold

18-karat = 18/24 = 0.75 = 75% pure gold

Step 2: Calculate pure gold needed

For 100g of 18-karat gold: 100 × 0.75 = 75g pure gold

Step 3: Calculate silver needed

Total weight = Pure gold + Silver

100g = 75g + Silver

Silver = 25g

Step 4: Verify the composition

Gold percentage: 75g / 100g = 0.75 = 75%

Silver percentage: 25g / 100g = 0.25 = 25%

Total: 75% + 25% = 100% ✓

75g pure gold + 25g silver
Final answer:

The jeweler needs 75 grams of pure gold and 25 grams of silver to create 100 grams of 18-karat gold.

Applied rules:

• Karat System: Karat ÷ 24 = Percentage of pure gold

• Mass Conservation: Total mass = Sum of component masses

• Percentage Calculation: Part ÷ Whole × 100%

Mixture & Concentration Exercises 4-5
4 Multiple Solution Mixing
Exercise 4
A lab technician needs to prepare 200 mL of a 12% acid solution. She has available 8% acid solution and 20% acid solution. How much of each should she mix to get the desired concentration?
Definition:

Multiple Solution Mixing: Combining two solutions with different concentrations to achieve a target concentration

Variables
x = 8% solution, y = 20% solution
Volume Equation
x + y = 200
Concentration Equation
0.08x + 0.20y = 0.12(200)
Step 1: Define variables

Let x = mL of 8% solution

Let y = mL of 20% solution

We know x + y = 200 (total volume)

Step 2: Set up the concentration equation

Amount of acid from 8% solution + Amount from 20% solution = Amount in 12% solution

0.08x + 0.20y = 0.12(200) = 24

Step 3: Solve the system of equations

From first equation: y = 200 - x

Substitute into second equation: 0.08x + 0.20(200 - x) = 24

0.08x + 40 - 0.20x = 24

-0.12x = -16

x = 133.33 mL of 8% solution

y = 200 - 133.33 = 66.67 mL of 20% solution

Step 4: Verify the solution

Amount of acid from 8%: 0.08 × 133.33 = 10.67 mL

Amount of acid from 20%: 0.20 × 66.67 = 13.33 mL

Total acid: 10.67 + 13.33 = 24 mL

Concentration: 24/200 = 0.12 = 12% ✓

133.33 mL of 8% + 66.67 mL of 20%
Final answer:

The technician should mix 133.33 mL of the 8% solution with 66.67 mL of the 20% solution to obtain 200 mL of 12% acid solution.

Applied rules:

• System of Equations: Use volume and concentration equations

• Solute Conservation: Total acid before = Total acid after

• Substitution Method: Solve systems of linear equations

5 Gasoline Mixture
Exercise 5
A gas station has regular gasoline (87 octane) and premium gasoline (93 octane). How many gallons of each should be mixed to create 120 gallons of 91 octane gasoline?
Definition:

Octane Mixture: Combining fuels with different octane ratings to achieve a desired rating

Variables
x = regular, y = premium
Volume Equation
x + y = 120
Octane Equation
87x + 93y = 91(120)
Step 1: Define variables

Let x = gallons of 87 octane (regular)

Let y = gallons of 93 octane (premium)

We know x + y = 120 (total volume)

Step 2: Set up the octane equation

Total octane value from regular + Total from premium = Total in final mixture

87x + 93y = 91(120) = 10,920

Step 3: Solve the system of equations

From first equation: y = 120 - x

Substitute into second equation: 87x + 93(120 - x) = 10,920

87x + 11,160 - 93x = 10,920

-6x = -240

x = 40 gallons of regular (87 octane)

y = 120 - 40 = 80 gallons of premium (93 octane)

Step 4: Verify the solution

Octane value from regular: 87 × 40 = 3,480

Octane value from premium: 93 × 80 = 7,440

Total octane value: 3,480 + 7,440 = 10,920

Average octane: 10,920 ÷ 120 = 91 ✓

40 gallons regular + 80 gallons premium
Final answer:

The gas station should mix 40 gallons of regular (87 octane) gasoline with 80 gallons of premium (93 octane) gasoline to create 120 gallons of 91 octane gasoline.

