How to Convert Mass-Based ppb into mg/L Using Density

A water sample contains 250 ppb of dissolved lead. What is that concentration in mg/L?

For a dilute aqueous solution with a density of approximately 1.00 kg/L, the answer is **0.250 mg/L**.

The conversion is usually straightforward:

\[
\boxed{\text{mg/L} \approx \frac{\text{ppb}}{1000}}
\]

But there's an important condition. This shortcut assumes that the ppb concentration is based on mass and that the solution has approximately the same density as water. If either assumption changes, the conversion needs more care.

Let's work through why the shortcut works, when it fails, and how to handle the different types of ppb concentration in HSC Chemistry calculations.

## 1. Why does dividing ppb by 1000 give mg/L?

Imagine you have one litre of very dilute salt solution. Its density is approximately 1.00 kg/L, so that litre has a total mass of approximately 1.00 kg.

Suppose the solution contains 250 ppb of a dissolved substance, measured by mass.

A mass-based concentration of 250 ppb means:

\[
250\text{ ppb} = 250\ \mu\text{g/kg}
\]

In other words, every kilogram of solution contains 250 micrograms of dissolved substance.

Since one litre of solution has a mass of approximately one kilogram, that litre contains approximately 250 micrograms of solute.

Now convert micrograms into milligrams:

\[
1000\ \mu\text{g} = 1\text{ mg}
\]

Therefore:

\[
250\ \mu\text{g/L}
= 0.250\text{ mg/L}
\]

That's the entire reason behind the shortcut. We're converting a mass fraction into a mass concentration by using the density of the solution.

[IMAGE: An annotated illustration of a one-litre transparent water sample beside a balance showing approximately 1.00 kg of solution. Magnify a tiny portion of the sample to represent dissolved solute, with labels showing 250 micrograms of solute per litre and the equivalent 0.250 mg/L. Make the distinction between the total solution mass and the dissolved solute mass clear.]

## 2. The important distinction: what does ppb actually measure?

Parts per billion (ppb) describes a ratio. It does not automatically tell you whether the concentration is measured by mass, volume, or amount of substance.

For a mass-based concentration:

\[
1\text{ ppb} = 1\ \mu\text{g/kg}
\]

This means one microgram of solute for every kilogram of total solution.

For a mass-per-volume concentration, a result might instead be reported directly in \(\mu\text{g/L}\).

These two measurements are not identical definitions. They become approximately numerically equivalent for dilute aqueous solutions because one litre of solution has a mass close to one kilogram.

| Concentration basis | Meaning of 1 ppb | What is needed to convert to mg/L? |
|---|---|---|
| Mass/mass (m/m) | 1 μg solute per kg solution | Solution density |
| Mass/volume (m/V) | 1 μg solute per L solution | Divide by 1000 |
| Volume/volume (V/V) | 1 μL solute per L solution | Density of the solute, if converting to mass per volume |

[IMAGE: A scientific comparison showing three labelled measuring arrangements: a balance comparing solute mass with total solution mass, a one-litre measuring vessel showing solute mass per solution volume, and a liquid-volume comparison showing microlitres of solute per litre of mixture. Clearly distinguish mass/mass, mass/volume, and volume/volume.]

For dissolved substances in dilute water samples, mass-based ppb is particularly useful. However, always check how the concentration has been defined in the question.

If the question gives ppb by mass, use density. If it explicitly gives ppb as μg/L, no density conversion is needed.

For a broader explanation of the relationship between these concentration units, see [ppm, ppb and mg/L: When Can You Convert Between Them?](https://tutorgum.com/learn/resources/ppm-ppb-and-mg-l-when-can-you-convert-between-them/).

One more basis is worth recognising: amount/amount concentration. For example, 1 ppb by amount means a mole fraction of \(1\times10^{-9}\), or 1 nmol of substance per mol of mixture.

This is common in trace-gas measurements. For an ideal gas, a volume fraction and a mole fraction have the same numerical value under the same temperature and pressure conditions.

However, amount-based ppb cannot be converted directly into mg/L using the mass-based formula. You would also need the molar mass and enough information to determine the amount of mixture per litre.

The main rule is simple: **identify what the ppb represents before choosing an equation.**

## 3. The conversion formula for ppb by mass into mg/L

For a mass-based ppb concentration, use:

\[
\boxed{
C_{\mathrm{mg/L}}=
\frac{C_{\mathrm{ppb}}\times\rho}{1000}
}
\]

Where:

- \(C_{\mathrm{mg/L}}\) = mass concentration in mg/L.
- \(C_{\mathrm{ppb}}\) = concentration in ppb by mass.
- \(\rho\) = density of the final solution in kg/L.
- 1000 = the conversion factor from micrograms to milligrams.

