Ways to Measure Solution Concentration in HSC Chemistry

Two glucose solutions each contain 4.0 g of dissolved glucose. One has a final volume of 100 mL; the other has a final volume of 400 mL. Before calculating, predict: which is more concentrated, and by what factor?

The 100 mL solution is four times as concentrated. It contains 4.0 g in every 100 mL, while the larger solution contains only 1.0 g in every 100 mL. The total glucose mass is the same, but it is spread through different volumes.

Concentration describes how much solute there is relative to a stated quantity of solution, not simply how much solute is in the container. The challenge is recognising what each description compares.

01First, separate solute, solvent, and solution

For glucose dissolved in water, glucose is the solute, water is the solvent, and the whole mixture is the solution. Once it is fully dissolved and mixed, equal-sized samples have the same composition. :chatgpt-content-reference{index=”0″}

Every concentration description below compares a quantity of solute with a quantity of the finished solution. The denominator might be the solution’s volume or its mass.

That gives you a useful habit: before touching the calculator, write what belongs above and below the fraction line.

02Mass concentration: grams per litre

Suppose you need to know how much dissolved solid a litre of solution contains. A mass-based description answers that directly.

Mass concentration is the mass of a specified solute divided by the volume of the solution. We will use the symbol \(C_m\): :chatgpt-content-reference{index=”1″}

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

Here, \(m\) is the solute mass in grams, \(\mathrm{g}\), and \(V\) is the final solution volume in litres, \(\mathrm{L}\). These units give \(C_m\) in \(\mathrm{g\,L^{-1}}\), meaning grams per litre.

The negative exponent just means “per”: \(\mathrm{g\,L^{-1}}\) is another way of writing \(\mathrm{g/L}\).

03Amount concentration: moles per litre

Mass concentration is useful, but grams do not directly count chemical particles. Equal masses of two different substances can contain different numbers of molecules.

A mole is a counting unit, rather like a dozen with a much larger count. One mole contains approximately \(6.022\times10^{23}\) specified entities, such as molecules or ions. The important limit of the dozen analogy is this: a fixed count does not imply a fixed mass. A dozen bowling balls and a dozen grapes would make very different shopping bags. :chatgpt-content-reference{index=”2″}

Molar mass, \(M\), tells you the mass of one mole of a substance. When \(M\) is in grams per mole, \(\mathrm{g\,mol^{-1}}\), dividing a mass in grams by \(M\) gives the amount of substance, \(n\), in moles, \(\mathrm{mol}\): \(n=m/M\).

Amount concentration, also called molar concentration or molarity, then compares those moles with the solution volume: :chatgpt-content-reference{index=”3″}

\[
c=\frac{n}{V}
\]

Here, \(c\) is amount concentration in \(\mathrm{mol\,L^{-1}}\), \(n\) is the amount of the specified solute in moles, and \(V\) is the final solution volume in litres.

The conversion between grams per litre and moles per litre

Both descriptions use the same solution volume. Only the numerator changes from grams to moles:

\[
c=\frac{n}{V}
=\frac{m}{MV}
=\frac{C_m}{M}
\qquad\text{and}\qquad
C_m=cM
\]

To convert \(\mathrm{g\,L^{-1}}\) to \(\mathrm{mol\,L^{-1}}\), divide by the solute’s molar mass in \(\mathrm{g\,mol^{-1}}\). To go back, multiply.

You need the solute’s identity, or its molar mass, for this conversion. A label saying “20 g of dissolved material per litre” is not enough to determine moles per litre.

04Percentage concentration: per 100 of what?

Now predict: does 5.0% (m/m) necessarily describe the same concentration as 5.0% (m/v)?

No. One refers to 100 g of solution; the other refers to 100 mL. Those quantities are not automatically interchangeable.

The letters beside the percentage tell you what is being compared. Here, \(m\) stands for mass and \(v\) stands for volume. :chatgpt-content-reference{index=”4″}

You may also encounter % (w/v) and % (w/w). In these labels, “weight” means mass, so they correspond to % (m/v) and % (m/m), respectively.