Applied rules:

• Weighted Average: Octane rating is a weighted average of components

• System of Equations: Use volume and property equations

• Property Conservation: Total property value remains constant

Mixture and Concentration Problem Solving Guide
\(\text{Mixture Equation: } C_1V_1 + C_2V_2 = C_fV_f\)
General Mixture Equation
Advanced Concepts:

Mass Balance: The total amount of solute remains constant in a closed system

Volume Additivity: When mixing solutions, volumes are generally additive

Weighted Averages: Final concentration is a weighted average of component concentrations

Conservation Laws: Mass and moles of solute are conserved during mixing

Mixture Problem Strategy:
  1. Identify components: List all solutions and their concentrations/volumes
  2. Define variables: Assign symbols to unknown quantities
  3. Set up equations: Use conservation of mass principle
  4. Solve systematically: Use algebraic methods
  5. Verify solution: Check against original conditions
Tip 1: Always draw a diagram showing initial and final states with concentrations and volumes.
Tip 2: Remember that the final concentration must be between the initial concentrations.
Tip 3: Convert percentages to decimals when doing calculations (15% = 0.15).
Tip 4: Check that your answer makes physical sense (no negative volumes).
Common Mistakes: Forgetting to conserve mass, using wrong base values, misplacing decimal points, not checking reasonableness.
Key Formulas: Mixture: C₁V₁+C₂V₂=C₃V₃, Dilution: C₁V₁=C₂V₂, Concentration: (Solute/Total)×100%
Critical Thinking Points:

• Physical Meaning: Understand what each variable represents in the real world

• Dimensional Analysis: Units must be consistent throughout calculations

• Verification: Always substitute back to check your answer

• Problem Context: Consider the practical constraints of the problem

Mixture Problem Types

🧪
Solution Mixing
C₁V₁ + C₂V₂ = C₃V₃
Example: 20% + 40% → 30%
Dilution
C₁V₁ = C₂V₂
Example: 15% → 10% by adding water
Alloy Creation
Mass conservation principle
Example: Pure metal + other metal
Property Mixtures
Weighted average properties
Example: Octane rating mixing

Questions & Answers

Question: In mixture problems, why do we multiply concentration by volume to find the amount of pure substance? Isn't concentration already a measure of how much substance is in the solution?

Answer: Yes, concentration is a measure of how much substance is in the solution, but it's a ratio, not an absolute amount. Concentration tells us the proportion of solute per unit volume.

For example:

  • A 20% solution means 20% of the volume is pure substance
  • But in 100 mL of 20% solution, you have 20 mL of pure substance
  • In 500 mL of 20% solution, you have 100 mL of pure substance

So concentration × volume gives you the actual amount of pure substance in that specific volume. This is crucial because when mixing solutions, we're combining the actual amounts of pure substances, not ratios.

The fundamental principle is conservation of mass: the total amount of pure substance before mixing equals the total amount after mixing.

Question: In dilution problems, why does the amount of solute stay the same when we add more solvent? Doesn't adding water somehow dilute the amount too?

Answer: Great question! When we add water (solvent) to a solution, we're only adding the diluting agent, not changing the amount of the substance we're interested in (solute).

Think of it like this:

  • You have 10 g of sugar dissolved in 100 mL of water (10% sugar solution)
  • When you add 100 mL of water, you now have 10 g of sugar in 200 mL of solution
  • The amount of sugar hasn't changed (still 10 g), but it's now distributed in more water
  • The concentration decreases (10 g / 200 mL = 5% sugar solution)

Adding solvent doesn't remove or add solute - it just spreads the same amount of solute over a larger volume, thereby reducing the concentration. This is the principle behind the dilution formula: C₁V₁ = C₂V₂.

Question: How do I know if my answer to a mixture problem makes sense? Sometimes I get answers that seem way too big or small.

Answer: There are several ways to check if your mixture problem answer makes sense:

1. Range Check: Your final concentration should be between the concentrations of the starting solutions.

2. Reasonableness Check: If you're mixing a 10% solution with a 50% solution to get a 25% solution, you should need more of the lower concentration solution.

3. Substitution Check: Plug your answers back into the original equations to verify they work.

4. Unit Check: Make sure your units are consistent and your answer has the correct units.

5. Reality Check: Volumes should be positive, concentrations should be between 0% and 100%, etc.

Example: If you mixed a 10% solution and a 50% solution and got 100% concentration, that's impossible and indicates an error.