PTCE Math: How to Master Dilution Calculations

At a glance.

• Dilutions are widely tested as part of the Order Entry and Processing knowledge domain of the PTCE.
• Dilution is the process of lowering a solution’s concentration by adding a diluent, such as water, without changing the amount of active ingredient present.
• Dilutions are used when a medication, disinfectant, or stock solution is more concentrated than needed for its intended use. Adding a measured amount of an appropriate diluent lowers the concentration while keeping the required amount of active ingredient in the final preparation.
• Percentage strength (w/v%, w/w%, and v/v%) are used to express the ratio of active ingredient to non-active ingredients in a medicine.
• A stock solution is a concentrated preparation used to make a weaker solution. Stock solution dilution calculations determine how much stock solution and diluent are needed to produce a required concentration and final volume.
• The formula C1.V1 = C2.V2 is used to help answer stock solution-related questions. In these questions, you are given three of the four of these variables, allowing you to plug in the values and find the required answer.

Dilutions as a topic of the PTCE

The latest PTCE outline includes dilution calculations as part of the Order Entry and Processing knowledge domain.

Pharmacy technicians are required to learn basic dilution procedures that feature:

• Dilution relationships – such as C1.V1 = C2.V2
• How to calculate the amount of stock solution required
• How to work with percentage strengths

Note: alligation calculations were removed as an examinable topic of the PTCE, though the three types of dilution calculation listed remain examinable.

Remember – complex dilution calculations will not be tested. Keep the scope of your exam preparation limited to basic dilution calculations, the type and style of which we cover throughout this guide.

What are dilutions?

Dilution is the process of lowering a solution’s concentration by adding a diluent, such as water, without changing the amount of active ingredient present.

As the word suggests, diluents are used to “dilute” the concentration.

For example:

• Liquid medication: adding purified water to 100 mL of a 10% solution to prepare a weaker 5% solution.
• Disinfectant: mixing a concentrated disinfectant with water to produce the lower concentration required for cleaning.
• Injectable medication: adding sterile diluent to a concentrated medication before administration to obtain the prescribed concentration.

Why are dilutions required in pharmacy?

Dilutions are used when a medication, disinfectant, or stock solution is more concentrated than needed for its intended use. Adding a measured amount of an appropriate diluent lowers the concentration while keeping the required amount of active ingredient in the final preparation.

This can make a product:

• safer to administer
• easier to measure accurately
• more suitable for a particular patient, route, or procedure

Dilutions must be calculated and prepared carefully because using too much or too little diluent can produce the wrong concentration and lead to ineffective treatment or patient harm.

How to work with percentage strengths

Percentage strength is a term that describes how much active ingredient is present in every 100 parts of a preparation.

There are three designations:

	Meaning	Example
w/w	weight per weight	Grams per 100 g
w/v	weight per volume	Grams per 100 mL
v/v	volume per volume	mL per 100 mL

For example:

• 5% w/w = 5 grams of active ingredient in 100 g of medicine
• 5% w/v = 5 grams of active ingredient in 100 mL of medicine
• 10% v/v = 10 mL of active ingredient in 100 mL of medicine

Let’s take a practical example:

• A 1% hydrocortisone cream contains 1 gram of hydrocortisone in every 100 grams of the finished cream. We can therefore write this as a 1% w/w percentage strength.

Remember, a medicine is made up of two parts – the active ingredient and the excipients (inactive ingredients).

In this case, we have 1 gram of active ingredient (hydrocortisone) and 99 grams of inactive ingredients (which includes diluent). Hypothetically then, if we were to add an additional 50 grams of inactive ingredient to this medicine, that would dilute the concentration of the hydrocortisone component. This means that, in each gram of cream, the medicine has a lower concentration and is therefore weaker.

We can visually represent this hypothetical dilution as follows:

Percentage strengths can be manipulated.