For mass/mass and volume/volume percentages, divide the solute quantity by the corresponding solution quantity, using matching units, then multiply by 100.

For mass/volume percentage, use grams of solute and millilitres of solution specifically. It is a conventional way of expressing grams per 100 mL, not a dimensionless mass fraction.

Mass/volume percentage is another mass-concentration label

The glucose solutions above contain 18.0 g per 1000 mL. Therefore, they contain 1.80 g per 100 mL and can be labelled 1.80% (m/v).

Using the rounded molar mass from the worked example, all three labels describe the same glucose concentration:

\[
18.0\ \mathrm{g\,L^{-1}}
\quad\longleftrightarrow\quad
1.80\%\ (m/v)
\quad\longleftrightarrow\quad
0.100\ \mathrm{mol\,L^{-1}}
\]

The numerical values differ because the units differ.

For a direct conversion between % (m/v) and \(\mathrm{g\,L^{-1}}\), multiply the percentage number by 10 to obtain grams per litre. Divide grams per litre by 10 to obtain the percentage number. No density is needed because both descriptions already use solution volume.

The most tempting mistake: using solvent instead of solution

A 5.0% (m/m) solution contains 5.0 g of solute in 100 g of solution. For a solution containing only that solute and water, the water mass is 95.0 g, not 100 g.

A 5.0% (v/v) solution uses 5.0 mL of pure liquid solute per 100 mL of final solution. Refer the volumes to the same temperature. Do not automatically conclude that exactly 95.0 mL of solvent was added: liquid volumes are not generally additive on mixing.

The practical rule is to establish the final solution volume, rather than assume it equals the volume of solvent added. This is central to preparing a standard solution.

05Density connects mass-based and volume-based descriptions

To convert % (m/m) to % (m/v), you need to connect the mass of the solution with its volume.

That connection is the solution’s density, \(\rho\), pronounced “rho”:

\[
\rho=\frac{m_{\text{solution}}}{V_{\text{solution}}}
\]

Here, \(m_{\text{solution}}\) is the total solution mass and \(V_{\text{solution}}\) is its volume. Using grams and millilitres gives density in \(\mathrm{g\,mL^{-1}}\), or grams per millilitre. Rearranging gives \(m_{\text{solution}}=\rho V_{\text{solution}}\). :chatgpt-content-reference{index=”5″}

Density and mass concentration have different numerators. Density uses the mass of the whole solution; mass concentration uses the mass of the specified solute.

Choose the density that matches the conversion

For % (m/m) to % (m/v), use the solution’s density. You are converting total solution mass to solution volume.

For % (v/v) to % (m/v), use the pure liquid solute’s density. You are converting the solute’s volume to its mass while keeping the same solution volume.

Treating 100 mL of solution as 100 g assumes a density of \(1.00\ \mathrm{g\,mL^{-1}}\). Use that only when it is given or justified as an approximation. “It contains water” is not sufficient justification.

06Questions and solutions

Question 1

A 75.0 mL solution sample contains 0.900 g of dissolved solid. Calculate its mass concentration in \(\mathrm{g\,L^{-1}}\) and its mass/volume percentage.

Solution 1

The mass concentration is \(12.0\ \mathrm{g\,L^{-1}}\), equivalent to 1.20% (m/v).

Mass concentration uses solute mass divided by final solution volume. Converting the volume gives \(75.0\ \mathrm{mL}=0.0750\ \mathrm{L}\).

\[
C_m=\frac{m}{V}
=\frac{0.900\ \mathrm{g}}{0.0750\ \mathrm{L}}
=12.0\ \mathrm{g\,L^{-1}}
\]

To determine the mass in a 100 mL sample:

\[
\frac{0.900\ \mathrm{g}}{75.0\ \mathrm{mL}}
\times100\ \mathrm{mL}
=1.20\ \mathrm{g}
\]

Therefore, the solution contains 1.20 g per 100 mL, giving 1.20% (m/v). These are two descriptions of the same ratio.

