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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 example

How 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.

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Dilution
How many mL of a 20% chlorhexidine gluconate stock solution are needed to prepare 110 mL of a 0.12% solution? Round your answer to two decimal places.
mL

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How to set up a dilution problem

Diluting a stock solution changes its volume and its concentration but not the amount of drug in it. That single fact is the whole method. Whatever quantity of drug is in the stock you measure out is the quantity in the final preparation, so concentration × volume before must equal concentration × volume after.

C₁ × V₁ = C₂ × V₂

Any three of the four solve for the fourth. The concentrations can be in any unit as long as both are in the same one (percent, mg/mL, ratio strength converted to percent), and the same goes for the volumes. The equation only works if the units match across the equals sign.

The usual question: how much stock

Prepare 500 mL of 0.5% chlorhexidine gluconate from a 4% stock.

V₁ = C₂ × V₂ ÷ C₁ = 0.5 × 500 ÷ 4 = 62.5 mL

Then the part of the answer that isn't a number: you measure 62.5 mL of the 4% stock and add water to bring the total to 500 mL. That means 437.5 mL of water, not 500. If the question asks how much diluent to add, the answer is V₂ − V₁.

Solving for final volume

How much 3% hydrogen peroxide can be made from 50 mL of a 30% solution?

V₂ = C₁ × V₁ ÷ C₂ = 30 × 50 ÷ 3 = 500 mL

The strength dropped by a factor of ten, so the volume went up by a factor of ten. Every dilution obeys that inverse relationship, and it's the fastest sanity check available: if the concentration fell and your volume also fell, the equation was set up upside down.

Solving for final strength

10 mL of 6% acetic acid is diluted to 240 mL with water. What is the resulting strength?

C₂ = C₁ × V₁ ÷ V₂ = 6 × 10 ÷ 240 = 0.25%

This one is alligation medial with a 0% diluent, and it gives the same answer either way. Use whichever setup you can write fastest.

Worked example: units that need converting first

How many mL of a 17% benzalkonium chloride stock are needed to make 1,000 mL of a 1:5,000 solution?

the target as a percent = 1 ÷ 5,000 × 100 = 0.02%

V₁ = 0.02 × 1,000 ÷ 17 = 1.176 → 1.18 mL

The conversion had to happen before C₁V₁ = C₂V₂, because 17 and 5,000 are not in the same units. A ratio strength in the question is a hint that the first step is a conversion, not the dilution equation.

Serial dilutions

When a question dilutes in stages, apply C₁V₁ = C₂V₂ to each stage in turn, or notice that the dilution factors multiply. Take 1 mL of a 10% solution to 10 mL (now 1%), then 1 mL of that to 10 mL (now 0.1%). Two 1-in-10 dilutions are a 1-in-100 dilution overall. The drug amount in the final 10 mL is 10 mg, and it came from 0.1 mL of the original stock.

Where students lose the points

10 Dilution practice problems with worked solutions

Every problem below is worked start to finish. Try it yourself first, then check the steps. For a fresh problem every time, use the practice box above. It generates unlimited variations of these same question types.

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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (0.2% × 490 mL) ÷ 10%.
  3. V₁ = 98 (%·mL) ÷ 10% = 9.8 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (1.5% × 700 mL) ÷ 30%.
  3. V₁ = 1050 (%·mL) ÷ 30% = 35 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (2% × 200 mL) ÷ 6%.
  3. V₁ = 400 (%·mL) ÷ 6% ≈ 66.67 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (0.5% × 940 mL) ÷ 5.25%.
  3. V₁ = 470 (%·mL) ÷ 5.25% ≈ 89.52 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (0.25% × 650 mL) ÷ 10%.
  3. V₁ = 162.5 (%·mL) ÷ 10% = 16.25 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (0.5% × 400 mL) ÷ 4%.
  3. V₁ = 200 (%·mL) ÷ 4% = 50 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (0.25% × 890 mL) ÷ 17%.
  3. V₁ = 222.5 (%·mL) ÷ 17% ≈ 13.09 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (0.02% × 730 mL) ÷ 5%.
  3. V₁ = 14.6 (%·mL) ÷ 5% = 2.92 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (3% × 90 mL) ÷ 30%.
  3. V₁ = 270 (%·mL) ÷ 30% = 9 mL.
  4. 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.
  5. 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
  1. Use C₁V₁ = C₂V₂.
  2. V₁ = (C₂ × V₂) ÷ C₁ = (0.2% × 530 mL) ÷ 10%.
  3. V₁ = 106 (%·mL) ÷ 10% = 10.6 mL.
  4. 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.
  5. 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.

Sources and how far they go

Where a convention on this page comes from a published standard, the standard is named. Where it comes from a textbook or from ordinary practice instead, the entry says so. That difference matters when a preceptor or an exam disagrees with you.

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