For example:

• 1% w/v = 1 gram of active ingredient in 100 mL of medicine
• We also know that 1 gram is equal to 1,000 mg
• If 100 mL of medicine contains 1,000 mg, then 1 mL contains 10 mg
• This can be represented as 10 mg/mL

Let’s work through a few more examples:

% Strength	How to convert to mg/mL
0.5% w/v	0.5 g in 100 mL 0.5 g = 500 mg 500 mg ÷ 100 mL = 5 mg/mL
2% w/v	2 g in 100 mL 2 g = 2,000 mg 2,000 mg ÷ 100 mL = 20 mg/mL
2.5% w/v	2.5 g in 100 mL 2.5 g = 2,500 mg 2,500 mg ÷ 100 mL = 25 mg/mL
4% w/v	4 g in 100 mL 4 g = 4,000 mg 4,000 mg ÷ 100 mL = 40 mg/mL
10% w/v	10 g in 100 mL 10 g = 10,000 mg 10,000 mg ÷ 100 mL = 100 mg/mL

In other words:

Percentage strength (w/v) × 10 = mg/mL

Take the percentage strength (such as 0.5% w/v), then multiply the figure by 10 to produce mg/mL (in this case 0.5 x 10 = 5), therefore 5 mg/mL.

Let’s work through the two other percentage strength designations (w/w and v/v):

• 2% w/w cream = 2 g of active ingredient in 100 g of medicine
• 2 g is the same as 2,000 mg
• 2,000 mg ÷ 100 g = 20 mg/g
• Shortcut: % w/w × 10 = mg/g

v/v is different and doesn’t follow the same shortcut:

Example: 5% v/v solution

5 mL of liquid ingredient in 100 mL of finished solution

5 mL ÷ 100 mL = 0.05 mL/mL

This means there are 0.05 mL of active ingredient in each 1 mL of solution.

Sample PTCE Dilution Question.

How many grams of sodium chloride are contained in 250 mL of a 0.9% w/v sodium chloride solution?

With this type of question, always start with the % strength solution.

We are told our solution has a percentage strength of 0.9% w/v. As we learned, this means 0.9 grams of medicine in each 100 mL of solution.

We can answer the question by multiplying this % by the target volume:

0.9% x 250 mL of solution = 2.25 grams

Alternatively, we can create a table to represent precisely what this means:

Final 250 mL Solution	Active ingredient
First 100 mL	0.9 grams
Second 100 mL	0.9 grams
Last 50 mL	0.45 grams
	2.25 grams of drug in total

This tabular form is useful because it visually represents the amount of drug in each 100 mL of solution. You can use this table format in your PTCE to help you double-check or simply just to answer the question directly.

What are stock solutions?

A stock solution is a concentrated preparation used to make a weaker solution. Stock solution dilution calculations determine how much stock solution and diluent are needed to produce a required concentration and final volume:

We are familiar with stock solutions in the home.

Think of orange concentrate. This is a highly concentrated orange solution (effectively, a stock solution). We dilute this solution with water to produce a palatable, useable concentration.

In pharmacy, stock solutions are no different. They must similarly be diluted to produce a target volume (mL) with a target concentration (% w/v, for example).

How to answer stock solution PTCE questions

The formula C₁V₁ = C₂V₂ is used to calculate how much of a concentrated stock solution is needed to prepare a weaker solution. Typically, a PTCE question will give you three of these four variables. You need to plug these values into the question and solve to determine the correct answer.

C1.V1 = C2.V2

• C₁ = concentration of the original stock solution
• V₁ = volume of stock solution needed
• C₂ = concentration of the final diluted solution
• V₂ = total volume of the final solution

Let’s review a sample question:

A pharmacy technician needs to prepare 250 mL of a 4% w/v solution using a 20% w/v stock solution. How many milliliters of the stock solution are required?

First, establish that the stock solution formula can be used. Here, we see that stock solution is mentioned in the question and that three variables have been included.

Let’s plug these values into the equation. Note that the first part is always about the original stock solution and the volume of stock solution required:

• C₁ = 20% — concentration of the original stock solution
• V₁ = unknown — volume of stock solution required

Always remember that the final part of the stock solution equation always focuses on final volume / concentration:

• C₂ = 4% — required final concentration
• V₂ = 250 mL — required final volume

Key points to consider:

• Identify each value first: write down C₁, V₁, C₂, and V₂ before inserting anything into the formula. This prevents the stock and final concentrations from being reversed.
• Keep the units consistent: both concentrations must use the same format, and both volumes must use the same unit. Convert units before calculating when necessary. For example: if the question asked for 0.5 L, you may need to convert this to 500 mL first.
• Remember what V₂ represents: V₂ is the final total volume, not the amount of diluent being added.
• Check whether the answer is reasonable: when making a weaker solution, the volume of concentrated stock should be smaller than the final volume.
• Follow the wording carefully: after calculating V₁, measure that amount of stock and add enough diluent to reach the required final volume. Do not automatically add the stated final volume as diluent.