Question 2

An empty bottle has a mass of 24.0 g. Filled with solution, it has a mass of 124.0 g. The solution contains 8.00 g of dissolved solute.

A student labels the bottle “8.00% (m/v)”. Which percentage description is supported by the measurements, and what additional measurement would allow the mass/volume percentage to be calculated?

Solution 2

The measurements support 8.00% (m/m), not necessarily 8.00% (m/v). The final solution volume is needed to calculate the mass/volume percentage.

First subtract the bottle’s mass:

\[
\begin{aligned}
m_{\text{solution}}
&=124.0\ \mathrm{g}-24.0\ \mathrm{g}
=100.0\ \mathrm{g}\\
\text{mass percentage}
&=\frac{8.00\ \mathrm{g}}{100.0\ \mathrm{g}}\times100\%
=8.00\%
\end{aligned}
\]

The denominator is 100.0 g of solution, so the correct description is 8.00% (m/m).

Nothing in the measurements establishes that the solution occupies 100 mL. Measuring its volume would resolve this. Alternatively, measuring its density would allow the volume to be calculated from the known solution mass.

Question 3

Solution A contains glucose at \(30.0\ \mathrm{g\,L^{-1}}\). Solution B contains urea, \(\ce{CO(NH2)2}\), at \(12.0\ \mathrm{g\,L^{-1}}\). Both solutes remain as molecules in water.

Use rounded molar masses of \(180\ \mathrm{g\,mol^{-1}}\) for glucose and \(60.0\ \mathrm{g\,mol^{-1}}\) for urea.

Which solution has the greater mass concentration, and which has the greater amount concentration? Explain how the rankings can differ.

Solution 3

Solution A has the greater mass concentration, but solution B has the greater amount concentration: \(0.200\ \mathrm{mol\,L^{-1}}\), compared with \(0.167\ \mathrm{mol\,L^{-1}}\) for A.

The mass-concentration ranking follows directly from the labels: \(30.0>12.0\).

For the amount concentrations, let \(c_A\) and \(c_B\) represent the concentrations of the respective solutes:

\[
\begin{aligned}
c_A
&=\frac{C_{m,A}}{M_A}
=\frac{30.0\ \mathrm{g\,L^{-1}}}{180\ \mathrm{g\,mol^{-1}}}
=0.167\ \mathrm{mol\,L^{-1}}\\
c_B
&=\frac{C_{m,B}}{M_B}
=\frac{12.0\ \mathrm{g\,L^{-1}}}{60.0\ \mathrm{g\,mol^{-1}}}
=0.200\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Here, \(C_{m,A}\) and \(C_{m,B}\) are the stated mass concentrations, and \(M_A\) and \(M_B\) are the corresponding molar masses.

A mole of urea has much less mass than a mole of glucose. Solution B can therefore contain more solute molecules per litre despite containing fewer grams of solute per litre.

The trap is treating “more concentrated” as complete information without specifying the basis of comparison.

Question 4

A solution is labelled 10.0% (v/v) of a liquid solute. The pure solute has a density of \(0.800\ \mathrm{g\,mL^{-1}}\). All volumes refer to the same temperature.

Determine the mass/volume percentage and the mass concentration in \(\mathrm{g\,L^{-1}}\). Does the label also prove that exactly 90.0 mL of water was used for every 100 mL of solution?

Solution 4

The solution is 8.00% (m/v), or \(80.0\ \mathrm{g\,L^{-1}}\). The label does not establish that 90.0 mL of water was used.

Consider 100 mL of solution. The label specifies 10.0 mL of pure liquid solute per 100 mL of finished solution.

Use the pure solute’s density, \(\rho_{\text{solute}}\), to convert its volume, \(V_{\text{solute}}\), to mass:

\[
\begin{aligned}
m
&=\rho_{\text{solute}}V_{\text{solute}}\\
&=(0.800\ \mathrm{g\,mL^{-1}})(10.0\ \mathrm{mL})
=8.00\ \mathrm{g}\\
C_m
&=\frac{8.00\ \mathrm{g}}{0.100\ \mathrm{L}}
=80.0\ \mathrm{g\,L^{-1}}
\end{aligned}
\]

There are 8.00 g of solute per 100 mL of solution, giving 8.00% (m/v).

