Dilution Practice Problems (C₁V₁ = C₂V₂)
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When you dilute a concentrated stock solution, the amount of drug stays constant while the volume increases, so the concentration falls proportionally. The dilution equation lets you solve for whichever variable is unknown, most often the volume of stock solution needed to hit a target concentration.
FormulaC₁ × V₁ = C₂ × V₂ → V₁ = (C₂ × V₂) ÷ C₁
Worked exampleHow many mL of a 20% stock are needed to make 120 mL of a 2% solution? V₁ = (2% × 120) ÷ 20% = 12 mL (then QS to 120 mL with diluent).
The drug amount is unchanged by dilution; only concentration and volume change. Remember to QS to the final volume when the question asks for it.
10 Dilution practice problems with worked solutions
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Problem 1: Dilution
How many mL of a 10% benzalkonium chloride stock solution are needed to prepare 490 mL of a 0.2% solution? Round your answer to two decimal places.
Answer 9.80 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.2% × 490 mL) ÷ 10%.
- V₁ = 98 (%·mL) ÷ 10% = 9.8 mL.
- If the question asks for the final preparation, QS with diluent up to 490 mL total. The 9.8 mL of stock is only part of the final volume.
- Rounded as the question asks: 9.80 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 2: Dilution
How many mL of a 30% hydrogen peroxide stock solution are needed to prepare 700 mL of a 1.5% solution? Round your answer to two decimal places.
Answer 35.00 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (1.5% × 700 mL) ÷ 30%.
- V₁ = 1050 (%·mL) ÷ 30% = 35 mL.
- If the question asks for the final preparation, QS with diluent up to 700 mL total. The 35 mL of stock is only part of the final volume.
- Rounded as the question asks: 35.00 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 3: Dilution
How many mL of a 6% acetic acid stock solution are needed to prepare 200 mL of a 2% solution? Round your answer to two decimal places.
Answer 66.67 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (2% × 200 mL) ÷ 6%.
- V₁ = 400 (%·mL) ÷ 6% ≈ 66.67 mL.
- If the question asks for the final preparation, QS with diluent up to 200 mL total. The 66.67 mL of stock is only part of the final volume.
- Rounded as the question asks: 66.67 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 4: Dilution
How many mL of a 5.25% sodium hypochlorite stock solution are needed to prepare 940 mL of a 0.5% solution? Round your answer to two decimal places.
Answer 89.52 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.5% × 940 mL) ÷ 5.25%.
- V₁ = 470 (%·mL) ÷ 5.25% ≈ 89.52 mL.
- If the question asks for the final preparation, QS with diluent up to 940 mL total. The 89.52 mL of stock is only part of the final volume.
- Rounded as the question asks: 89.52 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 5: Dilution
How many mL of a 10% benzalkonium chloride stock solution are needed to prepare 650 mL of a 0.25% solution? Round your answer to two decimal places.
Answer 16.25 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.25% × 650 mL) ÷ 10%.
- V₁ = 162.5 (%·mL) ÷ 10% = 16.25 mL.
- If the question asks for the final preparation, QS with diluent up to 650 mL total. The 16.25 mL of stock is only part of the final volume.
- Rounded as the question asks: 16.25 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 6: Dilution
How many mL of a 4% chlorhexidine gluconate stock solution are needed to prepare 400 mL of a 0.5% solution? Round your answer to two decimal places.
Answer 50.00 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.5% × 400 mL) ÷ 4%.
- V₁ = 200 (%·mL) ÷ 4% = 50 mL.
- If the question asks for the final preparation, QS with diluent up to 400 mL total. The 50 mL of stock is only part of the final volume.
- Rounded as the question asks: 50.00 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 7: Dilution
How many mL of a 17% benzalkonium chloride stock solution are needed to prepare 890 mL of a 0.25% solution? Round your answer to two decimal places.
Answer 13.09 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.25% × 890 mL) ÷ 17%.
- V₁ = 222.5 (%·mL) ÷ 17% ≈ 13.09 mL.
- If the question asks for the final preparation, QS with diluent up to 890 mL total. The 13.09 mL of stock is only part of the final volume.
- Rounded as the question asks: 13.09 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 8: Dilution
How many mL of a 5% potassium permanganate stock solution are needed to prepare 730 mL of a 0.02% solution? Round your answer to two decimal places.
Answer 2.92 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.02% × 730 mL) ÷ 5%.
- V₁ = 14.6 (%·mL) ÷ 5% = 2.92 mL.
- If the question asks for the final preparation, QS with diluent up to 730 mL total. The 2.92 mL of stock is only part of the final volume.
- Rounded as the question asks: 2.92 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 9: Dilution
How many mL of a 30% hydrogen peroxide stock solution are needed to prepare 90 mL of a 3% solution? Round your answer to two decimal places.
Answer 9.00 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (3% × 90 mL) ÷ 30%.
- V₁ = 270 (%·mL) ÷ 30% = 9 mL.
- If the question asks for the final preparation, QS with diluent up to 90 mL total. The 9 mL of stock is only part of the final volume.
- Rounded as the question asks: 9.00 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Problem 10: Dilution
How many mL of a 10% benzalkonium chloride stock solution are needed to prepare 530 mL of a 0.2% solution? Round your answer to two decimal places.
Answer 10.60 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.2% × 530 mL) ÷ 10%.
- V₁ = 106 (%·mL) ÷ 10% = 10.6 mL.
- If the question asks for the final preparation, QS with diluent up to 530 mL total. The 10.6 mL of stock is only part of the final volume.
- Rounded as the question asks: 10.60 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.