Let's see where this equation comes from.

Start with the definition:

\[
1\text{ ppb}=1\ \mu\text{g/kg}
\]

If the density of the solution is \(\rho\) kg/L, then one litre contains \(\rho\) kilograms of solution.

Therefore, a concentration of \(C_{\mathrm{ppb}}\) gives:

\[
C_{\mathrm{ppb}}\times\rho
\quad \mu\text{g/L}
\]

To convert micrograms into milligrams, divide by 1000:

\[
C_{\mathrm{mg/L}}=
\frac{C_{\mathrm{ppb}}\rho}{1000}
\]

That's the complete derivation.

[IMAGE: An annotated physical unit-conversion diagram showing a one-litre measuring vessel containing rho kilograms of solution. Beside it, show the relationship 1 ppb by mass = 1 microgram per kilogram, followed by multiplication by rho to obtain micrograms per litre and division by 1000 to obtain milligrams per litre. Use arrows connecting the actual solution vessel, mass measurement, and final concentration.]

### What if density is given in g/mL?

You can use the numerical value directly.

This is because:

\[
1.00\text{ g/mL}=1.00\text{ kg/L}
\]

For example:

\[
1.18\text{ g/mL}=1.18\text{ kg/L}
\]

You do not need to multiply or divide by another 1000.

**Important:** Use the density of the final solution, not automatically the density of pure water or the dissolved substance.

### Worked example 1: Converting ppb in dilute freshwater

A freshwater sample contains 180 ppb of a dissolved substance by mass. Its density is \(0.998\text{ kg/L}\).

Calculate its concentration in mg/L.

**Step 1**: Identify the concentration basis.

The question specifies ppb by mass.

Therefore:

\[
180\text{ ppb}=180\ \mu\text{g/kg}
\]

**Step 2**: Identify the solution density.

The density is:

\[
\rho=0.998\text{ kg/L}
\]

This means one litre of the solution has a mass of 0.998 kg.

**Step 3**: Apply the conversion formula.

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{C_{\mathrm{ppb}}\rho}{1000}\\
&=\frac{180\times0.998}{1000}\\
&=0.17964\text{ mg/L}
\end{aligned}
\]

**Step 4**: Round to three significant figures.

\[
\boxed{C=0.180\text{ mg/L}}
\]

The sample contains 0.180 mg of dissolved substance per litre of solution.

Notice that the answer is very close to 0.180 mg/L even if we approximate the density as exactly 1.00 kg/L. That's why the shortcut works well for dilute freshwater.

### Worked example 2: Converting ppb in a denser solution

A concentrated salt solution has a density of \(1.18\text{ g/mL}\). A trace metal impurity is present at 640 ppb by mass.

Calculate the concentration of the impurity in mg/L.

**Step 1**: Identify the concentration basis.

The concentration is mass-based:

\[
640\text{ ppb}=640\ \mu\text{g/kg}
\]

**Step 2**: Convert the density into kg/L.

We are given:

\[
\rho=1.18\text{ g/mL}
\]

Since the numerical values are equivalent:

\[
\rho=1.18\text{ kg/L}
\]

**Step 3**: Apply the conversion formula.

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{640\times1.18}{1000}\\
&=\frac{755.2}{1000}\\
&=0.7552\text{ mg/L}
\end{aligned}
\]

**Step 4**: State the final answer.

\[
\boxed{C=0.755\text{ mg/L}}
\]

The solution contains approximately 0.755 mg of the impurity per litre.

[IMAGE: A transparent one-litre sample vessel representing a concentrated salt solution, positioned beside a balance indicating the solution density of 1.18 g/mL. Include a magnified inset representing dissolved trace-metal particles distributed throughout the liquid. Label the given density and 640 ppb by mass, but do not display the calculated concentration.]

Notice something interesting: the answer is 0.755 mg/L, not 0.640 mg/L.

The solution is denser than water. Each litre contains 1.18 kg of solution, so it contains more solute per litre than a solution with the same ppb-by-mass concentration but a density of 1.00 kg/L.

The density has increased the mass concentration, even though the mass fraction of the impurity has stayed the same.

## 4. The common trap: ppb is not always the same as μg/L

Consider two laboratory reports.

Both contain the number 250, but their concentration units are different.