PTCE Dilution Practice Questions.

Below we put together eight sample PTCE dilution questions.

Attempt each of these questions first to test your own understanding of the concepts and formulae covered in this guide. Then, compare your answers and methods with the answer explanation section directly below.

Question 1.

How many grams of sodium chloride are contained in 500 mL of a 0.9% w/v solution?

a. 0.45 g

b. 4.5 g

c. 9 g

d. 45 g

Question 2.

What concentration in mg/mL is equivalent to a 2.5% w/v solution?

a. 0.25 mg/mL

b. 2.5 mg/mL

c. 25 mg/mL

d. 250 mg/mL

Question 3.

A 30 g container of cream contains 0.75 g of active ingredient. What is its percentage strength?

a. 0.025% w/w

b. 0.25% w/w

c. 2.5% w/w

d. 25% w/w

Question 4.

How many milliliters of alcohol are present in 350 mL of a 70% v/v preparation?

a. 24.5 mL

b. 70 mL

c. 245 mL

d. 280 mL

Question 5.

How many milliliters of a 10% stock solution are required to prepare 200 mL of a 2% solution?

a. 20 mL

b. 40 mL

c. 100 mL

d. 160 mL

Question 6.

How many milliliters of a 50 mg/mL stock solution are required to prepare 120 mL of a 10 mg/mL solution?

a. 24 mL

b. 36 mL

c. 60 mL

d. 96 mL

Question 7.

A technician has 80 mL of a 25% stock solution. To what final total volume should it be diluted to produce a 5% solution?

a. 100 mL

b. 320 mL

c. 400 mL

d. 500 mL

Question 8.

How many milliliters of a 30% stock solution are required to prepare 1.5 L of a 6% solution?

a. 95 mL

b. 200 mL

c. 280 mL

d. 300 mL

Answer Explanations.

Answer to Question 1 – b. 4.5 grams

A 0.9% w/v solution contains 0.9 g of sodium chloride in every 100 mL.

(0.9 / 100) x 500 = 4.5 g

Therefore, 500 mL contains 4.5 g of sodium chloride.

Answer to Question 2 – c. 25 mg/mL

A 2.5% w/v solution contains 2.5 g in every 100 mL. First convert grams to milligrams:

2.5 g x 1,000 = 2,500 mg

2,500 mg / 100 mL = 25 mg/mL

Answer to Question 3 – c. 2.5% w/w

Divide the weight of the active ingredient by the total weight of the cream, then multiply by 100:

(0.75 g / 30 g) x 100 = 2.5%

The cream has a percentage strength of 2.5% w/w.

Answer to Question 4 – c. 245 mL

A 70% v/v preparation contains 70 mL of alcohol in every 100 mL.

(70 mL / 100 mL) x 350 mL = 245 mL

Therefore, the preparation contains 245 mL of alcohol.

Answer to Question 5 – b. 40 mL

Establish the values:

• C₁ = 10%
• V₁ = unknown
• C₂ = 2%
• V₂ = 200 mL

Apply C1.V1 = C2.V2:

10% x V1 = 2% x 200 mL

10% x V1 = 400 mL

V1 = 400 mL / 10%

V1 = 40 mL

Use 40 mL of the 10% stock solution, then add enough diluent to make 200 mL total.

Answer to Question 6 – a. 24 mL

Establish the values:

• C₁ = 50 mg/mL
• V₁ = unknown
• C₂ = 10 mg/mL
• V₂ = 120 mL

Apply C1.V1 = C2.V2:

50 mg/mL x V1 = 10 mg/mL x 120 mL

50 mg/mL x V1 = 1,200 mL

V1 = 1,200 mL / 50

V1 = 24 mL

Use 24 mL of the 50 mg/mL stock solution and add enough diluent to make 120 mL total.