Subtracting \(10.0\ \mathrm{mL}\) from \(100\ \mathrm{mL}\) would assume that the separately measured liquid volumes add exactly on mixing. That assumption is not supplied, so the water volume cannot be determined from the label alone.

Question 5

A draft specification describes a solution as 120% (m/v), with a density of \(1.50\ \mathrm{g\,mL^{-1}}\). A reviewer rejects it because “a solution cannot contain more than 100% solute”.

Calculate the corresponding mass/mass percentage and assess the reviewer’s claim. Do the two stated numbers alone establish that a real solute could dissolve at this concentration?

Solution 5

The specification corresponds to 80.0% (m/m). It is not impossible merely because the mass/volume percentage exceeds 100%, but solubility information is still needed to establish physical feasibility.

Use a 100 mL solution basis. The label specifies 120 g of solute in that volume.

The total solution mass and mass percentage would be:

\[
\begin{aligned}
m_{\text{solution}}
&=\rho V_{\text{solution}}\\
&=(1.50\ \mathrm{g\,mL^{-1}})(100\ \mathrm{mL})
=150\ \mathrm{g}\\
\text{mass percentage}
&=\frac{120\ \mathrm{g}}{150\ \mathrm{g}}\times100\%
=80.0\%
\end{aligned}
\]

The remaining mass would be \(150-120=30\ \mathrm{g}\), available for solvent.

A mass/mass percentage cannot exceed 100%, because the solute mass cannot exceed the total solution mass. But % (m/v) compares grams with millilitres. A value of 120% (m/v) means 120 g per 100 mL, not 120 g per 100 g.

These numbers are not internally contradictory. However, they do not tell us whether a particular solute can dissolve in that amount of solvent under the stated conditions. That requires solubility data.

Question 6

A homogeneous solution in a sealed, flexible container expands by 5.00% on warming. Nothing enters or leaves, no reaction occurs, and all solute remains dissolved.

Which of these concentration descriptions change: \(\mathrm{g\,L^{-1}}\), \(\mathrm{mol\,L^{-1}}\), % (m/v), and % (m/m)? Calculate the percentage change for each description that changes. Has solvent necessarily been added because the amount concentration has decreased?

Solution 6

The values in \(\mathrm{g\,L^{-1}}\), \(\mathrm{mol\,L^{-1}}\), and % (m/v) all decrease by 4.76%. The mass/mass percentage stays unchanged. No solvent has been added.

The solute mass, solute amount in moles, and total solution mass all remain constant.

Each of the first three descriptions divides a fixed solute quantity by solution volume. The new volume is 1.050 times the original volume, so:

\[
\begin{aligned}
\frac{\text{new concentration}}{\text{original concentration}}
&=\frac{\text{original volume}}{1.050\times\text{original volume}}
=\frac{1}{1.050}
=0.95238\\
\text{percentage decrease}
&=(1-0.95238)\times100\%
=4.76\%
\end{aligned}
\]

A 5.00% increase in the denominator does not produce an exactly 5.00% decrease in the ratio. The relationship is reciprocal.

Mass/mass percentage has mass in both the numerator and denominator. Neither mass changes, so that percentage remains fixed.

The lower amount concentration does not prove that solvent entered. It shows that there are fewer moles per litre, which here results from expansion rather than dilution by added solvent.

07From a concentration to a usable quantity

A concentration becomes useful when you connect it to an actual sample. The relation \(n=cV\) gives the moles transferred, while \(m=C_mV\) gives the solute mass transferred. Use litres when the concentration is expressed per litre.

That is the starting point for solution dilution calculations: identify how much solute is present, then track what happens to the final solution volume. The unit tells you which quantity you are keeping track of.