**Report A:** 250 ppb by mass.

The solution has a density of \(1.20\text{ kg/L}\).

\[
C_{\mathrm{mg/L}}=
\frac{250\times1.20}{1000}
\]

\[
\boxed{C=0.300\text{ mg/L}}
\]

**Report B:** 250 μg/L.

This is already a mass-per-volume concentration.

Simply divide by 1000:

\[
250\ \mu\text{g/L}
=\boxed{0.250\text{ mg/L}}
\]

No density conversion is needed.

[IMAGE: Side-by-side comparison of two one-litre transparent sample containers. Container A is labelled 250 ppb by mass and density 1.20 kg/L. Container B is labelled 250 micrograms per litre. Show that both are measurements of dissolved material but use different concentration bases. Do not show the calculated mg/L results.]

The two answers are different because the original measurements describe different things.

- ppb by mass compares solute mass with total solution mass.
- μg/L compares solute mass with solution volume.

Some water-testing reports use ppb as shorthand for μg/L. This is a common convention for dilute water samples, but it is not a universal definition.

If a question explicitly gives μg/L, treat it as a mass-per-volume concentration. Do not automatically apply the mass-based ppb formula.

### A quick decision table

| What the question gives you | What to do |
|---|---|
| ppb by mass | Multiply by solution density, then divide by 1000 |
| μg/L | Divide by 1000 |
| ppb by volume | Obtain the solute density to convert its volume into mass |
| ppb by amount (mol/mol) | Use molar mass and the relevant amount-per-volume information |

For volume-based ppb, 1 ppb by volume means 1 nL of solute per litre of mixture. You need the density of the pure liquid solute to convert that volume into mass.

For a gas, additional information such as temperature, pressure, and molar mass may be needed.

The key is to avoid treating every concentration written in ppb as though it has the same physical meaning.

## 5. Questions and solutions

The following questions use illustrative data. They start with a straightforward conversion and gradually introduce density changes, dilution, and mixing.

### Question 1

A freshwater sample contains 72.0 ppb of dissolved zinc by mass.

Assume the density of the sample is \(1.00\text{ kg/L}\).

Calculate the concentration of zinc in mg/L.

### Solution 1

The concentration is **0.0720 mg/L**.

Since the concentration is given in ppb by mass, use:

\[
C_{\mathrm{mg/L}}=
\frac{C_{\mathrm{ppb}}\rho}{1000}
\]

Substitute:

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{72.0\times1.00}{1000}\\
&=0.0720\text{ mg/L}
\end{aligned}
\]

Therefore:

\[
\boxed{C=0.0720\text{ mg/L}}
\]

The result makes sense because the density is exactly \(1.00\text{ kg/L}\). The numerical value of the concentration in ppb is therefore 1000 times the numerical value in mg/L.

### Question 2

A chemical cleaning solution has a density of \(1.14\text{ g/mL}\) and contains 385 ppb of a dissolved impurity by mass.

A student calculates:

\[
385\div1000=0.385\text{ mg/L}
\]

Evaluate the student's answer and calculate the correct concentration.

### Solution 2

The student's answer is incorrect. The correct concentration is **0.439 mg/L**.

The student has assumed that the solution has a density of \(1.00\text{ kg/L}\). However, the question gives a different density.

First, convert the density:

\[
1.14\text{ g/mL}=1.14\text{ kg/L}
\]

Now apply the conversion formula:

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{385\times1.14}{1000}\\
&=\frac{438.9}{1000}\\
&=0.4389\text{ mg/L}
\end{aligned}
\]

To three significant figures:

\[
\boxed{C=0.439\text{ mg/L}}
\]

The student's calculation would be appropriate if the solution density were \(1.00\text{ kg/L}\).

Here, each litre contains 1.14 kg of solution. Since the concentration is expressed per kilogram, that additional solution mass must be included.

The important lesson is that the shortcut is an approximation based on density, not an exact conversion rule.

### Question 3

Two laboratory samples contain the same dissolved substance.

Both have a concentration of 210 ppb by mass. However, their densities differ.

- Sample A: \(0.940\text{ kg/L}\)
- Sample B: \(1.09\text{ kg/L}\)

Each sample has a volume of 1.00 L.

[IMAGE: Two identical one-litre laboratory sample vessels side by side. Label Sample A with density 0.940 kg/L and 210 ppb by mass, and Sample B with density 1.09 kg/L and 210 ppb by mass. Keep the vessels the same volume to emphasise that their total solution masses differ. Do not include calculated solute masses or answers.]