Answer to Question 7 – c. 400 mL

This question asks for the final volume rather than the stock volume:

• C₁ = 25%
• V₁ = 80 mL
• C₂ = 5%
• V₂ = unknown

Apply C1.V1 = C2.V2:

25% x 80 mL = 5% x V2

2,000 mL = 5% x V2

2,000 mL / 5% = V2

400 mL = V2

The 80 mL of stock solution should be diluted to a final total volume of 400 mL.

Answer to Question 8 – d. 300 mL

First convert the final volume:

1.5 liters x 1,000 = 1,500 mL

Establish the values:

• C₁ = 30%
• V₁ = unknown
• C₂ = 6%
• V₂ = 1,500 mL

Apply C1.V1 = C2.V2:

30% x V1 = 6% x 1,500 mL

30% x V1 = 9,000 mL

V1 = 9,000 mL / 30

V1 = 300 mL

Use 300 mL of the 30% stock solution, then add enough diluent to make a final volume of 1.5 L.

Topic Review.

• Dilution lowers the concentration without changing the amount of active ingredient. The ingredient is simply distributed throughout a larger final volume.
• Use C1.V1 = C2.V2 for stock solution dilution questions. This formula connects the initial stock concentration and volume (first half of the equation) with the required final concentration and volume (final half of the equation).
• Remember what each symbol means: C1 is the stock concentration, V1 is the stock volume needed, C2 is the final concentration, and V2 is the final total volume.
• Establish all four values before calculating. Writing them down first reduces the risk of reversing the initial and final concentrations.
• The stock solution should be stronger than the final solution. If C1 is lower than C2, ordinary dilution cannot produce the required strength.
• Keep corresponding units consistent. Concentrations should use the same format, while volumes should both be expressed in milliliters, liters, or another matching unit.
• Convert units before using the formula. For example, convert 1.5 L to 1,500 mL if the answer is required in milliliters.
• V2 represents the final total volume. It is not the amount of diluent that should be added.
• Add enough diluent to reach the final volume. If 50 mL of stock is required to prepare 250 mL, measure 50 mL and add diluent until the total volume reaches 250 mL.
• Use proportions for direct percentage-strength questions. For example, 0.9% w/v means 0.9 g in 100 mL, so the amount in another volume can be found using a proportion.
• Use the percentage-strength shortcut carefully. For w/v preparations, multiplying the percentage by 10 gives the concentration in mg/mL: 2.5% w/v equals 25 mg/mL.
• For w/w questions, compare weights. Divide the weight of the active ingredient by the total weight of the finished preparation and multiply by 100.
• For v/v questions, compare volumes. A 70% v/v preparation contains 70 mL of the liquid ingredient in every 100 mL of finished preparation.
• Check whether the answer is reasonable. When making a weaker solution, the stock volume should be smaller than the required final volume.
• Read what the question is actually asking for. It may ask for the stock volume, final volume, amount of active ingredient, concentration, or occasionally the amount of diluent.
• Do not use C1.V1 = C2.V2 for every percentage question. Use it when a solution is being diluted from one concentration to another; use a proportion when finding the amount already present in a preparation.

Frequently Asked Questions.

Can stock and final concentrations be written in different formats?

They can initially, but they must be converted into matching units before the equation is used. For example, convert 2% w/v to 20 mg/mL before comparing it with a concentration stated in mg/mL.

Should answers be rounded during each step?
Always keep the full calculated value until the final step, then round according to the question’s instructions or workplace policy. Rounding too early can produce an inaccurate final answer.

Are dilution and alligation the same calculation method?
No. Dilution usually combines a concentrated stock with a diluent, while alligation determines how much of two different strengths should be mixed to produce an intermediate strength.

How can I further improve my study preparation of PTCE math?

Our online test prep course is unique in offering a dedicated PTCE Math Module. This module covers dilutions in more detail as well as every other math topic tested on the PTCE exam. Full study guides, practice questions, worked examples, explainers, PDF notes, and test exams. Everything you need to succeed at PTCE math – all in one place.

Can C1.V1 = C2.V2 be used with concentrations expressed in mg/mL?
Yes. The equation works with mg/mL, percentage strengths, ratio strengths, and other concentration formats, provided that C1 and C2 use matching units.