(a) Calculate the concentration of each sample in mg/L.

(b) Which sample contains more dissolved substance per litre?

(c) Calculate the percentage by which the larger mass concentration exceeds the smaller one.

### Solution 3

Sample B has the larger mass concentration. It contains approximately 16.0% more dissolved substance per litre than Sample A.

**Part (a): Calculate the concentration of Sample A.**

Use:

\[
C_{\mathrm{mg/L}}=
\frac{C_{\mathrm{ppb}}\rho}{1000}
\]

Substitute:

\[
\begin{aligned}
C_A
&=\frac{210\times0.940}{1000}\\
&=0.1974\text{ mg/L}
\end{aligned}
\]

Therefore:

\[
\boxed{C_A=0.197\text{ mg/L}}
\]

**Calculate the concentration of Sample B.**

\[
\begin{aligned}
C_B
&=\frac{210\times1.09}{1000}\\
&=0.2289\text{ mg/L}
\end{aligned}
\]

Therefore:

\[
\boxed{C_B=0.229\text{ mg/L}}
\]

**Part (b): Compare the samples.**

Sample B has the larger concentration in mg/L.

Both samples have the same mass-based ppb value, but one litre of Sample B has a greater total mass.

In fact:

- One litre of Sample A has a mass of 0.940 kg.
- One litre of Sample B has a mass of 1.09 kg.

Since both contain 210 μg of solute per kilogram, Sample B contains more solute in the same volume.

**Part (c): Calculate the percentage difference.**

Use Sample A as the reference:

\[
\text{Percentage increase}
=
\frac{C_B-C_A}{C_A}\times100
\]

Substitute:

\[
\begin{aligned}
\text{Percentage increase}
&=\frac{0.2289-0.1974}{0.1974}\times100\\
&=15.96\%
\end{aligned}
\]

Therefore:

\[
\boxed{16.0\%}
\]

The tempting prediction is that equal ppb values must give equal mg/L values. That would only be true if the solution densities were also equal.

This question tests whether you understand what each concentration is measured against, rather than simply whether you can substitute numbers into a formula.

### Question 4

A laboratory has a 250.0 g sample of solution containing 480 ppb of a dissolved substance by mass.

A technician adds 750.0 g of pure water. No solute is lost, and no chemical reaction occurs.

The final diluted solution has a density of \(1.02\text{ kg/L}\).

Calculate:

(a) The mass of dissolved substance originally present.

(b) The new concentration in ppb by mass.

(c) The final concentration in mg/L.

[IMAGE: A laboratory dilution setup showing a 250.0 g sample of solution in one vessel and 750.0 g of pure water in a second vessel being combined into a larger mixing vessel. Label the initial solution mass, added water mass, and final solution density of 1.02 kg/L. Show the sequence clearly, without displaying any calculated concentrations.]

### Solution 4

The final concentrations are **120 ppb by mass** and **0.122 mg/L**.

The important detail is that dilution changes the total solution mass and volume, but the mass of dissolved substance remains constant.

**Part (a): Calculate the original solute mass.**

The original concentration is:

\[
480\text{ ppb}=480\ \mu\text{g/kg}
\]

Convert the original solution mass into kilograms:

\[
250.0\text{ g}=0.2500\text{ kg}
\]

Using the mass-based concentration:

\[
\begin{aligned}
m_{\mathrm{solute}}
&=480\ \mu\text{g/kg}\times0.2500\text{ kg}\\
&=120\ \mu\text{g}
\end{aligned}
\]

Therefore:

\[
\boxed{m_{\mathrm{solute}}=120\ \mu\text{g}}
\]

**Part (b): Calculate the new ppb concentration.**

The final mass of solution is:

\[
\begin{aligned}
m_{\mathrm{final}}
&=250.0+750.0\\
&=1000.0\text{ g}\\
&=1.000\text{ kg}
\end{aligned}
\]

The solute mass is still 120 μg.

Therefore:

\[
\begin{aligned}
C_{\mathrm{ppb}}
&=\frac{120\ \mu\text{g}}{1.000\text{ kg}}\\
&=120\ \mu\text{g/kg}
\end{aligned}
\]

Hence:

\[
\boxed{C_{\mathrm{ppb}}=120\text{ ppb}}
\]

The concentration has fallen from 480 ppb to 120 ppb because the total solution mass has increased by a factor of four.

**Part (c): Calculate the final mg/L concentration.**

This is where the density becomes important.

We know the final solution mass is 1.000 kg and its density is \(1.02\text{ kg/L}\).

Use:

\[
\rho=\frac{m}{V}
\]

Rearrange:

\[
V=\frac{m}{\rho}
\]

Substitute:

\[
\begin{aligned}
V
&=\frac{1.000\text{ kg}}{1.02\text{ kg/L}}\\
&=0.9804\text{ L}
\end{aligned}
\]

Convert the solute mass into milligrams:

\[
120\ \mu\text{g}=0.120\text{ mg}
\]

Now calculate the mass concentration:

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{0.120\text{ mg}}{0.9804\text{ L}}\\
&=0.1224\text{ mg/L}
\end{aligned}
\]

Therefore:

\[
\boxed{C_{\mathrm{mg/L}}=0.122\text{ mg/L}}
\]

Notice that the answer is not exactly 0.120 mg/L.

The final solution has a density of 1.02 kg/L, rather than exactly 1.00 kg/L. That small difference changes the conversion.

Also notice that we did not assume the final volume was the sum of the initial solution volume and the water volume. The supplied final density allowed us to calculate the actual final volume from the total mass.

### Question 5

Two solutions contain the same dissolved substance, X.

A chemist measures out the following equal-volume portions:

| Property | Solution A | Solution B |
|---|---:|---:|
| Volume | 0.500 L | 0.500 L |
| Density | 0.800 kg/L | 1.20 kg/L |
| Concentration | 300 ppb by mass | 900 ppb by mass |

The chemist mixes the two portions. Assume:

- No solute is lost during mixing.
- No chemical reaction occurs.
- The initial volumes are additive.

After mixing, the chemist adds 0.250 kg of pure water.

The resulting diluted solution has a measured density of \(0.980\text{ kg/L}\).

Calculate:

(a) The concentration of X in ppb by mass immediately after the first two solutions are mixed.

(b) The concentration of X in mg/L immediately after mixing.

(c) The concentration of X in ppb by mass after adding the water.

(d) The final concentration in mg/L.

(e) Explain why simply taking the arithmetic mean of the original 300 ppb and 900 ppb values gives an incorrect result.

[IMAGE: A three-stage laboratory mixing illustration. Stage one shows two separate 0.500 L vessels labelled A (0.800 kg/L, 300 ppb by mass) and B (1.20 kg/L, 900 ppb by mass). Stage two shows both portions poured into one mixing vessel, with the original volumes indicated as additive. Stage three shows 0.250 kg of pure water being added to the mixture, with the final measured density labelled 0.980 kg/L. Do not show calculated masses, final volumes, or resulting concentrations.]

### Solution 5

The final answers are:

- Immediately after mixing: **660 ppb by mass and 0.660 mg/L**.
- After adding water: **528 ppb by mass and 0.517 mg/L**.

The key is to track the actual mass of solute and the total mass of solution. Equal volumes do not necessarily mean equal solution masses.

**Part (a): Calculate the concentration immediately after mixing.**

First, determine the mass of each original solution.

For Solution A:

\[
\begin{aligned}
m_A&=\rho_A V_A\\
&=0.800\times0.500\\
&=0.400\text{ kg}
\end{aligned}
\]

For Solution B:

\[
\begin{aligned}
m_B&=\rho_B V_B\\
&=1.20\times0.500\\
&=0.600\text{ kg}
\end{aligned}
\]

Now calculate the mass of X in each solution.

For Solution A:

\[
\begin{aligned}
m_{X,A}
&=300\ \mu\text{g/kg}\times0.400\text{ kg}\\
&=120\ \mu\text{g}
\end{aligned}
\]

For Solution B:

\[
\begin{aligned}
m_{X,B}
&=900\ \mu\text{g/kg}\times0.600\text{ kg}\\
&=540\ \mu\text{g}
\end{aligned}
\]

The total solute mass is:

\[
\begin{aligned}
m_{X,\mathrm{total}}
&=120+540\\
&=660\ \mu\text{g}
\end{aligned}
\]

The total solution mass is:

\[
\begin{aligned}
m_{\mathrm{total}}
&=0.400+0.600\\
&=1.000\text{ kg}
\end{aligned}
\]

Therefore, the new mass-based concentration is:

\[
\begin{aligned}
C_{\mathrm{ppb}}
&=\frac{660\ \mu\text{g}}{1.000\text{ kg}}\\
&=660\ \mu\text{g/kg}
\end{aligned}
\]

Hence:

\[
\boxed{C_{\mathrm{ppb}}=660\text{ ppb}}
\]

**Part (b): Calculate the concentration immediately after mixing in mg/L.**

The question states that the initial volumes are additive.

Therefore:

\[
\begin{aligned}
V_{\mathrm{total}}
&=0.500+0.500\\
&=1.000\text{ L}
\end{aligned}
\]

The total solute mass is 660 μg, or 0.660 mg.

Therefore:

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{0.660\text{ mg}}{1.000\text{ L}}\\
&=0.660\text{ mg/L}
\end{aligned}
\]

Hence:

\[
\boxed{C_{\mathrm{mg/L}}=0.660\text{ mg/L}}
\]

The initial mixture has a density of:

\[
\rho=\frac{1.000\text{ kg}}{1.000\text{ L}}
=1.000\text{ kg/L}
\]

This explains why its numerical ppb value divided by 1000 happens to equal its mg/L value.

**Part (c): Calculate the new ppb concentration after adding water.**

Adding pure water increases the total solution mass but does not change the mass of X.

The new total solution mass is:

\[
\begin{aligned}
m_{\mathrm{final}}
&=1.000+0.250\\
&=1.250\text{ kg}
\end{aligned}
\]

The solute mass remains 660 μg.

Therefore:

\[
\begin{aligned}
C_{\mathrm{ppb}}
&=\frac{660\ \mu\text{g}}{1.250\text{ kg}}\\
&=528\ \mu\text{g/kg}
\end{aligned}
\]

Hence:

\[
\boxed{C_{\mathrm{ppb}}=528\text{ ppb}}
\]

Notice that the concentration has decreased because the same solute mass is now distributed throughout a greater total solution mass.

**Part (d): Calculate the final mg/L concentration.**

We know:

- Final solution mass = 1.250 kg.
- Final solution density = 0.980 kg/L.
- Solute mass = 660 μg = 0.660 mg.

First, calculate the final solution volume:

\[
\begin{aligned}
V_{\mathrm{final}}
&=\frac{m_{\mathrm{final}}}{\rho_{\mathrm{final}}}\\
&=\frac{1.250\text{ kg}}{0.980\text{ kg/L}}\\
&=1.276\text{ L}
\end{aligned}
\]

Now calculate the mass concentration:

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{0.660\text{ mg}}{1.276\text{ L}}\\
&=0.5172\text{ mg/L}
\end{aligned}
\]

To three significant figures:

\[
\boxed{C_{\mathrm{mg/L}}=0.517\text{ mg/L}}
\]

We can check this result using the ppb conversion formula:

\[
\begin{aligned}
C_{\mathrm{mg/L}}
&=\frac{528\times0.980}{1000}\\
&=0.51744\text{ mg/L}
\end{aligned}
\]

This agrees with the result above, allowing for rounding.

**Part (e): Explain why the arithmetic mean is incorrect.**

The arithmetic mean of the original concentrations is:

\[
\frac{300+900}{2}=600\text{ ppb}
\]

This answer is incorrect because the two solutions have equal volumes but different masses.

Solution A has a mass of 0.400 kg, while Solution B has a mass of 0.600 kg.

The original ppb values describe micrograms of X per kilogram of solution. Therefore, the concentrations must be weighted according to the mass of each solution, not simply averaged because their volumes are equal.

Solution B contributes more total solution mass and more dissolved substance than Solution A.

The correct mixed concentration is 660 ppb, not 600 ppb.

There is another important detail in the second stage. Adding water changes the mass-based concentration and the final solution density. You cannot assume that the numerical ppb value divided by 1000 will always give the correct mg/L value.

This problem brings together three ideas:

- Conservation of solute mass during mixing and dilution.
- Mass-based concentration depends on the total solution mass.
- Mass concentration in mg/L depends on the total solution volume.

Keeping those three quantities separate is what makes the calculation work.

## 6. Where to go next

Once you have converted a trace concentration into mg/L, you can use it in further HSC Chemistry calculations.

For example, if a question asks for the concentration in mol/L, you can convert the mass concentration into g/L and divide by the molar mass.

Read [How to Convert mg/L to mol/L in HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-convert-milligrams-per-litre-into-moles-per-litre/) for the next step.

Remember the central conversion:

\[
\boxed{
\text{mg/L}=
\frac{\text{ppb by mass}\times
\text{density (kg/L)}}{1000}
}
\]

For dilute aqueous solutions, density is often close to 1.00 kg/L, so dividing ppb by 1000 is a useful approximation. When the density differs, use the full equation.